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RESONANCE AIEEE_CENTRE OF MASS - 1

CENTRE OF MASS

CENTRE OF MASS

Every physical system has associated with it a certain point whose motion characterises the motion of

the whole system. When the system moves under some external forces, then this point moves as if theentire mass of the system is concentrated at this point and also the external force is applied at thispoint for translational motion. This point is called the centre of mass of the system.

CENTRE OF MASS OF A SYSTEM OF 'N' DISCRETE PARTICLES

Consider a system of N point masses m1, m

2, m

3, ................ m

n whose

position vectors from origin O are given by 1r

, 2r

, 3r

,...............

nr

respectively. Then the position vector of the centre of mass C of the

system is given by.

cmr =

n21

nn2211

m........mm

rm........rmrm

; cmr

=

n

1ii

n

1iii

m

rm

cmr

= M1

n

1iii rm

where, ii rm

is called the moment of mass of the particle w.r.t O.

M =

n

1iim is the total mass of the system.

Note: If the origin is taken at the centre of mass then

n

1iii rm

=0. hence, the COM is the point about which

the sum of �mass moments� of the system is zero.

POSITION OF COM OF TWO PARTICLES

Centre of mass of two particles of masses m1 and m

2 separated by a distance r lies in between the

two particles. The distance of centre of mass from any of the particle (r) is inversely proportionalto the mass of the particle (m)i.e. r 1/m

or2

1

rr

= 1

2

mm

or m1r

1 = m

2r

2

or r1 =

12

2

mm

mr and r

2 =

21

1

mm

mr

Here, r1 = distance of COM from m

1

and r2 = distance of COM from m

2

From the above discussion, we see thatr

1 = r

2 = 1/2 if m

1 = m

2, i.e., COM lies midway between the two particles of equal masses.

Similarly, r1 > r

2 if m

1 < m

2 and r

1 < r

2 if m

2 < m

1, i.e., COM is nearer to the particle having larger

mass.


Example 1. Two particles of mass 1 kg and 2 kg are located at x = 0 and x = 3 m. Find the position of theircentre of mass.

Solution :Since, both the particles lies on x-axis, the COM will also lie on x-axis. Let the COM islocated at x = x, thenr

1 = distance of COM from the particle of mass 1 kg = x

and r2 = distance of COM from the particle of mass 2 kg = (3 � x)

Using2

1

rr

= 1

2

mm

orx3

x

= 12

or x = 2 m

Thus, the COM of the two particles is located at x = 2 m. Ans.

Example 2. The position vector of three particles of masses m1 = 1 kg, m

2 = 2 kg and m

3 = 3 kg are

m)k�j�4i�(r1

, m)k�j�i�(r2

and m)k�2j�i�2(r3

respectively. Find the position vector

of their centre of mass.Solution :

The position vector of COM of the three particles will be given by

321

332211COM mmm

mmm

rrrr

Substituting the values, we get

321)k�2�j��i�3)(2()k�j�i�((2))k�j�4i�((1)

rCOM

m)k�j�i�3(21

Ans.

Example 3. Four particles of mass 1 kg, 2 kg, 3 kg and 4 kg are placed at the four vertices A, B, C and Dof a square of side 1 m. Find the position of centre of mass of the particles.

Solution :Assuming D as the origin, DC as x -axis and DA as y-axis, we have

m1 = 1 kg, (x

1, y

1) = (0, 1m)

m2 = 2 kg, (x

2, y

2) = (1m, 1m)

m3 = 3 kg, (x

3, y

3) = (1m, 0)

and m4 = 4 kg, (x

4, y

4) = (0, 0)

Co-ordinates of their COM are

xCOM

= 4321

44332211

mmmm

xmmmxmxm

= 4321

)0(4)1(3)1(2)0)(1(

=

105

= 21

m = 0.5 m


Similarly, yCOM

= 4321

44332211

mmmm

ymymymym

= 4321

)0(4)0(3)1(2)1)(1(

=

103

= 0.3 m

(xCOM

, yCOM

) = (0.5 m, 0.3 m) Ans.

Thus, position of COM of the four particles is as shown in figure.

Example 4. Consider a two-particle system with the particles having masses m1 and m

2. If the first particle

is pushed towards the centre of mass through a distance d, by what distance should thesecond particle be moved so as to keep the centre of mass at the same position?

Solution :Consider figure. Suppose the distance of m

1 from the centre of mass C is x

1 and that of m

2 from

C is x2. Suppose the mass m

2 is moved through a distance d towards C so as to keep the

centre of mass at C.

Then, m1x

1 = m

2x

2.........(i)

and m1(x

1 � d) = m

2 (x

2 � d). .........(ii)

Subtracting (ii) from (i)m

1d = m

2 d

or, d = 2

1

mm

d,

CENTRE OF MASS OF A CONTINUOUS MASS DISTRIBUTION

For continuous mass distribution the centre of mass can be located by replacing summation sign with anintegral sign. Proper limits for the integral are chosen according to the situation

xcm

=

dm

dmx, y

cm =

dm

dmy, z

cm =

dm

dmz

dm = M (mass of the body) x = x component of centre of mass of element

cmr

= M1

dmr

.

Note: If an object has symmetric mass distribution about x axis then y coordinate of COM is zero andvice-versa

CENTRE OF MASS OF A UNIFORM ROD

Suppose a rod of mass M and length L is lying along the x-axis with its one end at x = 0 and the

other at x = L. Mass per unit length of the rod = LM

Hence, dm, (the mass of the element dx situated at x = x is) = LM

dx

The coordinates of the element dx are (x, 0, 0). Therefore, x-coordinate of COM of the rod will be


xCOM

=

dm

dmxL

0

x=Lx=xx=0

dx

=M

dxLM

)x(L

0

= L

0 2L

dxxL1

The y-coordinate of COM is

yCOM

=

dm

dmy = 0

Similarly, zCOM

= 0

i.e., the coordinates of COM of the rod are

0,0,

2

L, i.e. it lies at the centre of the rod.

Example 5. A rod of length L is placed along the x-axis between x = 0 and x = L. The linear density (mass/length) of the rod varies with the distance x from the origin as = Rx. Here, R is a positive constant. Findthe position of centre of mass of this rod.

Solution :Mass of element dx situated at x = x is

dm = dx = Rx dxThe COM of the element has coordinates (x, 0, 0).Therefore, x-coordinate of COM of the rod will be

x=Lx=xx=0

dx

y

x

xCOM

=

dm

dmxL

0

=

L

0

L

0

dx)Rx(

dx)Rx)(x(

=

L

0

L

0

2

dxxR

dxxR

= 3L2

2x

3x

L

0

2

L

0

3

The y-coordinate of COM of the rod is yCOM

=

dm

dmy

= 0 (as y = 0)

Similarly, zCOM

= 0

Hence, the centre of mass of the rod lies at

0,0,

3

L2Ans.


CENTRE OF MASS OF A SEMICIRCULAR RING

Figure shows the object (semi circular ring). By observation we can say that the x-coordinate ofthe centre of mass of the ring is zero as the half ring is symmetrical abnout y-axis on both sidesof the origin. Only we are required to find the y-coordinate of the centre of mass.

Y

ycm

d y=Rsin

Rd

X

To find ycm

we use ycm

= M1

ydm ...(i)

Here for dm we consider an elemental arc of the ring at an angle from the x-direction of angularwidth d. If radius of the ring is R then its y coordinate will be R sin, here dm is given as

dm = R

M

× R d

So from equation ---(i), we have

ycm

= M1

RdR

M

0

(R sin) =

R

dsin0

ycm

=

R2...(ii)

CENTRE OF MASS OF SEMICIRCULAR DISC

Figure shows the half disc of mass M and radius R. Here, we are only required to find the y-coordinate of the centre of mass of this disc as centre of mass will be located on its half verticaldiameter. Here to find y

cm, we consider a small elemental ring of mass dm of radius x on the disc

(disc can be considered to be made up such thin rings of increasing radii) which will be integratedfrom 0 to R. Here dm is given as

dm = 2R

M2

( x)dx

Y

ycm

XR

dx

x

Now the y-coordinate of the element is taken as

x2, as in previous section, we have derived that

the centre of mass of a semi circular ring is concentrated at

R2

Here ycm

is given as ycm

= M1

R

0

dm

x2 =

M1

R

0

22

dxxR

M4y

cm =

3R4


1. Centre of mass of a uniform rectangular, square or circular plate lies at its centre. Axis of symme-try plane of symmetry.

2. For a laminar type (2-dimensional) body with uniform negligible thickness the formulae for findingthe position of centre of mass are as follows :

....tAtA....rtArtA

....mm....rmrm

r21

2211

21

2211COM

( m = At)

or ....AA....rArA

r21

2211COM

Here, A stands for the area,

3. If some mass of area is removed from a rigid body, then the position of centre of mass of the

remaining portion is obtained from the following formulae:

(i)21

2211COM mm

rmrmr

or 21

2211COM AA

rArAr

(ii)21

2211COM mm

xmxmx

or

21

2211COM AA

xAxAx

21

2211COM mm

ymymy

or

21

2211COM AA

yAyAy

and21

2211COM mm

zmzmz

or

21

2211COM AA

zAzAz

Here, m1, A

1, 1r

, x1, y

1 and z

1 are the values for the whole mass while m

2, AA

2, 2r

, 2x

, y2 and z

2 are the values

for the mass which has been removed. Let us see two examples in support of the above theory.

Example 6. Find the position of centre of mass of the uniform lamina shown in figure.

XaO

Y

Solution : Here,

A1 = area of complete circle = a2

A2 = area of small circle =

2

2a

=

4a2

(x1, y

1) = coordinates of centre of mass of large circle = (0, 0)


and (x2, y

2) = coordinates of centre of mass of small circle =

0,

2a

Using xCOM

= 21

2211

AAxAxA

we get xCOM

=

4a

a

2a

4a

22

2

=

43

81

a = � 6a

and yCOM

= 0 as y1 and y

2 both are zero.

Therefore, coordinates of COM of the lamina shown in figure are

0,

6

aAns.

CENTRE OF MASS OF SOME COMMON SYSTEMS

A system of two point masses m1 r

1 = m

2 r

2

The centre of mass lies closer to the heavier mass.

Rectangular plate (By symmetry)

xc =

2b

yc =

2L

A triangular plate (By qualitative argument)

at the centroid : yc =

3h

A semi-circular ring

yc =

R2x

c = 0

A semi-circular disc

yc =

3R4

xc = 0


A hemispherical shell

yc =

2R

xc = 0

A solid hemisphere

yc =

8R3

xc = 0

A circular cone (solid)

yc =

4h

A circular cone (hollow)

yc =

3h

Example 7. A uniform thin rod is bent in the form of closed loop ABCDEFA as shown in the figure. The y-coordinate of the centre of mass of the system is

(A)

r2(B)

23r6

(C)

r2

(D) Zero

Ans. (B)

Solution. The centre of mass of semicircular ring is at a distance

r2 from its centre. (Let = mass/

length)

Ycm

= 23

r6r2rrr

r4r2

r2r

MOTION OF CENTRE OF MASS AND CONSERVATION OF MOMENTUM :

Velocity of centre of mass of system

cmv

= M

dtrd

m..............dtrd

mdtrd

mdtrd

m nn

33

22

11

= M

vm..........vmvmvm nn332211

Here numerator of the right hand side term is the total momentum of the system i.e., summationof momentum of the individual component (particle) of the system


Hence velocity of centre of mass of the system is the ratio of momentum of the system to the mass of thesystem.

SystemP

= M cmv

Acceleration of centre of mass of system

cma

= M

dtvd

m..............dtvd

mdtvd

mdtvd

m nn

33

22

11

= M

am..........amamam nn332211

= M

systemonforceNet =

MForceernalintNetForceExternalNet

= M

ForceExternalNet

(action and reaction both of an internal force must be within the system. Vector summation willcancel all internal forces and hence net internal force on system is zero)

extF

= M cma

where extF

is the sum of the 'external' forces acting on the system. The internal forces which the

particles exert on one another play absolutely no role in the motion of the centre of mass.If no external force is acting on a system of particles, the acceleration of centre of mass of thesystem will be zero. If a

c = 0, it implies that v

c must be a constant and if v

cm is a constant, it

implies that the total momentum of the system must remain constant. It leads to the principal ofconservation of momentum in absence of external forces.

If 0Fext

then cmv

= constant

�If resultant external force is zero on the system, then the net momentum of the systemmust remain constant�.

Motion of COM in a moving system of particles:

(1) COM at rest :If F

ext = 0 and V

cm = 0, then COM remains at rest. Individual components of the system may

move and have non-zero momentum due to mutual forces (internal), but the net momentum ofthe system remains zero.(i) All the particles of the system are at rest.

(ii) Particles are moving such that their net momentum is zero.example:

(iii) A bomb at rest suddenly explodes into various smaller fragments, all moving in different directionsthen, since the explosive forces are internal & there is no external force on the system for explo-sion therefore, the COM of the bomb will remain at the original position and the fragment fly suchthat their net momentum remains zero.

(iv) Two men standing on a frictionless platform, push each other, then also their net momentumremains zero because the push forces are internal for the two men system.

(v) A boat floating in a lake, also has net momentum zero if the people on it changes their position,because the friction force required to move the people is internal of the boat system.

(vi) Objects initially at rest, if moving under mutual forces (electrostatic or gravitation)also have netmomentum zero.

(vii) A light spring of spring constant k kept compressed between two blocks of masses m1 and

m2 on a smooth horizontal surface. When released, the blocks acquire velocities in oppo-

site directions, such that the net momentum is zero.

(viii) In a fan, all particles are moving but COM is at rest


(2) COM moving with uniform velocity :If F

ext = 0, then V

cm remains constant therefore, net momentum of the system also remains

conserved. Individual components of the system may have variable velocity and momentum due tomutual forces (internal), but the net momentum of the system remains constant and COM contin-ues to move with the initial velocity.

(i) All the particles of the system are moving with same velocity.e.g.: A car moving with uniform speed on a straight road, has its COM moving with aconstant velocity.

(ii) Internal explosions / breaking does not change the motion of COM and net momentumremains conserved. A bomb moving in a straight line suddenly explodes into various smallerfragments, all moving in different directions then, since the explosive forces are internal & there isno external force on the system for explosion therefore, the COM of the bomb will continue theoriginal motion and the fragment fly such that their net momentum remains conserved.

(iii) Man jumping from cart or buggy also exert internal forces therefore net momentum of thesystem and hence, Motion of COM remains conserved.

(iv) Two moving blocks connected by a light spring on a smooth horizontal surface. If theacting forces is only due to spring then COM will remain in its motion and momentum willremain conserved.

(v) Particles colliding in absence of external impulsive forces also have their momentum con-served.

(3) COM moving with acceleration :If an external force is present then COM continues its originalmotion as if the external force is acting on it, irrespective ofinternal forces.Example:Projectile motion : An axe thrown in air at an angle with thehor izonta l wi l l per fo rm a compl ica ted m ot ion o frotation as well as parabolic motion under the effect ofgravitation

Hcom

= g2sinu 22

Rcom

= g2sinu2

T = gsinu2

Example:Circular Motion : A rod hinged at an end, rotates, than itsCOM performs circular motion. The centripetal force (F

c)

required in the circular motion is assumed to be acting onthe COM.

COM2

c RmùF

mg

mg

mg

Fc

Fc

Fc

mg

Fc

Rcom


Example 8. A projectile is fired at a speed of 100 m/s at an angle of 37º above the horizontal. At the highest point,

the projectile breaks into two parts of mass ratio 1 : 3, the lighter piece coming to rest. Find thedistance from the launching point to the point where the heavier piece lands.

Solution : Internal force do not effect the motion of the centre of mass, the centre of mass hits theground at the position where the original projectile would have landed. The range of theoriginal projectile is,

xCOM

= g

cossinu2 2

= m10

54

53

102 4

= 960 mThe centre of mass will hit the ground at this position. As the smaller block comes to restafter breaking, it falls down vertically and hits the ground at half of the range, i.e., at x =480 m. If the heavier block hits the ground at x

2, then

xCOM

= 21

2211

mmxmxm

960 = )m3m()x)(m3()480)(m( 2

x

2 = 1120 m Ans.

Momentum Conservation :

The total linear momentum of a system of particles is equal to the product of the total mass of the

system and the velocity of its centre of mass. P

= M cmv

extF

= dtdP

IfextF

= 0

dtdP

= 0 ; P

= constant

When the vector sum of the external forces acting on a system is zero, the total linear momentum ofthe system remains constant.

1P

+ 2P

+ 3P

+ ...............+nP

= constant.

Example 9. A shell is fired from a cannon with a speed of 100 m/s at an angle 60º with the horizontal (positive x-

direction). At the highest point of its trajectory, the shell explodes into two equal fragments. One ofthe fragments moves along the negative x-direction with a speed of 50 m/s. What is the speed of theother fragment at the time of explosion.

Solution :As we know in absence of external force the motion of centre of mass of a body remainsuneffected. Thus, here the centre of mass of the two fragments will continue to follow theoriginal projectile path. The velocity of the shell at the highest point of trajectory is

vM = ucos = 100 ×cos60º = 50 m/s.

Let v1 be the speed of the fragment which moves along the negative x-direction and the

other fragment has speed v2,. which must be along positive x-direction. Now from momen-

tum conservation, we have

mv = 2m

v1 +

2m

v2

or 2v = v2 � v

1

or v2 = 2v + v

1 = (2 × 50) + 50 = 150 m/s


Example 10. A man of mass m is standing on a platform of mass M kept on smooth ice. If the man starts movingon the platform with a speed v relative to the platform, with what velocity relative to the ice does theplatform recoil ?

Solution :Consider the situation shown in figure. Suppose the man moves at a speed w towards right andthe platform recoils at a speed V towards left, both relative to the ice. Hence, the speed of theman relative to the platform is V + w. By the question,

V + w = v, or w = v � V .............(i) w

V

/////////////////////////////////////////////////ice

Taking the platform and the man to be the system, there is noexternal horizontal force on the system. The linear momentumof the system remains constant. Initially both the man and theplatform were at rest. Thus,

0 = MV - mw or, MV = m (v � V) [Using (i)]

or, V = mM

mv

.

Example 11. A flat car of mass M is at rest on a frictionless floor with a child of mass m standing at its edge. Ifchild jumps off from the car towards right with an initial velocity u, with respect to the car, find thevelocity of the car after its jump.

Solution :Let car attains a velocity v, and the net velocity of the child with respect to earth will beu � v, as u is its velocity with respect to car.

m

M

///////////////////////////////////////////////////////////////////

uv M

m

///////////////////////////////////////////////////////////

Initially, the system was at rest, thus according to momentum conservation, momentumafter jump must be zero, as

m (u � v) = M v

v = Mm

mu

Example 12. In a free space a rifle of mass M shoots a bullet of mass m at a stationary block of mass Mdistance D away from it. When the bullet has moved through a distance d towards the blockthe centre of mass of the bullet-block system is at a distance of :

(A) ( )D d m

M m

from the block (B)

md M D

M m

from the rifle

(C) 2 d m DM

M m

from the rifle (D) (D d)

M

M m from the bullet

Ans. (A,B,D)

Solution.d x

D-d-xD

COM

Bullet of mass(m)

Block

M

Rifle

As; Mx = m(D � d � x)

x = mM

)dD(m

from the block

and x' = D � d � x

= mM

M)dD(

from the bullet.


Example 13. The centre of mass of two masses m & mmoves by distance 5x

when mass m is moved by

distance x and m is kept fixed. The ratio mm

is

(A) 2 (B) 4 (C) 1/4 (D) None of theseAns. (B)

Solution (m + m) 5x

= mx + mO

m + m = 5 m ; m = 4 m ;mm

= 4

Example 14. A uniform disc of mass �m� and radius R is placed on a smooth horizontal floor such that the plane

surface of the disc is in contact with the floor. A man of mass m/2 stands on the disc at its periphery.The man starts walking along the periphery of the disc. The size of the man is negligible as comparedto the size of the disc. Then the centre of disc.

(A) moves along a circle of radius 3R

(C) moves along a circle of radius 3R2

(C) moves along a circle of radius 2R

(D) does not move along a circle

Ans. (A)Solution. The centre of mass of man + disc shall always remain at rest. Since the man is always at periphery

of disc, the centre of disc shall always be at distance R/3 from centre of mass of two body system.Hence centre of disc moves in circle of radius R/3.

Example 15. A person P of mass 50 kg stands at the middle of a boat of mass 100 kg moving at a constantvelocity 10 m/s with no friction between water and boat and also the engine of the boat is shutoff. With what velocity (relative to the boat surface) should the person move so that the boatcomes to rest. Neglect friction between water and boat.

(A) 30 m/s towards right (B) 20 m/s towards right(C) 30 m/s towards left (D) 20 m/s towards left

Ans. (A)Solution. Momentum of the system remains conserved as no external force is acting on the system in

horizontal direction. (50 + 100) 10 = 50 × V + 100 × 0 V = 30 m/s towards right, as boat

is at rest. boatPV = 30 m/s

Example 16. Two men of masses 80 kg and 60 kg are standing on a wood plank of mass 100 kg, that hasbeen placed over a smooth surface. If both the men start moving toward each other with speeds1 m/s and 2 m/s respectively then find the velocity of the plank by which it starts moving.

Solution. Applying momentum conservation ;(80) 1 + 60 (� 2) = (80 + 60 + 100) v

v = 240

40 =

61

m/sec.


Example 17. Each of the blocks shown in figure has mass 1 kg. The rear block moves with a speed of2 m/s towards the front block kept at rest. The spring attached to the front block is light andhas a spring constant 50 N/m. Find the maximum compression of the spring. Assume, on afriction less surface

/////////////////////////////////////////////////////////////////

1kg1kgk=50N/m

Solution :Maximum compression will take place when the blocks move with equal velocity. As no netexternal horizontal force acts on the system of the two blocks, the total linear momentum willremain constant. If V is the common speed at maximum compression, we have,

(1 kg) (2 m/s) = (1 kg)V + (1 kg)Vor, V = 1 m/s.

Initial kinetic energy = 21

(1 kg) (2 m/s)2 = 2 J.

Final kinetic energy

= 21

(1 kg) (1m/s)2 + 21

(1 kg) (1 m/s)2 = 1 J

The kinetic energy lost is stored as the elastic energy in the spring.

Hence,21

(50 N/m) x2 = 2J � 1J = 1 J

or, x = 0.2 m.

Example 18. Figure shows two blocks of masses 5 kg and 2 kg placed on a frictionless surface and con-nected with a spring. An external kick gives a velocity 14 m/s to the heavier block towards thelighter one. Deduce (a) velocity gained by the centre of mass and (b) the separate velocities ofthe two blocks with respect to centre of mass just after the kick.

//////////////////////////////////////////////////////////////

5kg 2kg

Solution : (a) Velocity of centre of mass is

vcm

= 25

02145

= 10 m/s

(b) Due to kick on 5 kg block, it starts moving with a velocity 14 m/s immediately, butdue to inertia 2 kg block remains at rest, at that moment. Thus, velocity of 5 kgblock with respect to the centre of mass is v

1 = 14 � 10 = 4 m/s and the velocity

of 2 kg block w.r.t. to centre of mass is v2 = 0 � 10 = �10 m/s

Example 19. The two blocks A and B of same mass connected to a spring and placed on a smooth surface. Theyare given velocities (as shown in the figure) when the spring is in its natural length :

(A) the maximum velocity of B will be 10 m/s(B) the maximum velocity of B will be greater than 10 m/s(C) the spring will have maximum extension when A and B both stop(D) the spring will have maximum extension when both move towards left.

Ans. (A)


Solution. Suppose B moves with a velocity more than 10 m/s a should move at a velocity greater than 5m/s and increases the overall energy which is not possible since there is no external forceacting on the system. Hence B should move with a maximum velocity 10 m/s.Also both A and B can never stop so as to keep the momentum constant.Also both A and B can never move towards left simultaneously for momentum remaining con-served.Hence only (A) is correct.

IM PU LS E

Impulse of a force F

acting on a body for the time interval t = t1 to t = t

2is defined as :-

= 2

1

t

tdtF

= dtF

= dtvd

m

dt = vdm

= m( 12 v - v

) = PÄ

= change in momentum due to force F

Also, sRe

= 2

1

t

tsRe dtF

= PÄ

(impulse - momentum theorem)

Note: Impulse applied to an object in a given time interval can also becalculated from the area under force time (F-t) graph in the sametime interval.

Instantaneous Impulse :There are many cases when a force acts for such a short time that the effect is instantaneous,e.g., a bat striking a ball. In such cases, although the magnitude of the force and the time forwhich it acts may each be unknown but the value of their product (i.e., impulse) can be known bymeasuring the initial and final momenta. Thus, we can write.

if PPPdtF

Important Points :(1) It is a vector quantity.(2) Dimensions = [MLT�1](3) S unit = kg m/s(4) Direction is along change in momentum.(5) Magnitude is equal to area under the F-t. graph.

(6)

= dtF

= avF

dt = tFav

(7) It is not a property of a particle, but it is a measure of the degree to which an external forcechanges the momentum of the particle.

Example 20. The hero of a stunt film fires 50 g bullets from a machine gun, each at a speed of 1.0 km/s. Ifhe fires 20 bullets in 4 seconds, what average force does he exert against the machine gunduring this period.

Solution : The momentum of each bullet= (0.050 kg) (1000 m/s) = 50 kg-m/s.The gun has been imparted this much amount of momentum by each bullet fired. Thus, the rateof change of momentum of the gun

= s420)s/mkg50(

= 250 N.

In order to hold the gun, the hero must exert a force of 250 N against the gun.


Impulsive force :

A force, of relatively higher magnitude and acting for relatively shorter time, is called impulsive force.An impulsive force can change the momentum of a body in a finite magnitude in a very short timeinterval. Impulsive force is a relative term. There is no clear boundary between an impulsive and Non-Impulsive force.

Note: Usually colliding forces are impulsive in nature.Since, the application time is very small, hence, very little motion of the particle takes place.

Important points :1. Gravitational force and spring force are always non-Impulsive.2. Normal, tension and friction are case dependent.3. An impulsive force can only be balanced by another impulsive force.

1. Impulsive Normal : In case of collision, normal forces at the surface of collision are always impulsive

eg. Ni = Impulsive; N

g = Non-impulsive

N1

N2

/////////////////////////////

Both normals are Impulsive

//////////////////////////////////////////////////////////////////

N2

N3

N1

N1

N1, N

3 = Impulsive; N

2 = non-impulsive

Both normals are Impulsive


2. Impulsive Friction : If the normal between the two objects is impulsive, then the friction between thetwo will also be impulsive.

Friction at both surfaces is impulsive

Friction due to N2 is non-impulsive and due to N

3 and N

1 are impulsive

3. Impulsive Tensions : When a string jerks, equal and opposite tension act suddenly at eachend. Consequently equal and opposite impulses act on the bodies attached with the string in thedirection of the string. There are two cases to be considered.(a) One end of the string is fixed :

The impulse which acts at the fixed end of the string cannot change the momentum of thefixed object there. The object attached to the free end however will undergo a change inmomentum in the direction of the string. The momentum remains unchanged in a directionperpendicular to the string where no impulsive forces act.

(b) Both ends of the string attached to movable objects :In this case equal and opposite impulses act onthe two objects, producing equal and oppositechanges in momentum. The total momentum of thesystem therefore remains constant, although themomentum of each individual object is changed inthe direction of the string. Perpendicular to the stringhowever, no impulse acts and the momentum of eachparticle in this direction is unchanged.

///////////////////

All normal are impulsive but tension T is impulsive only for the ball A

TT is Impulsive

T is non-impulsive

T is non-impulsive

A

B

C

For this example:In case of rod, Tension is always impulsive and in case of spring, Tension is always non-impulsive.

Example 21. Two identical block A and B, connected by a massless string are placed on a frictionless horizontalplane. A bullet having same mass, moving with speed u strikes block B from behind as shown. If thebullet gets embedded into the block B then find :

A Bm u

C

mm

/////////////////////////////////////////////////////

(a) The velocity of A,B,C after collision.(b) Impulse on A due to tension in the string(c) Impulse on C due to normal force of collision.(d) Impulse on B due to normal force of collision.


Solution : (a) By Conservation of linear momentum v = 3u

(b)3

mudtT

(c)

u

3

umdtN =

3mu2

(d) dt)TN( = 3

mudtTNdt

3mu2

dtN

COLLISION OR IMPACT

Collision is an event in which an impulsive force acts between two or more bodies for a short time,which results in change of their velocities.

Note : (a) In a collision, particles may or may not come in physical contact.

(b) The duration of collision, t is negligible as compared to the usual time intervals of observationof motion.

(c) In a collision the effect of external non impulsive forces such as gravity are not taken into aaccount as due to small duration of collision (t) average impulsive force responsible for colli-sion is much larger than external forces acting on the system.

The collision is infact a redistribution of total momentum of the particles. Thus, law ofconservation of linear momentum is indispensable in dealing with the phenomenon of collisionbetween particles.

Line of Impact

The line passing through the common normal to the surfaces in contact during impact is called line ofimpact. The force during collision acts along this line on both the bodies.Direction of Line of impact can be determined by:(a) Geometry of colliding objects like spheres, discs, wedge etc.(b) Direction of change of momentum.If one particle is stationary before the collision then the line of impact will be along its motion aftercollision.

Classification of collisions

(a) On the basis of line of impact (i) Head-on collision : If the velocities of the colliding particles are along the same line before

and after the collision.

(ii) Oblique collision : If the velocities of the colliding particles are along different lines beforeand after the collision.

(b) On the basis of energy :

(i) Elastic collision : In an elastic collision, the colliding particles regain their shape and sizecompletely after collision. i.e., no fraction of mechanical energy remains stored as defor-mation potential energy in the bodies. Thus, kinetic energy of system after collision is equalto kinetic energy of system before collision. Thus in addition to the linear momentum, ki-netic energy also remains conserved before and after collision.


(ii) Inelastic collision : In an inelastic collision, the colliding particles do not regain their shapeand size completely after collision. Some fraction of mechanical energy is retained by thecolliding particles in the form of deformation potential energy. Thus, the kinetic energy ofthe particles after collision is not equal to that of before collision. However, in the absenceof external forces, law of conservation of linear momentum still holds good.

(iii) Perfectly inelastic : If velocity of separation along the line of impact just after collision be-comes zero then the collision is perfectly inelastic. Collision is said to be perfectlyinelastic if both the particles stick together after collision and move with same velocity,

Note : Actually collision between all real objects are neither perfectly elastic nor perfectly inelastic, its inelas-tic in nature.

Examples of l ine of impact and coll isions based on l ine of impact

(i) Two balls A and B are approaching each other such that their centres are moving along line CD.

Head on Collision

(ii) Two balls A and B are approaching each other such that their centre are moving along dotted lines as shownin figure.

Oblique Collision

(iii) Ball is falling on a stationary wedge.

Oblique Collision

COEFFICIENT OF RESTITUTION (e)

The coefficient of restitution is defined as the ratio of the impulses of reformation anddeformation of either body.


e = ndeformatioofpulseImnreformatioofpulseIm

=

dtF

dtF

d

r

= impactoflinealongapproachofVelocityimpactoflinealongseperationofVelocity

The most general expression for coefficient of restitution is

e = impact of line along contact of point of approach ofvelocity impact of line along contact of points of separation of velocity

Example for calculation of eTwo smooth balls A and B approaching each other such that their centres are moving along line CD inabsence of external impulsive force. The velocities of A and B just before collision be u

1 and u

2 respectively.

The velocities of A and B just after collision be v1 and v

2 respectively.

Line of impact

u1 u2

Just Before collision

C D A B

Line of impact

v1 v2 Just After collision

C D A B

ND

m1

ND

m2

u1

u2

NR

m1

NR

m2

vv

m1 m2

v1

v2

Deformation Reformation

u > u1 2 v < v1 2

Fext

= 0 momentum is conserved for the system. m

1u

1 + m

2 u

2 = (m

1 + m

2)v = m

1v

1 + m

2v

2

v = 21

2211

mmumum

=

21

2211

mmvmvm

.......(1)

Impulse of Deformation :J

D = change in momentum of any one body during deformation.

= m2 (v � u

2) for m

2

= m1 (�v + u

1) for m

1

Impulse of Reformation :J

R = change in momentum of any one body during Reformation.

= m2 (v

2 � v) for m

2

= m1 (v � v

1) for m

1

e = )(

)(

D

R

JnDeformatio of Impulse

JnReformatio of Impulse

= 21

12

uuvv

= impactoflinealongapproachofVelocity

impactoflinealongseparationofVelocity


Note : e is independent of shape and mass of object but depends on the material.The coefficient of restitution is constant for a pair of materials.(a) e = 1 Impulse of Reformation = Impulse of Deformation

Velocity of separation along the LOI = Velocity of approach along the LOI Kinetic energy of particles after collision may be equal to that of before collision.

Collision is elastic.(b) e = 0 Impulse of Reformation = 0

Velocity of separation along the LOI= 0

Kinetic energy of particles after collision is not equal to that of before collision.

Collision is perfectly inelastic .(c) 0 < e < 1 Impulse of Reformation < Impulse of Deformation

Velocity of separation along the LOI < Velocity of approach along the LOI Kinetic energy of particles after collision is not equal to that of before collision.

Collision is Inelastic.

Note : In case of contact collisions e is always less than unity. 0 e 1

Important Point :In case of elastic collision, if rough surface is present thenk

f < k

i(because friction is impulsive)

Where, k is Kinetic Energy.

/////////////////////////////////////Rough

A particle �B� moving along the dotted line collides with a rod also in state of motion as shown in the figure.

The particle B comes in contact with point C on the rod.To write down the expression for coefficient of restitution e, we first draw the line of impact. Then we resolvethe components of velocities of points of contact of both the bodies along line of impact just before and justafter collision.

Then e = x2x1

x1x2

uuvv

Coll ision in one dimension (Head on)

m1 m2

u1u2

(a)Before Collision

m1 m2

v1v2

(b)After Collision

u1 > u

2 v

2 > v

1

e = 21

12

uuvv

(u

1 � u

2)e = (v

2 � v

1)


By momentum conservation,m

1u

1 + m

2u

2 = m

1v

1 + m

2v

2

v2 = v

1 + e(u

1 � u

2)

and v1 =

21

2122211

mm)uu(emumum

v2 =

21

2112211

mm)uu(emumum

Special Case :(1) e = 0

v1 = v

2

for perfectly inelastic collision, both the bodies, move with same vel. after collision.

(2) e = 1and m

1 = m

2 = m,

we get v1 = u

2 and v

2 = u

1

i.e., when two particles of equal mass collide elastically and the collision is head on, they ex-change their velocities., e.g.

m m2m/s

Before Collision

v =01

(3) m1 >> m

2

m1 + m

2 m

1 and 0

mm

1

2

v1 = u

1No change

and v2 = u

1 + e(u

1 � u

2)

Now If e = 1v

2 = 2u

1 � u

2

Example 22. Two identical balls are approaching towards each other on a straight line with velocity 2 m/sand 4 m/s respectively. Find the final velocities, after elastic collision between them.

mm 4m/s2m/s

Solution :The two velocities will be exchanged and the final motion is reverse of initial motion for both.

m m4m/s 2m/s


Example 23. Three balls A, B and C of same mass �m� are placed on a frictionless horizontal plane in a

straight line as shown. Ball A is moved with velocity u towards the middle ball B. If all thecollisions are elastic then, find the final velocities of all the balls.

A B Cu

m m m

//////////////////////////////////////////

Solution :A collides elastically with B and comes to rest but B starts moving with velocity u

A B Cu

m m m

//////////////////////////////////////////

After a while B collides elastically with C and comes to rest but C starts moving with velocity u

A B Cu

m m m

//////////////////////////////////////////

Final velocitiesV

A = 0;

VB = 0 and V

C = u Ans.

Example 24. Four identical balls A, B, C and D are placed in a line on a frictionless horizontal surface. A andD are moved with same speed �u� towards the middle as shown. Assuming elastic collisions,

find the final velocities.

A B Cu

D/////////////////////////////////////////////////////

u

Solution :A and D collides elastically with B and C respectively and come to rest but B and C startsmoving with velocity u towards each other as shown

A B Cu

D/////////////////////////////////////////////////////

u

B and C collides elastically and exchange their velocities to move in opposite directions

A B Cu

D/////////////////////////////////////////////////////

u

Now, B and C collides elastically with A and D respectively and come to rest but A and D startsmoving with velocity u away from each other as shown

A B Cu

D/////////////////////////////////////////////////////

u

Final velocities VA = u ( ); V

B = 0; V

C = 0 and V

D = u ( ) Ans.

Example 25. Two particles of mass m and 2m moving in opposite directions on a frictionless surface collideelastically with velocity v and 2v respectively. Find their velocities after collision, also find thefraction of kinetic energy lost by the colliding particles.

m 2m2v v

Solution :Let the final velocities of m and 2m be v

1 and v

2 respectively as shown in the figure:

m 2mv2

By conservation of momentum:m(2v) + 2m(�v) = m(v

1) + 2m (v

2)

or 0 = mv1 + 2mv

2

or v1 + 2v

2 = 0 .........(1)


and since the collision is elastic:v

2 � v

1 = 2v �(�v)

or v2 � v

1 = 3v .........(2)

Solving the above two equations, we get,v

2 = v and v

1 = �2v Ans.

i.e., the mass 2m returns with velocity v while the mass m returns with velocity 2v in thedirection shown in figure:

m 2mv

The collision was elastic therefore, no kinetic energy is lost, KE loss = KEi - KE

f

or,

2222 v)m2(21

)v2(m21

)v)(m2(21

)v2(m21

= 0

Example 26. On a frictionless surface, a ball of mass m moving at a speed v makes a head on collision withan identical ball at rest. The kinetic energy of the balls after the collision is 3/4th of the original.Find the coefficient of restitution.

Solution :As we have seen in the above discussion, that under the given conditions :

By using conservation of linear momentum and equation of e, we get,

v2

e1'v1

and v

2e1

'v2

Given that if K43

K or21

mv1�2 +

21

mv2�2 =

43

2mv2

1

Substituting the value, we get

2

2e1

+

2

2e1

=

43

or e = 2

1Ans.

Example 27. A block of mass 2 kg is pushed towards a very heavy object moving with 2 m/s closer to theblock (as shown). Assuming elastic collision and frictionless surfaces, find the final velocitiesof the blocks.

2kg

veryheavyobject

10m/s

2m/s

/////////////////////////////////////////////////////////////////////

Solution :Let v

1 and v

2 be the final velocities of 2kg block and heavy object respectively then,

v1 = u

1 + 1 (u

1 � u

2) = 2u

1 � u

2 = �14 m/s

v2 = �2m/s


Example 28. A ball is moving with velocity 2 m/s towards a heavy wall moving towards the ball with speed1m/s as shown in fig. Assuming collision to be elastic, find the velocity of the ball immediatelyafter the collision.

Solution :The speed of wall will not change after the collision. So, let v be the velocity of the ballafter collision in the direction shown in figure. Since collision is elastic (e = 1),

separation speed = approach speedor v � 1 = 2 + 1

or v = 4 m/s Ans.

Example 29. Two balls of masses 2 kg and 4 kg are moved towards each other with velocities 4 m/s and2 m/s respectively on a frictionless surface. After colliding the 2 kg ball returns back withvelocity 2m/s.

2kg 4kg4m/s 2m/s

//////////////////////////////////////////// ////////////////////////////////////////

2kg2m/s

4kg v2

Just before collision Just after collision

Then find:(a) velocity of 4 kg ball after collision(b) coefficient of restitution e.(c) Impulse of deformation J

D.

(d) Maximum potential energy of deformation.(e) Impulse of reformation J

R.

Solution :(a) By momentum conservation,

2(4) � 4(2) = 2(�2) + 4(v2) v

2 = 1 m/s

(b) e = approach ofvelocity separation of velocity

= 63

)2(4)2(1

= 0.5

(c) At maximum deformed state, by conservation of momentum, common velocity is v = 0.J

D = m

1(v � u

1) = m

2(v � u

2) = 2(0 � 4) = �8 N -s

= 4(0 � 2) = � 8 N - s

or = 4(0 � 2) = � 8 N - s

(d) Potential energy at maximum deformed state U = loss in kinetic energy during deformation.

or U =

222

211 um

21

um21

� 21

(m1 + m

2)v2

=

22 )2(421

)4(221

� 21

(2 + 4) (0)2

or U = 24 Joule


(e) JR = m

1(v

1 � v) = m

2 (v � v

2) = 2 (�2 � 0) = �4 N-s

or = 4(0 � 1) = �4 N-s

or e = D

R

JJ

JR

= eJD

= (0.5) (�8) = �4 N-s

Coll ision in two dimension (obl ique)

1. A pair of equal and opposite impulses act along common normal direction. Hence, linear momen-tum of individual particles do change along common normal direction. If mass of the collidingparticles remain constant during collision, then we can say that linear velocity of the individualparticles change during collision in this direction.

2. No component of impulse act along common tangent direction. Hence, linear momentum or linearvelocity of individual particles (if mass is constant) remain unchanged along this direction.

3. Net impulse on both the particles is zero during collision. Hence, net momentum of both theparticles remain conserved before and after collision in any direction.

4. Definition of coefficient of restitution can be applied along common normal direction, i.e., alongcommon normal direction we can applyRelative speed of separation = e (relative speed of approach)

Example 30. A ball of mass m hits a floor with a speed v0 making an angle of incidence a with the normal. The

coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of theball.

Solution :The component of velocity v

0 along common tangential direction v

0 sin will remain un-

changed. Let v be the component along common normal direction after collision. Applying,Relative speed of separation = e (Relative speed of approach)along common normal direction, we get

v = ev0 cos

v sin0

v'

v (= ev cos )0

Thus, after collision components of velocity v� are v0 sin and ev

0 cos

20

20 )cosev()sinv('v Ans.

and tan =

cosev

sinv

0

0

or tan = e

tan Ans.

Note : For elastic collision, e = 1 v� = v

0and =


Example 31. A ball of mass m makes an elastic collision with another identical ball at rest. Show that if thecollision is oblique, the bodies go at right angles to each other after collision.

Solution :In head on elastic collision between two particles, they exchange their velocities. In thiscase, the component of ball 1 along common normal direction, v cos

becomes zero after collision, while that of 2 becomes v cos . While the componentsalong common tangent direction of both the particles remain unchanged. Thus, the com-ponents along common tangent and common normal direction of both the balls in tabularform are given below.

Ball

Before collision After collision Before collision After collision1 v sin v sin v cos

2 v cos

Component along common tangent direction

Component along common normal direction

From the above table and figure, we see that both the balls move at right angle aftercollision with velocities v sin and v cos .

Note : When two identical bodies have an oblique elastic collision, with one body at rest before collision, then

the two bodies will go in directions.

VARIABLE MASS SYSTEM :

If a mass is added or ejected from a system, at rate kg/s and relative velocity relv

(w.r.t. the system),

then the force exerted by this mass on the system has magnitude relv

.

Thrust Force ( tF

)

dtdm

vF relt

Suppose at some moment t = t mass of a body is m and its velocity is v

. After some time at t =

t + dt its mass becomes (m � dm) and velocity becomes vdv

. The mass dm is ejected with

relative velocity rv

. Absolute velocity of mass �dm � is therefore ( v

+ rv

). If no external forces are

acting on the system, the linear momentum of the system will remain conserved, or

fi PP

or m v

= (m � dm) ( v

+ d v

) + dm ( v

+ rv

)

or m v

= m v

+ md v

� (dm) v

� (dm) (d v

) + (dm) v

+ rv

dm


The term (dm) (d v

) is too small and can be neglected.

md v

= � rv

dm or

dtvd

m

=

dtdm

v r

Here,

dtvd

m

= thrust force tF

and �dtdm

= rate at which mass is ejecting or

dtdm

vF rt

Problems related to variable mass can be solved in following four steps1. Make a list of all the forces acting on the main mass and apply them on it.

2. Apply an additional thrust force tF

on the mass, the magnitude of which is

dtdm

v r

and direc-

tion is given by the direction of rv

in case the mass is increasing and otherwise the direction of

� rv

if it is decreasing.

3. Find net force on the mass and apply

dtvd

mFnet

(m = mass at the particular instant)

4. Integrate it with proper limits to find velocity at any time t.

Rocket propulsion :

Let m0 be the mass of the rocket at time t = 0. m its mass at any time t and v its velocity at that

moment. Initially, let us suppose that the velocity of the rocket is u.

Further, let

dtdm

be the mass of the gas ejected per unit time and vr the exhaust velocity of the

gases with respect to rocket. Usually

dtdm

and vr are kept constant throughout the journey of

the rocket. Now, let us write few equations which can be used in the problems of rocket propul-sion. At time t = t,

1. Thrust force on the rocket Ft = v

r

dtdm

(upwards)

2. Weight of the rocket W = mg (downwards)3. Net force on the rocket F

net = F

t � W (upwards)


or Fnet

= vr

dtdm

�mg

4. Net acceleration of the rocket a = mF

ordtdv

= mvr

dtdm

�g

or dv = mvr

dm � g dt

or v

urvdv

m

m0 mdm

�g t

0dt

Thus, v = u � gt + vr n

m

m0...(i)

Note : 1. Ft = v

r

dtdm

is upwards, as vr is downwards and

dtdm

is negative.

2. If gravity is ignored and initial velocity of the rocket u = 0, Eq. (i) reduces to v = vr ln

m

m0 .

Example 32. A rocket, with an initial mass of 1000 kg, is launched vertically upwards from rest under gravity.The rocket burns fuel at the rate of 10 kg per second. The burnt matter is ejected verticallydownwards with a speed of 2000 ms�1 relative to the rocket. If burning stopsafter one minute.Find the maximum velocity of the rocket. (Take g as at 10 ms�2)

Solution :Using the velocity equation

v = u � gt + vr ln

m

m0

Here u = 0, t = 60s, g = 10 m/s2, vr = 2000 m/s, m

0 = 1000 kg

and m = 1000 � 10 × 60 = 400 kg

We get v = 0 � 600 + 2000 ln

4001000

or v = 2000 ln 2.5 � 600

The maximum velocity of the rocket is 200(10 ln 2.5 � 3) = 1232.6 ms�1 Ans.

Example 33. A flat car of mass m0 starts moving to the right due to a

constant horizontal force F. Sand spills on the flat car froma stationary hopper. The rate of loading is constant andequal to kg/s . F ind the t ime dependence of thevelocity and the acceleration of the flat car in the processof loading. The friction is negligibly small.

F

m0

Solution : Initial velocity of the flat car is zero. Let v be its velocity at time t and m its mass at thatinstant. Then


At t = 0, v = 0 and m = m0 at t = t, v = v and m = m

0 + t

Here, vr = v (backwards)

dtdm

=

Ft = v

r

dtdm

= v (backwards)

Net force on the flat car at time t is Fnet

= F � Ft

or m dtdv

= F � v ....(i)

or (m0 + t)

dtdv

=F � v or

v

0 vFdv

=

t

0 0 tmdt

�

1 [n (F � v)]

0v =

1[n (m

0 + t)]

0t

n

vFF

= n

0

0

m

tm

vF

F

= 0

0

m

tm or v =

tmFt

0 Ans.

From Eq. (i),dtdv

= acceleration of flat car at time t

or = m

vF

a =

tm

tmtF

F

0

0 or a = 20

0

)tm(

Fm

Ans.

Example 34. A cart loaded with sand moves along a horizontal floor due to a constant force F coinciding indirection with the cart�s velocity vector. In the process sand spills through a hole in the bottom

with a constant rate kg/s. Find the acceleration and velocity of the cart at the moment t, if atthe initial moment t = 0 the cart with loaded sand had the mass m

0 and its velocity was equal

to zero. Friction is to be neglected.Solution : In this problem the sand spills through a hole in the bottom of the cart. Hence, the relative

velocity of the sand vr will be zero because it will acquire the same velocity as that of the

cart at the moment.v

r = 0

vF

v

m

Thus, Ft = 0

dt

dmvFas rt

and the net force will be F only. F

net = F

or m

dt

dv = F ....(i)

But here m = m0 � t (m

0 � t)

dtdv

= F


or v

0dv =

t

0 0 tmdtF

v = t00 )tm(n

F

or v =

Fln

tm

m

0

0Ans.

From eq. (i), acceleration of the cart

a = dtdv

= mF

or a = tmF

0 Ans.

LINEAR MOMENTUM CONSERVATION IN PRESENCE OF EXTERNAL FORCE.

extF

= dtPd

extF

dt = Pd

Pd

= extF

)mpulsive

dt

If extF

)mpulsive

= 0

Pd

= 0

or P

is constant

Note: Momentum is conserved if the external force present is non-impulsive. eg. gravitation or spring force

Example 35. Two balls are moving towards each other on a vertical line collides with each other as shown.Find their velocities just after collision.

3m/s

4m/s

2kg

4kg

Solution : Let the final velocity of 4 kg ball just after collision be v. Since,external force is gravitational which is non - impulsive, hence,linear momentum will be conserved.Applying linear momentum conservation:

2kg

4kg

v

4m/s

2(�3) + 4(4) = 2(4) + 4(v)

or v = 21

m/s


Example 36. A bullet of mass 50g is fired from below into the bob of mass450g of a long simple pendulum as shown in figure. The bulletremains inside the bob and the bob rises through a height of1.8 m. Find the speed of the bullet. Take g = 10 m/s2.

/////////////////////

v

Solution : Let the speed of the bullet be v. Let the common velocity of the bullet and the bob, after thebullet is embedded into the bob, is V. By the principle of conservation of the linear momentum,

V = kg05.0kg45.0v)kg05.0(

=

10v

The string becomes loose and the bob will go up with a deceleration of g = 10 m/s2. As it comesto rest at a height of 1.8 m, using the equation v2 = u2 + 2ax,

1.8 m = 2

2

s/m102

)10/v(

or, v = 60 m/s.

Problem 1. Three particles of masses 0.5 kg, 1.0 kg and 1.5 kg are placed at the three corners of a rightangled triangle of sides 3.0 cm, 4.0 cm and 5.0 cm as shown in figure. Locate the centre ofmass of the system.

Solution : 1.5 kg

0.5 kg1.0 kg(A) (B)

(C)

x

y

3cm5cm

4cm

taking x and y axes as shown.coordinates of body A = (0,0)coordinates of body B = (4,0)coordinates of body C = (0,3)

x - coordinate of c.m. = CBA

CCBBAA

mmmrMxmxm

= 5.10.15.0

05.140.105.0

= 34

kgcm

= cm = 1.33cm

similarly y - wordinates of c.m. = 5.10.15.0

35.100.105.0

=

35.4

= 1.5 cm

So, certre of mass is 1.33 cm right and 1.5 cm above particle A.

Problem 2. A block A (mass = 4M) is placed on the top of a wedge B ofbase length l (mass = 20 M) as shown in figure. When thesystem is released from rest. Find the distance moved by thewedge B till the block A reaches at lowest end of wedge. Assume all surfaces are frictionless.


Solution : Initial position of centre of mass

= BB

AABB

MMMXMX

=

M24M4.M20.XB

= 6

X5 B

Final position of centre of mass

= M24

Mx4M20)xX( B =

6x)xX(5 B

since there is no horizontal force on systemcentre of mass initially = centre of mass finally.5X

B + = 5X

B + 5x + x

= 6x

6

x

Problem 3. An isolated particle of mass m is moving in a horizontal xy plane, along x-axis. At a certainheight above ground, it suddenly explodes into two fragments of masses m/4 and 3m/4. Aninstant later, the smaller fragment is at y = + 15 cm. Find the position of heavier fragment atthis instant.

Solution : As particle is moving along x-axis, so, y-coordinate of COM is zero.

YM M =

4MY

4M

+ 4M3Y

4M3

0 × M = 15

4M

+ 4M3Y

4M3

cm54

Y M3

Problem 4. A shell at rest at origin explodes into three fragments of masses 1 kg, 2 kg and m kg. Thefragments of masses 1 kg and 2 kg fly off with speeds 12 m/s along x-axis and 8 m/s along y-axis respectively. If m kg flies off with speed 40 m/s then find the total mass of the shell.

Solution : As initial velocity = 0, Initial momentum = (1 + 2 + m) × 0 = 0

Finally, let velocity of M =

V . We know |

V | = 40 m/s.

Initial momentum = final momentum .

0 = 1 × 12 i� + 2 × 8 j� + m

V

V = �m

)j�16i�12(

|

V | = 2

22

m

)16()12( =

m1

22 )16()12( = 40 {given}

m = 40

)16()12( 22

= 0.5 kg

Total mass = 1 + 2 + 0.5 = 3.5 kg

Problem 5. A block moving horizontally on a smooth surface with a speed of 20 m/s bursts into two equalparts continuing in the same direction. If one of the parts moves at 30 m/s, with what speeddoes the second part move and what is the fractional change in the kinetic energy of thesystem.

Solution :


Applying momentum conservation ;

m × 20 = 2m

V + 2m

× 30 20 = 2V

+ 15

So, V = 10 m/s

initial kinetic energy = 21

m × (20)2 = 200 m

final kinetic energy = 21

. 2m

. (10)2 + 21

×2m

(30)2 = 25 m + 225 m = 250 m

fractional change in kinetic energy = E.K initial

)E K. initial()E K. final( =

m200m200m250

= 41

Problem 6. A block at rest explodes into three equal parts. Two parts starts moving along X and Y axesrespectively with equal speeds of 10 m/s. Find the initial velocity of the third part.

Solution :Let total mass = 3 m, initial linear momentum = 3m × 0

Let velocity of third part =

VUsing conservation of linear momentum :

m × 10 i� + m × 10 j� + m

V = 0

So,

V = (� 10 i� � 10 j� ) m/sec.

|

V | = 22 )10()10( = 10 2 , making angle 135o below x-axis

Problem 7. Blocks A and B have masses 40 kg and 60 kg respectively. They are placed on a smoothsurface and the spring connected between them is stretched by 1.5m. If they are releasedfrom rest, determine the speeds of both blocks at the instant the spring becomesunstretched.

Solution :

Let, both block start moving with velocity V1 and V

2 as shown in figure

Since no horizontal force on system so, applying momentumconservation

0 = 40 V1 � 60 V

2 21 V3V2 ........(1)

Now applying energy conservation, Loss in potential energy = gain in kinetic energy

21

kx2 = 21

m1V

12 +

21

m2V

22

21

× 600 × (1.5)2 = 21

× 40 × V1

2 + 21

× 60 × V2

2 .......(2)

Solving euation (1) and (2) we get,V

1 = 4.5 m/s, V

2 = 3 m/s.

Problem 8. Find the mass of the rocket as a function of time, if it moves with a constant accleration a,in absence of external forces. The gas escaps with a constant velocity u relative to therocket and its initial mass was m

0.


Solution : Using, Fnet = Vrel

dt

dm�

Fnet

= � udtdm

.......(1)

Fnet

= ma ......(2)Solving equation (1) and (2)

ma = � u dtdm

m

momdm

= t

ouadt�

nom

m =

uat

omm

= u/ate uat

emm 0

Ans.

Problem 9. A ball is approaching to ground with speed u. If the coefficient of restitution ise then find out: m

u

/////////////////////////

(a) the velocity just after collision.

(b) the impulse exerted by the normal due to ground on the ball.Solution :

e = approah of velocity

separation ofvelocity =

uv

(a) velocity after collision = V = eu ........(1)(b) Impulse exerted by the normal due to ground on the ball = change in momentum of ball.

= {final momentum} � {initial momentum}

= {m v} � {� mu}

= mv + mu = m {u + eu} = mu {1 + e} Ans.

RESONANCE NIT_CENTRE OF MASS - 36

PART - I : OBJECTIVE QUESTIONSHkkx - I : oLrqfu"B iz'u ¼OBJECTIVE QUESTIONS½

SECTION (A) : CALCULATION OF CENTRE OF MASS nzO;eku d sUnz d h x.kukA-1. The centre of mass of a body : oLrq d k nzO;eku d sUnz :

(1) Lies always at the geometrical centre (ges'kk T;kferh d sUnz ij fLFkr gksrk gSaA)

(2) Lies always inside the body (ges'kk oLrq d s vUnj gksrk gS)(3) Lies always outside the body (ges'kk oLrq d s ckgj gksrk gS)(4*) Lies within or outside the body (oLrq d s vUnj ;k ckgj gksrk gSA)

Sol. Centre of mass is a point which can lie within or outside the body.nzO;eku d sUnz og fcUnq gS t ks fd oLrq d s vanj ;k ckgj gks ld rk gSA

A-2. A uniform solid cone of height 40 cm is shown in figure. The distance of centre of mass of the cone frompoint B (centre of the base) is :,d leku Bksl 'kad q ft ld h Å ¡pkbZ 40 lseh- gS] fp=kkuqlkj fn[kk;k x;k gSA fcUnq B ls nzO;eku d sUnz d h nwjh¼vk/kkj d s d sUnz ls½ gksxh &

(1) 20 cm (2) 10/3 cm (3) 20/3 cm (4*) 10 cm

Sol. Centre of mass are rcm

= 4h

= 4

40 = 10 cm

nzO;eku d sUnz rcm

= 4h

= 4

40 = 10 cm

A-3. A body has its centre of mass at the origin. The x-coordinates of the particles(1) may be all positive(2) may be all negative(3) must be all non-negative(4*) may be positive for some particles and negative in other particles,d oLrq dk nzO;eku dsUnz ewy&fcUnq ij gSA d.kksa ds x�funsZ'kkad &(1) lkjs /kukRed gks ldrs gSA (2) lkjs _ .kkRed gks ldrs gSA(3) lkjs v_ .kkRed gksus gh pkfg,A (4*) dqN d.kksa ds fy;s /kukRed o dqN d.kksa ds fy;s _ .kkRed gks ldrs gSA

Sol. self explaintory


SECTION (B) : MOTION OF CENTRE OF MASS nzO;eku dsUnz dh xfrB-1. A bomb travelling in a parabolic path under the effect of gravity, explodes in mid air. The centre of mass

of fragments will:,d ce xq: Roh; izHkko esa ijoy; iFk ij xfr d jrk gS] chp gok esa foLQksfVr gks t krk gSA Vqd M+ks d k nzO;eku dsUnz(1) Move vertically upwards and then downwards

igys lh/ks Å ij d h vksj fQj uhps d h xfr d jsxk(2) Move vertically downwardslh/ks uhps d h vkSj xfr d jsxkA(3) Move in irregular pathvleku iFk ij xfr d jsxkA(4*) Move in the parabolic path which the unexploded bomb would have travelled.ml ijoy; iFk ij xfr d jsxk ft l ij ce QVus ls igys d j jgk gksrk gSA

B-2. If a ball is thrown upwards from the surface of earth and during upward motion :vxj iF̀oh d h lrg ls xsan Å ij d h vksj Qsad h t krh gS] rks Å ij d h vksj xfr d s nkSjku �(1) The earth remains stationary while the ball moves upwardsiF̀oh : d h gqbZ jgrh gS] t cfd xsan Å ij d h vksj xfr d jrh gSA(2) The ball remains stationary while the earth moves downwardsxasn : d h jgrh gS t cfd iF̀oh uhps d h vksj xfr d jrh gS(3) The ball and earth both moves towards each otherxsan rFkk iF̀oh nksuksa ,d nwljs d h vkSj xfr d jrs gSA(4*) The ball and earth both move away from each otherxsan rFkk iF̀oh nksuksa ,d nwljs ls nwj d h vksj xfr d jrs gSA

Sol. In abscence of external force both move away from each other to keep the centre of mass at rest.cká cy d h vuqifLFkfr esa nksuksa ,d nwljs ls nwj d h vksj xfr d jsxsa ft lls nzO;eku d sUnz fojke ij jgsA

B-3. Internal forces can change :

vkUrfjd cy ifjofrZr dj ldrk gS &(1) the linear momentum but not the kinetic energy of the system.(2*) the kinetic energy but not the linear momentum of the system.(3) linear momentum as well as kinetic energy of the system.(4) neither the linear momentum nor the kinetic energy of the system.(1) fudk; dk jSf[kd laosx fdUrq xfrt Å tkZ ugha (2*) fudk; dh xfrt Å tkZ fdUrq jSf[kd laosx ugha(3) fudk; dk jSf[kd laosx o lkFk gh xfrt Å tkZ Hkh (4) fudk; dk u rks jSf[kd laosx] u gh xfrt Å tkZ

Sol. Internal forces canot change velocity but can do work.vkarfjd cy osx d ks ifjofrZr ugh d jrs gSa ysfd u d k;Z d j ld rs gSA

B-4. If the external forces acting on a system have zero resultant, the centre of mass(1) must not move (2) must accelerate(3*) may move (4) may accelerate;fn fdlh fudk; ij yx jgs ckº; cyksa dk ifj.kkeh 'kwU; gks] nzO;eku dsUnz &(1) xfr ugha djsxkA (2) Rofjr gksxkA (3*) xfr dj ldrk gSA (4) Rofjr gks ldrk gSA

Sol. If initial velocity of system is not zero then centre of mass moves with constant velocity.;fn fud k; d k izkjfEHkd osx 'kwU; ugh gS rks nzO;eku d sUnz fu;r osx ls xfr d j ld rk gSA

B-5. Two balls are thrown in air. The acceleration of the centre of mass of the two balls while in air (neglectair resistance)(1) depends on the direction of the motion of the balls(2) depends on the masses of the two balls(3) depends on the speeds of the two balls(4*) is equal to gnks xsans ,d lkFk ok;q esa Qsadh x;h gaSA nksuksa xsanksa dk nzO;eku dsUnz dk Roj.k tc os ok;q esa gS &(1) xsanks dh xfr dh fn'kk ij fuHkZj djrk gSA (2) nksuksa xsanks ds nzO;ekuksa ij fuHkZj djrk gSA(3) nksuksa xsanksa dh pkyksa ij fuHkZj djrk gSA (4) g ds cjkcj gSA

Sol. acm

= 21

21

mmgmgm

= g


B-6. Two particles of mass 1 kg and 0.5 kg are moving in the same direction with speed of 2m/s and 6m/srespectively on a smooth horizontal surface. The speed of centre of mass of the system is :nks nzO;eku 1 kg rFkk 0.5 kg ,d gh fn'kk esa 2 m/sec rFkk 6 m/sec ls fpd us lrg ij xfr d j jgs gSA lewg d snzO;eku d sUnz d k osx gksxk&

(1*) 3

10 m/s (2)

710

m/s (3) 211

m/s (4) 3

12 m/s

Sol. vcm = sec/m3

102/11

621

21

.

B-7. Two particles having mass ratio n : 1 are interconnected by a light inextensible string that passes overa smooth pulley. If the system is released, then the acceleration of the centre of mass of the systemis :nks d .k ft ud k nzO;eku d k vuqikr n : 1 gS os ,d gYd h vforkU; jLlh ls t qM+s gq, gSa t ks fpd uh f?kjuh ls xqt jrhgSA vxj fud k; d ks NksM+ fn;k t krk gS rks lewg d s nzO;eku d sUnz d k Roj.k gksxkA

(1) (n � 1)2 g (2) g1n1n

2

(3*) g

1n1n

2

(4) g

1n

1n

Sol. a = mnm

)m�nm(

g

= )1n()1�n(

g

a1 = a2 = a

acm = )mnm(ma�nma 21

= a

)1n()1�n(

acm = g)1n(

)1�n(2

2

.

SECTION (C) : CONSERVATION OF LINEAR MOMENTUM js[kh; laosx laj{k.kC-1. Two particles A and B initially at rest move towards each other under a mutual force of attraction. The

speed of centre of mass at the instant when the speed of A is v and the speed of B is 2v is :[JEE - 89]

izkjEHk esa fLFkj A o B fi.M ijLij vkd "kZ.k cy ls ,d nwljs d h vksj xfr d jrs gSA t c A d h pky v o B d h pky2

v gS rks nzO;eku d sUnz d h pky gksxh \ [JEE - 89]

(1) v (2*) Zero 'kwU; (3) 2 v (4) 3

v

/2

Sol. Net external force is zero so net momentum will remain zero.d qy cká cy 'kwU; gSA blfy, d qy laosx 'kwU; gksxkA

C-2. If the KE of a body becomes four times its initial value, then the new momentum will be more than itsinitial momentum by;vxj fdlh oLrq dh xfrt Å tkZ izkjfEHkd ls pkj xquh gks tkrh gS rks u;k laosx izkjfEHkd laosx ls fdruk T;knk gksxkA(1) 50% (2*) 100% (3) 125% (4) 150%

Sol. 4kk

1

2 4v

v2

1

2

2

vv

1

2

Then rc 100p

p�p

1

12

= 100

mv

mv�mv

1

12

= 1001�

v

v

1

2

= 100%


C-3. A man of mass 'm' climbs on a rope of length L suspended below a balloon of mass M. The balloon isstationary with respect to ground. If the man begins to climb up the rope at a speed vrel (relative torope). In what direction and with what speed (relative to ground) will the balloon move?

(1*) downwards, Mm

mv rel

(2) upwards,

Mm

Mvrel

(3) downwards, M

mv rel (4) downwards, M

v)mM( rel

,d m nzO;eku d k vkneh L yEckbZ d h jLlh ij p<+ jgk gS] jLlh M nzO;eku d s xqCckjs ls yVd h gqbZ gSaA t ehu d slkis{k xqCckjk fLFkj gSA vxj vkneh jLlh ij mij dh vksj vrel (jLlh ds lkis{k) osx ls p<+uk 'kq: dj nsrk gS rks fdlfn'kk esa (t ehu d s lkis{k) rFkk fd l osx ls xqCckjk xfr d jsxkA

(1*) uhps] Mm

mv rel

(2) Å ij,

Mm

Mvrel

(3) uhps] M

mv rel (4) uhps M

v)mM( rel

Sol. we have ge t kurs gS Vm + Vb = Vrel Vm = Vrel � Vbby conservation of linear momentumjs[kh; laosx lj{k.k lsmVm � MVb = 0So vr% m (Vrel � Vb) � MVb = 0

Vb = Mm

mvrel

SECTION (D) : SPRING - MASS SYSTEM fLçax nzO;eku fudk;D-1. In the figure shown the change in magnitude of momentum of the block when it comes to its initial

position if the maximum compression of the spring is x0 will be :

fp=kkuqlkj CykWd d s losax esa ifjorZu gksxk t c CykWd nqckjk izkjfEHkd voLFkk esa vkrk gS ;fn fLizax d k vf/kd relEihM+u x

0 gSA

(1*) 2 k m x0

(2) k m x0

(3) zero (4) none of these buesa ls d ksbZ ugh

D-2. Two masses are connected by a spring as shown in the figure. One of the masses was given velocity v = 2k

, as shown in figure where 'k' is the spring constant. Then maximum extension in the spring will be

nks nzO;eku fp=kkuqlkj fLizax ls ca/ks gq;s gSaA ,d nzO;eku dks 2k osx nsrs gS tSlk fp=k esa n'kkZ;k x;k gS tgka k fLizax fu;rkadgSA rks fLizax esa vf/kdre f[kapko gksxk

(1) 2 m (2) m (3*) km2 (4) km3


Sol. by energy conservation 21

mv2 = 21

(2m)

2

2v

+

21

kx2

Å t kZ lja{k.k ls 21

mv2 = 21

(2m)

2

2v

+

21

kx2

x = mK2

SECTION (E) : IMPULSE vkosxE-1. A ball of mass 50 gm is dropped from a height h = 10 m. It rebounds losing 75 percent of its kinetic

energy. If it remains in contact with the ground for t = 0.01 sec., the impulse of the impact force is :,d 50 gm d h xsan 10 ehVj Å ¡pkbZ ls NksM+h t krh gSA ;g okfil 75 izfr'kr xfrt Å t kZ d h gkfu d s ckn mNyrhgSA vxj ;g /kjrh d s lkFk t = 0.01 lsd .M, d s fy, Li'kZ d jrh gS rks VDd j d s cy d k vkosx gksxkA(1) 1.3 N�s (2*) 1.05 N-s (3) 1300 N�s (4) 105 N�s

Sol. v1 = gh2 = 10102 = 210

k2 = 41

k1 v22 =

41

v12

v2 = 2v1 = 25

|P| = |�mv2 � (mv1)| = m|�v2 � v1|

|P| = 50 × 10�3 × 23

× 210 = 2

1015 2�

J = P = 1.05N-s.

E-2. The area of F-t curve is A, where 'F' is the force on one mass due to the other. If one of the collidingbodies of mass M is at rest initially, its speed just after the collision is :F-t oØ d k {ks=kQy A gS t gk¡ 'F' ,d nzO;eku ij nwljs d s dkj.k cy gSA vxj Vdjkus okyh oLrqvksa esa ls M nzO;ekuokyh izkjEHk esa fLFkj Fkh rks Vd jkus d s ckn bld k osx D;k gksxkA

(1*) A/M (2) M/A (3) AM (4) MA2

Sol.. Area of F-t curve = A = Impulse. F-t vkjs[k dk {ks=kQy = A = vkosxImpulse vkosax = dP = A = mv � 0

v = MA

.

E-3. The given figure shows a plot of the time dependent force Fx acting on a particle in motion along the x-

axis. What is the total impulse delivered by this force to the particle from time t = 0 to t = 2second?fn;k x;k fp=k x-v{k ds vuqfn'k xfr djrs gq;s ,d d .k ij dk;Z d jus okys le; ij fuHkZj cy F

x(t) dk xzkQ iznf'kZr

d jrk gSA le; t = 0 ls t = 2 sec rd bl cy }kjk d .k d ks fn;k x;k d qy vkosx D;k gS \

(1) 0 (2) 1 kg-m/s (3*) 2 kg-m/s (4) 3 kg-m/s


Sol. Impulse (vkosx) = dtF

= Area under curve (oØ ls ifjc) {ks=kQy )

= 21

(2) (2) = 2 kg-m/sec.

E-4. A mass of 100g strikes the wall with speed 5m/s at an angle as shown in figure and it rebounds with thesame speed. If the contact time is 2 × 10�3 sec., what is the force applied on the mass by the wall :100g nzO;eku dk dksbZ d.k 5m/s dh pky ls fdlh nhokj ls fp=k esa n'kkZ;s x, dks.k ij Vdjkrk gS] rFkk mlh pky lsokil ykSV vkrk gSA ;fn laidZ le; 2 × 10�3 sec gks rks d.k }kjk nhokj ij yxk, x, cy dk eku gS

60º

60º

100g

(1) 250 3 to right (2) 250 N to right (3*) 250 3 N to left (4) 250 N to left

(1) 250 3 nka;h vksj (2) 250 N nk;h vksj (3*) 250 3 N cka;h vksj (4) 250 N cka;h vksj

Sol. Fx = N3250�

102

)º60sinmVº�(60sinmV�

t

)Pi�Pf(

t

P3�

xxx

= 3250 N towards left cka;h rjQ

SECTION (F) : COLLISION VDd jF-1. A block moving in air explodes in two parts then just after explosion

(1*) the total momentum must be conserved(2) the total kinetic energy of two parts must be same as that of block before explosion.(3) the total momentum must change(4) the total kinetic energy must not be increasedgok esa xfr'khy ,d xqVdk nks Hkkxksa esa VwV tkrk gS rFkk VqdM+sa vyx&vyx gks tkrs gSa] rks foLQksV ds rqjUr ckn &(1) dqy laosx lajf{kr jgsxkA(2) nksuksa Hkkxksa dh dqy xfrt Å tkZ ogh jgsxh tks foLQksV ds igys xqVds dh Fkh(3) dqy laosx ifjofrZr gks tk;sxkA(4) dqy xfrt Å tkZ ugha c<+uh pkfg,A

F-2. In head on elastic collision of two bodies of equal masses, it is not possible :(1) the velocities are interchanged(2) the speeds are interchanged(3) the momenta are interchanged(4*) the faster body speeds up and the slower body slows down

leku nzO;eku okyh nks oLrqvksa dh lEeq[k çR;kLFk VDdj esa] ;g lEHko ugha gS & &(1) osx ijLij ifjofrZr gks tkrs gSA (2) pkysa ijLij ifjofrZr gks tkrh gSA(3) laosx ijLij ifjofrZr gks tkrs gSA (4*) rhozxkeh oLrq rst gks tk, rFkk /kheh oLrq /kheh pysaA

F-3. A bullet of mass m = 50 gm strikes a sand bag of mass M = 5 kg hanging from a fixed point, with a

horizontal velocity pv

. If bullet sticks to the sand bag then the ratio of final & initial kinetic energy of

the bullet is (approximately) :,d 50 gm d h xksyh ,d jsr d s yVd s gq, cSx ls (ft ld k nzO;eku 5 kg gSa) {ksfrt osx v ls Vd jkrh gSA vxj cwysVjsr d s cSx d s fpid t krh gS rks xksyh d h vfUre rFkk izkjfEHkd xfrt Å t kZ d k vuqikr D;k gksxk ¼yxHkx½( 1 )10�2 (2) 10�3 (3) 10�6 (4*) 10�4


Sol. 0.5 × vp + m × 0 = 5.05 v

i

f

vv

= 505.0

= 10�2

2

i

2f

)v(m21

)v(m21

= (10�2)2 = 10�4.

F-4. There are hundred identical sliders equally spaced on a frictionless track as shown in the figure.Initially all the sliders are at rest. Slider 1 is pushed with velocity v towards slider 2. In a collision thesliders stick together. The final velocity of the set of hundred stucked sliders will be :fp=kkuqlkj 100 CykWd ,d nwljs ls leku nwjh ij ?k"k.kZjfgr lrg ij mifLFkr gSA izkjEHk eaas lHkh fLFkj gSA igys CykWdd ks v osx ls nwljs dh rjQ /kDd k fn;k t krk gSA rFkk nksuksa ,d nwljs ds fpid t krs gS rks lHkh fpid s 100 CykWd d kvfUre osx D;k gksxk \

(1) 99v

(2*) 100

v(3) zero (4) v

Sol. by conservation of linear momentum Pi = Pf mv = (100 m) u u = v/100js[kh; laosx lj{k.k ls Pi = Pf mv = (100 m) u u = v/100

F-5. A massive ball moving with speed v collides head-on with a tiny ball at rest having a mass very lessthan the mass of the first ball. If the collision is elastic, then immediately after the impact, the secondball will move with a speed approximately equal to:,d Hkkjh xsan t ks v osx ls ,d NksVh xsan ¼ft ldk nzO;eku cgqr&cgqr d e gSA½ ls lEeq[k izR;kLFk VDd j d jrh gS rksVDd j d s ckn nwljh xsan d k osx yxHkx gksxkA(1) v (2*) 2v (3) v/2 (4) .

Sol. Velocity of heavy mass donot change after collisonVDd j d s ckn Hkkjh oLrq d k osx ifjofrZr ugh gksrk gSA

12

12

u�uv�v

= � e = � 1 v�0v�v2 = �1 v2 = 2v

F-6. A ball of mass 'm', moving with uniform speed, collides elastically with another stationary ball. Theincident ball will lose maximum kinetic energy when the mass of the stationary ball is,d 'm' nzO;eku d h xsan fu;r osx ls xfr dj jgh gSA ,d vU; fLFkj xsUn ls izR;kLFk VDdj djrh gSA vxj vkifrrxsan d h vf/kd re xfrt Å t kZ d h gkfu gksrh gS rks fLFkj xsan d k nzO;eku gksxkA [REE - 96](1*) m (2) 2m (3) 4m (4) infinity vuUr

Sol. If mass ;fn nzO;eku = mfirst ball will stop izFke xsan :d tk,xh v = 0so vr% k.e. = 0 (min U;wure)In other cases there will be some kinetic energy nwljs izdj.k esa leku gksxh(K.E. can't be negative _ .kkRed ugh gks ldrh)

F-7. During the head on collision of two masses 1 kg and 2 kg the maximum energy of deformation is 3

100J.

If before collision the masses are moving in the opposite direction, then their velocity of approachbefore the collision is :

nks nzO;ekuksa 1kg. rFkk 2kg d h ‘'kh"kkZfHkeq[k VDd j d s nkSjku laihMu d h egRre Å t kZ 3

100t wy gSA ;fn VDd j d s

igys nzO;eku foijhr fn'kk esa xfreku gSa ] rc mud k VDd j ls igys lkehI; osx gS &

(1*) 10 m/sec. (2) 5 m/sec. (3) 20 m/sec. (4) 10 2 m/sec.


Sol. U = )mm(mm

21

21

21

(V1 � V2)2 =

3100

(V1 � V2)2 × )m2m(2m.m2

=

3100

putting m = 1 kg j[kus ij:(V1 � V2) = 10 m/sec.

AlternateSolution:When deformation is maximum both the particles are moving with same velocity . So applying momentumconservation.tc nksuksa d.k leku osx ls xfr djsaxs rc laihM+u vf/kdre gksxkA vr% laosx lj{k.k lsm

1v

1 + m

2v

2 = m

1v

1� + m

2v

1�

v

1� =

21

2211

mmvmvm

Applying energy conservation: Å tkZ lj{k.k ls

21

m1v

12 +

21

m2v

22 =

21

(m1 + m2) (v1´ )2 + U deformation

U deformation

=21

21

21

mmmm

× (v1 � v2)2 = 3

100v

1 � v

2 = 10m/sec.

F-8. A block A of mass m moving with a velocity ' v

' along a frictionless horizontal track and a blocks of mass

m/2 moving with 2 v collides with block elastically. Final speed of the block A is :

,d m nzO;eku d k CykWd A, v

' osx ls ?k"kZ.k jfgr iFk ij py jgk gS vkSj ,d m/2 nzO;eku d k ,d CykWd B, 2

v

osx ls pyrs gq;s CykWd A,ls izR;kLFk : i ls Vd jkrh gS rks CykWd A d k vfUre pky gksxhA

(1) 5 v

3(2*) v (3)

2 v

3(4) none of these buesa ls d ksbZ ugha

Sol. Let the velocities of plank and body of mass 2m

ekuk Iykad vkSj 2m

nzO;eku dh oLrq fp=kkuqlkj VDdj ds ckn

move with speed v1 and v

2 after collision as shown.

Øe'k% osx v1 o v

2 ls xfr djrh gSA

From conservation of momentum.laosx lj{k.k ls

mv � 2m

2v = mv1 +

2m

v2

or 2v1 + v

2 = 0 ..........(1)

From equation of coefficient of restitution.izR;koLFkku xq.kkad

e = 1 = v2vvv 12

v

2 � v

1 = 3v ..........(2)

Solving 1 and 2 we get1 o 2 d ks gy d jus ij

v1 = �v


F-9. In a collision between two solid spheres, velocity of separation along the line of impact (assume noexternal forces act on the system of two spheres during impact) :nks Bksl xksyksa d h VDd j esa] VDd j js[kk d s vuqfn'k nwj t kus d k osx ¼;g ekfu;s fd nks xksyksa d s fud k; ij VDd jd s nkSjku d ksbZ ckº; cy d k;Z ugh d jrk gS½(1*) cannot be greater than velocity of approach

ikl vkus d s osx ls T;knk ugha gks ld rk(2) cannot be less than velocity of approachikl vkus d s osx ls d e ugha gks ld rk(3) cannot be equal to velocity of approachikl vkus d s osx d s cjkcj ugh gks ld rk(4) none of these (buesa ls d ksbZ ugha)

Sol. e = |approach ofvelocity ||separation of velocity|

For elastic collision e = 1|Velocity of separation| = |velocity of approach|For inelastic collision e < 1So |velocity of separation| < |velocity of approach|

gy% e = ||

||

kd osx dk lkisf{utnhd vkuskd osx dk lkisf{nwj gksus

izR;kLFk VDd j d s fy, e = 1

| nwj gksus d k lkisf{kd osx

| = |

ut nhd vkus d k lkisf{kd osx

|

vizR;kLFk VDd j d s fy, e < 1

vr% | nwj gksus d k lkisf{kd osx

| < |

ut nhd vkus d k lkisf{kd osx

|

F-10. In the figure shown the block A collides head on with another block B at rest. Mass of B is twice themass of A. The block A stops after collision. The co-efficient of restitution is :fp=kkuqlkj CykWd A, nwljs fLFkj CykWd B ls lEeq[k VDd j d jrk gSA B d k nzO;eku A ls nqxuk gSA VDd j d s ckn ACykd : d t krk gSA izR;kLFkrk xq.kkad gS %

(1*) 0.5 (2) 1 (3) 0.25 (4) it is not possible ;g lEHko ugha gSA

Sol. v1 =

21

22

21

121

mmu)e1(m

mmu)em�m(

=

m2m0)e1(m2

m2mu)m2e�m( 1

= 0

0 = m � e2m

e = 1/2

F-11. A particle of mass m moves with velocity v0 = 20 m/sec towards a wall that is moving with velocity v = 5 m/sec.

If the particle collides with the wall elastically, the speed of the particle just after the collision is :m nzO;eku d k ,d d .k v

0 = 20 m/sec ls xfr d jrk gqvk 5 m/sec ls xfr d jrh nhokj ls fp=kkuqlkj Vd jkrk gS

rks çR;kLFk VDd j d s ckn d .k d k osx D;k gksxkA

(1*) 30 m/s (2) 20 m/s (3) 25 m/s (4) 22 m/s


Sol. Sol.

V2 = Z

0

Vol. of Sep = Vel of approach ( elastic)nwj tkus dk osx = fudV vkus dk osx ( izR;kLFk) 20 + 5 = V � 5

V = 30 m/s Ans.

F-12. A super-ball is to bounce elastically back and forth between two rigid walls at a distance d from eachother. Neglecting gravity and assuming the velocity of super-ball to be v0 horizontally, the average forcebeing exerted by the super-ball on one wall is :,d lqij xsan] nks n<̀+ nhokjksa t ks d nwjh ij gS ] d s e/; vkxs ihNs xfr djrh gqbZ çR;kLFk : i ls mNyrh gSA xq: Roh;cy ux.; gS rFkk xsan d k {ksfrt osx v0 gS rks izR;sd nhokj ij lqij xsan }kjk yxk;k x;k cy gksxkA

(1) 21

d

mv20 (2*)

d

mv20 (3)

d

mv2 20 (4)

d

mv4 20

Sol. t = 0vd2

(time for succeesive collision Øekxr VDdjksa ds chp le;)

N × t = dP = mv0 � (�mv0)

N × 0vd2

= 2mv0

N = d

mv20

F-13. In the figure shown the block A collides head on with another block B at rest. Mass of B is twice themass of A. The block A stops after collision. The co-efficient of restitution is :fp=kkuqlkj CykWd A, nwljs fLFkj CykWd B ls lEeq[k VDd j d jrk gSA B d k nzO;eku A ls nqxuk gSA VDd j d s ckn ACykd : d t krk gSA izR;kLFkrk xq.kkad gS %

(1*) 0.5 (2) 1 (3) 0.25 (4) it is not possible ;g lEHko ugha gSA

F-14. A sphere of mass m moving with a constant velocity hits another stationary sphere of the same mass.If e is the coefficient of restitution, then ratio of speed of the first sphere to the speed of the secondsphere after collision will be :,d m nzO;eku d k xksyk fu;r osx ls xfr d jrk gqvk leku nzO;eku d s fLFkj xksys ls Vd jkrk gSA vxj e izR;kLFkrkxq.kkad gS rks VDd j d s ckn izFke xksys d s osx ,oa nwljs xksys d s osx vuqikr D;k gksxkA

(1*)

e1e1

(2)

e1e1

(3)

1e1e

(4)

1e1e

Sol. mu = mv1 + mv2 .......(i)u = v1 + v2 .......(i)

uv�v 12 = e ......(ii)

as solving have gy djus ij 2

1

vv

=

e1e�1

.


SECTION (G) : VARIABLE MASS ifjorZu'khy nzO;ekuG-1. If the force on a rocket which is ejecting gases with a relative velocity of 300 m/s, is 210 N. Then the rate of

combustion of the fuel will be :

;fn 300 m/s, osx ls xSlksa dks ckgj fudky jgs jkdsV ds }kjk xSalksa ij cy 210 N gS rks bZa/ku ds tyus dh nj gksxh :

(1) 10.7 kg/sec (2) 0.07 kg/sec (3) 1.4 kg/sec (4*) 0.7 kg/sec

Sol. F = dtdm

210 = 300 × dtdm

dtdm

= 0.7 kg/s.

PART - II : ASSERTION / REASONINGHkkx - II : [email protected] (ASSERTION/REASONING)

1. STATEMENT-1 : A sphere of mass m moving with speed u undergoes a perfectly elastic head on collisionwith another sphere of heavier mass M at rest (M > m), then direction of velocity of sphere of mass m isreversed due to collision [no external force acts on system of two spheres]STATEMENT-2 : During a collision of spheres of unequal masses, the heavier mass exerts more force onlighter mass in comparison to the force which lighter mass exerts on heavier mass.oDrO;-1 : ,d xksyk ft ld k nzO;eku m gS rFkk bld h pky u gS rFkk ;g nwljs Hkkjh xksys ls iw.kZr;k izR;kLFk lEeq[kVDd j d jrk gSA Hkkjh xksyk izkjEHk esa fojke esa gS rFkk Hkkjh xksys d k nzO;eku M gS tgk¡ (M > m) gSA VDd j d s d kj.km nzO;eku d s xksys d s osx d h fn'kk foijhr gks t krh gS [nksuksa xksyks d s fud k; ij d ksbZ ckº; cy d k;Zjr ugh gS]oDrO;-2 : vleku nzO;eku d s xksyksa d h VDd j d s nkSjku Hkkjh okyk xksyk gYd s okys xksys ij] gYd s okys xksys }kjkHkkjh okys xksys ij yxs cy d h rqyuk esa vf/kd cy yxkrk gSA(1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(3*) Statement-1 is True, Statement-2 is False(4) Statement-1 is False, Statement-2 is True(1) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA(3*) oDrO;-1 lR; gS] oDrO;-2 vlR; gS ;(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gS

Sol. Statement-2 contradicts Newton's third law and hence is false.oDrO;-2 U;wVu ds rhljs fu;e dk fojks/k djrk gS rFkk blfy;s vlR; gSA

2. STATEMENT-1 : In a perfectly inelastic collision between two spheres, velocity of both spheres justafter the collision are not always equal.STATEMENT-2 : For two spheres undergoing collision, component of velocities of both spheres alongline of impact just after the collision will be equal if the collision is perfectly inelastic. The componentof velocity of each sphere perpendicular to line of impact remains unchanged due to the impact.

oDrO; 1 : nks xksykasa d s chp ,d iw.kZ vizR;kLFk VDd j d s fy,] VDd j d s rqjUr ckn nksuksa xksysa d s osx ges'kk cjkcj

ugh gksaxsA

oDrO; 2 : VDd j d jrs gq;s nks xksysa d s fy,] VDd j d s rqjUr ckn VDd j d h js[kk d s vuqfn'k nksuksa xksyksa d s osxksa d s

?kVd cjkcj gksaxs ;fn VDd j iw.kZr;k vizR;kLFk gSA VDd j d h js[kk d s yEcor~ izR;sd xksys d s osx d k ?kVd VDd j

d s d kj.k vifjofrZr jgrk gSA(1*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.(2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1(3) Statement-1 is True, Statement-2 is False(4) Statement-1 is False, Statement-2 is True

(1*) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 d k lgh Li"Vhd j.k gSA

(2) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 d k lgh Li"Vhd j.k ugha gSA

(3) oDrO;-1 lR; gS] oDrO;-2 vlR; gS ;

(4) oDrO;-1 vlR; gS] oDrO;-2 lR; gS

mailto:[email protected]


Sol. (Easy) From statement-2, if the component of relative velocity normal to line of impact is non-zero, theyshall not have same velocity after collision. Hence statement-2 is correct explanation of statement-1.(Easy) oDrO;-2 ls, ;fn VDd j d h js[kk d s yEcor~ lkisf{kd osx d k ?kVd v'kwU; gS] rks VDd j d s ckn mud s lekuosx ugh gksaxsA blfy;s oDrO;-2 ] oDrO;-1 d k lgh Li"Vhd j.k gSA

3. Statement 1 : If the mass of the colliding particles remains constant, then the linear velocity of theindividual particles change during collision along common normal direction.Statement 2 : A pair of equal and opposite impulses act along common normal direction.(1*) Both statements 1 and 2 are true and statement 2 is the correct explanation of statement 1.(2) Both statements 1 and 2 are true but statement 2 is not correct explanation of statement 1.(3) Statement 1 is true but statement 2 is false(4) Both statements 1 and 2 are false.oDrO; 1 % vxj VDd j d j jgsa d .kksa d k nzO;eku fu;r jgs rks VDd j d s nkSjku izR;sd d .k d k viuk jsf[kd osxmHk;fu"B vfHkyEc fn'kk d s vuqfn'k ifjofrZr gksrk gSAoDrO; 2 % leku rFkk foijhr vkosx ;qXe mHk;fu"B vfHkyEc fn'kk d s vuqfn'k d k;Z d jrk gSA(1*) nksuksa oDrO; lR; gS rFkk oDrO; 2] oDrO; 1 d h lgh O;k[;k d jrk gSA(2) nksuks oDrO; lR; gS ijUrq oDrO; 2] oDrO; 1 d h lgh O;k[;k ugh d jrk gSA(3) oDrO; 1 lR; gS rFkk oDrO; 2 vlR; gSA(4) oDrO; 1 vlR; gS fd Urq oDrO; 2 lR; gSA

Ans. (1)

PART - I : OBJECTIVE QUESTIONSHkkx - I : oLrqfu"B iz'u ¼OBJECTIVE QUESTIONS½

Single choice type,dy fodYih izdkj

1. All the particles of a body are situated at a distance R from the origin. The distance of the centre ofmass of the body from the origin is

fdlh oLrq ds leLr d.k ewy fcUnq ls R nwjh ij fLFkr gSA oLrq ds nzO;eku dsUnz dh ewy fcUnq ls nwjh &[HCV - I](1) = R (2*) R (3) > R (4) R

2. A uniform thin rod of mass M and Length L is standing vertically along the y-axis on a smooth horizontalsurface, with its lower end at the origin (0,0). A slight disturbance at t = 0 causes the lower end to slipon the smooth surface along the positive x-axis, and the rod starts falling. The acceleration vector ofcentre of mass of the rod during its fall is : [JEE - 93]

[ R is reaction from surface],d le: i iryh M nzO;eku ,oa L yEckbZ d h NM+ y-v{k d s vuqfn'k fpd us {kSfrt ry ij Å /okZ/kj [kM+h gSA bld kfupyk fljk ewy fcUnq (0,0) ij gS y?kq fo{kksHk ls gh bldk fupyk fljk /kukRed x-v{k dh vksj fQlyrk gs rFkk NM+

fxjuk izkjEHk d jrh gS fxjrs le; bld s nzO;eku d sUnz d k Roj.k lfn'k gksxkA [ R lrg ls izfrfØ ;k cy gSA]

(1*) M

RgMaCM

(2)

MRgM

aCM

(3) RgMaCM

(4) None of these buesa ls dksbZ

ughaASol. Equation of motion xfr d s lehd j.k ls

cmaMRgM

so vr% cma

= M

RgM


3. A skater of mass m standing on ice throws a stone of mass M with a velocity of v in a horizontaldirection. The distance over which the skater will move back (the coefficient of friction between theskater and the ice is ) :,d Ld sVj (nzO;eku m) cQZ ij [kM+k gS] M nzO;eku dk iRFkj v osx ls {ksfrt fn'kk esa Qsd rk gSA LdsVj ihNs dh vksjft ruh nwjh r; d jsxk] oks gksxh\ (cQZ rFkk vkneh d s chp ?k"kZ.k xq.kkad gS) :

(1) gm2vM 22

(2)

gm2

Mv2

2

(3*)

gm2

vM2

22

(4)

gm2

vM22

22

Sol.

Pi = 0 ...(i) Pf = MV � mV1 ....(ii)

MV � mV1 = 0 v1 = mM

V..

using 02 = v12 � 2ax

v12 = 2gx

2

mMV

= 2g x.

x = gm2

VM2

22

4. In a vertical plane inside a smooth hollow thin tube a block of same mass as that of tube is released asshown in figure. When it is slightly disturbed it moves towards right. By the time the block reaches theright end of the tube then the displacement of the tube will be (where �R� is mean radius of tube).

Assume that the tube remains in vertical plane.Å /okZ/kj ry esa [kks[kyh ufyd k esa leku nzO;eku d k CykWd fp=kkuqlkj NksM+k t krk gS t c ;g ufyd k d s ,d fljsij igq¡prk gS rks ufyd k d k foLFkkiu gksxk (�R� ufyd k d h vkSlr f=kT;k gS) ekuks ufyd k Å /okZ/kj jgrh gSA

(1)

R2(2)

R4(3*)

2R

(4) R

Sol. Let the tube displaced by x towards left, then block will be displaced by (R � x) towards right ;

ekuk ufyd k ck;h rjQ x foLFkkfir gksrh gSA rc CykWd nk;h rjQ (R � x) foLFkkfir gksxkA

mx = m (R � x) x = 2R


5. A stationary body explodes into two fragments of masses m1 and m2. If momentum of one fragment isp, the minimum energy of explosion is,d fLFkj oLrq m1 rFkk m2 nzO;eku eas VwV t krh gSA vxj ,d Hkkx d k laosx p gS rks fo[k.Mu d h Å t kZ gksxh

(1) )mm(2p

21

2

(2)

21

2

mm2

p(3*)

21

212

mm2

)mm(p (4) )mm(2

p

21

2

Sol. use m1v1 = m2v2 =P

F.E. = 21

mv12 +

21

m2v22

= 21

m1

2

1mP

+

21

m2

2

2mP

= 21

21

122

mm)mm(P

.

6. A train of mass M is moving on a circular track of radius ' R ' with constant speed V. The length of thetrain is half of the perimeter of the track. The linear momentum of the train will be,d Vªsu ftldk nzO;eku M gS ,d leku osx V ls oÙ̀kkdkj iFk ftldh f=kT;k ' R ' gS eas xfr dj jgh gSA Vªsu dh yEckbZVªsd dh ifjf/k dh vk/kh gSA Vªsu dk js[kh; laosx gksxkA

(1) 'kwU; 0 (2*)

VM2(3) MVR (4) MV

Sol. If we treat the train as a ring of mass 'M' then its COM will be at a distance

R2 from the centre of the

circle. Velocity of centre of mass is :

;fn Vªsu d ks 'M' nzO;eku d k oy; ekus rks mld k nzO;eku d sUnz

R2 nwjh ij gksxkA nzO;eku d sUnz d k osx

VCM = RCM .

=

R2. =

R

V.

R2( =

RV

)

VCM =

V2 MVCM =

MV2

As the linear momentum of any system = MVCM

The linear momentum of the train =

MV2Ans.

fd lh Hkh fud k; d k js[kh; laosx = MVCM

vr% Vªsu dk js[kh; laosx =

MV2Ans.

7. Two particles approach each other with different velocities. After collision, one of the particles has amomentum p

in their center of mass frame. In the same frame, the momentum of the other particle is[REE - 98]

nks d .k ,d nwljs dh vksj fHkUu&fHkUu osx ls vk jgs gSaA VDd j ds ckn muesa ls ,d d.k dk muds nzO;eku d sUnz rU=kesa laosx p

gSA leku rU=k esa vU; d .k d k laosx gksxkA [REE - 98]

(1) 0 (2*) �p

(3) �p

/2 (4) � 2 p

Sol. Net momentum in centre of mass frame should be zeronzO;eku d sUnz d s funsZ'k ra=k esa d qy laosx 'kwU; gksxk

so vr% 'PP

= 0 P�'P


8. A particle of mass �m� and velocity � v � collides oblique elastically with a stationary particle of mass �m�.

The angle between the velocity vectors of the two particles after the collision is :,d �m� nzO;eku d k d .k t ks fd � v � osx ls xfr d j jgk gS ,d �m� nzO;eku d s fLFkj d .k ls fr;Zd izR;kLFk VDd jd jrk gSA VDd j d s ckn nksuksa d .kksa d s osx lfn'kksa d s e/; d ks.k gksxkA [REE - 97](1) 45° (2) 30° (3*) 90° (4) None of these buesa ls d ksbZ ugha

Sol.

be fore collision VDd j ls iwoZ after collision VDd j d s cknSo angle between velocity vectors is 90º

vr% osx lfn'kksa d s e/; d ks.k 90º gSA

9. Two homogenous spheres A and B of masses m and 2m having radii 2a and a respectively are placed intouch. The distance of centre of mass from first sphere is :nks lekax (homogenous) xksys A rFkk B ftuds Øe'k% nzO;eku m rFkk 2m o f=kT;k 2a rFkk a gS ,d nwljs ds ikl j[ksx;s gSA igys xksys ls nzO;eku dsUnz dh nwjh Kkr djksA(1) a (2*) 2a (3) 3a (4) none of these buesa ls dksbZ ugha

Sol. we have m1 r1 = m2 r2 mr = 2m (3a � r) r = 2a

10. A non�uniform thin rod of length L is placed along x-axis as such its one of ends at the origin. The linear

mass density of rod is = 0x. The distance of centre of mass of rod from the origin is :

L yEckbZ dh vleku NM+ ftldk js[kh; ?kuRo = 0x gSA ftls bl rjg j[kk x;k gS fd bldk ,d fljk ewy fcUnq ij

gSA x ewy fcUnq ls yEckbZ gSA ewy fcUnq ls nzO;eku dsUnz dh nwjh Kkr djksA(1) L/2 (2*) 2L/3 (3) L/4 (4) L/5

Sol.

3L2

xdx

dxx

dx

x)xd(

dm

xdmx

0

L

0

20

L

0cm

11. A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smoothsurface. The centre of mass :cUn ckWDls es a,d xsan bl rjg j[kh gqbZ gS fd ;g xsan ckWDl dh nhokjkssa ls Vdjkrh jgrh gSA ckWDls dks fpduh {kSfrt lrgij j[kk x;k gSA nzO;eku dsUnz(1) of the box remains constant (2*) of the box plus the ball system remains constant(3) of the ball remains constant (4) of the ball relative to the box remains constant(1) ckWDl dk fu;r jgsxkA (2*) xsan rFkk ckWDl fudk; dk fu;r jgsxkA(3) xsan dk fu;r jgsxkA (4) xsan dk ckWDl ds lkis{k fu;r jgsxkA

Sol. Net external force on box plus ball system is zero.xsan vkSj ckDl fud k; d qy cká cy 'kwU; gSA


12. A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless surface. The manwalks to the other end of the plank. If the mass of plank is M/3, the distance that the plank moves relative tothe ground is :M nzO;eku dk ,d O;fDr L yEckbZ ds r[rs tksfd ?k"kZ.kjfgr lrg ij j[kk gS ds ,d fljs ij [kM+k gSA O;fDr r[rs ds nwljsfljs dh vksj pyuk 'kq: djrk gSA ;fn r[rs dk nzO;eku M/3 gks rks tc O;fDr nwljs fljs ij igq¡prk gS rks r[rs }kjk r;nwjh gksxhA(1*) 3L/4 (2) L/4 (3) 4L/5 (4) L/3

Sol. m (L � x) + 3m

(�x) = 0

mL = 34

mx

13. Two blocks A and B are connected by a massless string (shown in figure) A force of 30 N is applied on blockB. The distance travelled by centre of mass in 2s starting from rest is :nks CykWd A rFkk B fp=kkuqlkj nzO;ekujfgr jLlh ls tqM+s gq, gSA F = 30 N dk cy B CykWd ij yxk;k tkrk gS rks xfr djusds 2s i'pkr~ nzO;eku dsUnz ds }kjk r; nwjh gksxh :

10kg 20kg F=30N

Smooth( )?k"kZ.kjfgr

AB

(1) 1m (2*) 2m (3) 3m (4) none of these buesa ls dksbZ ugha

Sol. acm

= )2010(30

= 1 ms2

S = 0 (2) + 21

(1) (2)2

= 2 m

14. The motion of the centre of mass of a system of two particles is unaffected by their internal forces :nks d.k fudk; dk nzO;eku dsUnz chp ds vkUrfjd cyksa ls vçHkkfor jgrk gS rks(1*) irrespective of the actual directions of the internal forces(2) only if they are along the line joining the particles(3) only if they are at right angles to the line joining the particles(4) only if they are obliquely inclined to the line joining the particles.(1*) vkUrfjd cy dh fdlh Hkh fn'kk ds fy,A(2) dsoy tc cy nksauks dks tksM+us okyh js[kk esa yxsA(3) dsoy tc cy nksauks dks tksM+us okyh js[kk ds yEcor~ yxsA(4) dsoy tc cy nksauks dks tksM+us okyh js[kk ls fr;Zd :i ls ysaxsA

Sol. vector sum of internal forces on system is zero.fud k; ij vkarfjd cyksa d s lfn'k ;ksx 'kwU; gksrk gSA

15. Two bodies of masses m and 4m are moving with equal linear momentum. The ratio of their kinetic energiesis :m nzO;eku 4m nzO;eku dh nks oLrq leku laosx ls xfr dj jgh gS] rks budh xfrt Å tkZvksa dk vuqikr gksxkA(1) 1 : 4 (2*) 4 : 1 (3) 1 : 1 (4) 1 : 2

Sol.2

1

kk

= 21

)m4(2/P

)m/(P2

2

= 4/1


16. If the momentum of a body increases by 20%, the percentage increase in its kinetic energy is equal to :;fn laosx 20% c<+k fn;k tk, rks oLrq dh xfrt Å tkZ esa çfr'kr ifjorZu (c<+uk) gksxk &

(1*) 44 (2) 88 (3) 66 (4) 20

Sol. we have 2.0p

p�p

1

12

2.1pp

1

2

so 1001�k

k100

k

k�k

1

2

1

12

=

1�

p

p21

22

× 100 (since k = m2

p2

)

= ((1.2)2 � 1) × 100 = 44 %

17. Two observers are situated in different inertial reference frames. Then :nks çs{kd vyx&vyx tM+Ro funsZ'k rU=k esa fLFkr gS rks(1*) the momentum of a body by both observers may be same(2) the momentum of a body measured by both observers must be same(3) the kinetic energy measured by both observes must be same(4) none of the above(1*) nksauks çs{kd ds vuqlkj oLrq dk laosx cjkcj gks ldrk gSA(2) nksauks çs{kd ds vuqlkj oLrq dk laosx cjkcj gh gksxkA(3) nksauks çs{kd ds vuqlkj oLrq dh xfrt Å tkZ cjkcj gh gksxkA(4) buesa ls dksbZ ugha

Sol. When velocity of both frames are same then momentum will be same else it will be diffrent accordingto diffrent observerst c nksuks funsZ'k rU=kks d k osx leku gS] rc laosx leku gksxk vU;kFkk vyx vyx iz{ksd d s lkis{k ;g vyx vyxgksxkA

18. A man is sitting in a moving train, then : ,d O;fDr xfr'khy jsyxkM+h (train) esa cSBk gqvk gS %(1) his momentum must not be zero (2) his kinetic energy is zero(3) his kinetic energy is not zero (4*) his kinetic energy may be zero(1) bldk laosx 'kwU; ugha gksxkA (2) bldh xfrt Å tkZ 'kwU; gksxhA(3) bldh xfrt Å tkZ 'kwU; ugha gksxhA (4*) bldh xfrt Å tkZ 'kwU; gks ldrh gSA

Sol. Velocity of man w.r.t. train is zero so kinetic energy of man w.r.t. train = 0jsyxkMh d s lkis{k O;fDr d k osx 'kwU; gS vr% jsyxkMh d s lkis{k O;fDr d h xfrt Å t kZ 'kwU; gksxh

19. A bomb dropped from an aeroplane explodes in air. Its total :ok;q;ku ls fxjk;k x;k ce gok esa fo?kfVr gks tkrk gS rks dqy(1) momentum decreases (2) momentum increases(3*) kinetic energy increases (4) kinetic energy decreases(1) laosx ?kVsxkA (2) laosx c<+sxkA(3*) xfrt Å tkZ c<+sxhA (4) xfrt Å tkZ ?kVsxhA

Sol. Kinetic energy increase due to internal energy.vkUrfjd Å t kZ d s d kj.k xfrt Å t kZ c<rh gS


20. Two blocks of masses m1 and m

2 are connected by a massless spring and placed on smooth surface. The

spring initially stretched and released. Then :nks CykWd m

1 rFkk m

2 dks nzO;ekujfgr fLçax dh lgk;rk ls tksM+dj fpduh lrg ij j[kk tkrk gSA fLçax dks çkjEHk esa [khapdj

NksM+ nsrs gSA rc :(1) the momentum of each particle remains constant separately(2) the momentums of each body are equal(3) the magnitude of momentums of each body are equal to each other(4) the mechanical energy of system remains constant(5*) both (3) and (4) are correct(1) çR;sd d.k dk vyx&vyx laosx fu;r jgsxkA (2) nksauks d.k dk laosx leku gksxkA(3) nksauks d.kksa dk laosx dk ifjek.k leku gksxkA (4) fudk; dh ;kaf=kd Å tkZ lajf{kr jgsxkA(5*) (3) rFkk (4) lgh gSA

Sol. 0pp 21

21 p�p

21 pp

21. A shell is fired from a cannon with a velocity v at an angle with the horizontal direction. At the highest pointin its path, it explodes into two equal pieces, one retraces its path to the cannon and the speed of the otherpiece immediately after the explosion is :,d xksys dks rksi ls {kSfrt ls dks.k ij v osx ls nkxk tkrk gSA blds iFk ds mPpre fcUnq ij ;g nks cjkcj Hkkxksa esa fo[kf.Mrgks tkrk gSA ,d Hkkx blh iFk dks rksi dh rjQ vuqlfjr djrk gS rks nwljs Hkkx dh pky VDdj ds rqjUr ckn gksxh &

(1*) 3 cos (2) 2 cos (3)

23

cos (4) 23 cos

Sol. P1 = P

f

mV cos = 2m

(�V cos + 2m

V)

V = 3V cos

22. The centre of mass of the shaded portion of the disc is : (The mass is uniformly distributed in theshaded portion) :pd rh d s Nk;kafd r gq;s Hkkx d k nzO;eku d sUnz gksxk % (Nk;kafd r Hkkx esa nzO;eku le: i forfjr gS)

A

(1*) 20R

to the left of AA (A ls ck;ha vksj 20R ij)

(2) 12R

to the left of A ( AA ls ck;ha vksj 12R

ij)

(3) 20R

to the right of AA (A ls nk;ha vksj 20R

ij)

(4) 12R

to the right of A (A ls nk;ha vksj 12R

ij)

Sol. A1 = R2 A2 = 16R2

x1 = 0 x2 = 4R3

xcen = 20R

�

16R

�R

4R3

16R

�0

22

2


23. A semicircular portion of radius �r� is cut from a uniform rectangular plate as shown in figure. The

distance of centre of mass 'C' of remaining plate, from point �O� is :

,d v)Zor̀h; Hkkx ft ld h f=kT;k r gS d ks fp=kkuqlkj ,d vk;rkd kj IysV ls cuk;k x;k gSA fcUnq O ls cph gqbZ IysVd s nzO;eku d sUnz C d h nwjh gksxhA

(1) )3(r2

(2) )4(2r3

(3) )4(r2

(4*) )4(3r2

Sol. A1 = 2r × r = 2r2

A2 = 2r2

x1 = 2r

x2 = 3r4

xcm =

2r

�r2

3r

2r

�2r

r2

22

22

= ]�4[3r2

2�4

r

32

�1r

2

3

24. In an elastic collision in absence of external force, which of the following is/are correct :[REE - 95]

,d izR;kLFk VDd j esa] ckg~; cy d h vuqifLFkfr esa fuEu esa ls d kSulk@ls d Fku lR; gSaA(1) The linear momentum is not conservedjs[kh; laosx lajf{kr ugha jgrk gSA(2) The potential energy is conserved in collisionVDd j esa fLFkfrt Å t kZ lajf{kr jgrh gSA(3) The final kinetic energy is less than the initial kinetic energyvfUre xfrt Å t kZ izkjfEHkd xfrt Å t kZ ls d e gksrh gSA(4*) The final kinetic energy is equal to the initial kinetic energyvfUre xfrt Å t kZ izkjfEHkd xfrt Å t kZ d s cjkcj gksrh gSA

25. A bag of mass M hangs by a long thread and a bullet (mass m) comes horizontally with velocity v andgets caught in the bag. Then for the combined system (bag + bullet) :,d M nzO;eku d k FkSyk ,d yEcs /kkxs ls yVd k gS rFkk ,d m nzO;eku d h xksyh v {kSfrt osx ls vkd j FkSys esa/kal t krh gSA rks FkSyk rFkk xksyh d s la;qDr fud k; d s fy, �(1) Momentum is mMv/(M + m) (2) KE is (1/2) Mv2

(3*) Momentum is mv (4) KE is m2v2/(M + m)(1) laosx mMv/(M + m) gksxkA (2) xfrt Å t kZ (1/2) Mv2 gksxhA(3*) laosx mv gksxkA (4) xfrt Å t kZ m2v2/(M + m) gksxhA


Sol. by conservation of linear momentum pi = pf

js[kh; laosx lehdj.k ls pi = pf

mv = (m + M) u u = )Mm(mv

so energy of system = 21

(m + M) × u2 = m2v2/2(M + m)

vr% fudk; dh Å tkZ = 21

(m + M) × u2 = m2v2/2(M + m)

26. A shell explodes in a region of negligible gravitational field, giving out n fragments of equal mass m.Then its total [REE - 97],d xksyk ,d ux.; xq:Roh; {ks=k esa foLQkksfVr gksrk gSA rFkk leku nzO;eku m ds n VqdM+s iznku djrk gSA rc bldhd qy &(1) Kinetic energy is smaller than that before the explosionxfrt Å t kZ foLQksfVr gksus d s igys okyh xfrt Å t kZ ls d e gksxhA(2) Kinetic energy is equal to the before the explosionxfrt Å t kZ foLQksfVr gksus d s igys okyh xfrt Å t kZ d s cjkcj gksxhA(3) Momentum and kinetic energy depend on nlaosx rFkk xfrt Å t kZ n ij fuHkZj d jsxhA(4*) Momentum is equal to that before the explosion.laosx foLQksfVr gksus d s igys okys laosx d s leku gksxkA

More than one choice typecgq fodYih izdkj27.* A system of particles has its centre of mass at the origin. The x-coordinates of all the particles

(A) may be positive(B) may be negative(C*) may be non-negative(D*) may be non-positive,d d.k fudk; dk nzO;eku dsUnz ewy&fcUnq ij gSA lHkh d.kksa ds x�funsZ'kkad &(A) /kukRed gks ldrs gSA (B) _ .kkRed gks ldrs gSA(C*) v&_ .kkRed gks ldrs gSA (D*) v&/kukRed gks ldrs gSA

28.* In which of the following cases the centre of mass of a system is certainly not at its centre ?(A*) A rod whose density continuously increases from left to right(B*) A rod whose density continuously decreases from left to right(C) A rod whose density decreases from left to right upto the centre and then increases(D) A rod whose density increases from left to right upto the centre and then decreasesfuEu es ls fdu fLFkfr;ksa ds fy;s fudk; dk nzO;eku dsUnz fuf'pr :i ls blds dsUnz ij ugha gksxk &(A) ,d NM ftldk ?kuRo cka;h ls nka;h vksj fujarj c<+rk jgsA(B) ,d NM ftldk ?kuRo cka;h ls nka;h vksj fujUraj de gksrk jgsaA(C) ,d NM ftldk ?kuRo cka;h ls nka;h vksj e/; fcUnq rd de gksrk jgs rRi'pkr~ c<+sA(D) ,d NM ftldk ?kuRo cka;h ls nka;h vksj e/; fcUnq rd c<+rk jgs rRi'pkr~ de gksrk jgsA

29.* If the net external force acting on a system is zero, then the centre of mass(A) must not move (B*) must not accelerate(C*) may move (D) may accelerate;fn fdlh fudk; ij yx jgs dqy ckº; cykssa dk ifj.kkeh 'kwU; gks] rks nzO;eku dsUnz &(A) xfr ugha djsxkA (B) Rofjr ugha gksxkA (C) xfr dj ldrk gSA (D) Rofjr gks ldrk gSA


30.* In an elastic collision in absence of external force, which of the following is/are correct :[REE - 95]

,d izR;kLFk VDd j esa] ckg~; cy d h vuqifLFkfr esa fuEu esa ls d kSulk@ls d Fku lR; gSaA(A*) The linear momentum is conservedjs[kh; laosx lajf{kr jgrk gSA(B) The potential energy is conserved in collisionVDd j esa fLFkfrt Å t kZ lajf{kr jgrh gSA(C) The final kinetic energy is less than the initial kinetic energyvfUre xfrt Å t kZ] izkjfEHkd xfrt Å t kZ ls d e gksrh gSA(D*) The final kinetic energy is equal to the initial kinetic energyvfUre xfrt Å t kZ, izkjfEHkd xfrt Å t kZ d s cjkcj gksrh gSA

31.* A small ball collides with a heavy ball initially at rest. In the absence of any external impulsive force, itis possible that(A) Both the balls come to rest(B*) Both the balls move after collision(C*) The moving ball comes to rest and the stationary ball starts moving(D) The stationary ball remains stationary, the moving ball changes its velocity.fojkekoLFkk esa j[kh ,d Hkkjh xsan ls ,d NksVh xsan Vdjkrh gSA fdlh Hkh cká vkosxh cy dh vuqifLFkfr esa ;g lEHko gS fd(A) nksuksa xsans fojkekoLFkk esa vk tk;saA(B) VDdj ds i'pkr~ nksuksa xasnss xfr'khy gks(C) xfr'khy xsan fLFkj gks tk;s rFkk fLFkj xsan xfr çkjEHk dj nsaA(D) fLFkj xsan fLFkj jgs] xfr'khy xsan dk osx ifjofrZr gks tk;sA

32.* A block moving in air explodes in two parts then just after explosion (neglect change in momentumduet to gravity)(A*) The total momentum of two parts must be equal to the momentum of the block before explosion.(B) The total kinetic energy of two parts must be equal as that of block before explosion.(C) The total momentum must change(D*) The total kinetic energy must increasegok esa xfr'khy ,d xqVdk nks Hkkxksa esa foLQksfVr gks tkrk gS rks foLQksV ds rqjUr ckn ¼xq:Ro ds dkj.k laosx ifjorZu ux.;gS½ &(A) nksuks Hkkxksa dk dqy laosx] foLQksV ds igys CykWd ds laosx ds cjkcj gksxk(B) nksuksa Hkkxksa dh dqy xfrt Å tkZ ogh jgsxh tks foLQksV ds igys xqVds dh Fkh(C) dqy laosx vko';d :i ls ifjofrZr gks tk;sxkA(D) dqy xfrt Å tkZ vko';d :i ls c<+ tk,sxhA

33.* Two bodies of same mass collide head on elastically then(A*) Their velocities are interchanged(B*) Their speeds are interchanged(C*) Their momenta are interchanged(D*) The faster body slows down and the slower body speeds up.

leku nzO;eku okyh nks oLrqvksa dh lEeq[k çR;kLFk VDdj esa &(A) buds osx ijLij ifjofrZr gks tkrs gSA (B) budh pkysa ijLij ifjofrZr gks tkrh gSA(C) buds laosx ijLij ifjofrZr gks tkrs gSA (D) rhozxkeh oLrq /kheh gks tkrh gS rFkk /kheh oLrq rst gks tkrh gSA


PART - II : COMPREHENSIONHkkx - II : vuqPNsn (COMPREHENSION)

Comprehension # 1vuqPNsn # 1

A 3kg block �A � moving with 4 m/sec on a smooth table collides inelastically and head on with an 8kg

block �B� moving with speed 1.5 m/sec towards �A �. Given e = 1/2

fpdus est ij 4 m/sec osx ls pyrk gqvk ,d 3kg nzO;eku dk CykWd �A � , 1.5 m/sec osx ls foifjr fn'kk esa A dh

vksj xfr'khy 8kg nzO;eku ds CykWd ds lkFk lEeq[k vçR;kLFk VDdj djrk gSA fn;k gS : e = 1/2

1. Final velocities of both the blocksnksuksa CykWd dk vfUre osx

(1*) VA = 2m/s, V

B = s/m

43

(2) VA = 1m/s, V

B = s/m

43

(3) VA = 2m/s, V

B = s/m

41

(3) VA = s/m

43

, VB = 2m/s

2. The impulse of reformation and deformationl:i.k rFkk fo:i.k dk vkosx(1) 12Ns, 6 Ns (2) 4Ns, 12 Ns (3*) 6Ns, 12 Ns (4) 12Ns, 4Ns

3. The maximum potential energy of deformationfo:i.k dh vf/kdre fLFkfrt Å tkZ(1) 34J (2) 35J (3) 30J (4*) 33J

4. Find out loss in kinetic energy of system.fudk; dh xfrt Å tkZ esa gkfu Kkr djksA

(1) ) J4

89(2*) ) J

499

(3) J4

79(4) zero

Ans. (1) VA = 2m/s, V

B = s/m

43

, (2) 6Ns, 12 Ns, (3) 33J (4) J4

99


Sol. using momentum conservation laosx laj{k.k ls

e = 21

3 × 4 � 8 × 1.5 = 3V1 + 8V

2

12 � 12 = 3 V1 + 8V

2

3 V1 + 8V

2 = 0 ....(1)

coffecient of restitution izR;koLFkku xq.kkad5.14V�V 12

=

21

....(2)

V2 � V

1 =

25.5

....(2)

reat in (i) 3 × 1 + 8

1V

25.5

= 0

3 × 1 + 22 + 8 × 1 = 0 V1 = �

1122

= � 2 m/sec

V2 = �

83

V1 = �

83

× (2) = 43

m/sec

(2) applying momentum conservation eqn. laosx lja{k.k lsm

1 V

1 + m

2V

2 = (m

1 + m

2)V V = 0 so

| PD

| = | m1 ( V � u

1) | = | m

1 u

1| = 3 × 4 = 12 Ns

| JR

| = |e. JD

| = 6 Ns

(3) P.E = 21

mu1 2 +

21

m2V

2 �

21

(m1 + m

2) V2.

= 21

× 3 × u2 + 21

× 8 × (1.5)2 � 0 = 335

(4) K = Ki � Kf = 33 �

22

4

38

2

1)2(3

2

1=

499

J.

Comprehension�2A smooth ball 'A' moving with velocity 'V' collides with another smooth identical ball at rest. Aftercollision both the balls move with same speed with angle between their velocities 60°. No external force

acts on the system of balls.,d fpd uh xsan 'A' pky 'V' ls nwljh fpd uh ,d leku xsan ls Vd jkrh gS t ks fd igys fojke esa iM+h FkhA VDd j d sckn nksuksa xsans leku pky ls pyrh gS o vc muds osxksa ds e/; dks.k 60° gSA xsanksa ds fudk; ij dksbZ ckº; cy dk;Zjrugha gksrk gSA

5. The speed of each ball after the collision isVDd j d s ckn izR;sd xsan d h pky &

(1) 2V

(2) 3V

(3*) 3

V(4)

3

V2

Sol. From conservation of momentumlaosx laj{k.k ls

mv = mv' cos30° + mv' cos30°

v' = 3

v30cos2

v


6. If the kinetic energy lost is fully converted to heat then heat produced is;fn xfrt Å t kZ esa gqbZ gkfu iw.kZr;k Å "ek esa ifjofrZr gksrh gS rks mRiUu Å "ek gksxh &

(1) 2mV

31

(2) 2mV

32

(3) 0 (4*) 2mV

61

Sol. Loss in kinetic energy xfrt Å t kZ esa gkfu

= 21 mv2 � 2 ×

22

mv61

3

vm

21

7. The value of coefficient of restitution isizR;ku;u xq.kkad (coefficient of restitution) d k eku gS &

(1) 1 (2*) 31

(3) 3

1(4) 0

Sol. Initially B was at rest, therefore line of impact is along final velocity of B.xsan B izkjEHk esa fojke esa Fkh vr% vkosx fØ ;k js[kk xsan B d s vfUre osx d s vuqfn'k gksxhA

e = 31

23

v

3

v21

30cosv60cos'v'v

centre of mass - dlpd.resonance.ac.in · resonance aieee_centre of mass - 2 example 1.two particles...

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