Chapter 9 1 Center of mass

Newton’s laws of motion are formulated for single particle. However, they can be extended without difficulty to systems of particles and bodies of finite dimensions.

We now consider two particles located on the x axis.

The position of the center of mass is defined by

21

2211

mm

xmxmxcm

where m1 and m2 are the masses of particle 1 and particle 2.

For the system with many particles,

the position of center of mass is given by

ii

iii

cm m

m

mmm

mmmr

rrrR

.....

.....

321

332211

where

iimM is the total mass of the system. Physically, the center of mass can be

interpreted as a weighted average position of the system of the particles. In the special case of a uniform gravitational field the center of mass coincides with the center of gravity.

For the continuous distribution of particles, the position of center of mass is given by

dm

dmzyx cmcmcmcm

rR ),,(

or

dm

zdmz

dm

ydmy

dm

xdmx

cm

cm

cm

2. Calculation of center of mass 2.1. Center of the mass for the uniform rod

22

2

0

0 L

L

L

dx

xdx

dm

xdmx L

L

cm

where is the line density (mass/length): = M/L 2.2 Center of the mass for the uniform triangle

32

32

12

1

3

2

2

3

2

2

32

0

2

22

0/

0

/

0

2

3

0

0/

0

/

0

b

a

baa

ab

dxa

bx

a

xbdx

dydx

ydydx

dxdy

ydxdy

dm

ydmy

a

a

baa

ba

dxa

bx

dxa

bxx

dydx

dyxdx

dxdy

xdxdy

dm

xdmx

a

a

abx

o

a

abx

o

a

cm

a

a

abx

o

a

abx

o

a

cm

where is the area density (mass/area). ((Geometry))

)3

,3

2()],()0,[(

3

1

23

2 babaa

OBOAOG

2.3 Center of mass for the half circle with radius a

((Solution))

0cmx from the symmetry.

aa

a

a

drdr

ddrr

rdrd

rdrdr

dm

ydmy a

a

cm 424.03

4

)2

(

3

12sin

sin2

3

00

00

2

3 Linear momentum 3.1 Definition of linear momentum

The linear momentum of the particle with a mass m and a velocity v, is defined by

p = mv The general statement of Newton’s second law

dt

dpF

or

vv

vFdt

dm

dt

dmm

dt

d )(

For a constant mass m, we have

av

F mdt

dm (conventional law)

3.2 System consisting of many particles

Now we apply the Newton’s second law to the motion of many particles shown in Fig.

Newton’s second law:

)(ei

jjii FFp

where jiF is the internal force applied on the i-th particle from the j-th particle (i ≠ j) and

)(eiF is the external force applied on the i-th particle.

________________________________________________________________________ We note that

ijji FF (Newton’s third law)

The forces two particles exert on each other are equal and opposite. (a) Weak law of action and reaction

(b) Definition for the strong law of action and reaction.

From the above equation, we have

)(

2

2

)( ei

jjiiii m

dt

dFFrp

or

i

ei

i

ei

j jji

iii

i

mdt

d

)(

)(

2

2

)(

F

FF

r

pP

or

)()(

2

2e

i

ei

iiim

dt

dFFrP ,

since ijji FF , where )(eF is the total external force. The center of mass RCM is the

average of the radii vectors, weighed in proportion to their mass,

CMi

ii Mm Rr

where M is the total mass defined by

i

imM .

Then we have

)(2

2

)( eCMM

dt

dFRP

,

or

)(2

2eCM

dt

dM F

RP

.

Motion of center of mass

x

y

O1 2 3 4

1

2

3

4

Fig. Typical motion of two bodies (with the same mass, red and blue circles) connecting by a uniform rod. The center of the mass is the midpoint of the two bodies. It shows a parabola-motion in the x-y plane.

Motion of center of mass

x

y

O1 2 3 4

1

2

3

4

Fig. Typical motion of two bodies (mass 2m for the red circles and mass m for blue

circles) connecting by a uniform rod. The center of the mass is deviated from the midpoint of the two bodies. It shows a parabola-motion in the x-y plane.

4. Conservation law of linear momentum If F(e) = 0, we have

0)( eFP

leading to

dt

dM CMR

P = conserved

This means that

0 if PPP (Momentum conservation law)

where Pi and Pf are the linear momentum in the initial state and the final state. ((Note))

When dt

d CMR= 0 at t = 0, RCM is independent of t.

5. Sample problems 5.1 Sample problem 9-2 A uniform metal plate (radius 2R) from which a disk of radius R is stamped out

We consider two circles (linear combination) Circle (radius 2R and the area density ) centered at the origin (0, 0) Circle (radius R and the area density -) centered at (-R,0).

0

3)(3

)(

)]([])2([

))](([0])2([2

2

22

22

cm

cm

y

R

R

RR

RR

RRRx

5.2 Typical problem Problem 9-16*** (SP-09) (10-th edition)

Ricardo, of mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30 kg canoe. When the canoe is at rest in the placid water, they exchange seats,

which are 3.0 m apart and symmetrically located with respect to the canoe’s center. If the canoe moves 40 cm horizontally relative to a pier post, what is Carmelita’s mass. ((Solution))

mR = 80 kg l = 2a = 3.0 m. M = 30 kg The position of the center of mass xcm does not change since there is no external force.

CR

RC

CR

CRcm mMm

axmMxaxm

mMm

axmMxaxmx

)()()()( 222111

or

)()()()( 222111 axmMxaxmaxmMxaxm RCCR

We put

yxx 21 Then we have

CR

CR

mMm

mmay

)(2

When y = 0.4m, we have mC = 57.6 kg.

5.3 Problem from Serway 8-39

Romeo (77.0 kg) entertains Juliet (55.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m away from Juliet who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo’s cheek. How far does the 80.0-kg boat move toward the shore it is facing? ((Solution))

mR = 77.0 kg: weight of Romeo mJ = 55.0 kg: weight of Juliet M = 80.0 kg: mass of the canoe l =2a = 2.70 m The position of the center of mass xcm does not change since there is no external force.

JR

JR

JR

RJcm mMm

axmmMx

mMm

axmMxaxmx

))(()()( 22111

or

))(()()( 22111 axmmMxaxmMxaxm JRRJ

We put

yxx 21

Then we have

mmMm

amy

CR

J 70.0212

0.557.22

6 Elastic collision (head-on collision): one-dimensional

In the absence of the external force, the total linear momentum is conserved before and after the collision of two particles with mass m1 and m2. We consider the elastic collision where the total energy is conserved before and after the collision.

Momentum conservation law

22112211 umumvmvm Energy conservation law (elastic collision)

222

211

222

211 2

1

2

1

2

1

2

1umumvmvm

or

21

221112

21

221211

)(2

2)(

mm

vmmvmu

mm

vmvmmu

((Mathematica))

Clear"Global`";

eq1 m1 v1 m2 v2 m1 u1 m2 u2;

eq2 1

2m1 v12

1

2m2 v22

1

2m1 u12

1

2m2 u22;

eq3 Solveeq1, eq2, u1, u2 Simplify

u1 v1, u2 v2, u1 m1 v1 m2 v1 2 m2 v2

m1 m2,

u2 2 m1 v1 m1 v2 m2 v2

m1 m2

7 Perfectly inelastic collision

Momentum conservation law;

umumvmvm 212211 The energy is not conserved in this case. 8 2D collisions

ffi mmm 221111 vvv (Two-dimensional vectors)

From the above figure

222111

22211111

sinsin

coscos

ff

ffi

vmvm

vmvmvm

The solution is

)cos(

sin

)cos(

sin

21

11

2

12

21

211

if

if

vm

mv

vv

Here we assume the case of elastic collision (the energy conservation law)

222

211

211 2

1

2

1

2

1ffi vmvmvm

When m1 = m2, this condition is rewritten as

221

((Note)) In the general case (including the case of inelastic collision), the energy is not conserved.

)]2sin(sin[sin)(csc2

2

1

2

1

2

1

212111212

2

211

211

222

211

mmm

vm

vmvmvmKKEEE

i

iffifif

When m1 = m2,

2121212

11

sinsin)csc()cot(2

2

1

ivm

E

Figure Plot of E/Ei as a function of 2, where m1 = m2. 1 is changed as a

parameter:1 = 10°, 20°, 30°, 40°, 50°, 60°, 70°, and 80°. E/Ei becomes zero for 1 + 2 = /2 (elastic scattering).

((Mathematica))

f eq5 . 1

180x1, 2

180x2;

PlotEvaluateTablef, x1, 10, 80, 10, x2, 0, 90,

PlotStyle TableHue0.12 i, Thick, i, 0, 7,

AxesLabel "2 degrees", "EEi",

PlotRange 0, 90, 0.6, 0

20 40 60 80q2 degrees

-0.6

-0.5

-0.4

-0.3

-0.2

-0.1

0.0DEEi

9 Elastic collision of particles with the same mass Problem 9-75** (SP-09) (10-th edition)

A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at 60° from the original direction. After the collisions, what are the speeds of (a) the target proton and (b) the projectile proton? ((Solution))

ffi 211 vvv 2

22

12

1 ffi vvv

ffffffffi 212

22

12

212

22

12

1 2)( vvvvvvvvv

or 021 ff vv

In other words, v1f and v2f are perpendicular to each other.

10 Impulse-linear momentum theorem 10.1 Theorem

f

i

t

t

if dt

dt

d

Fppp

pF

Definition of impulse J

tttdt avifav

t

t

if

f

i

FFFJ

pppJ

)(

or

avttF

pJ

,

with

f

i

t

tifav dt

ttFF

1

The average force avF can be evaluated from the values of pJ and the time interval

((Impulse-momentum theorem))

The change in the momentum of a particle is equal to the impulse of the net force acting on the particle. ((Note))

A moving particle has momentum and kinetic energy, but it does not carry with it a force. The force required to cause a particle to stop depends on how big the momentum change is and on how quickly the momentum change occurs. ((Example)) Impulse example The change in momentum of an object during a collision is equal to the product of the average force acting on an object and the time over which it acts. ((Example)) SmartPhysics p.144-146 A ball with a mass of m = 1 kg is released from rest from an initial height of (hi = 1 m) above the floor. It bounces back to half its original height (hf = 0.5 m). If we assume the ball is in contact with the floor for a time of t = 10 ms, what is the average force on the ball during the collision?

((Solution))

1ih m. 5.0fh m. m = 1 kg. 10t ms.

(i) v1 is the velocity of the ball which falls from the height hi and touchs on the floor

)2(0 21

2ighv , or ighv 21 = 4.43 m/s

(ii) v2 is the velocity of the ball which bounces back from the floor

)2(0 22

2fghv , or fghv 22 = 3.13 m/s

Impulse

)()( 2112 vvmmvmvtFpJ av 7.56 N s

The average force avF is obtained as

756)( 21

t

vvmFav N.

10.2 Example Problem 9-38** (10-th edition)

In the overhead view of Fig., a 300 g ball with a speed v of 6.0 m/s strikes a wall at an angle of 30° and then rebounds with the same speed and angle. It is in contact with the wall for 10 ms. In unit-vector notation, what are (a) the impulse on the ball from the wall and (b) the average force on the wall from the ball?

m = 0.3 kg, v = 6.0 m/s, = 30, t = 10 ms

tFpJ av

where (a)

NsmvpJ 8.1sin2 (b)

Nt

JFav 180

11 Systems with varying mass: A rocket 11.1 Formulation

We assume that we are at rest relative to an inertial reference frame (the Earth), watching a rocket accelerate through deep space with no gravitational or atmospheric drag force acting on it.

We consider a rocket plus fuel that is initially at rest. When fuel is ejected out the back of the rocket, it acquires momentum, and so the rocket must move forward to acquire opposite momentum to cancel the fuel’s momentum, since the total momentum of the system remains constant.

Suppose that a rocket of mass M is moving at the velocity v with respect to the Earth. Now a mass of fuel (- dM) is ejected with the velocity u with respect to the Earth. The rocket now moves forward with mass (M+dM) and velocity (v+dv)

Momentum conservation law:

))(()( dvvdMMudMMv . (1)

Note that

Vrocket-Earth = Vrocket-fuel + Vfuel-Earth where

Vrocket-Earth (= v + d v) is the velocity of rocket relative to the Earth. Vrocket-fuel (= vrel) is the velocity of rocket relative to the exhaust fuel. Vfuel-Earth (= u) is the velocity of the exhaust fuel relative to the Earth.

We get the relation

u = v + dv - vrel. (2) Substituting Eq.(2) for u into Eq.(1), we get

))(()( dvvdMMvdvvdMMv rel

or

dMdvvdMMdvMvdMvdMdvvdMMv rel .

or

MdvdMvrel 0 ,

or

dt

dvM

dt

dMvrel 0

Then we have

rel

rel

rel

Rvdt

dMv

dt

dMv

dt

dvM

)( (first rocket equation).

where

vrel: the fuel’s exhaust velocity relative to the rocket.

relRvT the thrust of the rocket engine (N)

R (=-dM/dt) the rate of fuel consumption Here we assume that vrel is constant. Then we have

M

dMvdv

dMvMdv

rel

rel

)ln(i

frel

M

M

relif M

Mv

M

dMvdvvv

f

i

,

or

)ln(f

irelif M

Mvvv

(second rocket equation)

We make a plot of rel

if

v

vv as a function of

f

i

M

M

10 20 30 40 50MiMf0

1

2

3

4vf-vive

11.2 The case for constant R (=-dM/dt) and vrel.

Here we assume that R (= -dM/dt) and vrel are independent of t. Then we have

dt

dMR

Rvdt

dvM rel

From the second equation, M is obtained as

RtMM 0

or

00

1M

Rt

M

M

where M0 is the initial mass at t = 0 and t<M0/R. Then we have

0

0

0 /1

/

MRt

MRv

RtM

Rv

M

Rv

dt

dv relrelrel

or

tM

Rvdt

MRt

MRvdvvv

o

rel

trel

1

1ln

/1

/

0 0

00

where 100

M

Rt.

(a) Plot of (v-v0)/vrel vs = Rt/M0.

)1

1ln(0

relv

vv

0.2 0.4 0.6 0.8 1.0x=RtM0

1

2

3

4

v-v0vrel

Note that in the limit of →0 (or t→0)

32)

1

1ln(

320

relv

vv

(b) Plot of M/M0 vs = Rt/M0.

10M

M

0.2 0.4 0.6 0.8 1.0x=RtM0

0.2

0.4

0.6

0.8

1.0

MM0

In conclusion:

The velocity of the rocket is proportional to the time in early stage. As the mass of rocket decreases in the late stage, the velocity of the rocket rapidly increases. ((Note)) The velocity of rocket

102 miles/h = 47.7 m/s 103 miles/h=0.447 km/s = 447 m/s 104 miles/h = 4.4704 km/s

The velocity for the orbit on the earth

91.7E

Eorbit R

GMv km/s = 1.77 x 104 miles/h

The escape velocity form the earth:

19.112

E

Eescape R

GMv km/s = 2.50 x 104 miles/h.

11.3 Problem 9-78 (10-th edition)

Consider a rocket that is in deep space and at rest relative to an inertial reference frame. The rocket’s engine is to be fired for a certain interval. What must be the rocket’s mass ratio (ratio of initial to final mass) over that interval if the rocket’s original speed relative to the inertial frame is to be equal to (a) the exhaust speed (speed of the exhaust products relative to the rocket) and (b) 2.0 times the exhaust speed? ((Solution)) Second rocket equation with vi = 0 and vf = v;

)exp(

)ln(0

relf

i

f

irelif

v

v

M

M

M

Mvvvv

(a) If v = vrel, we obtain 72.21 eM

M

f

i

(b) If v = 2 vrel,, we obtain 39.72 eM

M

f

i

11.4 Problem 9-79 (10-th edition)

A rocket that is in deep space and initially at rest relative to an inertial reference frame has a mass of 2.55 x 105 kg, of which 1.81 x 105 kg is fuel. The rocket engine is then fired for 250 s while fuel is consumed at the rate of 480 kg/s. The speed of the exhaust products relative to the rocket is 3.27 km/s. (a) What is the rocket’s thrust? After the 250 s firing, what are (b) the mass and (c) the speed of the rocket? ((Solution)) (a) The thrust of the rocket is given by T = Rvrel where R (=- dM/dt) is the rate of fuel consumption and vrel is the speed of the exhaust gas relative to the rocket. For this problem R = 480 kg/s and vrel = 3.27 103 m/s, so

T = R vrel =1.57 x 106 N (b) The mass of fuel ejected is given by Mfuel = R t , where t is the time interval of the burn. Thus, Mfuel = (480 kg/s)(250 s) = 1.20 105 kg. The mass of the rocket after the burn is

Mf = Mi – Mfuel = (2.55 – 1.20) x 105 kg = 1.35 x 105 kg. (c) Since the initial speed is zero, the final speed is given by the second rocket equation

)ln(f

irelf M

Mvv = 2.08 x 103 m/s

______________________________________________________________________ 11.5

Problem 9-76 (10-th edition) A 6090 kg space probe moving nose-first toward Jupiter at 105 m/s relative to the Sun fires its rocket engine, ejecting 80.0 kg of exhaust at a speed of 253 m/s relative to the space probe. What is the final velocity of the probe? ((Solution)) Second rocket equation;

)ln(f

irelif M

Mvvv

vf = 105 m/s + (253 m/s) ln(6090 kg/6010 kg) = 108 m/s.

12. Homework and SP-09 12.1 Problem 9-14 (SP-09) (10-th edition)

In Fig., two particles are launched from the origin of the coordinate system at time t = 0. Particle 1 of mass m1 = 5.00 g is shot directly along the x axis on a frictionless floor, with constant speed 10.0 m/s. Particle of mass m2 = 3.00 g is shot with a velocity of magnitude 20.0 m/s, at an upward angle such that it always directly above particle 1. (a) What is the maximum height Hmax reached by the c.o.m. of the two-particle system? In unit-vector notation, what are the (b) velocity and (b) acceleration of the c.o.m. when the c.o.m. reaches Hmax?

((Solution)) m1 = 5.00g, m2 = 3.00 g, v1 = 10.0 m/s, v2 = 20 m/s

)sin,cos(

),0,(

222

11

gtvv

v

v

v

r1 =(x1, y1) r2 =(x2, y2)

21

2211

21

2211

21

2211

,

,

mm

mm

mm

mm

mm

mm

com

com

com

aaA

vvV

rrR

12.2 Problem 9-37 (SP-09) (10-th edition)

A soccer player kicks a soccer ball of mass 0.45 kg that is initially at rest. The player’s foot is in contact with the ball for 3.0 x 10-3 s, and the force of the kick is given by

NtttF ])100.2()100.6[()( 296 for st 3100.30 , where t is in seconds. Find the magnitudes of (a) the impulse on the ball due to the kick, (b) the average force on the ball from the player’s foot during the period of contact, (c) the maximum force on the ball from the player’s foot during the period of contact, and (d) the ball’s velocity immediately after it loses contact with player’s foot.

((Solution))

0.5 1.0 1.5 2.0 2.5 3.0t ms

1000

2000

3000

4000

F N

a = 6.0 x 106, b = 2.0 x 109 t0 = 3.0 x 10-3 s, m = 0.45 kg The force is given by

2)( btattF The impulse J:

0

0

)(t

dttFJ

The average force Fav:

0t

JFav

12.3 Problem 9-70*** (SP-09) (10-th edition)

In Fig., puck 1 of mass m1 = 0.20 kg is sent sliding across a frictionless lab bench, to undergo a one-dimensional elastic collision with stationary puck 2. Puck 2 then slides off the bench and lands a distance d from the base of the bench. Puck 1 rebounds from the collision and slides off the opposite edges of the bench, landing a distance 2d from the base of the bench. What is the mass of puck2? (Hint: Be careful with signs).

((Solution)) m1 = 0.2 kg

The momentum conservation law:

2211211 0 umummvm (1) For the mass m1,

2

1

2

1gthy

tux

or

h

gdu

g

hudx

g

ht

gthy

22

22

2

02

10

1

1

2

For the mass m2,

h

gdu

g

hudx

g

ht

gthy

2

2

2

02

10

2

2

2

Then we have

2

1

22

2

1

2

h

gd

h

gd

u

u (2)

From Eqs.(1) and (2), we have

12

112

12

111

2

2

2

mm

vmu

mm

vmu

(3)

The energy conservation law:

222

211

211 2

1

2

1

2

1umumvm

The substitution of Eq.(3) into the energy conservation law leads to

02

)5(

21

11211

mm

vmmmm

Thus we get

12 5mm 12.4 Problem 9-64** (SP-09) (10-th edition) A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal (Fig.) At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless surface. The collision is elastic, find (a) the speed of the ball and (b) the speed of the block, both just after the collision. ((Solution))

M = 2.50 kg, m = 0.5 kg, R = 0.7 m. Energy conservation law

2

2

1mvmgR or smgRv /70.32

Momentum conservation law

mVMUmv The collision is elastic.

222

2

1

2

1

2

1mVMUmv

smMm

mvU

smvMm

MmV

/23.12

/47.2

13. Problems from Sewway 13.1 Serway Problem 8-51

A small block of mass m = 0.500 kg is released from rest at the top of a curve-shaped frictionless wedge of mass M = 3.00 kg, which sits on a frictionless, horizontal surface. When the block leaves the wedge, the velocity is measured to be v = 4.00 m/s to the right. (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height h of the wedge? ((Solution))

((Solution)) M = 3.00 kg m = 0.500 kg v = 4.00 m/s (a) The initial momentum of the system is zero, which remains constant through the motion (the momentum conservation). When the block of mass m leaves the wedge, we have

0 MVmv or

V = 0.667 m/s

(b) Energy conservation

Ei = Ef where

22

2

1

2

1MVmvE

mghE

f

i

Then we have h = 0.952 m. 13.2 Serway Problem 8-58

A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially stretched and with force constant k = 2.00 x 104 N/m. The cannon fires a 200-kg projectile at a velocity of 125 m/s directed 45.0° above the horizontal. (a) Assuming that the mass of the cannon and its carriage is 5000 kg, find the recoil

speed of the cannon. (b) Determine the maximum extension of the string. (c) Find the maximum force the spring exerts on the carriage. (d) Consider the system consisting of the cannon, carriage, and projectile. Is the

momentum of this system conserved during the firing? Why or why not?

((Solution)) k = 2.00 x 104 N/m mp = 200 kg; mass for the projectile vp = 125 m/s; velocity for the projectile = 45.0°. M = 5000 kg for the cannon and carriage

(a) Use conservation of the horizontal component of momentum for the system of the

shell, the cannon, and the carriage, from just before to just after the cannon firing.

045cos recoilpp Mvvm

or

vrecoil = -3.54 m/s (b) Use conservation of energy for the system of the cannon, the carriage, and the

spring from right after the cannon is fired to the instant when the cannon comes to rest.

Ef = Ei.

where

2max

2

2

12

1

kxE

mvE

f

recoili

Then we have

mvk

mx recoil 77.1max

(c) NkxF 4

maxmax 1054.3

(d) No. The rail exerts a vertical external force (the normal force) on the cannon and prevents it from recoiling vertically. Momentum is not conserved in the vertical direction. The spring does not have time to stretch during the cannon firing. Thus, no external horizontal force is exerted on the system (cannon, carriage, and shell) from just before to just after firing. Momentum of this system is conserved in the horizontal direction during this interval. ________________________________________________________________________ Appendix A.1 Ballistic pendulum

The ballistic pendulum is an apparatus used to measure the speed of a fast-moving projectile, such as a bullet. A bullet of mass m1 is fired into a large block of wood of mass m2 suspended from some light wires. The bullet embeds in the block, and the entire system swings through a height h. The speed of the bullet can be determined from the measurement of h.

(a) Momentum conservation law on the collision

ummmvm )(0. 2121 or

21

1

mm

vmu

, (1)

where v is the velocity of bullet before the collision and u is the velocity of bullet and block. _________________________________________________________________________

_______________________________________________________________________ (b) Energy conservation law: The kinetic energy at the point A is given by

221 )(

2

1ummK A .

The potential energy at the point C is given by

ghmmUB )( 21 Then we have

ghmmumm )()(2

121

221 , (2)

where h is the height between the point A and the point B. From Eqs.(1) and (2), we have

ghm

mmv 2

1

21

.

A.2 A variable-mass drop We consider a raindrop falling through a cloud of small water droplets Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force of the rain drop is

dt

dmv

dt

dvm

dt

dpFext .

Suppose the mass of the raindrops depends on the distance x that it has fallen. Then m = kx, where k is constant, and dm/dt = kv. This gives

)(kvvdt

dvmmg

or

2vdt

dvxxg

since Fext = mg. Reference: K.S. Krane, Am. J. Phys. 49, 113 (1981). We assume that v = dx/dt = v0 and x = x0 at t = 0. We solve this problem (nonlinear differential equation) using the Mathematica (NDSolve) numericallty.

0.1 0.2 0.3 0.4t

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

x

Fig. x vs t, where v0 = 0 and x0 = 0.01 – 0.20 (x0 = 0.02)

0.1 0.2 0.3 0.4t

0.5

1.0

1.5

2.0

u

Fig. dx/dt vs t, where v0 = 0 and x0 = 0.01 – 0.20 (x0 = 0.02).

Raindrop

Clear"Global`"g 9.8;

timexx0_, u0_, tmax_, opts__ :

Modulenumso1, numgraph,

numso1 NDSolvext ut ut2 g xt, xt ut, x0 x0, u0 u0,

xt, ut, t, 0, tmax;

numgraph PlotEvaluatext . numso11, t, 0, tmax, opts, DisplayFunction Identitytimelistx

timex, 0, 0.4, PlotStyle Hue4.5 0.01, Thick, AxesLabel "t", "x",

Background LightGray, PlotRange All, DisplayFunction Identity & Range0.01, 0.2, 0.02; Showtimelistx, DisplayFunction $DisplayFunction

0.1 0.2 0.3 0.4t

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

x

B. Center of mass frame and laboratory frame for elastic collision B.1 Momentum conservation and the energy conservation

m1 m2

v1 iL

v1 fL

v2 fL

q1L

q2L

Fig. Laboratory frame. The initial momentum is Liim 1v . The scattering angle of

the particle 1 with mass m1 is L1 .

In the laboratory frame, the velocities of the particles 1 and 2 before and after collision, are defined by

Li1v , L

i2v , Lf1v , L

f2v ,

where

02 Liv

since the particle with mass m2 is at rest before the collision. The center of mass velocity is given by

LiCM mm

m1

21

1 vv

In the center of mass frame

CMi1v , CM

i2v , CMf1v , CM

f2v ,

We have the relation

CMCMi

Li vvv 11 , CM

CMi

Li vvv 22 , (1a)

CMCM

fLf vvv 11 , CM

CMf

Lf vvv 22 , (1b)

where

LiCM

Li

CMi mm

m1

21

211 vvvv

, L

iCMLi

CMi mm

m1

21

122 vvvv

.

(i) The momentum conservation in the laboratory frame;

Lf

Lf

Li

Li mmmm 22112211 vvvv .

which means that the velocity of the center of mass remains unchanged before and after the collision. Using Eqs.(1a) and (1b), we get

)()( 221111 CMCM

fCMCM

fLi mmm vvvvv

or

0)( 21112211 CMLi

CMf

CMf mmmmm vvvv

or

02211 CMf

CMf mm vv , or CM

fCM

f m

m1

2

12 vv (2)

(ii) The energy conservation law;

222

2

11

2

22

2

11 2

1

2

1

2

1

2

1 Lf

Lf

Li

Li vmvmvmvm

Using Eqs.(1a) and (1b), we get

2112

22

2

11 2

1)(

2

1

2

1 LiCM

CMfCM

CMf mmm vvvvv

or

2112

2112112

222

11 2

1))((

2

1)()(

2

1)(

2

1 LiCMCM

CMf

CMf

CMf

CMf mmmmmmm vvvvvvv

or

212

21

2

112

222

11 2

1))((

2

1

2

1)(

2

1)(

2

1 LiCM

Li

CMf

CMf mmmmm vvvvv (3)

where is the reduced mass and is given by

21

21

mm

mm

.

Using Eqs.(2) and (3),

212

12

12

211 2

1)(

2

1)(

2

1 Li

CMf

CMf m

mmm vvv

or

2121

2121

2

212

11 2

1)(

2

1)(

2

1 Li

CMf

CMf mm

mm

m

mm vvv

or

21

121 mm

vmv

LiCM

f , and

21

112 mm

vmv

LiCM

f

and

Li

LiCM

fCMf v

mm

vmmvv 1

21

12121

)(

in magnitudes.

In conclusion we have

CMf

CMf m

m1

2

12 vv . L

iCMi mm

m1

21

12 vv

.

with

21

1211 mm

vm LiCM

iCM

f vv .

21

1122 mm

vm LiCM

iCM

f vv .

m1 m2

v1 iCM v2 i

CM

v1 fCM=v1 i

CM

v2 fCM=v2 i

CM

qCM

Fig. The center of mass frame. 21

1211 mm

vmvv

LiCM

fCMi

,

CM

LiCM

fCMi v

mm

vmvv

21

1122 . L

iCMi

CMi

CMf

CMf vvvvv 12121 . The scattering

angle of the particle 1 with mass m1 is CM . Note that

Li

Li

Li

CMf

CMf

CMCM

fCMCM

fLf

Lf

vmm

vm

mm

vm1

21

12

21

11

12

1212

||

)()(

vv

vvvvvv

Li

Li

Li

CMi

CMi

CMCMiCM

CMi

Li

Li

v

mm

m

mm

m

1

121

21

21

1

12

1212 )()(

vv

vv

vvvvvv

That is, the rate at which two objects approach each other before an elastic collision is the same at the rate at which they separate afterward. We can use this result to identify elastic collisions in any inertial reference frame. Namely, the relative velocity of two objects at a given time, that is, the difference in the velocity vectors of the objects must be the same in all inertial reference frames. B.2 Scattering angles CM , L

1 and L2

The y and z components of CMfv1 and CM

fv2 ;

CMLiCMCM

fy

CMf mm

vmv sinsin

21

1211

v

CMLiCMCM

fz

CMf mm

vmv coscos

21

1211

v

CMLiCMCM

fy

CMf mm

vmv sinsin

21

1122

v

CMLiCMCM

fz

CMf mm

vmv coscos

21

1122

v

where the z axis is in the horizontal direction and the y axis is in the vertical direction. Using Eq.(1b), we have

CMCM

fLf vvv 11 , CM

CMf

Lf vvv 22 ,

or

CMLiCMCM

fyCMy

CMfy

Lf mm

vmv sinsin

21

12111

vvv

Li

CM

LiCM

Li

CMCMCM

f

zCMz

CMfz

Lf

vmm

mm

mm

vm

mm

vm

vv

121

21

21

11

21

12

1

11

)cos(

cos

cos

vvv

CMLi

yCMy

CMfy

Lf mm

vm sin21

1122

vvv

Li

CM

LiCM

Li

zCMz

CMfz

Lf

vmm

mm

mm

vm

mm

vm

121

11

21

11

21

11

22

)cos(

cos

vvv

where

LiCM mm

m1

21

1 vv

.

The scattering angle in the laboratory scheme is obtained as

CM

CM

z

Lf

y

LfL

mm

m

cos

sintan

21

2

1

1

1

v

v

2cot

cos1

sin

cos

sintan

11

1

2

2

2

CM

CM

CM

CM

CM

z

Lf

y

LfL

mm

m

v

v

When m1 = m2, we have

2

tan

2cos2

2cos

2sin2

cos1

sintan

21

CM

CM

CMCM

CM

CML

.

Then we have

221

LL

since

1tantan 21 LL .

vCM

v2 fCM=vCM v1 f

CM

v1 iL

v1 fCM

qCM qL

m2m1 =2

Fig. The relation between scattering angle in the laboratory frame and in the

center of mass frame. 21

1211 mm

vmvv

LiCM

fCMi

, CM

LiCM

fCMi v

mm

vmvv

21

1122 .

CMf

CMf

Li vvv 211 . CM

iCMi

Li vvv 211 . In this figure, we assume that m2/m1 = 2.

APPENDIX C Energy of a system of many particles The kinetic energy of many particles is given by

i

iiimK )(2

1vv

The velocity in the laboratory frame iv is given by

CMiCMi vvv

where vCM is the velocity of the center of mass and CM

iv is the velocity in the center of

mass frame. Then K can be rewritten as

CMrel

i

CMiiCM

iCM

CMii

i

CMiiCM

i

CMiCM

CMiCMi

i

CMiCM

CMiCMi

KK

mM

mmM

m

mK

22

22

22

)(2

1

2

1

)()(2

1

2

1

)2[2

1

)()(2

1

vv

vvvv

vvvv

vvvv

where

0)( CMCMi i

CMiiCMii MMmm vvvvv

Then K consists of the kinetic energy of the center of mass (KCM) and the kinetic energy of the objects as viewed in the center of mass frame (Krel).