msu physics 231 fall 2015 1 physics 231 topic 6: momentum and collisions alex brown october 7 2015

MSU Physics 231 Fall 2015 1

Physics 231Topic 6: Momentum and Collisions

Alex BrownOctober 7 2015


MomentumF = m a Newton’s 2nd lawF = m v/t a=v/tF = m (vfinal - vinital)/t

Define p = mv momentum (kg m/s)

F= (pfinal - pinitial)/tF= p/t = m v/t p = (pfinal - pinitial) = m v impulse

The net force acting on an object equalsthe change in momentum (p) in time period (t). Since velocity is a vector, momentum is alsoa vector, pointing in the same direction as v.


ImpulseF=p/t Force=change in (mv) per time

period (t).p=Ft The change in momentum equals

the force acting on the object timeshow long you apply the force.

Definition: p = Impulse

What if the force is not constant within the timeperiod t?

ts s s

F1

F2

F3

p = (F1 t + F2 t + F3 t)

= area under F vs t plot


car hitting haystack and stoping

car hitting wallThe change in momentum (impulse) is the same, but the forcereaches a much higher value when the car hits a wall!


Some examplesA tennis player receives a shot approaching him (horizontally) with 50m/s and returns the ball in the opposite direction with 40m/s. The mass of the ball is 0.060 kg.A) What is the impulse delivered by the ball to the racket?B) What is the change in kinetic energy of the ball?

A) Impulse=change in momentum (p). p = m (vfinal - vinitial) = 0.060(-40-50) = -5.4 kg m/s

B) KEi = 75 J KEf = 48 J (no PE!) Enc = KEi - KEf = 27 J (energy lost to heat)


Child safetyA friend claims that it is safe to go on a car trip with yourchild without a child seat since he can hold onto your 12kg child even if the car makes a frontal collision (lasting 0.05s and causing the vehicle to stop completely) at v=50 km/h(about 30 miles/h). Is he to be trusted?

F = p/t force = impulse per time period = m (vf-vi) /t vf = 0 and vi = 50 km/h = 13.9 m/s m = 12kg t=0.05s

F = 12(-13.9)/0.05 = -3336 N

This force corresponds to lifting a mass of 340 kg!!!


Question (assume t is the same for both cases)

The velocity change is largest in case: A B

The acceleration is largest in case: A B

The momentum change is largest in case: A B

The impulse is largest in case: A B


F21 t = p1 = p1f - p1i

F12 t = p2 = p2f – p2i

Newton’s 3rd law: F12 = - F21

F12 t = - F21 t

p1f - p1i = - p2f + p2i

Rewrite: p1i + p2i = p1f + p2f

CONSERVATION OF MOMENTUM(only for a closed system)

Conservation of Momentum

m1m2

p1i p2i

m1m2

Force of 2 on 1 = F21 F12

m1

p1f m2

p2f

p1 = m1 v1

p2 = m2 v2


questionAn object is initially at rest in outer space (no gravity).It explodes and breaks into two pieces, not of equal size.The magnitude of the momentum of the larger piece is …a) Greater than that of the smaller pieceb) Smaller than that of the smaller piecec) The same as that of the smaller piece

Initial momentum is zero0 = plarge,final + psmall,final

So plarge,final = - psmall,final

Magnitude is the same, direction opposite.


QuizAn object is initially at rest in outer space (no gravity).It explodes and breaks into three pieces, not of equal size.The momentum of the largest piece:a) Must be larger than that of each of the smaller piecesb) Must be less than that of each of the smaller piecesc) Must be equal in magnitude but opposite in direction to the

sum of the smaller pieces.d) Cannot be determined from the momenta of the smaller

piecesThe conservation of momentum must hold, even with more than 2 objects. Although the momenta of each piece could be pointing in a different direction, the sum must be equal to zero after the explosion. So, the momentum of one of the pieces must always be opposite in direction, but equal in magnitude to the sum of the momenta of the other two pieces.


Moving in spaceAn astronaut (100 kg) is drifting away from the spaceship with v=0.2 m/s. To getback he throws a wrench (2 kg) in the direction away from the ship. With whatvelocity does he need to throw the wrenchto move with v=0.1 m/s towards the ship?

Initial momentum: mavai + mwvwi =100*0.2+2*0.2 = 20.4 kg m/sAfter throw: mavaf + mwvwf = 100*(-0.1) + 2*vwf kg m/s

Conservation of momentum: mavai + mwvwi = mavaf + mwvwf

20.4 = -10 + 2*vwf vwf=15.7 m/s

(a) 0.1 m/s (b) 0.2 m/s (c) 5 m/s (d) 16 m/s(e) this will never work?


Perfectly inelastic collisions Conservation of momentum: m1v1i+m2v2i=m1v1f+m2v2f

After the collision m1 and m2 form one new object with massM = m1 + m2 and one velocity vf =v1f = v2f

m1v1i + m2v2i = vf (m1 + m2)

vf = (m1v1i + m2v2i)/ (m1 + m2)

before after

Demo: perfect inelastic collision on airtrack


Perfectly inelastic collisions Conservation of momentum: m1v1i+m2v2i=m1v1f+m2v2f


m1v1i + m2v2i = vf (m1 + m2)vf = (m1v1i + m2v2i)/ (m1 + m2)

Let c = m2/m1 then

vf = (v1i + c v2i)/ (1+c)

If m1 = m2 (c=1) then vf = (v1i + v2i)/ 2

If c=1 and v2i = 0 then vf = v1i/2


Perfect inelastic collision: an example

A car collides into the back of a truck and their bumpers getstuck. What is the ratio of the mass of the truck and the car?(mtruck/mcar = m2/m1 =c) What is the fraction of KE lost?

50 m/s 20 m/s 25 m/s

Before collision: KEi = ½m1(50)2 + ½5m1(20)2

After collision: KEf = ½6m1(25)2

Ratio: KEf/KEi= [6*(25)2]/[(50)2+5*(20)2] = 0.8317% of the KE is lost (damage to cars!)

vf = [v1i + cv2i] / (1+c) solve for c

c = [v1i - vf] / [vf – v2i] = (50-25)/(25-20) = 5


Types of collisionsInelastic collisions Elastic collisions

•Momentum is conserved

•Some energy is lost in the collision: kinetic energy is not conserved

•Perfectly inelastic: the objects stick together.

•Momentum is conserved

•No energy is lost in the collision: kinetic energy is conserved

KEi - KEf = Enc KEi - KEf = 0


Inelastic Solution

vf = [m1 v1i + m2v2i] / (m1+m2)

With m2 = c m1

vf = [v1i + cv2i] / (1+c)

Special case m2 = m1 (c=1)

vf = [v1i + v2i] / 2

Conservation of momentum: m1v1i + m2v2i = m1v1f + m2v2f


m1v1i + m2v2i = vf (m1 + m2)KEi - KEf = Enc


Elastic collisionsConservation of momentum: m1v1i+m2v2i=m1v1f+m2v2f

Conservation of KE: ½m1v1i

2+½m2v2i2=½m1v1f

2+½m2v2f2

Two equations and two unknowns – after some algebra (see book)


Elastic Solution

v1f = [(m1 –m2) v1i + 2m2v2i] / (m1+m2)

v2f = [(m2 –m1) v2i + 2m1v1i] / (m1+m2)

For m2 = c m1

v1f = [(1-c) v1i + 2cv2i] / (1+c)

v2f = [(c-1) v2i + 2v1i] / (1+c)

Special case m2 = m1 (c=1)

v1f = v2i

v2f = v1i

KEi - KEf = 0


Given m2=m1 and v1i=0

What are the velocities of m1 and m2 after the collision in terms of the initial velocity of m2 if m1 is originally at rest?

c = 1v1f = v2i

v2f = v1i = 0

m1 m2

m1 m2

Elastic collisions of equal mass

For m2 = c m1

v1f = [(1-c) v1i + 2cv2i] / (1+c)

v2f = [(c-1) v2i + 2v1i] / (1+c)


Collisions in one dimension given

given m1 and m2 or c = m2 /m1

v1i and v2i

Elastic (KE conserved)

v1f = [(1-c) v1i + 2c v2i] / (1+c)

v2f = [(c-1) v2i + 2v1i] / (1+c)

Inelastic (stick together, KE lost)

vf = [v1i + c v2i] / (1+c)

Momentum p = mvImpulse p = pf - pi = FtConservation of momentum (closed system) p1i + p2i = p1f + p2f


inelastic collision

With m2 = c m1

vf = [v1i + cv2i] / (1+c)


elastic collisions

For m2 = c m1

v1f = [(1-c) v1i + 2cv2i] / (1+c)

v2f = [(c-1) v2i + 2v1i] / (1+c)


Clicker Quiz

In a system with 2 moving objects when a collision occursbetween the objects:a) the total kinetic energy is always conservedb) the total momentum is always conservedc) the total kinetic energy and total momentum are always conservedd) neither the kinetic energy nor the momentum is conserved

For both inelastic and elastic collisions, the momentum isconserved.


Clicker QuizBA

Consider a closed system of objects A and B.Object A is thrown at an object B that was originally at rest, as shown

in the figure. It is not known if A is heavier or lighter than B. After A collides with B:

a) Object A must go left (back in the direction it came from)b) Object A must go rightc) Object A could go left or rightd) Object B could go lefte) The total momentum will be different than before the collision

If the collision is perfectly inelastic. A & B stick together and must go right (conservation of momentum). If the collision is elastic and mA<<mB A will go left. If mA>>mB both will go right. So A could go left or right. B can not go left (A would have to go left to, which violates momentum conservation.


Ballistic Pendulumblockm2=1 kgv2i = 0

bulletm1 = 0.1 kgvb = v1i = ? h=0.2m

What is the speed of the bullet?

There are 2 stages:• The collision• The swing up

The swing of the block Use conservation of Mechanical energy. (M g y + ½Mv2)bottom = (M g y + ½Mv2)top M = m1 +m2

0 + ½M(vf)2 = ½M(vb)2/(1+c)2 = Mgh gives (vb)2 = 2 g h (1+c)2

The collision The bullet gets stuck in the block (perfect inelastic collision). Use conservation of momentum.

c = (m2/m1) = 10 vf = (v1i + c v2i)/(1+c) = vb/(1+c)

vb = 21.8 m/s


Clicker Quiz

The kinetic energy of an object is quadrupled. Its momentum will change by a factor of…

a) 0b) 2c) 4d) 8e) 16


Impact of a meteoriteEstimate what happens if a 2.8x1013 kg meteorite collides with earth:

a) Is the orbit of earth around the sun changed?b) how much energy is released?Assume: meteorite has same density as earth, the collision isinelastic and the meteorite’s v is 10 km/s (relative to earth)

A) Earth’s mass: me = 6x1024kg mm = 2.8x1013 kg ve=0 vm=1x104 m/s

Conservation of momentum: c = me/mm= 2.14 x 1011

vf = vm /(1+c) vf=4.7x10-8 m/s (very small)B) Energy=Kinetic energy loss: (½meve

2+½mmvm2) - (½mmevme

2)

0 + 0.5 (2.8x1013) (1x104)2 - 0.5 (6x1024) (4.7x10-8)2 = 1.4x1021 J

Largest nuclear bomb existing: 100 megaton TNT = 4.2x1017 JEnergy release: 3300 nuclear bombs!!!!!


problemm1=5 kg m2=10 kg. M1 slides down and makesan elastic collision with m2. How high doesm1 go back up?

ME = mgh + (1/2)mv2

Step 1. What is the velocity of m1 just before it hits m2?Conservation of ME: MEtop= MEbottom = m1gh + 0 = 0 + 0.5m1v2

So v1 = 9.9 m/s

Step 2. Elastic collision with c = 2, v1i = 9.9 and v2i = 0

so v1f = -3.3

Step 3. m1 moves back up; use conservation of ME again.MEtop= MEbottom = m1gh’ + 0 = 0 + 0.5 m1 v1f

2

So h’ = 0.55 m

h=5m

m1

m2


Playing with blocksm1=0.5 kg collision is elasticm2=1.0 kgh1=2.5 mh2=2.0 m

A) determine the velocity of the blocks after the collisionB) how far away from the bottom of the table does m2 land


Determine the velocity of the blocks after the collision

m1=0.5 kg collision is elasticm2=1.0 kgh1=2.5 mh2=2.0 m

Step 1: determine velocity of m1 at the bottom of the slideConservation of ME (mgh+½mv2)top =(mgh+½mv2)bottom

0.5*9.81*2.5 + 0 = 0 + 0.5*0.5*v2

so: v1=7.0 m/sStep 2: Elastic collision equation with c = 2, v1i = 7 and v2i = 0

Gives v1f = -2.33 and v2f = 4.67


How far away from the table does m2 land?

m1=0.5 kg collision is elasticm2=1.0 kgh1=2.5 mh2=2.0 mv1f=-2.3 m/s v2f=4.7 m/s

This is a parabolic motion with initial horizontal velocity.Horizontal vertical

x = vxot y = yo + vyot - ½gt2

x = 4.7 t 0 = 2.0 + 0 - 0.5*9.81*t2

= 2.96 m t=0.63 s


atvv yy 0

221 attvy oy

221 attvy y

tvvy yoy )(21

)( 20

2

21

yya vvy

(A)

(B)

(C)

(D)

(E)

with a = -g


at the top all have vy = 0 (why?) at the top and bottom vx (Q) < vx (N) (why?)at the top vx (P) = vx (R) = 0 (why?)For all (PE+KE) (bottom) = (PE+KE) (top) mvo

2/2 = m(vox2 + voy

2)/2 = mgh + [mvx2/2]

For Q and N get time from (Eq. C)


a) Speed at X compared to V ? b) Velocity at X compared to P ? c) Size of total force at R compared to T ? d) Size of total force at Q compared to T ?


a) Speed at X compared to V ? X larger than Vb) Velocity at X compared to P ? samec) Size of total force at R compared to T ? T larger than Rd) Size of total force at Q compared to T ? Q larger than T


Remember at the a given time Power = P = F v = m a v


questionA girl (g) is standing on a plank (p) that is lying on ice (assume frictionless surface between ice and board). She starts to walk with 1 m/s, relative to the plank. What is her speed relative to the ice? Use mg=70 kg mp=100 kgChoose the ice as a frame of referenceInitial: both she and the plank are at rest.Final: vgp = vg – vp =1 vgp = velocity of girl relative to plank vg = velocity of girl relative to icevp = vg - vgp vp = velocity of plank relative to ice

Conservation of momentum: Initial (not moving): =0Final: mg vg+ mp vp = mg vg + mp(vg-vgp) = 70 vg + 100(vg-1) = 0

Solve for vg = 0.59 m/s

msu physics 231 fall 2015 1 physics 231 topic 6: momentum and collisions alex brown october 7 2015

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