Physics 141. Lecture 15. No lecture on Tuesday 10/28. I...

Frank L. H. Wolfs Department of Physics and Astronomy, University of Rochester, Lecture 15,

Physics 141.Lecture 15.

No lecture on Tuesday 10/28.I will be across the Atlantic.



• Course information:• Homework set # 7 is due on Friday 11/4/2015.• Lab report # 3 is due on Friday 10/28/2015.

• Collisions (Chapter 10):• Elastic collisions of macroscopic objects.• Inelastic collisions of macroscopic objects.



No lecture on Thursday 10/27.I will be in the Netherlands.


This is where I grew up.This is where I used to ride my bike.


The Personal Response System (PRS).Quiz.


Chapter 10: Exploring collisions.Macroscopic to microscopic.

Motion of the center-of-mass of theball + bird system is not changed asa result of their interaction.


Collisions.The collision force.

• During a collision, a strong forceis exerted on the colliding objectsfor a short period of time.

• The collision force is usuallymuch stronger then any externalforce.

• The result of the collision force isa change in the linear momentumof the colliding objects.

• The change in the momentum ofone of the objects is equal to

p f −pi = dp

pi

p f

∫ =F t( )dt

ti

t f

∫


Collisions.The collision impulse.

• If we measure the change in thelinear momentum of an object wewill obtain information about theintegral of the force acting on it:

• The integral of the force is calledthe collision impulse J:

p f −pi = dp

pi

p f

∫ =F t( )dt

ti

t f

∫

Ji = dp

pi

p f

∫ =F t( )dt

ti

t f

∫



• Consider you are involved in acollision: you first move with 55mph and after the collision youare at rest.

• The change in momentum is thusfixed and the collision impulse isalso fixed:

• What happens to you depends onthe magnitude of the force! Anincrease in time Δt results in areduction of the force.

Ji =

F t( )dt

ti

t f

∫



• Interactions between sub-atomicparticles are usually studied bycomparing their momenta beforeand after an interaction.

• The change in their momentaprovides us with informationabout the collision impulse.

• Determining the force from thecollision impulse required aknowledge of the timedependence of the interaction.

BeforeAfter

Collision between protons.


Elastic and inelastic collisions.

• If we consider both collidingobject, then the collision forcebecomes an internal force and thetotal linear momentum of thesystem must be conserved if thereare no external forces acting onthe system.

• Collisions are usually dividedinto two groups:

• Elastic collisions: kinetic energyis conserved.

• Inelastic collisions: kineticenergy is NOT conserved.


Elastic and inelastic collisions.

• Kinetic energy does not need tobe conserved during the timeperiod that the collision force isacting on the system. The kineticenergy may become 0 J for ashort period of time.

• During the time period that thecollision force is non-zero, someor all of the initial kinetic energymay be converted into potentialenergy (for example, the potentialenergy associated withdeformation).


3 Minute 00 Second Intermission.

• Since paying attention for 1 hourand 15 minutes is hard when thetopic is physics, let’s take a 3minute 00 second intermission.

• You can:• Stretch out.• Talk to your neighbors.• Ask me a quick question.• Enjoy the fantastic music.• Solve a WeBWorK problem.


Elastic collisions in one dimension.

• Consider the elastic collisionshow in the Figure.

• Conservation of linearmomentum requires that

• Conservation of kinetic energyrequires that

• Two equations with twounknown!

V1iV2i = 0

Vcm

V1fV2f

(a)

(b)

(c)

m1v1i = m1v1 f + m2v2 f

12

m1v1i2 =

12

m1v1 f2 +

12

m2v2 f2



• The solution for the final velocityof mass m1 is:

• The solution for the final velocityof mass m2 is:

V1iV2i = 0

Vcm

V1fV2f

(a)

(b)

(c)

v1 f =

m1 − m2

m1 + m2

v1i

v2 f =

m1

m2

v1i − v1 f( ) = 2m1

m1 + m2

v1i


Elastic collisions in one dimension.Special cases.

• m1 = m2:

• v1f = 0 m/s• v2f = v1i

• m2 >> m1:

• v1f = - v1i• v2f = (2m1/m2) v1i

• m1 >> m2:

• v1f = v1i• v2f = 2v1i

V1iV2i = 0

Vcm

V1fV2f

(a)

(b)

(c)

Note: the motion of the centerof mass is not changed due to the collision.



• We can use an air track to study elastic collisions.

• The velocity of the carts can be determined if we measure the length oftime required to pass through the photo gates: velocity = length/time, orwe can use a photo gate to study the motion.

TEL-Atomic, http://www.telatomic.com/at.html



• Let us focus on specific examples where one cart (cart 2) is at rest:

• M1 = M2: v1f = 0 m/s, v2f = v1i

• M1 = 2M2: v1f = (1/3)v1i, v2f = (4/3) v1i

• 2M1 = M2: v1f = -(1/3)v1i, v2f = (2/3) v1i



Inelastic collisions in one dimension.

• In inelastic collisions, kineticenergy is not conserved.

• A special type of inelasticcollisions are the completelyinelastic collisions, where the twoobjects stick together after thecollision.

• Conservation of linearmomentum in a completelyinelastic collision requires that

m1vi = (m1+m2)vf

after

Vf

m1 + m2

Vi

m1 m2

before



• The final velocity of the system is equal to

• The final kinetic energy of the system is equal to

• Note: not all the kinetic energy can be lost, even in acompletely inelastic collision, since the motion of the centerof mass must still be present. Only if our reference frame ischosen such that the center-of-mass velocity is zero, will thefinal kinetic energy in a completely inelastic collision bezero.

v f =

m1

m1 + m2

vi

K f =

12

m1 + m2( )v f2 =

12

m1 + m2( ) m1

m1 + m2

vi

⎛

⎝⎜⎞

⎠⎟

2

=m1

m1 + m2

Ki



• Let us focus on one specific example of a procedure to measure thevelocity of a bullet:

• We shoot a 0.3 g bullet into a cart

• The final velocity of the cart is measured and conservation of linearmomentum can be used to determine the velocity of the bullet:

vi = (M1 + M2)vf / M1



Done for today! This is why I do not fly American Airlines.

Physics 141. Lecture 15. No lecture on Tuesday 10/28. I...

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