momentum

Physics 211: Lecture 14, Pg 1

Physics 211: Lecture 14Physics 211: Lecture 14

Today’s AgendaToday’s Agenda

Motion of the center of mass, momentum conservation, F=ma

Collisions: you gotta conserve momentum! elastic or inelastic (energy conserving or not)

Inelastic collisions in one dimension and in two dimensions

Explosions

Comment on energy conservation

Ballistic pendulum

sum of all theexternal forces

acceleration ofthe center of mass


Center of Mass Motion: ReviewCenter of Mass Motion: Review

We have the following law for CM motion:

This has several interesting implications:

It tells us that the CM of an extended object behaves like a simple point mass under the influence of external forces: We can use it to relate FF and AA like we are used to doing.

It tells us that if FFEXT = 0, the total momentum of the system does not change. The total momentum of a system is conserved if there are

no external forces acting.

F P AEXT CMddt

M


(2)

Lecture 14, Lecture 14, Act 1Act 1Center of Mass MotionCenter of Mass Motion

Two pucks of equal mass are being pulled at different points with equal forces. Which experiences the bigger acceleration? (Puck 2 has many meters of rope wrapped around it.)

FFM

M

AA1

T

T

AA2

(a)(a) AA1 AA2 (b)(b) AA1 AA2 (c)(c) AA1 = AA2

(1)

Pucks


Linear Momentum:Linear Momentum:

Only the total external force matters!

m1

m3

m2

FF1,EXT

EXTNETi EXTidtd

,, FFP

CMEXTNET Mdtd APF ,

Which is the same as:

Newton’s 2nd law applied to systems!


We have just shown that MAA = FFEXT

Linear Acceleration depends only on external force, not on where it is applied!

Expect that AA1 and AA2 will be the same since F1 = F2 = T = F / 2

The answer is (c) AA1 = AA2.

So the final CM velocities should be the same!

Lecture 14, Lecture 14, Act 1Act 1SolutionSolution


The final velocity of the CM of each puck is the same! Notice, however, that the motion of the particles in each of the pucks is different (one is

spinning).

VV

VV

This one has morekinetic energy(rotation)

Lecture 14, Lecture 14, Act 1Act 1SolutionSolution


Momentum ConservationMomentum Conservation

The concept of momentum conservation is one of the most fundamental principles in physics.

This is a component (vector) equation. We can apply it to any direction in which there is no external

force applied. You will see that we often have momentum conservation (FEXT=0)

even when mechanical energy is not conserved. Elastic collisions don’t lose mechanical energy In inelastic collisions mechanical energy is reduced We will show that linear momentum must still be conserved, F=ma

F PEXT

ddt

ddtP 0 FEXT 0


Elastic vs. Inelastic CollisionsElastic vs. Inelastic Collisions

A collision is said to be elastic when kinetic energy as well as momentum is conserved before and after the collision. Kbefore = Kafter

Carts colliding with a spring in between, billiard balls, etc.

vvi

A collision is said to be inelastic when kinetic energy is not conserved before and after the collision, but momentum is conserved. Kbefore Kafter

Car crashes, collisions where objects stick together, etc.


Inelastic collision in 1-D: Example 1Inelastic collision in 1-D: Example 1

A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V :

What is the initial speed of the bullet v? What is the initial energy of the system? What is the final energy of the system? Is kinetic energy conserved?

v

before

V

after

x


Example 1...Example 1...

Consider the bullet & block as a system. After the bullet is shot, there are no external forces acting on the system in the x-direction. Momentum is conserved in the x direction!Momentum is conserved in the x direction!

Px, i = Px, f mv = (M+m)V

vV

initial final

v M mm

V

x



Now consider the kinetic energy of the system before and after:

Before:

After:

So

E mv m M mm

V M mm

M m VB

12

12

12

22

2 2

E M m VA 12

2

Kinetic energy is NOT conserved!Kinetic energy is NOT conserved! (friction stopped the bullet)However, momentum was conserved, and this was useful.

E mM m

EA B

v M mm

V


Inelastic Collision in 1-D: Example 2Inelastic Collision in 1-D: Example 2

M + m

vv = ?

M

m

VV v = 0ice

(no friction)



Before the collision:

Use conservation of momentum to find v after the collision.

After the collision:

P Vi M m ( )0 P vf M m ( )

Conservation of momentum:P Pi f

M M mV v ( )

Vv)mM(

M

vector equation

Air track



Now consider the K.E. of the system before and after:

Before:

After:

So E MV M M mM

v M mM

M m vBUS

12

12

12

22

2 2

E M m vA 12

2

E MM m

EA B

Kinetic energy is NOT conservedKinetic energy is NOT conserved in an inelastic collision!

Vv)mM(

M


Lecture 14, Lecture 14, Act 2Act 2Momentum ConservationMomentum Conservation

Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface.

The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck. Which box ends up moving faster?

(a)(a) Box 1 (b)(b) Box 2 (c)(c) same

1 2



Since the total external force in the x-direction is zero, momentum is conserved along the x-axis. In both cases the initial momentum is the same (mv of ball). In case 1 the ball has negative momentum after the collision, hence the box must have more positive momentum if the total is to be conserved. The speed of the box in case 1 is biggest!

1 2

x

V1 V2



1 2

xV1 V2

mvinit = MV1 - mvfin

V1 = (mvinit + mvfin) / M

mvinit = (M+m)V2

V2 = mvinit / (M+m)

V1 numerator is bigger and its denominator is smaller than that of V2.

V1 > V2


Inelastic collision in 2-DInelastic collision in 2-D

Consider a collision in 2-D (cars crashing at a slippery intersection...no friction).

vv1

vv2

VV

before after

m1

m2

m1 + m2


Inelastic collision in 2-D...Inelastic collision in 2-D...

There are no net external forces acting. Use momentum conservation for both components.

vv1

vv2

m1

m2

VV = (Vx,Vy)

m1 + m2

P Px i x f, , m v m m Vx1 1 1 2

V mm m

vx

1

1 21X:

P Py i y f, , m v m m Vy2 2 1 2

V mm m

vy

2

1 22y:



So we know all about the motion after the collision!

VV = (Vx,Vy)

Vx

Vy

V m

m mvx

1

1 21

V m

m mvy

2

1 22

1

2

11

22

x

y

pp

vmvm

VV

tan



We can see the same thing using momentum vectors:

tan pp

2

1

PP

pp1

pp2

PP

pp1

pp2


Explosion (inelastic un-collision)Explosion (inelastic un-collision)

Before the explosion:M

m1 m2

v1 v2

After the explosion:


Explosion...Explosion...

No external forces, so PP is conserved.

Initially: PP = 0

Finally: PP = m1vv1 + m2vv2 = 0

m1vv1 = - m2vv2 M

m1 m2

vv1 vv2

Rocket

Bottle


Lecture 14, Lecture 14, Act 3Act 3Center of MassCenter of Mass

A bomb explodes into 3 identical pieces. Which of the following configurations of velocities is possible?

(a)(a) 1 (b)(b) 2 (c)(c) both

m mvv VV

vv

m

m mvv vv

vv

m

(1) (2)



m mvv vv

vv

m

(1)

No external forces, so PP must be conserved. Initially: PP = 0 In explosion (1) there is nothing to balance the upward momentum

of the top piece so PPfinalfinal 0.

mmvvmmvv

mmvv



No external forces, so PP must be conserved. All the momenta cancel out.

PPfinalfinal = 0.

(2)

m mvv vv

vv

mmmvv

mmvv

mmvv


Comment on Energy ConservationComment on Energy Conservation

We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved. Mechanical Energy is lost: Where does it go??

» Heat (bomb)» Bending of metal (crashing cars)

Kinetic energy is not conserved since work is done during the collision!

Momentum along a certain direction is conserved when there are no external forces acting in this direction. In general, momentum conservation is easier to satisfy

than energy conservation.


Ballistic PendulumBallistic Pendulum

H

L LL L

m

M

A projectile of mass m moving horizontally with speed v strikes a stationary mass M suspended by strings of length L. Subsequently, m + M rise to a height of H.

Given H, what is the initial speed v of the projectile?

M + mv

V

V=0


Ballistic Pendulum...Ballistic Pendulum...

Two stage process:

1. m collides with M, inelastically. Both M and m then move together with a velocity V (before having risen significantly).

2. M and m rise a height H, conserving K+U energy E. (no non-conservative forces acting after collision)


Ballistic Pendulum...Ballistic Pendulum... Stage 1: Momentum is conserved

Stage 2: K+U Energy is conserved

in x-direction: mv m M V ( ) V mm M

v

Eliminating V gives: gH2mM1v

( )E EI F12

2( ) ( )m M V m M gH V gH2 2


Ballistic Pendulum DemoBallistic Pendulum Demo

H

L LL L

m

M

In the demo we measure forward displacement d, not H:

M + mv

d

L

d H

L-H L d L H

H L L d

2 2 2

2 2


Ballistic Pendulum Demo...Ballistic Pendulum Demo...

L

d H

L-H

Ld

LdLL

LdLL

dLLH

2211

2

2

2

2

2

22

dL

1for

gH2mM1v

Lgd

mM1v

for d << L

Let’s see who can throw fast...

Ballistic

pendulum


Recap of today’s lectureRecap of today’s lecture

Inelastic collisions in one dimension (Text: 8-6) Inelastic collisions in two dimensions (Text: 8-6)

Explosions

Comment on energy conservation

Ballistic pendulum (Ex. 8-14)

Look at textbook problems Look at textbook problems Chapter 8: # 21, 50, 69, 81, Chapter 8: # 21, 50, 69, 81, 87 87

momentum

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