physics 151: lecture 18, pg 1 physics 151: lecture 18 today’s agenda l topics çreview of momentum...

Physics 151: Lecture 18, Pg 1

Physics 151: Lecture 18Physics 151: Lecture 18

Today’s AgendaToday’s Agenda

TopicsReview of momentum conservation2-D CollisionsSystems of particles


Lecture 17 Lecture 17 ACT 4ACT 4

The law of conservation of momentum applies to a collision between two bodies if:

a. they exert forces on each other respectively proportional to their masses.

b. they exert forces on each other respectively proportional to their velocities.

d. their accelerations are proportional to their masses.

e. they exert equal and opposite forces on each other.


Lecture 18Lecture 18Review problem: Review problem: qualitative qualitative

Two boys in a canoe toss a baseball back and forth. What effect will this have on the canoe? Neglect (velocity-dependent) frictional forces with water or air.

a. None, because the ball remains in the canoe.

b. The canoe will drift in the direction of the boy who throws the ball harder each time.

c. The canoe will drift in the direction of the boy who throws the ball with less force each time.

d. The canoe will oscillate back and forth always moving opposite to the ball.

e. The canoe will oscillate in the direction of the ball because the canoe and ball exert forces in opposite directions upon the person throwing the ball.


Lecture 17, Lecture 17, ACT 3ACT 3Momentum 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 fastest ?

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

1 2


Lecture 17Lecture 17Review problemReview problem

A 3.0-kg mass is sliding on a horizontal frictionless surface with a speed of 3.0 m/s when it collides with a 1.0-kg mass initially at rest as shown in the figure. The masses stick together and slide up a frictionless circular track of radius 0.40 m. To what maximum height, h, above the horizontal surface will the masses slide?


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

VV = (Vx,Vy)

m1

m2

m1 + m2

b,xa,x PP m v m m Vx1 1 1 2

Vm

m mvx

1

1 21

b,ya,y PP m v m m Vy2 2 1 2

Vm

m mvy

2

1 22


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

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

V

Vtan PP

pp1

pp2

PP

pp1

pp2


Elastic CollisionsElastic Collisions

Elastic means that energy is conserved as well as momentum.

This gives us more constraints.We can solve more complicated problems !!Billiards (2-D collision).The colliding objects

have separate motionsafter the collision as well as before.

Start with a simpler 1-D problem.

Before After

See text: 9.4


Elastic Collision in 1-DElastic Collision in 1-D

v1,b v2,b

before

x

m1m2

See text: 9.4

v1,av2,a

afterm1

m2



v1,b v2,b

v1,av2,a

before

after

x

m1 m2

Conserve PX

m1v1,b + m2v2,b = m1v1,a + m2v2,a

Conserve Energy

1/2 m1v21,b + 1/2 m2v2

2,b = 1/2 m1v21,a + 1/2 m2v2

2,a

Suppose we know v1,b and v2,b

We need to solve for v1,a and v2,a

Should be no problem 2 equations & 2 unknowns !

See text: 9.4

m1v1,b + m2v2,b = m1v1,a + m2v2,a



After some moderately tedious algebra, (see text book Chapter 9, section3) we can derive the following equations for the final velocities,

See text: 9.4

BBAv

mmm

vmmmm

v2

21

2

1

21

21

1)

2()(

BBAv

mmmm

vmmm

v2

21

12

1

21

1

2)()

2(

1) m1v1,b + m2v2,b = m1v1,a + m2v2,a

2) v1,b - v2,b = - (v1,a - v2,a)

In general:


Example - Elastic CollisionExample - Elastic Collision

Suppose I have 2 identical bumper cars. One is motionless and the other is approaching it with velocity v1. If they collide elastically, what is the final velocity of each car ?

Note that this means,

m1 = m2 = m

v2B = 0

See text: 9.4

Animation


Lecture 18,Lecture 18, ACT 3ACT 3Elastic CollisionsElastic Collisions

I have a line of 3 bumper cars all touching. A fourth car smashes into the others from behind. Is it possible to satisfy both conservation of energy and momentum if 2 cars are moving after the collision?All masses are identical, elastic collision.

A) Yes B) No

Before

After? Animation


Example of 2-D elastic collisions:Example of 2-D elastic collisions:BilliardsBilliards

If all we know is the initial velocity of the cue ball, we don’t have enough information to solve for the exact paths after the collision. But we can learn some useful things...

See text: Ex. 9.11


BilliardsBilliards

Consider the case where one ball is initially at rest.

ppa

ppb

FF PPa

before afterthe final direction of the

red ball will depend on

where the balls hit.

vvcm

See Figure 12-14

See text: Ex. 9.11


We know momentum is conserved: ppb = ppa + PPa

We also know that energy is conserved:

Comparing these two equations tells us that:

BilliardsBilliards

pb2 = (ppa + PPa )2 = pa

2 + Pa2 + 2 ppa PPa

m2

P

m2

p

m2

p 2a

2a

2 b

ppa PPa = 0

and must therefore be orthogonal!

Or … one momentum must be zero.

ppa

ppb

PPa

Ppp 2a

2a

2 b

See text: Ex. 9.11


BilliardsBilliards

The final directions are separated by 90o.

ppa

ppb

FF PPa

before after

vvcm

See text: Ex. 9.11


Lecture 18 –Lecture 18 – ACT 4ACT 4Pool SharkPool Shark

Can I sink the red ball without scratching ?Ignore spin and friction.

See text: Ex. 9.11

A) Yes B) No C) More info needed


Lecture 18Lecture 18 – ACT 4 – ACT 4Pool SharkPool Shark

From above, after the collision the two balls move off at right angles.

Thus if the red ball goes toward a pocket, so does the cue ball

See text: Ex. 9.11

B) No


Billiards.Billiards.

More generally, we can sink the red ball without sinking the white ball – fortunately.

See text: Ex. 9.11

Animation


Recap of today’s lectureRecap of today’s lecture

Momentum and Collisions Ch. 9.1-9.4 (part of 9.4)

physics 151: lecture 18, pg 1 physics 151: lecture 18 today’s agenda l topics çreview of momentum...

Documents

masses slide

vv1vv1 vv2vv2 v v

m1m1 m2m2 slide

inelastic collision

v y m1m1 m2m2

opposite forces

d cars

x m1m1 m2m2