Linear Momentum and Second Newton’s Law Definition of momentum: Change in momentum: 2 nd Newton’s Law: Definition of acceleration: We can write 2 nd Newton’s.

Linear Momentum and Second Newton’s Law

Definition of momentum: Change in momentum:

2nd Newton’s Law: Definition of acceleration:

amF

t

va

vmp

We can write 2nd Newton’s Law as:

1. Second Newton’s Law in terms of momentum:

t

pF

t

vmF

vmp

1

We can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time.

same change in momentum

This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete.

F

2

A. It decreases the change in momentumB. It decreases the force on the egg

Example: In an egg-tossing contest, two people toss a raw egg back and forth. After each successful toss, each person takes a step back. Catching the egg without breaking it becomes harder and harder. Usually the trick is moving your hand down with the egg when you receive it. This works better because:

• If the flying egg has speed v , the change in momentum is:

Δp = 0 – mv = –mv (independent of how you catch it)

• By moving your hand with the egg, you are increasing the time interval over which this Δp must take place. So the average force on the egg

•Catching the egg is harder and harder because its speed becomes larger (and the required change in momentum, too), so exerting a small force becomes harder as well.

tpFev / decreases.

3

2. System of particles. Conservation of momentum

exti

toti

FF

PP

t

PF tot

ext

constPPF tottotext

00

total momentum:

total external force:

The total momentum of an isolated system of objects remains constant!

1. Increases2. Does not change3. Decreases

Example: Suppose rain falls vertically into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating water, the speed of the cart

Mass is increasing P = mv must be conserved (Fext = 0)

Speed must decrease 4

1 2 1 2i i f fp p p p

12 1

2

75 kg0.2 m/s 0.17 m/s

90 kgf f

mv v

m

Compare changes in linear momentum: 1 1 1 1 (75 kg)(0.2 m/s - 0) 15 kg m/sf ip m v v

1 1 2 20 0 f fm v m v

2 2 2 2 (90 kg)( 0.17 m/s - 0) 15 kg m/sf ip m v v

Example: Two guys of masses m1 =75kg and m2=90kg pull on both ends of a rope on an ice rink. After a couple of seconds,the thin one is moving at 0.2 m/s. What is the speed of the big one?

No friction, no net vertical force Fext = 0 ptotal is conserved

75.0 kg

5

A BAv Bv

vvv BfAf

vmmvmvm

vmvmvmvm

BABBAA

BfBAfABBAA

)(

BA

BBAA

mm

vmvmv

Collisions (and explosions)

•Momentum is conserved in all collisions.•Collisions in which kinetic energy is conserved as well are called elastic collisions, and those in which it is not are called inelastic.•With inelastic collisions, some of the initial kinetic energy is lost to thermal or potential energy. It may also be gained during explosions, as there is the addition of chemical or nuclear energy.

1. Completely inelastic collisions(The objects stick together after collision, so there is only one final velocity)

6

Example: A ball of mass, m1 = 2 kg has a horizontal velocity, of v1 = 7 m/s. The ball collides into a cart full of sand, as shown below. The cart has a mass, m2 = 8 kg and a horizontal velocity v2 = 1 m/s. The ball and the cart are moving towards each other. Find the velocity of the cart and the ball after the ball collides with the cart and gets stuck in the sand.

m2

m1

v1

v2

m1 = 2 kgv1 = 7 m/sm2 = 8 kg v2 = 1 m/s v - ?

)(

)(

21

2211

212211

mm

vmvmv

vmmvmvm

smkgkg

smkgsmkgv /6.0

)82(

/18/72

7

A. The big block is initially at rest

B. The small block is initially at rest

C. The speed is the same in both cases

Example (Big block, small block): Consider the following two collisions between two blocks of masses m and M (> m). In both cases, one of the blocks is initially moving with speed v and the other is at rest. After the collision, they move together. The final speed of the two objects is larger when:

fMm vMmMvmv Mm

Mvmvv Mm

f

A.

Mm

mvv

v

vv

fA

M

m

0

B.

fAfB

M

m

vMm

Mvv

vv

v

0

8

p

p’ p+p’

rubber

p

putty

Example: You want to knock down a large bowling pin by throwing a ball at it. You can choose between two balls of equal mass and size. One is made of rubber and bounces back when it hits the pin. The other is made of putty and sticks to the pin. Which ball do you choose?

A. The rubber ballB. The putty ballC. It makes no difference

9

Example: Ballistic Pendulum A simple device for measuring the speed of projectile

M

01 U

h

;2

21

1

mvK ghmMU )(3

mM

;03 K

)()( 3322 UKUK ghmMvMm

)(2

)( 22

ghm

Mmv 2

)(1

1,vm mM

p1=mv1

2. Immediately after collision:

p2=(m+M)v2

;2

)( 22

2

vMmK

02 U

1. Before collision: 3. At the highest point:

p3=0

p1=p2 mv1 = (m+M)v2 21

)(v

m

Mmv

2v

10

2a. Elastic Collisions in 1D

2212

212

212

21

BfBAfABBAA

BfBAfABBAA

vmvmvmvm

vmvmvmvm

(1)

(2)

BfAfBA vvvv )( (2a)

Velocity of A relative to B after the collision

Velocity of A relative to B before the collision

Conservation of momentum:

Conservation of energy:

2. Elastic Collisions

The kinetic energy of the system is conserved:

after the collision it is the same as that before

11

Example: Carefully place a small rubber ball (mass m) on top of a much bigger basketball (mass M>>m) and drop these from some height h. What is the velocity of the smaller ball after the basketball hits the ground, reverses direction, and then collides with small rubber ball?

Remember that relative speed has to be equal before and after collision! Relative velocity has opposite direction.Before the collision, the basketball bounces up with v and the rubber ball is coming down with v, so their relative velocity is –2v. After the collision, it therefore has to be +2v!!This means that, after collision, the velocity of the smaller ball after is 3v.

v

v

m

M v

vv

3vV?

Relative speed is 2v Here too!

12

Example: A steel ball with mass m and initial speed v0 collides head-on with another ball of mass 2m that is initially at rest. What are the final speeds of the balls?

1) Conservation of momentum:

2a) Relative velocity:

Velocity of 1 relative to 2 before the collision

210 0 vvv

Velocity of 1 relative to 2 after the collision

210 20 mvmvmv

Adding these equations: 20 32 vv 032

2 vv 031

1 vv

13

A. One ball rises on the right, but higher than h

B. Two balls rise on the right at height h

C. A ball will rise on each side to height h

If two balls on the left are pulled to a certain height h and released, what happens?

Example (Newton’s cradle - a row of adjacent steel-ball )

pi = pf 2mvi = mvf vf = 2vi ,

but then kinetic energy would not be conserved:

Ki = 2½mvi2 = mvi

2 Kf = ½mvf2 = 2mvi

2

pi = pf 2mvi = 2mvf vf = vi.

Ki = 2 ½mvi2 = mvi

2 Kf = 2 ½mvf2 = mvi

2 Ok!

Same height means |vi| = |vf| (from conservation of energy), but this

violates conservation of momentum: pi = 2mvi ; pf = mvi – mvi = 0

No!

No!

14

3. What happens to the total kinetic energy?

Special case: Perfectly inelastic collisions When the objects stick together Example: Pin and putty

1. K is constant

2. K decreases

3. K increases

One body breaks into a number of parts. The explosion mechanism provides the extra energy.

Explosions

Superelastic collisions

Some internal energy is transformed into kinetic energy because of a collision.Example: An excited atom hits another atom and drops to a lower state without radiation.

Whenever a deformation is involved. Example: Most macroscopic collisions.

Inelastic collisions

When internal forces are conservative or objects are “hard”. Examples: Elementary particles, billiard balls.

Elastic collisions

15