Transcript lecture08

Linear Momentum and Second Newton’s Law

2 nd Newton’s Law:

F

m

a

Definition of acceleration: We can write 2 nd

F

m

Newton’s Law as:

v

 

t

a

  

v t

Definition of momentum:

p

m v

 

p

m

v

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

F

  

p

t

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.

F

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.

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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: A. It decreases the change in momentum B. It decreases the force on the egg • If the flying egg has speed

v

, the change in momentum is: Δ

p =

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 Δ

p

must take place. So the average force on the egg

F ev

 

p

/ 

t

decreases.

•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.

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2. System of particles. Conservation of momentum

F ext

  

P tot

t

total momentum: total external force:   

P i

F i

  

P tot

F ext

F

ext

 0   

P

tot

 0  

P

tot

const

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

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 1.

2.

3.

Increases Does not change Decreases Mass is increasing

P

=

mv

must be conserved (

F

ext = 0) Speed must decrease 4

75.0 kg Example:

Two guys of masses

m

1 =75kg and

m

2 =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

p

1

i

p

2

i

p

1

f

p

2

f F ext = 0

p

total

is conserved

m v

1 1

f

m v

2 2

f v

2

f

 

m m

2 1

v

1

f

  75 kg 90 kg 0.2 m/s   0.17 m/s Compare changes in linear momentum:   1

m v

1 1

f

v

1

i

  (75 kg)(0.2 m/s - 0)   2 2  2

f i

   15 kg m/s   15 kg m/s 5

Collisions (and explosions)

v v A B A B

•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)

v

Af

 

v Bf

v

m m A v

A v

A A

 

m B m B v

v

B B

 

m

(

m A v

Af A

 

m B m B

)

v

v

Bf

v

m A

v A m A

 

m B

v B m B

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Example:

A ball of mass,

m

1 = 2 kg has a horizontal velocity, of v 1 = 7 m/s. The ball collides into a cart full of sand, as shown below. The cart has a mass,

m

2 = 8 kg and a horizontal velocity v 2 = 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.

m

1

= 2 kg v

1

= 7 m/s m

2

= 8 kg v

2

= 1 m/s v - ?

m 1 v 1 v 2 m 2

m

1

v

1

v

 

m

1

m

2

v

2

v

1  (

m

1 

m

2

v

2

m

 2 ) (

m

1 

m

2 )

v v

 2

kg

 7

m

/

s

 8

kg

( 2

kg

 8

kg

)  1

m

/

s

 0 .

6

m

/

s

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Example (Big block, small block):

between two blocks of masses

m

Consider the following two collisions 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: A. The big block is initially at rest B. The small block is initially at rest

mv m

Mv M

 

m

M

v f

C. The speed is the same in both cases

v f

mv m

m

 

Mv M

M

A.

v m v M

v

 0

v fA

 

m mv

M

B.

v m v M

 0 

v v fB

 

m Mv

M

 

v fA

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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 ball B. The putty ball C. It makes no difference

p

putty

p p’ p+p’

rubber

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

m

,

v

1

M

M

m

v

2

M

m h

1. Before collision: 2. Immediately after collision: 3. At the highest point:

p

1

=mv

1 K

1 

mv

1 2 ; 2

U

1  0

p

2

=(m+M)v

2 K

2  (

m

M

)

v

2 2 ; 2

U

2  0

p

3

=0

K

3  0 ;

U

3  (

M

m

)

gh

p

1

=p

2

mv

1

(

K

2 

U

2 )  (

K

3 

U

3 )

= (m+M)v

2 v

1  (

m

m M

)

v

2 (

m

M

)

v

2 2 2  (

M

m

)

gh v

1  (

m

M

)

m

2

gh

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2. Elastic Collisions

The kinetic energy of the system is conserved: after the collision it is the same as that before

2a. Elastic Collisions in 1D Conservation of momentum: Conservation of energy:

m A v A

m B v B

m A v Af

m B v Bf

1 2

m A v

2

A

 1 2

m B v B

2  1 2 2

m A v Af

 1 2

m B

2

v Bf

(1) (2)

(

v A

v B

)   

v Af

v Bf

Velocity of A relative to B before the collision Velocity of A relative to B after the collision (2a)

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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?

m v

Relative speed is 2v Here too!

V?

3v

M v v v v

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 coming down with

v

, so their relative velocity is

v

and the rubber ball 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.

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Example:

A steel ball with mass

m

and initial speed

v

0 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:

mv

0

v

0  0 

mv

1  2

mv

2  0   

v

1 

v

2 

Velocity of 1 relative to 2 before the collision Velocity of 1 relative to 2 after the collision

Adding these equations

:

2

v

0  3

v

2

v

2  2 3

v

0

v

1   1 3

v

0 13

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

If two balls on the left are pulled to a certain height

h

and released, what happens?

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

h

p

i

= p

f

2mv

i

= mv

f

v

f

= 2v

but then kinetic energy would not be conserved:

i ,

K

i = 2½

mv

i 2 =

mv

i 2

K

f = ½

mv

f 2 = 2

mv

i 2 B. Two balls rise on the right at height

h

p

i

= p

f

K

i

= 2 ½mv

i 2

2mv = mv

i i 2

= 2mv

f

v

f

= v

i.

K

f

= 2 ½mv

f 2

= mv

i 2

No!

Ok!

C. A ball will rise on each side to height

h

Same height means

|v

i

| = |v

f

|

(from conservation of energy), but this violates conservation of momentum:

p

i

= 2mv

i

; p

f

= mv

i

– mv

i

= 0

14 No!

3. What happens to the total kinetic energy?

1.

K

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

2.

K

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

Special case: Perfectly inelastic collisions When the objects stick together Example: Pin and putty Explosions 3.

K

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

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.

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