Conservation of Momentum

A corner stone of physics is the conservation of momentum. This can be seen in all types of collisions.

2

Momentum Review Question:

Imagine a rubber and steel bullet each with the same mass and velocity. They each hit a wood block. The rubber bullet bounces off, while the steel bullet burrows into the block. Which one

moves the wood block more?

steel rubber

3

The steel bullet burrows into the block transferring all of its momentum to the block.

ΔP= mv It moves the block.

The rubber bullet bounces off transferring more momentum. If it bounces at the same

speed, but opposite direction, ΔP = 2mv. Thus, the block moves twice as much.

4

Conservation of Momentum:In all collisions or interactions, momentum of a

system is always conserved.

You may have previously learned about conservation of mass or energy from chemistry

class...

5

Since momentum is a vector quantity, direction must be taken into account to see that

momentum truly is conserved.

6

Rifle and Bullet ExampleThe rifle and bullet can be considered a system. Before

firing, they are both motionless and have a total momentum of 0.

After firing, the total momentum still equals 0. The rifle has momentum to the left, the bullet to the right. The rifle has a much larger mass so its velocity is less, but their momentum is still conserved.

mv mv

7

Types of Collisions:Elastic collision: momentum is conserved. The objects colliding aren’t deformed or smashed, thus no kinetic energy is lost. Ex: billiard ball

collisions

8

Inelastic collision: momentum is still conserved. Kinetic energy is lost. This often happens when

object interlock or stick together. The objects are also often deformed or crunched. Ex: car crash

9

Conservation of Momentum Problems:

When solving problems involving the conservation of momentum, the most important

thing to consider is:

Total momentum before collision

Total momentum after collision=

10

This cannon recoils quite a bit! The momentum of the projectile flying forwards must be equaled by the cannon itself recoiling backwards.

The movable parts of the cannon help reduce some of this effects by increasing the time of the recoil. Thus, the force is lessened.

P projectile P cannon

11

Explosion Sample Problem:

A 300 kg cannon fires a 10 kg projectile at 200 m/s. How fast does the cannon recoil

backwards?

BOOM!

12

Solution Set upThe momentum of the projectile must be

equal in size to the momentum of the cannon.

They must be equal since they must cancel each other out, initial momentum is 0!

BOOM

13

Calculation

P after = P before

mcannonvcannon + mprojvproj = 0

(300 kg) (vcannon) + (10kg) (200m/s) = 0

300kg

2000kgm/svcannon

vcannon = -6.67 m/s

Before firing, velocity = 0m/s.

Negative sign indicates the cannon moves in the opposite direction

to the projectile

14

Q: Why does the cannon move so much slower compared to the projectile?

A: It is much more massive, more inertia.

Q: What does the negative sign indicate?

A:The cannon moves in the opposite direction compared to the projectile.

15

Hit and Stick Sample ProblemJoe has a mass of 70kg and is running at 7 m/s with a football. He slams into 110kg Biff who was initially motionless. During this collision, Biff holds onto and tackles Joe. This type of event may be called a “hit and stick” collision. What is their resulting velocity after the collision?

16

Since they stick together, add their

masses.

Biff’s initial velocity is zero, so this term

drops out.

Hit and Stick Solution

afterbefore PP

BiffJoeBiffJoe PPP

3212211 )vm(mvmvm

32111 )vm(mvm

3110kg)v(70kg70kg(7m/s)

3v110kg)(70kg

70kg(7m/s)

2.7m/sv3

Do math carefully

Since all velocities were in the same direction, no – signs are needed here.

17

Collisions do not always take place in a nice neat line:

Often, collisions take place in 2 or 3 dimensions:

18

One ball collides into another. By using momentum vector components, you can predict

the result:

Another Example:

Before impact:

Total P before

After impact:

Y components cancel out

X components add up to previous P

19

It’s easiest to break the momentum into X and y components. Since momentum is always

conserved:

after Xbefore X PP

after Ybefore Y PP

20

Sample Problem:Two pool balls, each 0.50kg collide. Initally, the

first moves at 7 m/s, and the second is motionless. After the collision, the first moves 40o to the left of its original direction, the second moves 50o to the right of its original direction. Find both velocities

after the collision.

BA

Before Collision

A

B

After Collision

40o

50o

21

The X and Y components of momentum are both conserved. You can visualize this several ways:

A B

A

B

After the collision, the sum of the X components equals the original momentum. The y components cancel out

since there was no momentum in that direction originally.

22

Without using components, it can also be noticed that both momentum vectors after the collision

add up to the original momentum vector:

A

B

Remember that vectors can be moved anywhere as long as their magnitude and

relative direction are unchanged.

B

A B

23

Problem Solution:

A B

3.5kgm/s.5kg(7ms/)

mvPbefore

A

B

Diagram NOT to scale!

40o

kgm/s 3.5

B40sin o

m/s 4.5v

vkg .5kgm/s 2.25

mvP

kgm/s 2.25B

Use trig to find the momentum of ball B. Then find its velocity…

24

Now find the velocity of ball A: 5.36 m/s

Notice how the velocities of the balls don’t add up to the original velocity. Also, when added as scalars the momentums don’t add

up either. Only as vectors do the momentum vectors seem to be conserved.

25

Questions???

Homework:Page 208

Problems #39,53,56,62