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

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

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.

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...

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

Rifle and Bullet Example The 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. The 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

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

Inelastic collision: momentum is still conserved 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

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 =

This cannon recoils quite a bit 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

Explosion Sample Problem: A 300 kg cannon fires a 10 kg projectile at 200 m/s. How fast does the cannon recoil backwards? BOOM! A 300 kg cannon fires a 10 kg projectile at 200 m/s. How fast does the cannon recoil backwards?

Solution Set up The 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! The 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. P before = P after BOOM

Calculation P after = P before mcannonvcannon + mprojvproj = 0 Before firing, velocity = 0m/s. P after = P before mcannonvcannon + mprojvproj = 0 (300 kg) (vcannon) + (10kg) (200m/s) = 0 After setting the momentum before equal to the momentum after, substitute known values and solve for the velocity of the cannon. Negative sign indicates the cannon moves in the opposite direction to the projectile vcannon = -6.67 m/s

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.

Hit and Stick Sample Problem Joe 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? Joe 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?

Hit and Stick Solution Biff’s initial velocity is zero, so this term drops out. Since they stick together, add their masses. Do math carefully After setting the momentum before equal to the momentum after, substitute known values and solve for the velocity of the football player combination. Since all velocities were in the same direction, no – signs are needed here.

Collisions do not always take place in a nice neat line: Often, collisions take place in 2 or 3 dimensions:

Another Example: One ball collides into another. By using momentum vector components, you can predict the result: After impact: Before impact: Total P before Y components cancel out X components add up to previous P

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

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. A After Collision 40o A B B 50o Before Collision

The X and Y components of momentum are both conserved 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.

Without using components, it can also be noticed that both momentum vectors after the collision add up to the original momentum vector: A B A B B Remember that vectors can be moved anywhere as long as their magnitude and relative direction are unchanged.

Problem Solution: Diagram NOT to scale! Use trig to find the momentum of ball B. Then find its velocity… A 40o B A B

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.

Questions??? Homework: Page 208 Problems #39,53,56,62