Conservation of Momentum GOALS: Define MOMENTUM Understand how MOMENTUM is conserved in a collision or explosion. Use the principle of Conservation of Momentum to predict what happens during an interaction between objects.

Anything that is moving has MOMENTUM. MOMENTUM is defined as MASS times Velocity. The symbol for MOMENTUM is p p = m v

MOMENTUM is a VECTOR and so, just like velocity, MOMENTUM has direction. V=5 m/s M = 2 kg What is the MOMENTUM of this object? The number of the answer is 10. The units of the answer will be kg m/s The direction of the answer will be “to the right”.

DEMO: Newton’s Cradle From King’s Toy to Child’s Play Read the descriptions on the linked page (and watch the demos, as possible) for the 1-ball and 2-ball Newton’s Cradle ptr/chptr3_energy.htm

Some good online demos for collisions os/avimov/bychptr/chptr3_energy.htm The above link is the same as with the Newton’s Cradle discussions. Look at the demos describing the silver lab-car collisions (there is 1 at the top “Linear Momentum” and several in the middle). These might help with visualizations for the two major types of collisions.

M=2 kg M=5 kg V=10 m/s V=-3 m/s What is the total MOMENTUM of this “system” of two objects moving toward a collision? ANSWER: We have 20 kg m/s to the right and -15 kg m/s to the left The TOTAL MOMENTUM of this system is 5 kg m/s to the right.

M=2 kg M=5 kg V=10 m/s V=-3 m/s When two objects collide, they can either stick together in an “Inelastic Collision”, or… They can bounce off each other in an “Elastic Collision”.

M=2 kg M=5 kg V=10 m/s V=-3 m/s Suppose now these two objects collide, and stick together in an “Inelastic Collision”. What will the speed of these two objects be after the collision?

M=2 kg M=5 kg V=10 m/s V=-3 m/s “Inelastic Collision”. The combined mass will be 7 kg and the MOMENTUM will still be +5 kg m/s (to the right). This makes the velocity= p/m or m/s.

M=2 kg M=5 kg V=? m/s V=+2.25 m/s Suppose the more massive 5 kg object is turned around by an elastic collision and is moving to the right at m/s. What is the smaller 2 kg mass going to do after the “elastic collision”?

M=2 kg M=5 kg V=? m/s V=2.25 m/s Elastic Collision The final MOMENTUM is still going to be +5 kg m/s. The new MOMENTUM of the 5 kg mass is kg m/s, which is MORE than the total we started with.

M=2 kg M=5 kg V=? m/s V=+2.25 m/s So in order to have a total MOMENTUM of 5 kg m/s we have to add –6.25 kg m/s of momentum due to the 2 kg mass. The new velocity of the 2 kg mass is m/s, which is back to the left. P = kg m/s P = kg m/s

Impulse and Momentum (How we derive the “Impulse- Momentum Relationship”) F = ma F = m  v /  t F  t = m  v Impulse = momentum change [N-s] equivalent to [kg-m/s]

Ponder the following… a.k.a. “How would you look in a car crash if your airbag was a dashboard”?

Suppose this is a graph of the force of a tennis racket on a tennis ball during a serve. The force is not constant, but a momentum change will occur.

The area under the graph is the total impulse. In this case the impulse is given as 12.6 N-s. The tennis ball will therefore have a momentum change of 12.6 kg-m/s. If the ball has a mass of kg, what will the change in speed of the ball be?

The impulse and momentum change is 12.6 N-s.  v =  p / m  v = 12.6 N-s /.06 kg  v = 210 m/s If the ball has a mass of kg, what will the change in speed of the ball be?

Since we have cut this chapter short, you are responsible for only a small portion of Ch. 6 material. (Never fear—the underlying principles are very similar to what we’ve already discussed in all of the previous chapters). I would like you to be able to carry out the simple calculations behind momentum (p=mv). I would also like you to know the basic difference between elastic and inelastic collisions. Finally, you should also realize the idea of “conservation of momentum” has the same consequences of “conservation of energy” from Ch. 5 (a.k.a. “Total Momentum is BORING”)