# Moving Reference Frames and elastic collision

1. Sep 1, 2007

### fudawala

1. The problem statement, all variables and given/known data
Consider a head-on, elastic collision between two bodies whose masses are m and M, with m << M. It is well known that if m has speed v0 and M is initially at rest, m will bounce straight back with its speed unchanged, while M will remain at rest (to an excellent approximation). Use this fact to predict the final velocities if M approaches with speed v0 and m is initially at rest.

2. Relevant equations
u = u' + v (the classical velocity addition formula)
Newton's Second Law: F = ma & F' = m'*a' (The two laws for the two fixed reference frames S and S')

3. The attempt at a solution
Basically, the way I would solve this problem is think that m and M are both masses. Since u = u' + v and u' = u - v, using Newton's First Law, small m is isolated from all outside forces so then the velocity u is constant relative to the lab. Then the velocity of M is going to be v0.

2. Sep 1, 2007

### Staff: Mentor

But what's the final velocity of the small mass?

Hint: View things from a frame in which M is at rest, then you can apply the given fact. Once you solve the collision in that frame, transform your answer back to the original frame.

3. Sep 1, 2007

### fudawala

Moving Reference Frame

I don't understand the Hint that you gave me. Can you give an example.

4. Sep 1, 2007

### Staff: Mentor

The principle of (galilean) relativity says that we can view things from any inertial frame we want. So pick a frame in which we know the answer. In the original frame of reference, m is at rest while M moves with velocity v0. Instead, view things from a frame in which M is at rest. In that frame, what is the initial velocity of m? The final velocity of m?

5. Sep 1, 2007

### fudawala

I think the initial velocity of m is going to be v0 when M is the fixed reference frame. The final velocity will be unchanged when m is bounced straight back without changing the velocity so I would think that the final velocity is constant.

Last edited: Sep 1, 2007
6. Sep 1, 2007

### Staff: Mentor

OK. Let's say that initially m was at rest (in the original frame) and M was moving at speed v0 to the right. Transforming to a moving frame in which M is at rest (just add -v0 to all velocities) gives m a velocity of -v0. In other words: m moves to the left with speed v0.
m does not come to a halt after colliding elastically with M: it bounces straight back with the same speed.

Remember, you are told this:

You are expected to use that fact.

7. Sep 1, 2007

### Staff: Mentor

The speed is unchanged (in the moving frame), but the velocity is not constant. Using what I said in the last post, what are the initial and final velocities of m in the moving frame?