# Homework Help: Quick Collision Problem

1. May 18, 2010

### bon

1. The problem statement, all variables and given/known data

5. A ball of mass M travelling at non-relativistic speed v elastically collides with a
stationary ball of mass m. Show that the maximum speed which the second ball can
have after the collision is

2M/(M+m) v

What is the minimum speed?

2. Relevant equations

3. The attempt at a solution

Ok well i know it's an elastic collision so I know that momentum and energy are conserved... So if i label the final velocities of the two masses v1 and v2 I can write;

Mv = Mv1+ mv2

and 1/2Mv^2 = 1/2 M v1^2 + 1/2 mv2^2

But how do i now work out the maximum and minimum possible values for v2?

I know that the minimum speed would imply they stick together, but then the collision would no longer be elastic..So is it asking for maximum and minimum speeds GIVEN the collision is elastic - in which case I don't understand why there should be a range of possible speeds? Surely the system is determined..? Confused :S

2. May 18, 2010

### bon

anyone able to help with this?

3. May 18, 2010

### phyzguy

The speeds will depend on the angle between the two outgoing velocity vectors. You're assuming the collision is head-on, which it isn't necessarily. Try writing the final velocities as a function of this angle and find which angles give the largest and smallest speeds for m after the collision.

4. May 18, 2010

### bon

Ahh absolutely. Thanks

So I've tried solving this in the centre of mass frame, as it should be quicker..

i worked out Vcm = Mv/M+m

so I know that the magnitude of v2' = Mv/(M+m)..

I now need to choose the direction so as to maximise (and then minimise the speed in LABF) - So i can see that when v2' is parallel to Vcm the max speed will be what they stated, but my analysis also suggests that the minimum speed should be zero - which is counterintuitive..am i wrong?

I worked out it should be zero as follows:

Imagine we're in cmf - draw an arrow at some angle to horizontal to represent v2'...draw a horizontal arrow to represent Vcm - these are of the same magnitude..we need to "vector" add them to get v2 - but this means the possible values of v2 trace out a circle...and the minimum is 0..

where have i gone wrong? thanks!

5. May 18, 2010

### phyzguy

I think you're right - the minimum is zero.

6. May 18, 2010

### bon

Cool - how does that work then? Physically speaking...

(Also - am I right in thinking that there is no maximum scattering angle in CMF?)

Thanks!

7. May 18, 2010

### phyzguy

Physically the minimum of zero corresponds to an angular deflection of the incoming mass of zero degrees. If you think of two billiard balls, this is the limiting case where the incoming ball just barely grazes the stationary ball and imparts basically zero momentum. Yes, in the CM frame, the two masses are always oppositely directed, but can go out at any angle.