# Kinetic Energy

1. Aug 5, 2009

### AntiStrange

1. The problem statement, all variables and given/known data
A particle with mass 'm' and kinetic energy 'K' collides with a stationary particle of mass M. Then it says to find the total kinetic energy in the center of mass system 'Kcm'.
$$K_{cm} = \frac{M}{M + m}K$$

2. Relevant equations
conservation of momentum: total initial momentum = total final momentum
KE = (1/2)mv^2

3. The attempt at a solution
I tried using conservation of momentum:
$$m(v_{1}) = (m + M)v_{2}$$
where v1 is the velocity of the particle with mass 'm' and v2 is the velocity after the collision of both particles moving together (assuming this happens).

Then I just solved for v2 and tried to find the kinetic energy now which I assumed would be Kcm:
$$K_{cm} = \frac{m}{2(m + M)}K$$
which looks pretty close except it's supposed to be a M on top, not a m, and there is no 2 in the denominator.

2. Aug 5, 2009

### tms

Your attempted solution assumes that both bodies move with the same speed after the collision. That will not generally be the case. You will need to use both the momentum and energy equations in the lab frame, find the motion of the CM in the lab frame, then transform to the CM frame and calculate the energies there.

3. Aug 6, 2009

### prob_solv

@AntiStrange : I have calculated the equation, and I find the my answer is the same as the answer key assuming that the collision is inelastic ( tow particles moving together ). My hypothesis is that you made a mathematics mistake ( equation mistake ) .

@tms : We can't do this question without assuming that it's an inelastic collision. Because if it's an elastic collision, the Kcm must be the same as K ( no energy is gone )

4. Aug 6, 2009

### kuruman

The problem says nothing about calculating the energy after the collision. Just calculate the kinetic energy of the center of mass before the collision starting from

$$K_{CM}=\frac{1}{2}(M+m)V^{2}_{CM}$$

and expressing the VCM in terms of K.

After the collision, if the collision is elastic, the CM energy will be the same. If the collision is perfectly inelastic, the CM energy will be zero.

5. Aug 7, 2009

### tms

First, if the collision is inelastic, there is no way of telling, from the information given, how much energy is lost, so the problem can't be solved.

Second, kinetic energy is not an invariant under the Galilean transformation. Just consider two different masses, and look at the kinetic energy in the two frames in which one or the other mass is at rest.

Last edited: Aug 7, 2009