# I can't get started can someone help. Kinetic Energy in term of momentum

1. Sep 16, 2008

### Husker70

1. The problem statement, all variables and given/known data
The question is this:
Expressing the kinetic energy in terms of momentum (K=1/2mv2 = p2/2m),
prove using symbols, not numbers, that the fractional loss during the collision is
equal to M/(m+M)

2. Relevant equations

3. The attempt at a solution
I'm not sure where to even begin as I don't understand it
Thanks,
Kevin
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 16, 2008

### Dschumanji

"During the collision"

You have to be more specific, what kind of collision occured?

3. Sep 16, 2008

### Husker70

This question was asked from a lab for a ballistic pendulum experiment
Thanks,
Kevin

4. Sep 16, 2008

### ritwik06

Hi Kevin..
Thats relatively simple.
This is a case of a Ballistic Pendulum, therefore I take it as a complete;y inelastic collision, i.e. the bullet gets embedded.
Conserve the momentum:
mv=(M+m)(v1)
Let v1 b the velocity of the pendulum when the bullet gets embedded.
v1=mv/(M+m)
now find the change in kinetic energy:
$$\frac{1}{2}mv^{2}- \frac{1}{2}\frac{(M+m)m^{2}v^{2}}{(M+m)^{2}}$$

Simplify this to get:

Loss in KE = $$\frac{mMv^{2}}{2(M+m)}$$

Divide this by initial Kinetic energy to get the fraction loss:

=M/(M+m)

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