A bullet embedded in a sliding block

[Identity]

Homework Statement

A bullet of mass m is fired with horizontal velocity u at a block of mass M. The bullet embeds itself inside the block. The force of resistance of the block is constant.
Find the depths of penetration in the cases:
a) The block is held fixed.
b) The block is free to move on a smooth horizontal surface

Homework Equations

Conservation of momentum? Kinematic equations? Conservation of energy? I don't know.

The Attempt at a Solution

Using simple kinematics for a), with the resistive force $$F = -k$$ of the block on the bullet, I get the depth, x, as

$$x = \frac{mu^2}{2k}$$

However, I'm having more trouble with part b)

I've tried a conservation of energy/conservation of momentum approach:

$$mu = (M+m)v \Rightarrow v = \frac{m}{M+m}u$$

Now, the work done by the bullet on the block is $$Fx$$, which translates into $$\frac{1}{2}Mv^2$$.

$$Fx = \frac{1}{2}Mv^2$$

$$kx = \frac{1}{2}Mv^2$$

$$x = \frac{Mv^2}{2k}=\frac{M}{2k}\left(\frac{M}{M+m}u\right)^2$$

However, this isn't correct, what's the proper way to do this?

Thanks

 

[Doc Al]

However, I'm having more trouble with part b)

I've tried a conservation of energy/conservation of momentum approach:

$$mu = (M+m)v \Rightarrow v = \frac{m}{M+m}u$$

This is good; v is the final speed of the 'bullet + block'.

Hint: Compare the initial and final kinetic energy of the system. (How much is 'lost'?)

 

[Identity]

Ah right!
$$Fx = \frac{1}{2}mu^2 - \frac{1}{2}(M+m)v^2$$

Thanks Doc Al :)