Kinetic Energy of a Block-Bullet System A bullet of mass $m$ is fired towards a wooden block of mass $M$. At a particular instant of time when the bullet is inside the block, the speed of the block is $V$ and the speed of the bullet, relative to the block, is $v$. I would like to find the total kinetic energy of the system at this point.
Considering the bullet and block as separate entities, it should be $$\frac 12 MV^2 +\frac 12 m(v+V)^2$$
But I could also look at the bullet and block as one body with velocity $V$, and then add the extra velocity of the bullet which has not been accounted for: $$\frac 12 (M+m)V^2 +\frac 12 m v^2 $$
Which one of these is correct?
 A: The kinetic energy of system depends on the choice of reference frame.  To compare the energy of different objects or use conservation of energy, each energy must be defined in the same reference frame.
The most common choice would be to define a "lab frame."  This is your reference frame, standing at rest next to the experiment happening.  In this case all velocities used in the kinetic energy equations should be measured relative to you.
The total kinetic energy is the sum of the individual energy of each particle
$$ K = \frac{1}{2}MV^2 + \frac{1}{2}mu^2,$$
where $V$ is the speed of the block relative to you and $u$ is the speed of the bullet relative to you.
Instead of knowing the bullet's speed relative to you directly, you know the bullet's speed relative to the block.  We can rewrite the bullet's speed relative to you in terms of the variables you care about
$$u = v + V.$$
Putting it all together, the correct way to account for the kinetic energy of the system would be
$$K = \frac{1}{2}MV^2 + \frac{1}{2}m(v+V)^2$$
