# Final speed of a two block system [closed]

Suppose a large block $$M$$ is moving in a smooth surface with speed $$v_o$$. Then a smaller block $$m$$ is carefully placed on it. Find the final speed

I tried two methods which gave two different answers

1. Kinetic energy conservation (Assuming no heat loss)
2. Linear momentum conservation, since net force on the system is zero.

Kinetic energy:

Initially smaller block $$m$$ is at rest, block $$M$$ has speed $$v_o$$ Finally both move with speed $$v$$ $$\frac{1}{2}M v_o^2 = \frac{1}{2}(M+m) v^2$$ $$v=\sqrt{\frac{M}{M+m}}v_o$$ Linear momentum: $$Mv_o=(M+m)v$$ $$v=\frac{M}{M+m}v_o$$ What did I do wrong to have two different results?

• @ACB yes apparently the momentum approach is the right one. Aug 11 at 11:11
• This situation is an example of an inherently inelastic process, which means that energy is never conserved no matter how ideal the experiment is. Can you see where the energy loss occurs in this case? Aug 11 at 12:34
• @Vincent Thacker Yes, I did not know that this would be considered an inelastic process. Aug 11 at 12:39

Why would you assume that kinetic energy was conserved? If the blocks stick together and move with a final speed of $$v$$, it follows naturally that there was friction involved between the two blocks. The internal energy of the system would increase (by a magnitude $$f_kd$$) becuase there is an increase in temperature of the two blocks (This effect is not negligible ; neglecting it is equivalent to assuming that the friction was absent, and consequently the block wouldn't stick to it)