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?

  • $\begingroup$ @ACB yes apparently the momentum approach is the right one. $\endgroup$
    – Eyy boss
    Aug 11 at 11:11
  • 1
    $\begingroup$ 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? $\endgroup$ Aug 11 at 12:34
  • $\begingroup$ @Vincent Thacker Yes, I did not know that this would be considered an inelastic process. $\endgroup$
    – Eyy boss
    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)

This does not mean that the conservation of energy is violated. You will have to include the transfer of energy by heat and simply kinetic energy alone is not conserved. This however is impossible to calculate from the information given

The isolated system for momentum only works in the x direction because there are no external forces acting in the x direction:

  • Friction was internal to the system
  • The only external force was possibly while putting the upper block on the lower one. It acts in the y-direction and will not affect the momentum in the x direction.

Hope this helps


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