I was solving a homework question:

A body of mass M (Fig. 1.43) with a small disc of mass m placed on it rests on a smooth horizontal plane. The disc is set in motion in the horizontal direction with velocity v. To what height (relative to the initial level) will the disc rise after breaking off the body M? The friction is assumed to be absent.

diagram for given question

I was trying to solve this using conservation of energy, and relative velocities. I was thinking that the position of the centre of mass would be constant (at least horizontally) so the disc and body will acquire different velocities, respectively $\frac{Mv}{M + m}$ and $\frac{-mv}{M+m}$.

So my equation looked like: $$\frac{1}{2}mv^2 = \frac{1}{2}m(\frac{Mv}{M + m})^2 + \frac{1}{2}M(\frac{-mv}{M+m})^2 + mgh$$

With this I obtain $h = \frac{mv^2}{2g(M+m)}$, while the answer has $M$ in the numerator instead of $m$.

I looked at the solution, and it uses conservation of momentum like this: $$mv = (m+M)v_x$$

Can someone explain why the two bodies will move with the same velocity?

  • 1
    $\begingroup$ I do not see how you get $\frac{Mv}{M+m}$ and $\frac{-mv}{M+m}$ in the first place; their sum, if you want to talk about how fast the centre of mass should be moving, should be $\frac{mv}{M+m}$, which is what the answer key has. And they should be having the same horizontal velocity because the track ends vertically upwards, i.e. their relative motion there is purely vertical, hence no relative horizontal motion. $\endgroup$ Commented Aug 17, 2023 at 5:19

1 Answer 1


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Let's say there's a boat moving towards the left at velocity $v$, You're on the boat, and jump vertically into the air. Does the boat leave you behind?

It should be clear the answer is "no". That's because you and the boat are both moving at the same speed.

The same applies to the system in the question: when the little mass $m$ leaves the system, it is moving at the same speed as big mass $M$.


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