# Relative velocity, momentum conservation

I am having difficulty in conserving momentum for objects moving with relative velocities. For example:

A block of mass $$m$$ with a semicircular track of radius $$r$$ rests on a frictionless horizontal surface. A ball of radius $$R$$ and mass $$M$$ is released from the top point. What is the speed of the block when the ball has reached the bottom of the track?

My approach:

The ball falls from a height of $$r$$ to $$(r-R)$$. Let the velocities of block and ball be $$v_2$$ and $$v_1$$ resp (from the ground frame).

By energy conservation: $$Mg(r-R)=\left(\frac12mv_2^2+\frac12Mv_1^2\right)$$

By momentum conservation: $$0=mv_2+Mv_1$$

Am I wrong here?

Second approach:

Let final velocity of ball wrt to block be v. Let final velocity of block be v2(wrt ground). by energy conservation(wrt block):

$$Mg(r-R)=\frac12Mv^2$$

By momentum conservation: $$0=mv_2+M(v+v_2)$$

Will the answer be same, or am I wrong somewhere?

• I don’t understand your description. A picture would help.
– Dale
Aug 25 '19 at 12:41

Kinetic energy does not have to be conserved. You don't know if it is or not until you calculate the final velocities. Internal forces (between ball and track) can change kinetic energy. But if you assume KE is conserved then you need both equations (momentum and KE conservation) to calculate final velocities. But your expression for heights seems wrong. Only a picture will clarify the setup.

In respect to the block there is no momentum of the block. The momentum of the ball is not conserved. Neither its KE. So your second approach does not really work.

• the momentum of the block is conserved in the second case as its written from ground frame of reference and no external force is acting in x direction.I will provide a picture. Aug 25 '19 at 17:09
• In the reference frame of the block the block's velocity is zero and so is block's momentum. And this is so at any time.
– nasu
Aug 25 '19 at 17:44

You are almost right with the first approach. You miss one aspect which is that the ball of mass $$M$$ will probably be rolling and so there is a rotational kinetic energy which has to do with its angular velocity, adding some term like $$\frac15 M (v_1 -v_2)^2$$ or so. It would help to instead imagine something more like a toy racecar on a racecar track?

The second approach has got to be wrong. It takes the first approach and puts it into a reference frame moving with velocity $$v_2$$, defining $$v=v_1-v_2$$. That aspect is actually fine, but you have to be careful with this reference frame because you are saying that there is no kinetic energy to start with, and that part is wrong.

Instead for the energy balance one has (skipping $$v$$ for the moment to illustrate how it must all turn out),$$Mg(r-R) +\frac12 (M+m) v_2^2 = \frac12 M (v_1-v_2)^2$$ Note that after expansion you have two different expressions that must equal the same number $$Mg(r-R)$$! This means that we must have,$$M (v_1-v_2)^2 - (M + m) v_2^2 = M v_1^2 + m v_2^2$$ it turns out that those two expressions are not symbolically equal, so this is not a trivial equation $$0=0$$. So this equation is indeed telling you that energy balances only work between reference frames if a certain criterion holds. So let me give you this exercise: If you expand this out, what do you get? And: suppose that instead of choosing a reference frame moving at $$v_2$$, what if it were just some unknown velocity $$u$$, does that change anything?