How do velocities become equal when potential energy is maximum? [closed]

Question

I was given this diagram and asked to find the maximum potential energy during the subsequent motion if one of the blocks is projected with a kinetic energy K initially.

So I did reach upon the answer by using the concept of conservation of momentum and conservation of total mechanical energy. BUT, while apply conservation of momentum I assumed the motion of both the blocks to be of equal velocity during the instant when the potential energy between them will be maximum (my lecturer said something like that in class the other day and also somewhat intuitive. )

So i wanted to know why both velocities would be equal at the inatant PE is max. Could someone ptovide me the mathematical derivation please.

Answer 1

Initially the left hand block is moving to the right with the right hand block at rest.
As time goes on the left hand block will slow down due to force on it due to the right hand block and the right hand block will speed up. During this time the separation of the blocks decreases and so the electric potential energy of the system is increasing.
This will continue until there is a time when the velocities of the blocks are the same and at that instant the relative velocity between the blocks is zero.
At this instant the separation of the blocks is a minimum and the electric potential energy is a maximum because after that instant the speed of the right hand block to the right will be greater than the speed of the left hand block to the right, ie their separation will now be increasing.

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The electric potential energy $U(r)=\dfrac {kq_1q_2}{r}$, where $r$ is the separation of the two bodies.

$\dfrac{dU(r)}{dt} = -\dfrac{kq_1q_2}{r^2}\cdot \dfrac{dr}{dt} $.

When $\dfrac{dr}{dt}=0$, ie the rate of separation of the two bodies is zero when they have equal velocities, the rate of change of potential energy with time is zero, ie at a turning point.

Answer 2

Your diagram shows the experiment in the lab frame i.e. for the observer standing in the lab watching the masses the left mass is moving with velocity $K$ and the right mass is stationary:

But suppose we switch to the centre of mass frame i.e. the frame in which the total momentum is zero. We can do this by moving at velocity $K/2$ relative to the lab, which is the same as subtracting $K/2$ from all velocities. In this frame the left mass has velocity $K/2$ and the right mass has velocity $-K/2$. This is shown in the middle diagram.

Now in the COM frame the velocities of the masses are always equal and opposite, so the moment of closest approach is when both masses have come to a halt, as shown in the bottom diagram. At this moment the velocities of the masses are both zero, so the velocities of the masses are both the same.

And finally to get back to the lab frame we just add back $K/2$ to all velocities, so in the lab frame at the moment of closest approach both velocities have the same value of $K/2$.

Answer 3

In this system potential energy is at a maximum when the distance between the bodies is minimal. And since velocity difference is the (time) derivative of distance ... ;)

Edit: Or, more explicitly (you still need to fill in the blanks...): Call $x_1$ and $x_2$ the positions of the blocks, $v_1$ and $v_2$ their velocities. Minimum distance $d=x_2 - x_1$ requires that its time derivative equals $$ \dot{d} = ...? $$ Now how is that related to the velocities?