Two blocks A and B of mass $m_{A}$ and $m_{B}$ are connected together by means of a spring and are resting on a horizontal frictionless table. The blocks are then pulled apart so as to stretch the spring and then released. Show that the ratio of their kinetic energies at any instant is in the inverse ratio of their masses.

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Solution

As the intial momentum of the system is zero (as they are relased from rest), and no external force acts on the system, the net momentum of the system should be $0$ at any instant in time.

Let us say v1 and v2 are the speeds of the mass $1 K g$ and $2 K g$ respectively,

So $1 \times v 1 = 2 \times v 2$

or $\frac{v 1}{v 2} = \frac{2}{1}$

And ratio of their momentum is $1$ .

Ratio of $K E = \frac{1 \times v 1^{2}}{2 \times v 2^{2}} = 2.$

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