Question

Two bodies of same mass tied with an inelastic string of length l lie together. One of them is projected vertically upwards with velocity root 6gl . Find the maximum height up to which the centre of mass of system of the two masses rises.

Answer 1

Answer on Question#56940 - Physics – Mechanics | Relativity

Two bodies of same mass tied with an inelastic string of length $l$ lie together. One of them is projected vertically upwards with velocity $v_{i} = \sqrt{6gl}$. Find the maximum height up to which the centre of mass of system of the two masses rises.

Solution:

Applying the law of conservation of energy to the initial moment and the moment when the first body reaches the height $l$ we obtain:

$$\frac{m v_{i}^{2}}{2} = \frac{m v^{2}}{2} + m g l,$$

Where $v$ – is the speed of the first body at height $l$.

Thus

$$v = \sqrt{v_{i}^{2} - 2 g l} = \sqrt{6 g l - 2 g l} = 2 \sqrt{g l}$$

At this moment both bodies start to move as one piece with its centre of mass at height $l/2$. According to the law of conservation of energy the new speed of the system (mass is twice the mass of one body) is given by (the process is inelastic)

$$v' = \frac{v}{2} = \sqrt{g l}$$

After this moment the centre of mass of the system is elevated by the height $\Delta h$, which is given by

$$\Delta h = \frac{v'^{2}}{2 g} = \frac{g l}{2 g} = \frac{l}{2}$$

Thus the maximum elevation of the centre of mass is

$$h_{\max} = \frac{l}{2} + \Delta h = \frac{l}{2} + \frac{l}{2} = l$$

Answer: $l$.

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