Question

A bob of mass m attached with a string of length l tied to a point on ceiling is released from a position when its string is horizontal. At the bottom most point of its motion, an identical mass m gently stuck to it. Find the angle from the vertical to which it rises.

Answer 1

Answer on Question#56937 - Physics – Mechanics | Relativity

A bob of mass $m$ attached with a string of length $l$ tied to a point on ceiling is released from a position when its string is horizontal. At the bottom most point of its motion, an identical mass $m$ gently stuck to it. Find the angle from the vertical to which it rises.

Solution:

Applying the law of conservation to the initial moment and the moment right before the collision with the second bob, we obtain:

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

Where $v$ – is the speed of the first bob right before the collision.

Therefore

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

Since the collision is inelastic, then according to the law of conservation of momentum the new speed $v'$ of the system is

$$v' = \frac{m v}{2 m} = \frac{1}{2} \sqrt{2 g l} = \sqrt{\frac{g l}{2}}$$

According to the law of conservation of energy the final elevation of the system is given by

$$h = \frac{v'^2}{2 g} = \frac{g l / 2}{2 g} = \frac{l}{4}$$

Thus two bobs at moment of maximum elevation are $l - h = \frac{3}{4} l$ below the ceiling.

The required angle is given by

$$\theta = \arccos \frac{l - h}{l} = \arccos \frac{3}{4} \approx 41.41{}^\circ$$

Answer: $\arccos \frac{3}{4} \approx 41.41{}^\circ$.

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