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Q.

The bob of a pendulum of length 980cm is released from rest with its string making an angle 60° with the vertical. It collides with a ball of mass 20g resting on a smooth horizontal surface at the position of rest of pendulum. If the collision is perfectly elastic, with what velocity will the bob move immediately after the collision? Mass of the bob is 10g

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a

490 {cms}^{- 1}

b

\frac{2}{3} \times 980 {cms}^{- 1}

c

\frac{980}{4} {cms}^{- 1}

d

\frac{980}{3} {cms}^{- 1}

answer is D.

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Detailed Solution

Question Image

$m_{1} = 10 kg, m_{2} = 20 kg$

$u_{1} = ?, u_{2} = 0$

$v_{1} = \frac{2 m_{2} u_{2}}{m_{1} + m_{2}} + \frac{(m_{1} - m_{2}) u_{1}}{m_{1} + m_{2}}$

$v_{1} = (\frac{10 - 20}{10 + 20}) u_{1} = (\frac{- 10}{30}) u_{1} \to (1)$

$u_{1} = \sqrt{2gh}$

$u_{1} = \sqrt{2 \times 9.8 \frac{980}{2} \times 100}$

$u_{1} = \sqrt{960400}$

$u_{1} = 980 cm/s$

Substitute in equation(1)

$v_{1} = \frac{- 1}{3} \times 980 = \frac{980}{3} cm/s$

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The bob of a pendulum of length 980cm is released from rest with its string making an angle 60° with the vertical. It collides with a ball of mass 20g resting on a smooth horizontal surface at the position of rest of pendulum. If the collision is perfectly elastic, with what velocity will the bob move immediately after the collision? Mass of the bob is 10g