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Question: The bob A of a pendulum of mass m released from horizontal to the vertical hits another bob B of the...

The bob A of a pendulum of mass m released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in figure. If the length of the pendulum is 1 m, what is the speed with which bob B starts moving:

(Neglect the size of the bobs and assume the collision to be elastic)(Take g=10ms−2g = 10ms^{- 2})

A

4.47ms−14.47ms^{- 1}

B

5.47ms−15.47ms^{- 1}

C

6.47ms−16.47ms^{- 1}

D

3.47ms−13.47ms^{- 1}

Answer

4.47ms−14.47ms^{- 1}

Explanation

Solution

As the collision is elastic and two balls have the same mass, therefore ball A transfers its entire momentum to the ball B and ball A does not rise at all. After collision ball A comes to rest and ball B moves with speed of ball A.

∴\thereforeThe speed with which bob B starts moving is

v=2gh=2×10ms−2×1m=20ms−1v = \sqrt{2gh} = \sqrt{2 \times 10ms^{- 2} \times 1m} = \sqrt{20}ms^{- 1}

=4.47 m s-1