Question

A simple pendulum of length 1 m has a wooden bob of mass $ 1 \,kg $ . It is struck by a bullet of mass $ 10^{-2}\, kg $ moving with a speed of $ 2 \times 10^2\, ms^{-1} $ . The height to which the bob rises before swinging back is (Take $ g = 10 \,ms^{-2} $ )

• A. $ 0.2\, m $
• B. $ 0.6\, m $
• C. $ 8\, m $
• D. $ 1\, m $

Answer 1

The Correct Option is A

Momentum of bullet $ = 10^{-2} \times 2 \times 10^2 = 2\, kg\, ms^{-1} $ . Let the combined velocity of the bob $ + $ bullet $ = v $ . Momentum of bob $ + $ bullet $ = (10^{-2} + 1)v = 1.01\, v $ . By conservation of momentum, $ 1.01\, v = 2\, kg\, ms^{-1} $ or $ v = \frac{2}{1.01}=1.98\,ms^{-1} $ . By conservation of energy $ \frac{1}{2}\left(M + m\right)v^{2} = \left(M + m\right)gh $ or $ = h = \frac{v^{2}}{2\,g} = \frac{\left(1.98\right)^{2}}{2 \times 10} $ $ = 0.19 \simeq 0.2\,m $