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 × 10² ms^–1 . The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g = 10 ms^-2 )

• A. 0.30 m
• B. 0.20 m
• C. 0.35 m
• D. 0.40 m

Answer 1

The Correct Option is B

Applying the conservation of momentum:

$mu = (M + m)V$

Substituting the given values:

$10^{-2} \times 2 \times 10^2 = (1 + 10^{-2}) \times V$

$V = 2 \, \text{ms}^{-1}$

The height to which the bob rises can be calculated using:

$h = \frac{V^2}{2g}$

$h = \frac{2^2}{2 \times 10} = 0.2 \, \text{m}$