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 )

Updated On: Nov 15, 2024
  • 0.30 m
  • 0.20 m
  • 0.35 m
  • 0.40 m
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The Correct Option is B

Solution and Explanation

Applying the conservation of momentum:

mu=(M+m)Vmu = (M + m)V

Substituting the given values:

102×2×102=(1+102)×V10^{-2} \times 2 \times 10^2 = (1 + 10^{-2}) \times V

V=2ms1V = 2 \, \text{ms}^{-1}

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

h=V22gh = \frac{V^2}{2g}

h=222×10=0.2mh = \frac{2^2}{2 \times 10} = 0.2 \, \text{m}

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