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

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Conservation of Momentum

A simple pendulum of length 1 m has a wooden bob of mass 1 kg . It is struck by a bullet of mass

1 0^{- 2} kg

moving with a speed of

2 \times 1 0^{2} ms^{- 1}

. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use

g = 10 m / s^{2}

)

(A)0.30 m
(B)0.20 m✓ Correct
(C)0.35 m
(D)0.40 m

Answer: (B)

Solution

By conservation of momentum during the perfectly inelastic collision (bullet embeds in bob):

m u = (M + m) V ⟹ 1 0^{- 2} \times 2 \times 1 0^{2} = (1 + 1 0^{- 2}) V ⟹ V \approx 2 m/s

By conservation of energy as the bob-bullet system swings up:

h = \frac{V ^{2}}{2 g} = \frac{4}{20} = 0.20 m

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Single correct option

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