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 m s⁻¹. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g = 10 m/s²) (a) 0.20 m (b) 0.40 m (c) 0.30 m (d) 0.35 m

Answer: (a). Momentum conservation: v = mu/m+M = 10⁻² × 2001.01 ≈ 1.98 m/s. Energy conservation after collision: h = v²/2g = ((1.98)²)/(20). h ≈ 0.196 m ≈ 0.20 m. So the answer is (a) 0.20 m.

A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is… | JEE Main 2024 (1 Feb, Shift 1)

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Simple Pendulum

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

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Answer: (a)

Momentum conservation: $v = \frac{mu}{m+M} = \frac{10^{-2} \times 200}{1.01} \approx 1.98\text{ m/s}$ .
Energy conservation after collision: $h = \frac{v^2}{2g} = \frac{(1.98)^2}{20}$ .
$h \approx 0.196\text{ m} \approx 0.20\text{ m}$ .
So the answer is (a) $0.20\text{ m}$ .

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