A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed $10\,\mathrm{ms}^{-1}$ gets embedded in it, then loss of kinetic energy will be

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Conservation of Momentum

A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200 g travelling with speed

10 ms^{- 1}

gets embedded in it, then loss of kinetic energy will be

(A)4.9 J
(B)9.8 J✓ Correct
(C)14.7
(D)19.6 J

Answer: (B)

Solution

By conservation of momentum (no external horizontal force):

m_{b} v_{b} = (m_{b} + m_{s}) v_{f}

0.2 \times 10 = (0.2 + 9.8) v_{f}

v_{f} = \frac{2}{10} = 0.2 m/s

Initial kinetic energy:

K E_{i} = \frac{1}{2} m_{b} v_{b}^{2} = \frac{1}{2} \times 0.2 \times 100 = 10 J

Final kinetic energy:

K E_{f} = \frac{1}{2} (m_{b} + m_{s}) v_{f}^{2} = \frac{1}{2} \times 10 \times 0.04 = 0.2 J

Loss in kinetic energy:

Δ K E = 10 - 0.2 = 9.8 J

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option

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