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 (c) 14.7 J (d) 19.6 J

Answer: (b). Momentum: 0.2 × 10 = 10 v ⇒ v = 0.2 m/s Ki = 1/2(0.2)(10)² = 10 J, Kf = 1/2(10)(0.2)² = 0.2 J Loss = 10 - 0.2 = 9.8 J So the answer is (b) 9.8 J.

A bag of sand of mass 9.8 kg is suspended by a rope. A bullet of 200… | JEE Main 2022 (25 Jul, Shift 2)

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Collisions

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:

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Answer: (b)

Momentum: $0.2 \times 10 = 10\,v \Rightarrow v = 0.2\text{ m/s}$
$K_i = \frac{1}{2}(0.2)(10)^2 = 10\text{ J}$ , $K_f = \frac{1}{2}(10)(0.2)^2 = 0.2\text{ J}$
Loss $= 10 - 0.2 = 9.8\text{ J}$
So the answer is (b) 9.8 J.

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