A bullet of mass 200 g having initial kinetic energy 90 J is shot inside a long swimming pool as shown in the figure. If its kinetic energy reduces to 40 J within 1 s, the minimum length of the pool, the bullet has to travel so that it completely comes to rest is (a) 45 m (b) 90 m (c) 125 m (d) 25 m

Answer: (a). 1/2(0.2)v² = K: K = 90 J ⇒ v₁ = 30 m/s; K = 40 J ⇒ v₂ = 20 m/s. Taking constant retardation: a = (30 - 20)/(1) = 10 m/s² Distance to stop: s = (v₁²)/(2a) = 900/20 = 45 m So the answer is (a) 45 m.

A bullet of mass 200 g having initial kinetic energy 90 J is shot… | JEE Main 2022 (28 Jul, Shift 2)

Q14JEE Main 2022 · 28 JUL · SHIFT 2Medium

Work-Energy Theorem

A bullet of mass $200\text{ g}$ having initial kinetic energy $90\text{ J}$ is shot inside a long swimming pool as shown in the figure. If its kinetic energy reduces to $40\text{ J}$ within $1\text{ s}$ , the minimum length of the pool, the bullet has to travel so that it completely comes to rest is

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Answer: (a)

$\frac{1}{2}(0.2)v^2 = K$ : $K = 90\text{ J} \Rightarrow v_1 = 30\text{ m/s}$ ; $K = 40\text{ J} \Rightarrow v_2 = 20\text{ m/s}$ .
Taking constant retardation: $a = \frac{30 - 20}{1} = 10\text{ m/s}^2$
Distance to stop: $s = \frac{v_1^2}{2a} = \frac{900}{20} = 45\text{ m}$
So the answer is (a) 45 m.

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