Work Energy Power

JEE Main level, single correct, medium

80 of 183

Question 80 of 183

A bullet of mass 200g200g having initial kinetic energy 90J90J is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to 40J40J within 1s1s, 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) 45m45m

Let us assume that the motion of the bullet is of constant retardation inside the water.
Initial kinetic energy 12mv02=90J\dfrac{1}{2}m{v}_{0}^{2}=90J
⇒v0=30ms−1\Rightarrow {v}_{0}=30m{s}^{-1}
Let the retardation is aa, then after 1s1s
12m(30−a×1)2=40J\dfrac{1}{2}m{\left(30-a\times 1\right)}^{2}=40J
⇒30−a=20\Rightarrow 30-a=20
⇒a=10ms−2\Rightarrow a=10m{s}^{-2}
Stopping distance for the constant retardation is given by,
l=u22a=30220=45ml=\dfrac{{u}^{2}}{2a}=\dfrac{{30}^{2}}{20}=45m.
Therefore, the minimum length of the pool required is 45m45m.

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