JEE MAIN - Physics (2022 - 28th July Evening Shift - No. 5)
A bullet of mass $$200 \mathrm{~g}$$ having initial kinetic energy $$90 \mathrm{~J}$$ is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to $$40 \mathrm{~J}$$ within $$1 \mathrm{~s}$$, the minimum length of the pool, the bullet has to travel so that it completely comes to rest is
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45 m
90 m
125 m
25 m
Explanation
$${1 \over 2}m{x^2} = 90$$
$$ \Rightarrow {1 \over 2} \times 0.2 \times {x^2} = 90$$,
$${x^2} = 900$$
$$x = 30$$ m/s
$${1 \over 2}m{v^2} = 40 \Rightarrow v = {2 \over 3} \times 30 = 20$$ m/s
$$20 = 30 - a \times 1 \Rightarrow a = - 10$$ m/s2
$$0 - {x^2} = 2as$$
$$s = {{{x^2}} \over { - 2a}} = {{30 \times 30} \over {2 \times 10}}$$
$$ = 45$$ m
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