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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

Option: 1
$45 \mathrm{~m}$

Option: 2
$90 \mathrm{~m}$

Option: 3
$125 \mathrm{~m}$

Option: 4
$25 \mathrm{~m}$

Answers (1)

best_answer

Assuming that the bullet undergoes constant retardation

$\mathrm{K E_i=90 \mathrm{~J}=\frac{1}{2} m u^2}$

$\mathrm{K E_f=40 \mathrm{~J}=\frac{1}{2} m v^2}$

$\mathrm{\frac{u}{v}=\frac{3}{2} \rightarrow(1)}$

Also, $\mathrm{90=\frac{1}{2}(0.2) u^2}$

$\mathrm{u=30 \mathrm{~m} / \mathrm{s}}$

$\mathrm{\therefore v=20 \mathrm{~m} / \mathrm{s} \quad (from\: eqn \: 1 )}$

$\mathrm{from \: t=0 \: to \: t=1 \mathrm{~s},}$

$\mathrm{ v=u+a t}$

$\mathrm{20=30+a(1)}$

$\mathrm{a=-10 \mathrm{~m} / \mathrm{s}^2}$

The time at which bullet stops,

$\mathrm{\therefore \quad v=0 =30+(-10) t }$

$\mathrm{ t=3 s }$

$\mathrm{S =u t+\frac{1}{2} a t^2 }$

$\mathrm{=30(3)+\frac{1}{2}(-10)(3)^2}$

$=90-45=45\mathrm{m}$

Hence 1 is correct option

Posted by

Rakesh

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