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Q.

At a certain height, a body at rest explodes into two equal fragments with one fragment receiving a horizontal velocity of $10 \sqrt{3} m s^{- 1}$ . The horizontal distance between the two fragments, when their displacement vectors are inclined at $60^{0}$ relative to each other is $(g = 10 m s^{- 2})$

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a

$40 \sqrt{3} m$

b

$240 \sqrt{3} m$

c

$60 \sqrt{3} m$

d

$480 \sqrt{3} m$

answer is C.

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Detailed Solution

Question Image

When bomb explodes at rest

${\vec{P}}_{i} = {\vec{P}}_{f}$

$O = m u_{1} \hat{i} + {\vec{P}}_{2}$

$\Rightarrow {\vec{P}}_{2} = - m u_{1} \hat{i}$

$\Rightarrow m {\vec{u}}_{2} = - m u_{1} \hat{i}$

$∴ {\vec{u}}_{2} = - u_{1} \hat{i}$

$\Rightarrow u_{2} = 10 \sqrt{3} m / s$

From $Δ O A C :$ $\tan 30 = \frac{u t}{y}$

$\frac{1}{\sqrt{3}} = \frac{u t}{\frac{1}{2} g t^{2}} \Rightarrow t = \frac{2 \sqrt{3} u}{g}$

$∴ A B = u_{1} t + u_{2} t = 2 u t$

$= 2 \times 10 \sqrt{3} \times \frac{2 \sqrt{3} \times 10 \sqrt{3}}{10}$

$∴ A B = 120 \sqrt{3} m$

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