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A particle of mass m is dropped from height 80 m after 2 sec it explodes into two fragments of mass ratio 2 : 3. If the lighter mass stops momentarily then velocity of heavier mass is:

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$20 {ms}^{- 1}$

$\frac{100}{3} {ms}^{- 1}$

$\frac{20}{3} {ms}^{- 1}$

$\frac{50}{3} {ms}^{- 1}$

answer is D.

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Detailed Solution

Velocity of particle after $20 \sec = u - gt$

$= - 20 \hat{j}$

Momentum will be conserved before and after explosion $P_{i} = P_{f}$

$\Rightarrow m \times (- 20) \hat{i} = \frac{2 m}{5} \times v_{1} + \frac{3 m}{5} v$

$\Rightarrow v_{2} = \frac{- 100}{3} \hat{i}$

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