A bomb moving with velocity $(40 \hat{i} + 50 \hat{j} - 25 \hat{k}) m / s$ explodes into two pieces of mass ratio $1 : 4$ . After explosion the smaller piece moves away with velocity $(200 \hat{i} + 70 \hat{j} + 15 \hat{k}) m / s$ . The velocity of larger piece after explosion is

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$(45 \hat{i} - 35 \hat{j})$

$(45 \hat{j} - 35 \hat{k})$

$(- 35 \hat{i} + 45 \hat{k})$

$(45 \hat{k} - 35 \hat{j})$

answer is A.

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Detailed Solution

Let the mass of the unexploded bomb be $5 m$ . m. It explodes into the two pieces of masses $m a n d 4 m$ respectively.

Initial momentum of the unexploded bomb $= 5 (40 \hat{i} + 50 \hat{j} - 25 \hat{k)}$

After explosion, momentum of the smaller piece $= m \vec{v_{1}}$

$= m (200 \hat{i} + 70 \hat{j} + 15 \hat{k)}$

and momentum of the larger piece $= 4 m \vec{v_{2}}$

where $\vec{v_{1}} a n d \vec{v_{2}}$ are the velocities of the two pieces respectively. According to the law of conservation of momentum, we get

$5 m (40 \hat{i} + 50 \hat{j} - 25 \hat{k)} = m (200 \hat{i} + 70 \hat{j} + 15 \hat{k)} + 4 m \vec{v_{2}} 4 m \vec{v_{2}} = 5 m (40 \hat{i} + 50 \hat{j} - 25 \hat{k}) - m (200 \hat{i} + 70 \hat{j} + 15 \hat{k)} \vec{v_{2}} = \frac{1}{4} (180 \hat{i} - 140 \hat{k}) = (45 \hat{j} - 35 \hat{k)}$