A bomb of mass $1 k g$ initially at rest, explodes and breaks into three fragments of masses in the ratio $1 : 1 : 3$ . The two pieces of equal mass fly off perpendicular to each other with a speed $15 m / s$ each. The speed of heavier fragment will be:

5 m / s

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15 m / s

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45 m / s

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5 \sqrt{2} m / s

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Solution

The correct option is D $5 \sqrt{2} m / s$
The correct option is D

Masses of fragments: $\frac{1}{5} \times 1 k g$ , $\frac{1}{5} \times 1 k g$ , $\frac{3}{5} \times 1 k g$

ie., $0.2 k g, 0.2 k g, 0.6 k g$

So, The linear momentum of that fragments:

$0.2 \times 15 m / s = 3 k g m / s, 3 k g m / s, 0.6 \times v k g m / s$

The vector sum of the first two: $3 \sqrt 2 k g - m / s$ as they two linear momenta are perpendicular. This will be in North East direction, if the two pieces fly off in North and East directions respectively.

So linear momentum of the large piece is in the direction of South West direction as the Linear momentum of all three pieces is 0 (as before explosion).

Velocity of the third piece = $\frac{3 \sqrt{2} k g m / s}{0.6 k g} = 5 \sqrt{2} m / s$
in South West direction.

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