Work Energy and Power - Result Question 68

71. A body of mass $5 kg$ explodes at rest into three fragments with masses in the ratio $1: 1: 3$. The fragments with equal masses fly in mutually perpendicular directions with speeds of $21 m /$ $s$. The velocity of heaviest fragment in $m / s$ will be

[1989]

(a) $7 \sqrt{2}$

(b) $5 \sqrt{2}$

(c) $3 \sqrt{2}$

(d) $\sqrt{2}$

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Answer:

Correct Answer: 71. (a)

Solution:

  1. (a) Momentum of Ist part $=m \times 21=21 m$

Momentum of 2nd part $=m \times 21=21 m$

Resultant momentum, $=\sqrt{(m 21)^{2}+(m 21)^{2}}$

Resultant momentum $=$ momentum of $3 rd$ part

$\Rightarrow 21 \sqrt{2} m=3 m v$

$\Rightarrow v=7 \sqrt{2} m / sec$

Masses of the pieces are 1, 1,3 kg. Hence $(1 \times 21)^{2}+(1 \times 21)^{2}=(3 \times V)^{2}$

That is, $V=7 \sqrt{2} m / s$