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:
- (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, $V=7 \sqrt{2} m / s$
