Question 14 of 130

A body of mass $M$ at rest explodes into three pieces, in the ratio of masses $1:1:2$. Two smaller pieces fly off perpendicular to each other with velocities of $30{ms}^{-1}$ and $40{ms}^{-1}$ respectively. The velocity of the third piece will be

A$35m{s}^{-1}$B$50m{s}^{-1}$C$25m{s}^{-1}$D$15m{s}^{-1}$

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Answer: (C) $25m{s}^{-1}$

Let the masses of the three parts be $m',m'$ and $2m'$.
Momentum of first part ${p}_{1}=m'\times 30=30m'$ and momentum of second part ${p}_{2}=m'\times 40=40m'$.
Third part will fly in a direction opposite to the direction of resultant of ${p}_{1}$ and ${p}_{2}$, so that total momentum of the system after explosion remains zero.
Momentum of the third part is ${p}_{3}=\sqrt{{p}_{1}^{2}+{p}_{2}^{2}}=\sqrt{{\left(30m'\right)}^{2}+{\left(40m'\right)}^{2}}=50m'$
Velocity of third part, $v=\dfrac{50m'}{2m'}=$$25m{s}^{-1}$