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A particle of mass $4 \mathrm{~m}$ explodes into three pieces of masses $m, m$ and $2 m$. The equal masses move along $\mathrm{X}$-axis and $\mathrm{Y}$-axis with velocities $4 \mathrm{~ms}^{-1}$ and $6 \mathrm{~ms}^{-1}$ respectively. The magnitude of the velocity of the heavier mass is
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Verified Answer
The correct answer is:
$\sqrt{13} \mathrm{~ms}^{-1}$
Let third mass particle $(2 m)$ moves making angle $\theta$ with $X$-axis.
The horizontal component of velocity of
$2 m$ mass particle $=u \cos \theta$
And vertical component $=u \sin \theta$
From conservation of linear momentum in $X$-direction
$m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}$ or $\quad 0=m \times 4+2 m(u \cos \theta)$
or $-4=2 u \cos \theta$ or $-2=u \cos \theta...(i)$
Again, applying law of conservation of linear momentum in $Y$-direction
$$
\begin{array}{l}
0=m \times 6+2 m(u \sin \theta) \\
\Rightarrow-\frac{6}{2}=u \sin \theta \text { or }-3=u \sin \theta...(ii)
\end{array}
$$
Squaring Eqs. (i) and (ii) and adding,
$$
\begin{aligned}
(4)+(9) &=u^{2} \cos ^{2} \theta+u^{2} \sin ^{2} \theta \\
&=u^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right) \\
\text { or } \quad 13 &=u^{2} \\
\therefore \quad u &=\sqrt{13} \mathrm{~ms}^{-1}
\end{aligned}
$$
The horizontal component of velocity of
$2 m$ mass particle $=u \cos \theta$
And vertical component $=u \sin \theta$
From conservation of linear momentum in $X$-direction
$m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}$ or $\quad 0=m \times 4+2 m(u \cos \theta)$
or $-4=2 u \cos \theta$ or $-2=u \cos \theta...(i)$
Again, applying law of conservation of linear momentum in $Y$-direction
$$
\begin{array}{l}
0=m \times 6+2 m(u \sin \theta) \\
\Rightarrow-\frac{6}{2}=u \sin \theta \text { or }-3=u \sin \theta...(ii)
\end{array}
$$
Squaring Eqs. (i) and (ii) and adding,
$$
\begin{aligned}
(4)+(9) &=u^{2} \cos ^{2} \theta+u^{2} \sin ^{2} \theta \\
&=u^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right) \\
\text { or } \quad 13 &=u^{2} \\
\therefore \quad u &=\sqrt{13} \mathrm{~ms}^{-1}
\end{aligned}
$$
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