Question

A particle of mass $4 \,m$ explodes into three pieces of masses $m, m$ and $2 \,m$. The equal masses move along $X$-axis and $Y$-axis with velocities $4\, ms ^{-1}$ and $6\, ms ^{-1}$ respectively. The magnitude of the velocity of the heavier mass is

• A. $\sqrt{17}\, ms^{-1}$
• B. $2 \sqrt{13}\,ms^{-1}$
• C. $\sqrt{13}\,ms^{-1}$
• D. $\frac{ \sqrt{13}}{2} \,ms^{-1}$

Answer 1

Answer: (C)

$\sqrt{13}\,ms^{-1}$

Answer 2

Let third mass particle $(2m)$ 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 $0 = m\times4 + 2m \left( u \cos\theta\right)$
or $-4 = 2u \cos\theta -2 = u \cos\theta$ ....(i)
Again, applying law of conservation of linear momentum in Y-direction
$0 = m \times6+2m \left(u \sin\theta\right)$
$\Rightarrow - \frac{6}{2} = u \sin\theta$ or $-3 = u \sin \theta$ ....(ii)
Squaring Eqs. (i) and (ii) and adding,
$\left(4\right) + \left(9\right) = u^{2} \cos^{2} \theta + u^{2} \sin^{2} \theta$
$= u^{2} \left(\cos^{2} \theta + \sin^{2} \theta\right)$
or $13 = u^{2}$
$\therefore u = \sqrt{13} ms^{-1}$