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 $4m{s}^{-1}$ and $6m{s}^{-1}$ respectively. The magnitude of the velocity of the heavier mass is

• A. $\overline{1}7m{s}^{-1}$
• B. 2 $\overline{1}3m{s}^{-1}$
• C. $\overline{1}3m{s}^{-1}$
• D. $\frac{\overline{1}3}{2}m{s}^{-1}$

Answer 1

Answer: (C)

$\overline{1}3m{s}^{-1}$

Let third mass particle (2 m) moves making angle 8 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 4 + 2m (u $\cos \theta$ )
or -4 = 2u $\cos \theta$
or -2 = u $\cos \theta$ (i)
Again, applying law of conservation of linear momentum in y-direction.
0 = m 6 + 2m(u $\sin \theta$ )
$\Rightarrow -\frac{6}{2}=u\sin \theta$
or -3 = u $\sin \theta$ (ii)
Squaring Eqs. (i) and (ii) and adding, we get
(4) + (9) = $u^{2}co{s}^2\theta +u^2{\sin }^2\theta$

$=u^2({\cos }^2\theta +{\sin }^2\theta )$

or 13 = $u^2$

or u = $\overline{1}3m{s}^{-1}$