Question

A body of mass $M$ at rest explodes into three pieces, two of them of mass $M/4$ each, are thrown off in perpendicular directions with velocities of $3m/s$ and $4m/s$ respectively. The third piece will be thrown off with a velocity of

• A. 1.5 m/s
• B. 2 m/s
• C. 2.5 m/s
• D. 3 m/s

Answer 1

Answer: (C)

Let one mass piece of mass $M/4$ be thrown off in the horizontal direction with a speed of $3m/s$, and the other mass piece of mass $M/4$ be thrown in the vertical direction with a speed of $4m/s$ and the remaining mass $M/2$ be thrown with a velocity $vm/s$, making an angle $\theta$ with the horizontal direction.
So, from conservation of momentum,
Along horizontal direction
$\frac{M}{4}\times 3+\frac{M}{2}\times v\cos \theta =0\dots$ (i)
Along vertical direction
$\frac{M}{4}\times 4+\frac{M}{2}\times v\sin \theta =0\dots$ (ii)
So, from Eqs. (i) and (ii), we get
$v\cos \theta =-\frac{3}{2}\dots$ (iii)
and $v\sin \theta =-2\dots$ (iv)
So, $v=\sqrt{(3/2{)}^2+(2{)}^2}$
$=\sqrt{9/4+4}=\frac{5}{2}$