Question

A body of mass $10 \,kg$ at rest explodes into two pieces of masses $7\, kg$ and $3 \,kg$. If the total increase in kinetic energy due to explosion is $1680 \,J$, the magnitude of their relative velocity in m/s, after explosion is

• A. 40
• B. 50
• C. 70
• D. 80

Answer 1

Answer: (A)

From law of conservation of momentum,
Final momentum = initial momentum
$m_{1} v_{1}+m_{2} v_{2}=m v$
$7 v_{1}+3 v_{2}=0$
$\therefore v_{2}=-\frac{7 v_{1}}{3}$
So, the two pieces are moving in opposite directions. Increase in
$KE =1680$ ie,
$\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}-0=1680$
$\frac{1}{2} \times 7\left(v_{1}\right)^{2}+\frac{1}{2} \times 3\left(-\frac{7 v_{1}}{3}\right)^{2}=1680$
$7 v_{1}^{2}+3 \times \frac{49 v_{1}^{2}}{9}=1680 \times 2$
$7 v_{1}^{2}+\frac{49 v_{1}^{2}}{3}=3360$
$v_{1}^{2}=\frac{3360 \times 3}{70}=144$
$\therefore v_{1}=\sqrt{144}=12 \,m / s \therefore $
$v_{2}=-7 \times \frac{12}{3}=-28\, m / s$
Relative velocity $=v_{1}-v_{2}$ $=12-(-28)$
$=40\,m / s$