Question

A body of mass $10kg$ at rest explodes into two pieces of masses $7kg$ and $3kg$. If the total increase in kinetic energy due to explosion is $1680J$, 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
$m_1{v}_1+m_2{v}_2=mv$
$7v_1+3v_2=0$
$\therefore v_2=-\frac{7v_1}{3}$
So, the two pieces are moving in opposite directions. Increase in
$KE=1680$ ie,
$\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2-0=1680$
$\frac{1}{2}\times 7{\left(v_1\right)}^2+\frac{1}{2}\times 3{\left(-\frac{7v_1}{3}\right)}^2=1680$
$7v_1^2+3\times \frac{49v_1^2}{9}=1680\times 2$
$7v_1^2+\frac{49v_1^2}{3}=3360$
$v_1^2=\frac{3360\times 3}{70}=144$
$\therefore v_1=\sqrt{144}=12m/s\therefore$
$v_2=-7\times \frac{12}{3}=-28m/s$
Relative velocity $=v_1-v_2$ $=12-(-28)$
$=40m/s$