Question

A bomb of mass $m$ is moving in $x-$ direction with velocity $u$. Bomb splits into masses of $\frac{1}{3}m$ and $\frac{2}{3}m$ moving horizontally in the same plane. If an additional energy of $4m{u}^2$ is generated, the relative speed of two masses is :

• A. 3u
• B. 4u
• C. 6u
• D. 8u

Answer 1

Answer: (C)

6u

By conservation of momentum
$mu=\frac{m{u}_1}{3}+\frac{2}{3}m{u}_2$
or $3u=u_1+2u_2$ .....(i)
also
additional energy= change in kinetic energy
or $4m{u}^2=\frac{m{u}_1^2}{6}+\frac{2m{u}_2^2}{6}-\frac{1}{2}m{u}^2$
or $24m{u}^2=m{u}_1^2+2m{u}_2^2-3m{u}^2$
or $27m{u}^2=m{u}_1^2+2m{u}_2^2$
Solving (i) and (ii)
$u_1=5u$ and $u_2=-u$
$\therefore$ Relative velocity $=5u-(-u)=6u$