Question

A bomb of mass $m$ initially at rest, on the ground suddenly explodes in to two fragments. The fragment of mass $\frac{2m}{3}$ moves out with a velocity $v$. With the total energy released during the explosion, the unexploded bomb can be raised to what height above the ground?

• A. $\frac{v^2}{2g}$
• B. $\frac{2v^2}{g}$
• C. $\frac{v^2}{g}$
• D. $\frac{v^2}{4g}$

Answer 1

The correct option is C $\frac{v^2}{g}$

Using linear momentum conservation since these is low internal force:

$\Rightarrow \frac{m}{3}\times v^{\prime }=\frac{2m}{3}.v$

$\Rightarrow v^{\prime }=2v$

So, total K.E. after explosion $=K{E}_A+K{E}_B$

$=\frac{1}{2}\frac{m}{3}(2v{)}^2+\frac{1}{2}\times \frac{2m}{3}\times v^2$

$=m{v}^2$
Now for raising it to a certain height this energy must be converted to P. E:
$m{v}^2=mgh$

$\Rightarrow h=\frac{v^2}{g}$