A shell is fired from a cannon with velocity v m/s at an angle $\mathrm{\theta }$ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed in m/s of the other piece immediately after the explosion is:

In case of projectile motion as the highest point

${\left(\mathrm{v}\right)}_{\mathrm{vertcial}} = 0 \mathrm{and} {\left(\mathrm{v}\right)}_{\mathrm{horizontal}} = \mathrm{v} \cos \mathrm{\theta }$

the initial linear momentum of the system will be mv $\cos \mathrm{\theta }$. Now as fore of blasting is internal and force of gravity is vertical, so linear momentum of the system along horizontal is conserved, i.e.,

   ${\mathrm{P}}_1 + {\mathrm{P}}_2 = \mathrm{mv} \cos \mathrm{\theta } \mathrm{or} {\mathrm{m}}_1{\mathrm{v}}_1 + {\mathrm{m}}_2{\mathrm{v}}_2 = \mathrm{mv} \cos \mathrm{\theta }$

But it is given that ${\mathrm{m}}_1 = {\mathrm{m}}_2 = \frac{\mathrm{m}}{2} \mathrm{and} \mathrm{as} \mathrm{one} \mathrm{part} \mathrm{retraces} \mathrm{its} \mathrm{path}, {\mathrm{v}}_1 = -\mathrm{v} \cos \mathrm{\theta }$

$\frac{1}{2}\mathrm{m}(-\mathrm{vcos\theta })+\frac{1}{2}{\mathrm{mv}}_2 = \mathrm{mv} \cos \mathrm{\theta }$

Solving, we get: ${\mathrm{v}}_2 = 3\mathrm{v} \cos \mathrm{\theta }$