A shell is fired from a cannon with a speed $100m{s}^{-1}$ at an angle $60°$with the horizontal (positive $x$-direction). At the highest point of its trajectory, the shell explodes into two equal fragments. One of the pieces moves along $x$-direction with a speed of $50m{s}^{-1}$Determine the speed of the other piece at the time of explosion.

Explanation:

As we know that in the absence of external force the motion of the center of mass of a body remains unaffected. Thus, here the center of mass of the two fragments will continue to follow the original projectile path. The velocity of the shell at the highest point of trajectory is
$$\begin{gathered} v_m=u\cos \theta \\ \Rightarrow v_m=u\cos \left(60°\right) \\ \Rightarrow v_m=100\times \frac{1}{2} \\ \Rightarrow v_m=50m{s}^{-1} \end{gathered}$$

Let $v_1$be the speed of the fragment which moves along the negative $x$-direction and the other fragment has speed $v_2$ which must be along the positive $x$-direction.
Now, from the conservation of momentum, we get
$$\begin{gathered} m{v}_m=-\frac{m}{2}{v}_1+\frac{m}{2}{v}_2 \\ \Rightarrow 2v_m=v_2-v_{1 \\ } \\ \Rightarrow v_2=2v_M+v_{1 \\ } \\ \Rightarrow v_2=2\times 50+50 \\ \Rightarrow v_2=100+50 \\ \Rightarrow v_2=150m{s}^{-1} \end{gathered}$$

Hence, the speed of the other piece at the time of the explosion is $150m{s}^{-1}$