Question

A cannon ball is fired with a velocity 200 m/sec at an angle of ${60}^{\circ }$ with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically upwards with a velocity 100 m/sec, the second one falling vertically downwards with a velocity 100 m/sec. The third fragment will be moving with a velocity

• A. 100 m/s in the horizontal direction
• B. 300 m/s in the horizontal direction
• C. 300 m/s in a direction making an angle of 60° with the horizontal
• D. 200 m/s in a direction making an angle of 60° with the horizontal

Answer 1

The correct option is B

300 m/s in the horizontal direction

Momentum of ball (mass m) before explosion at the highest point = $mv\overrightarrow{i}=mucos{60}^{\circ }\hat{i}$

$=m\times 200\times \frac{1}{2}\hat{i}=100m\hat{i}kgm{s}^{-1}$

Let the velocity of third part after explosion is V

After explosion momentum of system =

${\overrightarrow{P}}_1+{\overrightarrow{P}}_2+{\overrightarrow{P}}_3$

$=\frac{m}{3}\times 100\hat{j}-\frac{m}{3}\times 100\hat{j}+\frac{m}{3}\times V\hat{i}$

By comparing momentum of system before and after the explosion

$\frac{m}{3}\times 100\hat{j}-\frac{m}{3}\times 100\hat{j}+\frac{m}{3}V\hat{i}=100m\hat{i}\Rightarrow V=300m/s$