• # question_answer 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 directionB)  300 m/s in the horizontal direction                 C)  300 m/s in a direction making an angle of $60{}^\circ$ with the horizontalD)  200 m/s in a direction making an angle of $60{}^\circ$ with the horizontal

Momentum of ball before explosion at the highest point $=mv\hat{i}=mu\cos {{60}^{o}}\hat{i}$             $=m\times 200\times \frac{1}{2}\hat{i}=100\,m.\hat{i}\,kg\,.m/s$ Let the velocity of third part after explosion is V. After explosion momentum of system             $=\overrightarrow{{{P}_{1}}}+\overrightarrow{{{P}_{2}}}+\overrightarrow{{{P}_{3}}}$             $=\frac{m}{3}\times 100\hat{j}-\frac{m}{3}\times 100\,\hat{j}+\frac{m}{3}\times V\hat{i}\,100m\hat{i}\,$             or, $V=300\,m/s$ Here, the correction option is (b).