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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 \[60{}^\circ \] with the horizontal

D)  200 m/s in a direction making an angle of \[60{}^\circ \] with the horizontal

Correct Answer: B

Solution :

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).