question_answer

A cannon ball is fired with a velocity 200 m/sec at an angle of 60° 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 [NCERT 1983; AFMC 1997]

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 (mass m) before explosion at the highest point\[=mv\hat{i}=mu\cos 60{}^\circ \hat{i}\]               = \[m\times 200\times \frac{1}{2}\hat{i}\]= \[100\ m\hat{i}\ kgm{{s}^{-1}}\]                    Let the velocity of third part after explosion is V               After explosion momentum of system = \[{{\vec{P}}_{1}}+{{\vec{P}}_{2}}+{{\vec{P}}_{3}}\]               = \[\frac{m}{3}\times 100\hat{j}-\frac{m}{3}\times 100\hat{j}+\frac{m}{3}\times V\hat{i}\]               By comparing momentum of system before and after the explosion          \[\frac{m}{3}\times 100\hat{j}-\frac{m}{3}\times 100\hat{j}+\frac{m}{3}V\hat{i}=100m\hat{i}\] Þ\[V=300\,m/s\]