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\[V=300\,m/s\]You need to login to perform this action.
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