Work Energy And Power Question 82
Question: 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]
Options:
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
Show Answer
Answer:
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} $
therefore $ V=300m/s $
BETA
