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
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
300 m/s in a direction making an angle of 
with the horizontal
200 m/s in a direction making an angle of 
with the horizontal
The correct option is B 300 m/s in the horizontal direction
Momentum of ball (mass m) before explosion at the highest point =mv^i=mu cos 60∘^i
=m×200×12^i=100 m^i kgms−1
Let the velocity of third part after explosion is V
After explosion momentum of system = =→P1+→P2+→P3
=m3×100^j−m3×100^j+m3×V^i
By comparing momentum of system before and after the explosion
m3×100^j−m3×100^j+m3V^i=100 m^i⇒V=300 m/s
300 m/s in the horizontal direction
Momentum of ball (mass m) before explosion at the highest point =mv^i=mu cos 60∘^i
=m×200×12^i=100 m^i kgms−1
Let the velocity of third part after explosion is V
After explosion momentum of system = =→P1+→P2+→P3
=m3×100^j−m3×100^j+m3×V^i
By comparing momentum of system before and after the explosion
m3×100^j−m3×100^j+m3V^i=100 m^i⇒V=300 m/s
