A cannot ball is fired with a velocity 200m / 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

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° with the horizontal
D. 200 m/s in a direction making an angle of `60^(@)` with the horizontal

Answer

B

Explanation

Momentum of ball (mass m) before explosion at the highest point = `mv hat(i)= m u cos 60^(@) hat(i)`
`m xx 200 xx (1)/(2) hat(i) = 100 m hat(i) kgms^(-1)`

Let the velocity of third part after explosion is V
velocity of third part after explosion is V After explosion momentum of system `= vec(P)_(1) + vec(P)_(2) + vec(P)_(3)`
`=(m)/(3) xx 100 hat(j) - (m)/(3) xx 100 hat(j) + (m)/(3) xx V hat(i)`
By comparing momentum of system before and after the explosion
`(m)/(3) xx 100 hat(j)- (m)/(3) xx 100 hat(j) + (m)/(3) V hat(i) = 100 m hat(i) rArr V= 300 m//s`