A cannon ball is fired with a velocity 200 m s-1 at an angle of 60° with the horizontal. At the highest point of its flight, it explodes into 3 equal fragments, one falling vertically downwards with a velocity 100 m s-1. The second is going vertically upwards with a velocity 100 m s-1, the third fragment will be moving with the velocity

Question

A cannon ball is fired with a velocity 200 m s-1 at an angle of 60° with the horizontal. At the highest point of its flight, it explodes into 3 equal fragments, one falling vertically downwards with a velocity 100 m s-1. The second is going vertically upwards with a velocity 100 m s-1, the third fragment will be moving with the velocity (A) 100 m s-1 in horizontal direction (B) 300 m s-1 in horizontal direction (C) 300 m s-1in a direction making an angle of 60° with the horizontal (D) 200 m s-1 in a direction making an angle of 60° with the horizontal.

Tardigrade Admin · Accepted Answer

Let 3m be the mass of the cannon ball before explosion. At the highest point before explosion, the velocity of the ball has horizontal component = u cos 60°. If upsilon′ is the velocity of the third fragment after explosion, then according to law of conservation of linear momentum 3mu cos 60° = mv cos 90° + m(-v cos 90°) + mv′ or v′ = 3u cos 60° = 3 × 200 × 1/2 = 300 m s-1 along horizontal direction.

$100 m s^{- 1}$ in horizontal direction

$300 m s^{- 1}$ in horizontal direction

$300 m s^{- 1}$ in a direction making an angle of 60° with the horizontal

$200 m s^{- 1}$ in a direction making an angle of 60° with the horizontal.

100ms−1 in horizontal direction

300ms−1 in horizontal direction

300ms−1in a direction making an angle of 60° with the horizontal

200ms−1 in a direction making an angle of 60° with the horizontal.

$100 m s^{- 1}$ in horizontal direction

$300 m s^{- 1}$ in horizontal direction

$300 m s^{- 1}$ in a direction making an angle of 60° with the horizontal

$200 m s^{- 1}$ in a direction making an angle of 60° with the horizontal.