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  4. 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

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Q. 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

JIPMERJIPMER 2015Laws of Motion

Solution:

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 $v′ = 3u \,\cos\, 60° = 3 \times 200 \times 1/2$
= $300\, m \,s^{-1}$ along horizontal direction.