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. $300\, m\, s^{-1}$in 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.

Answer 1

Answer: (B)

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

Answer 2

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.