A cannon ball is fired with a velocity of $200$ m/sec at an angle of $60^{o}$ with the horizontal. At the highest point of the flight it explodes into $3$ equal fragments. One goes vertically upwards with a velocity $100$ m/sec, the second one falling vertically downwards with a velocity of $100$ m/sec. The third fragment will be moving with a velocity?

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Solution

The horizontal component of the ball =Vcos $60^{0}$ --(1) and
the vertical component= Vsin $60^{0}$
Let the mass of the canon before fragmenting be M

From law of conservation of momentum we get,

Momentum of vertical component before explosion=Momentum of vertical component after explosion

Now at the highest point of flight vertical component will be zero due the action of gravity.

0 $= \frac{M}{3} (100) + \frac{M}{3} (V) v = - 100$

where v is the velocity of the fragment falling downwards

Momentum of horizontal component before explosion=Momentum of horizontal component after explosion.

Velocity of horizontal component= $\frac{200}{2} = 100 m / s$ from (1)
M $\times$ 100= $\frac{M}{3} \times V_{H} V_{H} = 300 m / s$

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