First slide

Law of conservation of momentum and it's applications

Question

A cannon ball is fired with a velocity 200 m/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

Moderate

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

Solution

Momentum of ball (mass m) before explosion at the highest point $= m v \hat{i} = m u \cos 60 ° \hat{i}$
$= m \times 200 \times \frac{1}{2} \hat{i} = 100 m \hat{i} {kgms}^{- 1}$
Let the velocity of third part after explosion is V
After explosion momentum of system = ${\vec{P}}_{1} + {\vec{P}}_{2} + {\vec{P}}_{3}$
$= \frac{m}{3} \times 100 \hat{j} - \frac{m}{3} \times 100 \hat{j} + \frac{m}{3} \times V \hat{i}$
By comparing momentum of system before and after the explosion
$\frac{m}{3} \times 100 \hat{j} - \frac{m}{3} \times 100 \hat{j} + \frac{m}{3} V \hat{i} = 100 m \hat{i} \Rightarrow V = 300 m / s$

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