A bomb is thrown in a horizontal direction with a velocity of $50 m / s$ . It explodes into two parts of masses $6 K g$ and $3 K g$ . The heavier fragment continues to move in the horizontal direction with a velocity of $80 m / s$ . Calculate the velocity of the lighter fragment.

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Solution

Given,

Total Mass, $M_{t o t a l} = 9 k g$

Initial velocity, $V_{i n i t i a l} = 50 m s^{- 1}$

Lighter Mass $= M_{1}$

Final velocity of $(M_{2} = 6 k g) p a r t$ , $V_{2} = 80 m s^{- 1}$
from conservation of momentum

Final momentum = initial momentum

$M_{2} V_{2} + M_{1} V_{1} = M_{t o t a l} V_{i n i t i a l}$

$6 \times 80 + 3 \times V_{1} = 9 \times 50$

$V_{1} = - 10 m s^{- 1}$

Final velocity of lighter mass is $- 10 m s^{- 1}$

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