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Standard IX

Physics

Law of Conservation of Mometum

Ball A of mas...

Question

Ball A of mass $5 k g$ travelling at $6 m s^{- 1}$ collides with another ball B of mass $2 k g$ travelling at $2 m s^{- 1}$ in the same direction. After collision, the velocity of ball A reduces to $2 m s^{- 1}$ . Calculate the velocity of ball B after the collision.

$10 m s^{- 1}$

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$12 m s^{- 1}$

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$6 m s^{- 1}$

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$8 m s^{- 1}$

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Solution

The correct option is B
$12 m s^{- 1}$

Given:
Mass of ball A $(m_{1})$ = $5 k g$
Mass of ball B $(m_{2})$ = $2 k g$
Initial velocity of ball A $(u_{1})$ = $6 m s^{- 1}$
Initial velocity of ball B $(u_{2})$ = $2 m s^{- 1}$
Final velocity of ball A $(v_{1})$ = $2 m s^{- 1}$
According to the law of conservation of momentum,
Initial momentum of the system = Final momentum of the system
$\Rightarrow$ $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$\Rightarrow$ $5 \times 6 + 2 \times 2 = 5 \times 2 + 2 \times v_{2}$
$\Rightarrow$ $v_{2} = 12 m s^{- 1}$

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Ball A of mass 5 kg travelling at 6 ms−1collides with another ball B of mass 2 kg travelling at 2 ms−1 in the same direction. After collision, the velocity of ball A reduces to 2 ms−1. Calculate the velocity of ball B after the collision.

Ball A of mass $5 k g$ travelling at $6 m s^{- 1}$ collides with another ball B of mass $2 k g$ travelling at $2 m s^{- 1}$ in the same direction. After collision, the velocity of ball A reduces to $2 m s^{- 1}$ . Calculate the velocity of ball B after the collision.