Two small glass spheres of masses 10 g and 20 g are moving in a straight line in the same direction with velocities of $3 m s^{- 1}$ and $2 m s^{- 1}$ respectively. They collide with each other. After collision, glass sphere of mass 10 g moves with a velocity of $2.5 m s^{- 1}$ . Find the velocity of the second ball after collision.

2.25 m s^{- 1}

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5.5 m s^{- 1}

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2.75 m s^{- 1}

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7.5 m s^{- 1}

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Solution

The correct option is B $2.25 m s^{- 1}$
Here, $m_{1} = 10 g = \frac{10}{1000} = 0.01 k g$

$m_{2} = 20 g = 0.02 k g$

$u_{1} = 3 m s^{- 1}; u_{2} = 2 m s^{- 1}$

$v_{1} = 2.5 m s^{- 1}; v_{2} = ?$

Total momentum of both the spheres before collision $= m_{1} u_{1} + m_{2} u_{2}$

$= 0.01 \times 3 + 0.02 \times 2 = 0.07 k g m s^{- 1}$

Total momentum of both the spheres after collision

$= m_{1} v_{1} + m_{2} v_{2}$

$= 0.01 \times 2.5 + 0.02 v_{2} = 0.025 + 0.02 v_{2}$

Now, according to the law of conservation of momentum,

Total momentum after collision $=$ Total momentum before collision

$∴ 0.025 + 0.02 v_{2} = 0.07$

or $0.02 v_{2} = 0.07 - 0.025 = 0.045$

or $v_{2} = \frac{0.045}{0.02} = 2.25 m s^{- 1}$

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