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A steel sphere of mass 100 gm moving with a velocity of $4 m/s$ collides with a dust Particle elastically moving in the same direction with a velocity of $1 m/s$ . The velocity of the dust particle after the collision is

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8 m/s

9 m/s

7 m/s

6 m/s

answer is B.

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Detailed Solution

m_{1} = 100 gm, m_{2} = 0 gm

$u_{1} = 4 m/s, u_{2} = 1 m/s$

$v_{2} = \frac{(m_{2} - m_{1}) u_{2}}{(m_{1} + m_{2})} + \frac{{2m}_{1} u_{1}}{(m_{1} + m_{2})}$

$= \frac{- 100 \times 1}{100} + \frac{2 \times 100 \times 4}{100}$

$= - 1 + 8$

$= 7 m/s$

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