A man of mass 60 kg and a boy mass 30 kg are standing together on frictionless ice surface. If they push each other apart, man moves away with a speed of $0.4\ {ms}^{1}$ relative to ice after 5 s. They will be away from each other at a distance of

Question

Dr. Abubacker · Accepted Answer

(c) As external force is zero
$m_{1}v_{1} + m_{2}v_{2} = 0$
$v_{2} = \frac{- m_{1}v_{1}}{m_{2}} = - \frac{60 	imes 0.4}{30} = - 0.8\ {ms}^{- 1}$
It means that boy moves with speed ${0.8\ ms}^{- 1}\ $opposite to velocity of man.
So, relative velocity of man and boy is
$v_{r} = 0.8 + 0.4 = 1.2\ {ms}^{- 1}$
Hence, separation $d = v_{r}t = 1.2 	imes 5 = 6$m

Q. A man of mass 60 kg and a boy mass 30 kg are standing together on frictionless ice surface. If they push each other apart, man moves away with a speed of $0.4\ {ms}^{1}$ relative to ice after 5 s. They will be away from each other at a distance of

Explanation

Reviewed by Dr. Abubacker

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