A man of mass 60 kg sitting on ice pushes a block of mass of 12 kg on ice horizontally with a speed of 5 ms^-1.The coefficient of friction between the man and ice and between block and ice is 0.2. If g =10 ms^-2,the distance between man and the block, when they come to rest is

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6 m

6.5 m

3 m

7 m

answer is B.

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Detailed Solution

$S_{1} = \frac{v^{2}}{2 μg},$ according to law of conservation of momentum,

$S_{2} = \frac{v^{1^{2}}}{2 μg};$

find $v^{1};$ $60 v' + 12 \times 5 = 0$

$v' = - 1 m / s$

$S = S_{1} + S_{2}$

$= \frac{v^{2}}{2 μg} + \frac{v^{1^{2}}}{2 μg}$ $= \frac{1}{2 μg} (5^{2} + 1^{2}) = 6.5 m$

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