A person of mass M = 90 kg standing on a smooth horizontal plane of ice throws a body of mass m = 10 kg horizontally on the same surface. If the distance between the person and body after 10 seconds is 10 metres, the K.E of the person (in Joules) is

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0.90

4.5

zero

0.45

answer is A.

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Detailed Solution

$V_{1} + V_{2} = \frac{10}{10} = 1 m/s$

From L.C.L.M

$M_{1} V_{1} = M_{2} V_{2}$

${90V}_{1} = {10V}_{2}$

$V_{2} = {9V}_{1}$

$∴ V_{1} + 9 V_{1} = 1$

$V_{1} = \frac{1}{10} m/s$

KE of Man

$= \frac{1}{2} M_{1} V_{1}^{2}$

$= \frac{1}{2} \times 90 \times {(\frac{1}{10})}^{2}$

$= 0.45 J$

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