Question

# A running man has half the kinetic energy that a boy of half his mass has.The man speeds up by 1ms−1 and then has the same kinetic energy as that of the boy. The original speeds of man and boy in ms−1 are:

A
(2+1),(21)
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B
(2+1),2(2+1)
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C
2,2
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D
(2+1),2(21)
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Solution

## The correct option is B (√2+1),2(√2+1)Let, m be the mass of the man, m′ be the mass of the boy,v be the velocity of the man,v′ be the velocity of the boy.As per the question,12mv2=12(12m′v′2)12mv2=14m′v′212mv2=14m2v′2v=v′2v′=2vIn later case12m(v+1)2=12m′v′212m(v+1)2=12(m2)(2v)212m(v+1)2=mv2v2+2v+1=2v2v2−2v−1v2−2v=1v2−2v+1=2(v−1)2=2v−1=√2v=√2+1 and v′=2vv′=2(√2+1)Therefore, the speed of man is √2+1 and the speed of boy is 2(√2+1).

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