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Question: A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by...

A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by 1 m/s so as to have same K.E. as that of boy. The original speed of the man will be

A

2m/s\sqrt { 2 } m / s

B

(2−1)m/s( \sqrt { 2 } - 1 ) m / s

C

1(2−1)m/s\frac { 1 } { ( \sqrt { 2 } - 1 ) } m / s

D

12m/s\frac { 1 } { \sqrt { 2 } } m / s

Answer

1(2−1)m/s\frac { 1 } { ( \sqrt { 2 } - 1 ) } m / s

Explanation

Solution

Let m = mass of the boy, M = mass of the man, v = velocity of the boy and V = velocity of the man

Initial kinetic energy of man 12mvt1t2\frac { 1 } { 2 } m \frac { v } { t _ { 1 } } t ^ { 2 }

=12[12(M2)v2]= \frac { 1 } { 2 } \left[ \frac { 1 } { 2 } \left( \frac { M } { 2 } \right) v ^ { 2 } \right]

⇒ V2=v24V ^ { 2 } = \frac { v ^ { 2 } } { 4 } .....(i)

When the man speeds up by 1 m/s ,

12M(V+1)2=12mv2=12(M2)v2\frac { 1 } { 2 } M ( V + 1 ) ^ { 2 } = \frac { 1 } { 2 } m v ^ { 2 } = \frac { 1 } { 2 } \left( \frac { M } { 2 } \right) v ^ { 2 } ⇒ (V+1)2=v22( V + 1 ) ^ { 2 } = \frac { v ^ { 2 } } { 2 }

⇒ V+1=v2V + 1 = \frac { v } { \sqrt { 2 } } .....(ii)

From (i) and (ii) we get speed of the man .