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Question

A man who is running has half the kinetic energy of the boy of half his mass. The man speeds up by 1 m/s and then has the same kinetic energy as the boy. The original speed of the man was

A
2 m/s
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B
(21) m/s
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C
2 m/s
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D
(2+1) m/s
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Solution

The correct option is D (2+1) m/s
Given mass of boy is half that of man, and kinetic energy is twice.
Let M be mass of the man, M2 be mass of the boy
Vboy is velocity of the boy.
Vman is velocity of the man.

2K.Eman=K.E.boy
2×12M×V2man=12.M2V2boy
Vman=Vboy2......(i)


Now, If velocity of man is increased by 1 m/s, K.Eman=K.E.boy
12M(Vman+1)2=12.M2V2boy
(Vman+1)2=V2boy2
from (i)
(Vman+1)2=4V2man2=2V2man
(Vman+1)=2Vman
1=(21)Vman
Vman=121
Vman=(2+1) m/sec

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