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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/sGiven 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=(√2−1)Vman ⇒Vman=1√2−1 ⇒Vman=(√2+1) m/sec

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