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Question: A running man has half the kinetic energy than a boy of half his mass has. The man speed up by \(1.0...

A running man has half the kinetic energy than a boy of half his mass has. The man speed up by 1.0 ms−11.0 \mathrm {~ms} ^ { - 1 } and then he has the same energy as the boy. The original speeds of the man and boy respectively are :

A

2.4 ms−1,1.2 ms−12.4 \mathrm {~ms} ^ { - 1 } , 1.2 \mathrm {~ms} ^ { - 1 }

B

1.2 ms−1,4.4 ms−11.2 \mathrm {~ms} ^ { - 1 } , 4.4 \mathrm {~ms} ^ { - 1 }

C

2.4 ms−1,4.8 ms−12.4 \mathrm {~ms} ^ { - 1 } , 4.8 \mathrm {~ms} ^ { - 1 }

D

4.8 ms−1,2.4 ms−14.8 \mathrm {~ms} ^ { - 1 } , 2.4 \mathrm {~ms} ^ { - 1 }

Answer

2.4 ms−1,4.8 ms−12.4 \mathrm {~ms} ^ { - 1 } , 4.8 \mathrm {~ms} ^ { - 1 }

Explanation

Solution

As the kinetic energy of the man is half the kinetic energy of the boy

Or 12MV2=1212(M2)v2\frac { 1 } { 2 } \mathrm { MV } ^ { 2 } = \frac { 1 } { 2 } \frac { 1 } { 2 } \left( \frac { \mathrm { M } } { 2 } \right) \mathrm { v } ^ { 2 } (∵m=M2)\left( \because \mathrm { m } = \frac { \mathrm { M } } { 2 } \right) Or

Here, M and V for man and m and v for boy.

Now when the velocity of the man is increased to (V+1) then kinetic energies become equal

12M(V+1)2=12(M2)v2=14M(2 V)2\frac { 1 } { 2 } \mathrm { M } ( \mathrm { V } + 1 ) ^ { 2 } = \frac { 1 } { 2 } \left( \frac { \mathrm { M } } { 2 } \right) \mathrm { v } ^ { 2 } = \frac { 1 } { 4 } \mathrm { M } ( 2 \mathrm {~V} ) ^ { 2 }

v2+2 V+1=2 V2\mathrm { v } ^ { 2 } + 2 \mathrm {~V} + 1 = 2 \mathrm {~V} ^ { 2 }

Or V2−2 V−1=0\mathrm { V } ^ { 2 } - 2 \mathrm {~V} - 1 = 0

And