A man who is running has half the kinetic energy of a boy of half his mass. The man speeds up by 1 `ms^(-1)` and then has the same kinetic energy as the boy. The original speed of the boy was:

Options

A. `2.4 m s^(-1), 1.2 m s^(-1)`
B. `1.2 m s^(-1), 4.4 m s^(-1)`
C. `2.4 m s^(-1), 4.8 m s^(-1)`
D. `4.8 m s^(-1), 2.4 m s^(-1)`

Answer

C

Explanation

As the kinetic energy of the man is half the kinetic energy of the boy
`therefore 1/2MV^2=1/2(1/2mv^2)`
or `1/2MV^2=1/2 1/2(M/2)v^2 " " (because m=M/2)`
or `v^2=4V^2 "or " v=2V`
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
`1/2M(V+1)^2 =1/2(M/2)v^2=1/4M(2V)^2`
`V^2+2V+1=2V^2 "or " V^2-2V-1=0`
`V=sqrt2+1 =2.4 m s^(-1)`
and `v=2(sqrt2+1)=4.8 m s^(-1)`