Question

The kinetic energy of a man is half the kinetic energy of a body of half of his mass. If the man increases his speed by $1m/s$ then his kinetic energy becomes equal to that of the boy. The ratio of the velocity of the boy and that of the man is:

• A. $\frac{2}{1}$
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $\frac{4}{3}$

Answer 1

Answer: (A)

$\frac{2}{1}$

Let,

$K$ and $K^{\prime }$ be the kinetic energies of man and a boy respectively. Similarly, $m$ and $m^{\prime }$, $v$ and $v^{\prime }$ be the masses and velocities of man and a boy respectively.

Given that

$K=\frac{1}{2}{K}^{\prime }$

$m=2m^{\prime }$

$v=v+1=v^{\prime \prime }$(say) $\Rightarrow K=K^{\prime }$

$\frac{1}{2}m{v}^2=\frac{1}{2}\frac{1}{2}{m}^{\prime }{v}^{\prime 2}$

$\Rightarrow \frac{v^2}{v^{\prime 2}}=\frac{m^{\prime }}{2m}$

$\therefore \frac{v^2}{v^{\prime 2}}=\frac{m^{\prime }}{4m^{\prime }}$

$\therefore \frac{v}{v^{\prime }}=\frac{1}{2}$