Question

A running man has half the kinetic energy than a boy of half his mass has. The man speed up by $1.0m{s}^{-1}$ and then he has the same energy as the boy. The original speeds of the man and boy respectively are

• A. $2.4m{s}^{-1},1.2m{s}^{-1}$
• B. $1.2m{s}^{-1},4.4m{s}^{-1}$
• C. $2.4m{s}^{-1},4.8m{s}^{-1}$
• D. $4.8m{s}^{-1},2.4m{s}^{-1}$

Answer 1

Answer: (C)

$2.4m{s}^{-1},4.8m{s}^{-1}$

Let $M$ and $m$ be the man of man and boy respectively

The velocities man and boy are $v_1$ & $v_2$ respectively

mass of boy in half of man of man

$m=\frac{M}{2}$

$K.E$ of boy $=K.E$ of man

$K.E_{man}=\frac{1}{2}K.E_{boy}$ [As per date]

$\frac{1}{2}m{v}_1^2=\frac{1}{2}\left(\frac{m}{2}\right)v_2^2$

$v_2=2v_1.....\left(1\right)$

When the speed of man increases by $1m/s$ then energy of man and boy are same

$L=\frac{1}{2}m(v_1+1{)}^2=\frac{1}{2}\frac{m}{2}{v}_2^2$

$\therefore =2(v_1+1{)}^2=v_2^2....(2)$

Solving $\left(1\right)$ & $\left(2\right)$

$velocityofman{v}_1=2.41m/s$

$velocityofboy{v}_2=2v_1=4.82m/s$

Answer is $C2.4148$