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. $\sqrt{2}m/s$
• B. $(\sqrt{2}-1)m/s$
• C. $2m/s$
• D. $(\sqrt{2}+1)m/s$

Answer 1

The correct option is D $(\sqrt{2}+1)m/s$

Let v₁ is the velocity of man and v₂ is the velocity of boy. It is given that a man who is running has half the kinetic energy of the boy of half his mass. So, initially the case is as follows

$K{E}_m=\frac{1}{2}K{E}_b$

$\frac{1}{2}m{v_1}^2=\frac{1}{2}\frac{1}{2}\frac{m}{2}{v_2}^2$

$v_1=\frac{1}{2}{v}_2...........\left(1\right)$

Second case

Now, the man speeds up by 1 m/s and has the same kinetic energy as the boy.

$\frac{1}{2}m{(v_1+1)}^2=\frac{1}{2}\frac{m}{2}{v_2}^2$

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

$v_2=\sqrt{2}(v_1+1)...............\left(2\right)$

From equation (1) and (2)

$v_1=\frac{1}{2}\left(\sqrt{2}\right(v_1+1\left)\right)$

$2v_1=\sqrt{2}{v}_1+\sqrt{2}$

$v_1=\frac{\sqrt{2}}{2-\sqrt{2}}$

$v_1=(\sqrt{2}+1)m/s$