Question

A running man has half the kinetic energy of a running boy of half his mass. The man speeds up by $1m/s$ and then has $K.E.$ as that of the boy. What were the original speeds of man and the boy?

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

Answer 1

Answer: (C)

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

Let the man's initial speed and mass is v and m the boy's speed is u and mass $\frac{m}{2}$

According to the data given,

$\frac{1}{2}m{v}^2=\left(\frac{1}{2}\right(\frac{m}{2}\left)u^2\right)/2$

$4v^2=u^2$

$u=2v$..(1)

Second case:$\frac{1}{2}m(v+1{)}^2=\frac{1}{2}(m/2)u^2$

Using (1),$\frac{1}{2}m(v+1{)}^2=\frac{1}{2}(m/2\left)\right(2v{)}^2$

$(v+1{)}^2=2v^2$

$v^2+2v+1=2v^2$

$v^2-2v-1=0$

$v=\frac{2\pm \sqrt{4+4}}{2}=(1+\sqrt{2})$

so,$u=2v=2(1+\sqrt{2})$