Question

A running man has half the kinetic energy that a boy of half his mass has the man speeds up by $1.0$ meter/sec and then has the same kinetic energy as the boy .What were the original speeds of man and boy ?

• A. $\text{4.8 m/sec, 2.4 m/sec}$
• B. $\text{2.5 m/sec , 5 m/sec}$
• C. $\text{4.2 m/sec, 8.4 m/sec}$
• D. $\text{8.4 m/sec , 4.2 m/sec}$

Answer 1

The correct option is B $\text{2.5 m/sec , 5 m/sec}$

$m_{man}=2m_{boy}$ ---(1)

man $K.E_m=\frac{1}{r}{m}_m{u}^2$

boy $K.E_b=\frac{1}{2}{m}_b{v}^2$

& K.Km=$\frac{1}{2}K.Eb=\frac{1}{2}\times \frac{1}{2}mb{v}^2=\frac{1}{4}mb{v}^2$

but $m_b=\frac{m_m}{2}$

so K.Em=$\frac{1}{4}\frac{mm}{2}.v^2$

but $K.E_m=\frac{1}{2}{m}_m{u}^2$

so $\frac{1}{2}{m}_m{u}^2=\frac{1}{8}{m}_m{v}^2$

$\Rightarrow u^2=\frac{u^2}{4}=\frac{v^2}{2^2}$

where $u\to u+1$

K.E'=$\frac{1}{2}{m}_m(u+1{)}^2$

But K.E'=K.Eb so $\frac{1}{2}{m}_m(4+1{)}^2=\frac{1}{2}mb{v}^2$

so $\frac{1}{2}{m}_m(u+1{)}^2=\frac{1}{2}(\frac{mm}{2}).4u^2$

we have $(u+1{)}^2=2u^2\Rightarrow u\simeq 2.5m/s$

so $v=2u\simeq 5m/s$