Question

A particle of mass $m$ has half the kinetic energy of another particle of mass $\frac{m}{2}$. The ratio of the speeds of the lighter and heavier particles is

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $1:4$

Answer 1

The correct option is C $2:1$

For the particle $\left(1\right)$,
given mass $m_1=m$, let speed be $v$
Kinetic energy $=K{E}_1=\left(\frac{1}{2}{m}_1{v}^2\right)$


For the particle $\left(2\right)$,
given mass $m_2=\frac{m}{2}$, let speed be $v^{\prime }$
Kinetic energy $=K{E}_2=\left(\frac{1}{2}{m}_2{v}^{\prime 2}\right)$

Since, particle of mass $m$ has half the kinetic energy of another particle of mass $\frac{m}{2}$, we get

$K.E_1=\frac{K.E_2}{2}$

$\Rightarrow \frac{1}{2}{m}_1{v}^2=\frac{1}{2}\times \frac{1}{2}{m}_2{v}^{\prime 2}$
Given that $(m_1=m)$ and $(m_2=\frac{m}{2})$

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

$v^2=\frac{v^{\prime 2}}{4}$, hence the ratio of speed of lighter particle and that of heavier particle is given by,
$\Rightarrow \frac{v^{\prime }}{v}=\frac{2}{1}$

Option C is the correct answer.