Question

Two particles $A$ and $B$ with particle $A$ having mass $m$, are moving towards each other under their mutual force of attraction. If their speeds at certain instant are $v$ and $2v$ respectively, then find the K.E of the system

Answer 1

Since the particles are moving under their mutual attraction, net force acting on the system $(m_1+m_2)$ is equal to ${\overrightarrow{F}}_1+{\overrightarrow{F}}_2=0$
Therefore, the lnear momentum of the system is constant
$\Rightarrow |m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2|=$ constant
Since, initially the particles were released from rest, $P_i=0$
$\Rightarrow p_f=m_1{\overrightarrow{v}}_1+m_2{\overrightarrow{v}}_2=0\Rightarrow m_2=\frac{m_1{v}_1}{v_2}\Rightarrow m_2=\frac{mv}{2v}=\frac{m}{2}$ (where $m_1=m,v_1=v,v_2=2v$)
The KE of the system $KE=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=\frac{1}{2}m{v}^2+\frac{1}{2}\left(\frac{m}{2}\right){\left(2v\right)}^2=\frac{3}{2}m{v}^2$