Question

Kinetic energy of the system just after the collision is

• A. $\frac{8G{m}^2}{3R}$
• B. $\frac{2G{m}^2}{3R}$
• C. $\frac{4G{m}^2}{3R}$
• D. cannot be determined

Answer 1

Answer: (B)

$\frac{2G{m}^2}{3R}$

Equating the gravitational potential energies and the kinetic energy we get
$\frac{G{m}_1{m}_2}{d}=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}^2{v}_2^2$
and using conservation of momentum we get
$m_1{v}_1=m_2{v}_2$
or
$v_2=\frac{m_1{v}_1}{m_2}$
Substituting this we get
$v_1=\sqrt{\frac{G{m}_2^2}{d(m_1+m_2)}}$
Similarly we get
$v_2=\sqrt{\frac{G{m}_1^2}{d(m_1+m_2)}}$
thus the relative velocity is given as
$v_1-(-v_2)=v_1+v_2=\sqrt{\frac{2G(m_1+m_2)}{d}}$(velocities have opposite direction)
this is the relative velocity before collision
thus relative velocity after collision will be $\frac{v_1+v_2}{2}$(coefficient of restitution is 1/2)
Thus we get the final kinetic energy as
$\frac{1}{2}\mu v^2$($\mu$ is the reduced mass and is equal to $\frac{m_1{m}_2}{m_1+m_2}$)
or
$\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}\frac{2G(m_1+m_2)}{4d}$
Substituting $m_1=m$ and $m_2=8m$ we get the answer as
$\frac{2G{m}^2}{3r}$(distance between the two bodies is r+2r)