Question

Two particles having masses $1kg$ and $7kg$ respectively attract each other. Initially they are at rest and infinite separation.The velocity of approach of the particles are at a separation of $1m$ is (G=universal gravitational constant)

• A. $\sqrt{2}Gm/s$
• B. $4\sqrt{G}$ $m/s$
• C. $\sqrt{\frac{G}{2}}m/s$
• D. $\frac{G}{4}m/s$

Answer 1

The correct option is B $4\sqrt{G}$ $m/s$

Potential energy when they are at $\infty$ distance is $P{E}_1=0$

Now its value when they are at a distance $r=1m$ will be $P{E}_2=-G\frac{m_1{m}_2}{r}=-7G$

so the value of change in PE is $7G$

It should be equal to $increase$ in Kinetic energy., as the initial velocity was $zero$ so the velocity of approach will be

given by $\frac{\mu v^2}{2}=7G$

where $\mu$ is the $reduced$ mass of the $system$ of masses, $\mu =\frac{m_1{m}_2}{m_1+m_2}=\frac{7}{8}kg$

putting above value in the equation for KE, we get $v=\sqrt{\frac{2\times 7G\times 8}{7}}=4\sqrt{G}$ $m/s$