JEE Main & Advanced Sample Paper JEE Main Sample Paper-4

  • question_answer
    Two particles of mass m & M are initially at rest and infinitely separated from each other. Due to mutual interaction (gravitational) they approach each other. Their relative velocity of approach when they are at a distance d is

    A) \[\sqrt{\frac{2Gd}{M+m}}\]                       

    B) \[\sqrt{\frac{2G(M+m)}{d}}\]

    C) \[\sqrt{\frac{2GMm}{(M+m)d}}\]                           

    D) \[\sqrt{\frac{2Gm}{d}}\]

    Correct Answer: B

    Solution :

    1st method- Let us say the spheres are moving with velocities \[{{v}_{1}}\And {{v}_{2}}\] when they are at a separation of d. Then from momentum conservation, \[m{{v}_{1}}=M{{v}_{2}}\] From energy conservation, \[\left( \frac{mv_{1}^{2}}{2}+\frac{Mv_{1}^{2}}{2} \right)-0=\left[ -\frac{GMm}{d}-0 \right]\] After solving above equation we get \[{{v}_{1}}+{{v}_{2}}=\sqrt{\frac{2G(M+m)}{d}}\] II2nd  Method : Using C frame or reduced mass concept. Let \[{{v}_{r}}\] be the relative velocity when they are at a separation of d. then,\[\left( \frac{mM}{m+M} \right)\frac{v_{r}^{2}}{2}=-\left[ -\frac{GMm}{d}-0 \right]\] \[\Rightarrow \]\[{{v}_{r}}=\sqrt{\frac{2G(M+m)}{d}}\]

You need to login to perform this action.
You will be redirected in 3 sec spinner