Question

Two stars each of mass M and radius R are approaching each other for a head-on collision. They start approaching each other when their separation is r > > R. If their speeds at this separation are negligible, the speed v with which they collide would be

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

Answer 2

Since the speeds of the stars are negligible when they are at a distance r, hence the initial kinetic energy of system is zero. Therefore, the initial total energy of the system is

$\mathrm { E } _ { \mathrm { i } } = \mathrm { KE } + \mathrm { PE } = 0 + \left( - \frac { \mathrm { GMM } } { \mathrm { r } } \right) = - \frac { \mathrm { GM } ^ { 2 } } { \mathrm { r } }$

Where M represent the mass of each star and r is initial separation between them.

When two stars collide their centres will be at a distance twice the radius of a star i.e, 2R.

Let v be the speed with which two stars collide. Then total energy of the system at the instant of their collision is given by

$\mathrm { E } _ { \mathrm { f } } = 2 \times \left( \frac { 1 } { 2 } \mathrm { Mv } ^ { 2 } \right) + \left( - \frac { \mathrm { GMM } } { 2 \mathrm { R } } \right) = \mathrm { Mv } ^ { 2 } - \frac { \mathrm { GM } ^ { 2 } } { 2 \mathrm { R } }$

According to law of conservations of mechanical energy

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