Two - body collision energy and velocity

[dean barry]

Here is the problem :
Two heavenly bodies (m1 and m2) (non orbiting) in free space at a distance D are released from rest and allowed to approach each other under gravitational influence to a final distance d.

This is the way I've worked the problem :

The total potential energy difference between the two positions
= ( ( G * m1 * m2 ) / d ) ) - ( ( G * m1 * m2 ) / D ) )
(Joules)

This also equals the final KE of the two bodies (call it KEt)

Find the final KE of both bodies (using mass ratio) :
KE (m1) = ( m2 / ( m1 + m2 ) ) * KEt
KE (m2) = ( m1 / ( m1 + m2 ) ) * KEt

To find the final velocity of each :
v (m1) = sqrt ( ( KE (m1) ) / ( ½ * m1 ) )
v (m2) = sqrt ( ( KE (m2) ) / ( ½ * m2 ) )

The momentum of each body is equal at position D

Comments please :

 

[sankalpmittal]

Here is the problem :
Two heavenly bodies (m1 and m2) (non orbiting) in free space at a distance D are released from rest and allowed to approach each other under gravitational influence to a final distance d.

This is the way I've worked the problem :

The total potential energy difference between the two positions
= ( ( G * m1 * m2 ) / d ) ) - ( ( G * m1 * m2 ) / D ) )
(Joules)

This also equals the final KE of the two bodies (call it KEt)

Find the final KE of both bodies (using mass ratio) :
KE (m1) = ( m2 / ( m1 + m2 ) ) * KEt
KE (m2) = ( m1 / ( m1 + m2 ) ) * KEt

To find the final velocity of each :
v (m1) = sqrt ( ( KE (m1) ) / ( ½ * m1 ) )
v (m2) = sqrt ( ( KE (m2) ) / ( ½ * m2 ) )

The momentum of each body is equal at position D

Comments please :

Hint : Use conservation of mechanical energy approach.
You can also use "lost mass" approach if you know what it is.

Make the second equation via conservation of linear momenta.

Mods, move this thread to homework section.

 

[dauto]

Where did

KE (m1) = ( m2 / ( m1 + m2 ) ) * KEt
KE (m2) = ( m1 / ( m1 + m2 ) ) * KEt

come from? You can't just drop those equations without any justification. BTW, why not use K₁, K₂, and K instead of KE (m1), KE (m2), and KEt? Makes for much cleaner equations.