Deriving momentum transfer in elastic collisions between two particles

[subho123]

transfer of momentum...help needed ?

Homework Statement

two particles of masses m1 & m2 are moving along a straight line with velocities u & v respectively. they collide to each other..if the collision between them is an elastic collision then show that momentum transferred from first particle to second particle is 2m1*m2(u - v)/m1 + m2...

Homework Equations

I am trying to use the condition of an elastic collision...their linear momentum & kinetic energy will be conserved..but the proof is not coming out...please help me with it...

if the velocity of first particle after collision is u1 & the same of the 2nd particle is v1 then we know, u1 = (m1 - m2)*u/(m1 + m2) + 2m2*v/(m1 + m2)

v1 = 2m1*u/(m1 + m2) - (m1 - m2)*v/(m1 + m2)

The Attempt at a Solution

 

[ehild]

Show your work.

ehild

 

[subho123]

v1 = 2m1*u/(m1 +m2) - (m1 - m2)*v/(m1+m2)

momentum of the 2nd particle after collision

m2v1 = 2m1*m2*u/(m1+m2) - (m1-m2)*m2*v/(m1+m2)

transfer of momemtum

m2v1 - m2v = 2m1*m2*u/(m1+m2) - (m1-m2)*m2*v/(m1+m2) - m2v

[2m1*m2*u - (m1-m2)*m2*v - (m1+m2)*m2v]/(m1+m2)

[2m1*m2*u - (m1-m2)*m2*v - m1*m2*v - m2^2v]/(m1+m2)

[2m1*m2*u - m1*m2v + m2^2v - m1*m2*v - m2^2v]/(m1+m2)

[2m1*m2*u - 2m1*m2*v]/(m1+m2)

2m1*m2(u-v)/(m1+m2)


it is sloved

BTW thanks

 

[ehild]

Good job!

ehild