Question

Two particles with masses 4m and 3m are moving toward each other along the x axis with the same initial speeds v_i. The particle with mass 4m is traveling to the left, and particle 3m is traveling to the right. They undergo a head-on elastic collision and each rebounds along the same line as it approached. Find the final speeds of the particles.

particle 4m

Answer 1

The momentum of 4m before collision = -4m vi

The momentum of 3m before collision = + 3m vi
(Negative sign because it is moving toward left). (Any convention can be chosen)
Total momentum before collision = -4m vi + 3m vi = = -mvi.
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Let w1 be the velocity of 4m after collision .
Let w2 be the velocity of 3m after collision .
Total momentum of both after collision is
4mw1+3mw2

In the left and right direction
Total momentum of both is -mvi
Initial momentum in this direction was
-mvi.
Hence
-mvi. =4mw1+3mw2

-vi=4w1+3w2-----------------------------(1)

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Considering the kinetic energies of the particles
Initial k.e = (1/2)4mvi^2+(1/2)3mvi^2 = (1/2)*7m.vi^2 .
Final k.e (1/2)4m w1^2 +(1/2) 3m w2^2
Hence
(1/2)*7m vi^2 = (1/2)m (4w1^2 + 3 w2^2)
7vi^2 = 4w1^2 + 3 w2^2
Eliminating w2 using ---------------(1)
7vi^2 = 4w1^2 + 3 {((-vi-4w1)/3)^2}
7vi^2 = 4w1^2 + (16/3)w1^2 + (1/3)vi^2 + 8viw1
6.67 vi^2 = 9.33 w1^2 +8viw1

(Solving the quadratic we get) (other solution of quadratic =-1.37 vi is rejected because lighter particle will travel towards right)
w1 = 0.519 vi(towards right)
From 1
-vi=4w1+3w2

w2=(-vi-4w1)/3 =-1.025 vi (hence towards left)