Is Momentum Conserved in a Perfectly Elastic Non-Central Collision?

[John O' Meara]

A small steel ball moving with speed v0 in the positive x-direction makes a perfectly elastic, non central collision with an identical ball originally at rest. After impact, the first ball moves with speed v1 in the first quadrant at an angle (theta1) with the x-axis and the second with speed v2 in the fourth quadrant at an angle (theta2) with the x-axis. (a) Write down equations expressing conservation of linear momentum in the x-direction, and in the y-direction. (b)Square these equations and add them. (c) At this point, introduce the fact that the collision is perfectly elastic. (d) Prove that (theta1) + (theta2) = pi/2
My attempt:
momentum before = momentum after.
Po + P'o = P + P'
(a) m1v0 = m1v1cos(theta1) + m2v2cos(theta2)
0 = m1v1sin(theta1) + m2v2sin(theta2)
(b) Squaring
m1^2v1^2 = m1^2xv1^2cos(theta1)^2 + m2^2xv2^2cos(theta2)^2
0 = m1^2xv1^2sin(theta1)^2 + m2^2xv2^2sin(theta2)^2.
But m1=m2 and after adding we get:
v0^2=v1^2 +v2^2.
(c) m1v0^2 = m1v1^2 + m2v2^2,
but m1=m2, therefore
v0^2=v1^2 + v2^2.
(c) give us no more new info.
(d)

 

[Doc Al]

My attempt:
momentum before = momentum after.
Po + P'o = P + P'
(a) m1v0 = m1v1cos(theta1) + m2v2cos(theta2)
0 = m1v1sin(theta1) + m2v2sin(theta2)

First problem: Since theta2 is below the x-axis, that second term in the second equation should be negative:
0 = m1v1sin(theta1) - m2v2sin(theta2)

(b) Squaring
m1^2v1^2 = m1^2xv1^2cos(theta1)^2 + m2^2xv2^2cos(theta2)^2
0 = m1^2xv1^2sin(theta1)^2 + m2^2xv2^2sin(theta2)^2.

Second problem: Take the square properly!
$$(a + b)^2 \ne a^2 + b^2$$

 

[John O' Meara]

Thanks very much for the corrections.After the corrections this is what I got for (b)
v0^2 = v1cos(theta1)^2 + v1v2cos(theta1)cos(theta2)+v1v2cos(theta2)cos(theta1) + v2^2cos(theta2)^2 ...(iii)

0 = v1^2sin(theta1)^2 - v1v2(sin(theta1)sin(theta2) + sin(theta2)sin(theta1)) + v2^2sin(theta2)^2 ...(iv)

Which gives on adding (iii) & (iv)

v0^2 = v1^2 + v2^2 + 2v1v2cos((theta1) + (theta2)) ...(v)

Part (c) is as before
v0^2 = v1^2 + v2^2

(d) When cos((theta1) + (theta2)) = cos(pi/2)=0. Then,
v0^2 = v1^2 + v2^2.
which proves that
(theta1) + (theta2) = pi/2, for a perfectly elastic non central collision.