A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical
stationary ball as shown. If the first ball moves away with angle 30 ° to the
original path, determine
a. the speed of the first ball after the collision.
b. the speed and direction of the second ball after the collision.
m1v1 + m2v2 = m1v1' + m2v2'
1/2m1v1^2 + 1/2m2v2^2 = 1/2m1v1'^2 + 1/2m2v2'^2
The Attempt at a Solution
m1 = m2 = 2.0kg
v1 = 3.0m/s
v2 = 0m/s
theta1 = 30
In the x direction
m1v1 + m2v2 = m1v1'cos30 + m2v2'cos(theta2)
(2.0)(3.0) + (2.0)(0) = (2.0)v1'cos30 + (2.0)v2'cos(theta2)
masses cancel out to leave;
3.0 = v1'cos30 + v2'cos(theta2)
In the y direction
m1(0) + m2(0) = m1v1'sin30 + m2v2'sin(theta2)
masses cancel to leave:
-v1'sin30 = v2'sin(theta2)
Now here is where I am getting stuck. I know that I have to somehow combine the x and y equations as well as the conservation of NRG eqn to find the three unknowns, but I am lost as to how the algebra will work out. I think I have to square them somehow to eventually get cos^2(theta2) + sin^2(theta2) = 1 to get around the unknown theta... but everytime I have tried, it doesn't work out. Please help!
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