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.
p = mv
Conservation of momentum
maVa + mbVb = ma'Va' + mb'Vb'
Conservation of KE
1/2maVa^2 + 1/2mbVb^2 = 1/2maVa'^2 + 1/2mbVb'^2
The Attempt at a Solution
I have looked at a few solutions on here and elsewhere, but the concept is just not quite clicking yet... I'm not really sure how to mess around with the equations to get what I need. I understand that you have to break it down into components, so:
ma = mb
p(x) = mVa = mVa' + mVb'
mVa = m(Va' + mVb')
Va = Va' + Vb' (and the horizontal components of the velocities are *cosθ)
Va = Va' *cosθ + Vb' *cosθ (the image included with the question shows ball B going below the x axis at an angle = θ, so I am wondering if the second θ is negative...)
I am not too sure what to do with this information ... maybe solve for Va' or Vb' and leave it there for now? So it would be:
Va' = (Va - Vb' *cosθ)/cosθ
Vb' = (Va - Va' *cosθ)/cosθ
p(y) = mVa + mVb = mVa' + mVb'
0 = Va' + Vb'
0 = Va' *sinθ + Vb' *sinθ (again, not sure if the second θ is negative, since it is below the x axis)
Va' *sinθ = -Vb' *sinθ
Also, from the conservation of energy:
0.5m(Va^2) + 0 = 0.5m(Va'^2 + Vb'^2)
Va^2 = Va'^2 + Vb'^2
I'm stuck after this. I don't know what I'm looking for after this point... Any help is appreciated! Thanks!