Question

Two ice pucks (one orange and one blue) of equal mass are involved in a perfectly elastic glancing collision as shown in the figures below. The orange puck is initially moving to the right at

v_oi = 7.30 m/s, strikes the initially stationary blue puck, and moves off in a direction that makes an angle of θ = 36.0° with the horizontal axis while the blue puck makes an angle of ϕ = 54.0° with this axis as in the second figure.

Note that for an elastic collision of two equal masses, the separation angle

θ + ϕ = 90.0°.

Determine the speed of each puck after the collision in meters per second.

V0f= m/s

Vbf= m/s

Answer 1

m1v1 = m1v1' + m2v2 (conservation of momentum); however since m1 = m2
v1 = v1' + v2 (where v are vectors, not scalars)

Therefore
Vo = Vo'*cos(th) + Vb*cos(ph), (x direction) and
Vo'*sin(th) = Vb*sin (ph) or Vb = Vo'* (sin(th)/sin(ph)) (y direction; basically the two vertical vectors must cancel one another out since the original vector had no vertical vector component)
Substituting
Vo = Vo'*cos(th) + Vo'* (sin(th)/sin(ph))*cos(ph)
7.30m/s = Vo'*(cos36° + sin36°/tan54°), or

7.30m/s = Vo'*(cos36° +0.4270)
Vo' = 5.906m/s
Vb = 5.906m/s*(sin36°/sin54°) = 4.290m/s