物理既問題, urgent

1. Using Law of Conservation of Momentum
(a) Momentum along x-direction = mu
(b) Momentum along y-direction = 0

2. In direction along the original direction of the ball A, using conservation of momentum,
mu = m(va).cos(theta-a) + m(vb).cos(theta-b)
i.e. u = (va).cos(theta-a) + (vb).cos(theta-b)

In direction perpendicular to original direction of ball A,
0 = m(va).sin(theta-a) - m(vb).sin(theta-b)
i.e. (va).sin(theta-a) = (vb).sin(theta-b)

3. (a) momentum
(b) Since the two balls are of the same mass and the collision is elastic, hence, 60 + (theta) = 90
i.e. (theta) = (90 - 60) degrees = 30 degrees

(c) Using equations obtained from Q2 above, we have,
2 = (va).cos(60) + (vb).cos(30) ---------------- (1)
(va).sin(60) = (vb).sin(30) ---------------- (2)
solving (1) and (2) for (va) and (vb) gives
va = 1 m/s and vb = 1.732 m/s