Question

A 2.0-kg ball moving at 10 m/s makes an off-center collision with a 3.0-kg ball that is initially at rest. After the collision, the 2.0-kg ball is deflected at an angle of 30° from its original direction of motion and the 3.0-kg ball is moving at 4.0 m/s. Find the speed of the 2.0-kg ball and the direction of the 3.0-kg ball after the collision (angle from original direction of motion)

Answer 1

m1=2 kg, v = 10 m/s, m2 = 3 kg, α =30º, u2 = 4 m/s. Find β and u1.
Take the x axis to be the original direction of motion of 2kkg ball. The equations expressing the conservation of momentum for the components in the x and y directions separately are
p(x): m1•v = m1•u1•cos α + m2•u2•cos β,
p(y): 0 = m1•u1•sin α - m2•u2•sin β.

cos β = (m1•v - m1•u1•cos α)/ m2•u2,
sin β = m1•u1•sin α/ m2•u2,

sin β/ cos β = tan β = m1•u1•sin α/ m1•v(1-cos α) =
=sin α/ (1-cos α) = 0.5/(1-0.866) = 3.73,
α = arctan 3.73 = 75º.

u1 = m2•u2•sin β / m1•sin α = 3•4•sin75º/2•sin30º = 11.6 m/s.

Answer 2

Apparently the answer is 5.3 at 26 degrees but I don't understand why?