Question

Two balls of equal mass undergo a collision. Ball one is initially travelling horizontally with a speed of 10 m/s, ball two is initially at rest. After the collision, ball one moves away with a velocity of 4.7 m/s at an angle of 60 degrees from it's original path and ball two moves at an unknown angle. Determine the magnitude and direction of velocity of ball two after the collision.

Answer 1

before collision
x momentum = 10m
y momentum = 0

after collision
x momentum = 10m = 4.7cos 60 m + Vx m

y momentum = 0 = 4.7sin60m + Vy m

then
v = sqrt(Vx^2+Vy^2)
tan A = Vy/Vx
note A is in quadrant IV, below x axis if ball one went northeast

Answer 2

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