Question

A 5.5 kg bowling ball moving at 9.55 m/s collides with a 0.875kg bowling pin, which is scattered at an angle of theta = 84.5 degrees from the initial direction of the bowling ball, with a speed of 17m/s

a) calculate the direction, in degrees, of the final velocity of the bowling ball. This angle should be measured in the same way that theta is.

b) calculate the magnitude of the final velocity, in meters per second, of the bowling ball.

Answer 1

Let's assume the initial direction was along the x-axis, or "horizontal."

a) Conserve vertical momentum:
0 = 0.875kg * 17m/s * sin84.5º - 5.50kg * v * sinΘ
where Θ is measured clockwise from the +x axis.
Then v*sinΘ = 2.7 m/s ← #1

Conserve horizontal momentum:
5.50kg * 9.55m/s = 0.8750kg * 17m/s * cos84.5º + 5.50kg * v * cosΘ
v*cosΘ = 9.3 m/s ← #2

Divide #1 by #2:
v*sinΘ / v*cosΘ = tanΘ = 2.7/9.3 = 0.2903
Θ = 16.2º ◄ to the initial direction

v * cos16.2º = 9.3 m/s
v = 9.68 m/s ◄ velocity