A 6.95-kg bowling ball moving at 10.0 m/s collides with a 1.60-kg bowling pin, s
ID: 1436478 • Letter: A
Question
A 6.95-kg bowling ball moving at 10.0 m/s collides with a 1.60-kg bowling pin, scattering it with a speed of 8.00 m/s and at an angle of 38.5° with respect to the initial direction of the bowling ball.
(a) Calculate the final velocity (magnitude and direction) of the bowling ball.
(b) Ignoring rotation, what was the original kinetic energy of the bowling ball before the collision?
J
(c) Ignoring rotation, what is the final kinetic energy of the system of the bowling ball and pin after the collision?
Explanation / Answer
(a) Let’s assume that the direction of the ball’s initial velocity is the positive x direction. Let’s determine its initial momentum.
M = 6.95 * 10 = 69.5
For the pin, final x momentum = 1.6 * 8 * cos 38.5 = 12.8 * cos 38.5
For the pin, final y momentum = 1.6 * 8 * sin 38.5 = 12.8 * sin 38.5
To determine the x component of the ball’s final momentum, subtract 12.8 * cos 38.5 from 69.5.
x = 69.5 – 12.8 * cos 38.5 = 59.48
vx = 59.48 ÷ 6.95 = 8.56 m/s
Since the ball was initially moving in the positive x direction, it had no y momentum. For the sum of the y momentums to be 0, its y momentum must be -12.8 * sin 38.5.
vy = -12.8/6.95 * sin 38.5
This is approximately negative 2.96 m/s
To determine the magnitude of its final velocity, use the following equation.
Speed = (x^2 + y^2)
x^2 = 8.56^2
y^2 = (-12.8/6.95 * sin 38.5)^2 = (-2.96)^2
Speed = 82.035
This is approximately 9.06 m/s.
Tan = y ÷ x = -12.8/6.95 * sin 38.5 ÷ 8.56
The angle is approximately negative 83.9. This is 83.9 clockwise from the direction of the ball’s initial velocity.
(b) For the ball, original KE = ½ * 6.95 * 10^2 = 347.5 J
(c) For the ball, final KE = ½ * 6.95 * 82.035 = 285.07 J
For pin, final KE = ½ * 1.6 * 8^2 = 51.2 J