A 725kg bowling ball moving at 100 ms collides with a 160kg

A 7.25-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 34.0° 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? (c) Ignoring rotation, what is the final kinetic energy of the system of the bowling ball and pin after the collision?

Solution

Here we will assume that the direction of the ball’s initial velocity is the positive x direction. Let’s determine its initial momentum.

M = 7.25 * 10 = 72.5
For the pin, final x momentum = 1.6 * 8 * cos 34.0 = 12.8 * cos 34.0
For the pin, final y momentum = 1.6 * 8 * cos 34.0 = 12.8 * sin 34.0
To determine the x component of the ball’s final momentum, subtract 12.8 * cos 34.0 from 72.5.

x = 72.5 – 12.8 * cos 34.0 = 61.9
vx = 61.9 / 7.25 = 8.54 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 34.0.

vy = -12.8/7.25 * sin 34.0 = - 0.99 m/s.
(a) For determine the magnitude of its final velocity, use the following equation.

Final velocity = (x^2 + y^2) = sqrt [8.54^2 + (-0.99)^2] = 8.60 m/s
For direction -
Tan = y ÷ x = -0.99 / 8.54

=> = - 6.61^o   
This is 90^o - 6.61^o = 83.39^o clockwise from the direction of the ball’s initial velocity.

(b) For the ball, original KE = ½ * 7.25 * 10^2 = 362.5 J
(c) For the ball, final KE = ½ * 7.25 * 8.60^2 = 268.10 J
For pin, final KE = ½ * 1.6 * 8^2 = 51.2 J

So, the final KE of the system = 268.10 + 51.20 = 319.30 J