Question

A 0.40 kg soccer ball is moving to the right with speed of 25m/s when it collides with a 0.60 kg basketball moving to the left with speed of 20 m/s. The basketball bounces off with speed of 12 m/s at an angle of 32.0° relative to its initial path as shown if the figure.

At what angle, φ, relative to its original path does the soccer ball move after the collision? What is the speed of the soccer ball after the collision? Is this collision elastic?

Answer 1

m1 = 0.4 kg, m2 =0.6 kg, u1 =25 m/s, u2 = -20 m/s

v2 = 12 m/s

from conservation of momentum long x direction

m1u1x +m2u2x = m1v1x +m2v2x

(0.4*25) - (0.6*20) = (0.4*v1x) +(0.6*12*cos(32))

v1 =-20/26 m/s

From conservation of momentum along y direction

m1u1y +m2u2y = m1v1y +m2v2y

0+0 = (0.4*v1y) +(0.6*12*sin(32))

v1y =-9.54 m/s

v1 = (20.26^2 +9.54^2)^0.5 = 22.4 m/s

tan(theta) = 9.54/20.26

theta = 25.2 degrees

φ = 180 +25.2 = 205 degrees

Ki-Kf = (1/2)m1u1^2 +(1/2)m2u2^2 -(1/2)m1v1^2 - (1/2)m2v2^2

= (0.5*0.4*25^2)+(0.5*0.6*20^2)-(0.5*0.4*22.4^2)-(0.5*0.6*12^2)

Ki - Kf = 101.5 J

Kinetic energy is not conserved , so it is inelastic collision