A 0.075 kg ball hits a frictionless, rigid, horizontal surface with a speed of 1.2 m/s at the angle q1 = 70 deg. The angle of rebound is q2 = 62 deg. Compute:
a) the speed of the ball immediately after rebound.
b) the resultant impulse acting on the ball during its time of contact with the surface.
See attachment for diagram.© BrainMass Inc. brainmass.com March 4, 2021, 6:17 pm ad1c9bdddf
Momentum is conserved in the x direction for this collision the initial velocity is a vector vx is 1.2 m/s cos(70) Vy is -1.2 m/s sin(70) The final velocity vector is v_final cos (62) + v_final sin(62)
the y component of the velocity ...
The solution provides written explanation for each step of calculations performed to find the speed and impulse of a bouncy ball during its rebound off a horizontal surface.