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.
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.