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.

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 ...

Solution Summary

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.

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