Question

A 1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.020 s, what is the magnitude of the average force on the floor from by the ball? Neglect air resistance.

Answer 1

The speed of the ball just before the bat floor v2 = u ²+2gh = 0+2x9.8 X 32 = 627.2 I for velocity V=25.04 m/s 7 + up (positive) to down (negating @ We know that impulse of the ball I= DP = mi - my Here m= 1.2 kg 12 = 10ml I-mer-v) V = - 25.04 ms I= 12 (10-(-25.04)) I = 1.2x 35.04 I = 42.048 Kg m/s Using Newtors and law Here F = AP AP = 42.048 Ns. At = 0.02os At 42.048 F= 0.020 F = 2102-4N