Question

6. A 0.3 kg rubber ball is dropped from the window of a building. It strikes the sidewalk below at 30 m/s and rebounds up at 20 m/s. The magnitude of the impulse due to the collision with the sidewalk is:

Answer 1

The magnitude of the impulse due to the collision of a rubber ball with the sidewalk

$$\Delta p=m|{\bf v_2-v_1}|=m(v_2-(-v_1))\\=0.3\:\rm kg*(20+30)\: m/s=15\: kg*m/s$$