JEE Main Physics question
A 1.5 kg ball drops vertically onto a floor, hitting it with a speed of 25 ms⁻¹. It rebounds with an initial speed of 15 ms⁻¹. If the ball was in contact with the floor for only 0.03 seconds, the force exerted on the floor by the ball is
Answer: (A)
Solution
Taking downward as positive, Δp = − = 1.5(25 − (−15)) = 1.5 × 40 = 60 kg·m/s. F = Δp/Δt = 60/0.03 = 2000 N.