JEE Main Physics question
A player kicks a football of mass 0.5 kg and the football begins to move with a velocity of 10 m/s. If the contact between the leg and the football lasts for 1/50 sec, then the force acted on the football should be
Answer: (C)
Solution
Impulse = change in momentum = mΔv = 0.5×10 = 5 kg·m/s. Average force F = Impulse/Δt = 5/(1/50) = 5×50 = 250 N.