Question

A soccer player hits a 0.20 kg ball which is incoming with a speed 5 m/s. After the kick, the ball changes its direction by 90 degrees and keeps its original speed. What is the average force exerted by the player if the impact time is 0.02 seconds?

Answer 1

Average force is given by,
Favg = dP/dt

here, dP = change in momentum = Pf - Pi = m*(Vf - Vi)

dt = time interval = 0.02 s

m = mass = 0.20 kg

Vi = initial speed = 5 m/s i (Let, initially ball comes in +x direction)

Vf = final speed = 5 m/s j (As, bal changes it's direction by 90 deg)

then, Favg = 0.20*(5 j - 5 i)/0.02

Favg = -50 i + 50 j

then, magnitude of force will be:

|Favg| = sqrt(50^2 + 50^2) = 50*sqrt(2)

|Favg| = 70.7 N

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