What is the average force during the interaction between the player's foot and the ball?

Question

A 0.28-kg soccer ball is rolling at 7.0 m/s toward a player. The player kicks the ball back in the opposite direction and gives it a velocity of -14 m/s. What is the average force during the interaction between the player's foot and the ball if the interaction lasts 3.0E-2 s?
_____ N

PHYSICS FINAL REVIEW, PART 1
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Answer 1

a = v/t = 21/0.03 = 700

F = 0.28 * 700 = 196 N

Answer 2

There are two ways to do this, they are equivalent, but you might find one easier. The first is to use the change in momentum. F=dP/dt, or if you don't have calculus, this more or less means the change in momentum divided by the change in time. The initial momentum was (0.25kg)*(6m/s), the final was (0.25kg)*(-14 m/s). Subtracting the initial from the final, you get that the change in momentum was -5 kg*m/s. The change in time was 1/50 s, so -5 kg*m/s divided by 1/50 s gives you an average force of -250 kg*m/s^2, or -250 Newtons. The second method uses the formulation that most are more familiar with, F=ma. To find the average force, find the average acceleration. To find the average acceleration use the vormula dv/dt=a, or the change in velocity divided by the change in time equals the average acceleration. dv=-14-6=-20 m/s. Again the change in time is 1/50 s so your average acceleration is -1000 m/s. Plugging this value for acceleration and the known mass of the ball into F=ma, you get the same answer of -250 Newtons

Answer 3

The answer is simple. F = m x a.

a = (change in velocity)/(change in time) = dv/dt.

So your acceleration is -21(m/s)/3.0E-2s = -700 m/s^2.

Force = m x a = (0.28kg)(-700m/s^2) = -196 N.

If the question is asking for magnitude of force, it's 196N.