<p>A 0.25-kg soccer ball is rolling at 6.0 m/s toward a player. The player kicks the ball back in the opposite direction and gives it a velocity of - 14m/s. What is the average force during the interaction between the player's foot and the ball if the interaction lasts 2.0 x 1/100 s?</p>
<p>f=ma
v=vo+at
get the a
average force then is
f=ma</p>
<p>Force is equal to the change in momentum over time.</p>
<p>so force = (-14)(0.25) - (6)(0.25) all over (the time)</p>
<p>PS: i took the december physics II exam....i dont remember this question being on it....</p>
<p>bleh my way is better</p>
<p>actually mavman100, this has to be done through momentum. If you work it out both methods give different results. This is a typical momentum problem.</p>
<p>Thank you so much for your explanation!</p>
<p>avelez89 if you can think outside the box .. it shows more physics skizzles .. no doubt it is a momentum problem i just think my way shows how all of mechanichs fit togethor ..your way is easier though.. but i feel mine is better</p>
<p>we had almost the exact same problem on a test today...strange.</p>
<p>i think I got it wrong :(</p>
<p>Isn't Faverage = (Mass)(Velocity)/(Change in time)?</p>
<p>tribe567: Force(average) = Change in Momentum over change in time. very close though.</p>
<p>oh yes, but isn't the change in momentum equal to delta p which is equal to pf - pi which is equal to the (mass x final velocity - mass x initial velocity) which is almost the same thing as mass x velocity? almost!</p>
<p>Mavman, I do not believe your way is correct. What you need here is impulse. Impulse = Force * Time = Change in Momentum. Find the final momentum minus initial momentum, then divide by the time. This IS the correct way to do this problem, I've done several problems EXACTLY like this and that is how it works.</p>