Rubber Throw Down
How to find the magnitude of the average force?
Alex throws a 0.15-kg rubber ball down onto the floor. The ball's speed just before impact is 6.5 m/s, and just after is 3.5 m/s. If the ball is in contact with the floor for .025 s, what is the magnitude of the average force applied by the floor on the ball?
Using the equation F= (mv(final)-mv(initial))/(change in) time
((.15*3.5) - (.15*6.5))/0.25
I get 18N as the magnitude, but that's not an answer out of the four answers I can choose from. What am I doing wrong?
Some of the numbers are not consistent throughout your question BUT:
lets assume the following:
mass(m)= 0.15
initial velocity(u) = -6.5
final velocity(v) = 3.5
time(t) = 0.025
The formula you have stated is correct:
F = (mv-mu)/t
The key thing to remember is that velocity and hence momentum are VECTORS that means they need a direction as well as a magnitude(size), in this example I will take the downward direction to be negative (ie) any velocity/momentum acting downwards will have a negative sign before it so u=-6.5 because the ball is going downwards)
So:
plug the numbers into the formula except this time, put a negative in front of initial velocity(u) and you get:
F = ((0.15*3.5)-(0.15*-6.5)/0.025 = 60N (the fact that the force is positive means that it acts upwards which makes sense, force is a vector meaning you usually have to give a written direction as well as a magnitude but since you are asked for a magnitude only this should be ok).
PS: The time is not clear in the question and if the time is in fact 0.25s, the answer will be 6N)
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