The impulse momentum theorum?

Question

A baseball (m=149g) approaches a bat horizontally at a speed of 40.2 m/s and is hit straight back at a speed of 45.6 m/s. If the ball is in contact with the batfor a time of 1.10ms, what is the average force exerted on the ball by the bat? Neglect the weight of the ball since it is so much less than the force of the bat. Choose the direction of the incoming ball as the positive direction.

so..would you use
f=m(v2-v1)/t

.149kg(45.6-40.2)/1.1x10^3s
=7.31x10-4N?

this doesnt seem right..is that too small of a force from the bat? im confused about if i did this correct...and what would be my direction? away from the bat? in the negative direction?

i

Answer 1

Your equation is correct. You signs are wrong. Also you have a scientific notation error.

V1= 40.2
V2= -45.6 note the negative, since it is in the opposite direction
t=1.1x10^-3 note the negative again.

If you set the problem up correctly, the sign of your answer should take care of itself, but it is good to think about it: which direction is the force acting? In the opposite direction of the incoming ball, therefore it will be negative.

Answer 2

It is way too small by six orders of magnitude: 1.10ms = 1.10×10^-3 s, not 1.10×10^3 s. And then it is also too small by an additional factor of about 16 because you aren't paying attention to your signs.

You're told to choose the direction of the incoming ball as the positive direction, so v1 = +40.2 m/s and v2 = -45.6 m/s. So the force is 0.149 (-45.6 - 40.2) / (1.10×10^-3)
= -1.16×10^4 N.