Maybe you got the signs wrong? Remember, the incoming ball, and the ball leaving the bat will have velocities of different signs.
A baseball (m = 149 g) approaches a bat horizontally at a speed of 43.3 m/s (97 mi/h) and is hit straight back at a speed of 49.5 m/s (111 mi/h).
If the ball is in contact with the bat for a time of 1.10 ms, 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.
i got 836.4N which is incorrect!
i first solved for J = m * vf - m * v0 which gives the answer in m/s
but then i used: J / chng time to find the avg. Force.
I think i'm getting confused because of the units.
You have to keep in mind that when it's hit back, it's final velocity is actually NEGATIVE 49.5 m/s. The change in velocity is vf-v0, or (-49.5)-(43.3)= -92.8m/s.
The change in the ball's momentum is mass * change in velocity or 0.149Kg * 92.8m/s, which equals 13.8Kg*m/s.
Impulse is force times time, and impulse equals change in momentum, so 13.8Kg*m/s = F*0.0011s. Solving for F gives 12545N. Is that what your book/teacher says is the right answer?
I quickly tried 49.5m/s-43.3m/s. Solving it using that delta_v gives you the answer you got, so your problem is in the signs.
Perhaps the problem is that the force is in the negative direction, so the component of the force in the positive direction is -12570.2 N?
topsquark is right, I should not have rounded an intermediate answer. If I were actually doing this problem, I'd solve it for force symbolically then let my calculator handle all the arithmetic.