# find force when baseball is hit

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• November 14th 2007, 02:55 PM
rcmango
find force when baseball is hit
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.

please help.
• November 15th 2007, 08:15 AM
qspeechc
Maybe you got the signs wrong? Remember, the incoming ball, and the ball leaving the bat will have velocities of different signs.
• November 15th 2007, 05:11 PM
Pulsar06
Quote:

Originally Posted by rcmango
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).

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.

Pulsar
• November 15th 2007, 08:33 PM
rcmango
no, i see how you got 12545.45 N's, however this is not the correct answer.

I think i must use the impulse momentum theorem of (m * v f) - (m * v0)

to get the impulse somehow, but i'm not sure how to set it up.

thanks so far.

Still need help.
• November 16th 2007, 07:24 AM
topsquark
Quote:

Originally Posted by rcmango
no, i see how you got 12545.45 N's, however this is not the correct answer.

I think i must use the impulse momentum theorem of (m * v f) - (m * v0)

to get the impulse somehow, but i'm not sure how to set it up.

thanks so far.

Still need help.

Pulsar06's method is absolutely correct. The only thing I would change is to not round the numbers in the middle of the problem. That gives me a slightly different answer of 12570.2 N. (This is almost the same as Pulsar06's answer to the correct number of significant digits, but if you are expected to input a decimal answer into the computer it might not get the number of sig. digs. correct. I have seen this kind of problem before.)

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?

-Dan
• November 16th 2007, 01:57 PM
Pulsar06
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.

Quote:

Originally Posted by rcmango
I think i must use the impulse momentum theorem of (m * v f) - (m * v0)

to get the impulse somehow, but i'm not sure how to set it up.

Right. This is essentially what I was trying to say. "(m * v f) - (m * v0)" is the change in the momentum. Since mass * velocity = momentum, and you're subtracting the initial momentum from the final momentum, the "impulse momentum theorem" as you called it should give 13.8(ish) KgM/s. Once you have that, use the fact that $FT=dP$ to solve for F.
• November 16th 2007, 11:03 PM
rcmango
Okay, i figured it out. The problem set up is just fine for my understanding.