How long does it take a 390 N force to stop a 0.15 kg baseball?

[JMarino]

Homework Statement

A 0.15 kg baseball moving at +26 m/s is slowed to a stop by a catcher who exerts a constant force of 390 N. How long does it take this force to stop the ball? How far does the ball travel before stopping?

Homework Equations

The Attempt at a Solution

For P=mV = 0.15Kg x 26M/s = 3.9Kgm/s. Then what equation?
I = F deltaT = m deltaV
deltaT=m deltaV / F = 3.9/390 = .01s Is that correct?
What about last question of travel distance. Is speed constant?

 

[denverdoc]

first part looks good.

The speed cannot be constant, after all its 26m/s one instant, and zero the next.

there are several approaches to second part, have you learned about work at all?

 

[JMarino]

somewhat, where do you suggest I look specifically?

 

[JMarino]

work = force x distance moved. There are two variables I do not know in this equation...

 

[denverdoc]

well there's a nifty thing known as work-energy theorum, which would suggest that right before the ball hits the mitt, it has a total energy, and when it comes to rest it has less, and that the difference would we the dot product of force and distance, in this case just the simple product of the two, or force*distance. That distance is the one you seek.

 

[denverdoc]

work = force x distance moved. There are two variables I do not know in this equation...

see above as well. I'm not trying to be confusing, its just that the energy can be a combination of different types such as potential and kinetic. I would assume the difference in potential energies are negligible, so look at the kinetic energies.

 

[JMarino]

Thanks for the help!

 

[JMarino]

W = KE-f - KE-0
W = 1/2 m(V-f)2 - 1/2m(V-0)2
W = 0 - 1/2 * .15 * (26)2 = 50.7

D = -50.7/F = 50.7/350

?

 

[denverdoc]

looks fine except the force was 390. If you remember kinematics, there is also an eqn;
Vf^2-Vi^2=2a*x where a would be 390/0.15

see the similarity with what you have done?