Baseball thrown against a wall

  • Thread starter Thread starter link2440
  • Start date Start date
  • Tags Tags
    Baseball Wall
AI Thread Summary
To determine the force per square inch that a baseball would hit a wall when thrown at 50 mph, one must calculate the change in momentum over a specified time, such as 0.001 seconds, when the ball comes to a stop. The average force can be derived from this change in momentum divided by the time interval. However, the pressure exerted on the wall depends on the contact area between the ball and the wall, which varies during impact. High-speed cameras could provide more accurate data on the force dynamics and contact area changes. Overall, the complexities of physics and material behavior make precise calculations challenging without experimental data.
link2440
Messages
2
Reaction score
0
How do I determine the force per square inch that a baseball would hit a wall?

Assume the baseball was thrown at 50 mph? What would be the force? and does that translate into a weight or pressure?

Obviously the ball would stop immediately. If we assumed it stopped like a baseball bat of .001 seconds. Is that all we would need?

Thanks!

Jeff
 
Physics news on Phys.org
link2440 said:
Obviously the ball would stop immediately. If we assumed it stopped like a baseball bat of .001 seconds. Is that all we would need?

Maybe you mistyped... Obviously it is impossible for it to stop immediately. But it could stop in .001 seconds. Force is the derivative of momentum with respect to time. If you don't know what that means we can say that the average force is the average change in momentum over time. We know the time, .001 s. The change in momentum is the final momentum minus the initial momentum. Momentum is mass times velocity.

The final momentum would be the mass of the ball (use kilograms) and the final velocity of the ball (use meters per second). If the ball comes to a rest, its final velocity is zero and thus its final momentum is zero. If the ball bounces back then its final velocity would be negative because its going in a different direction that it started in. We know the original velocity (convert your miles per hour to meters per second to make it easier). Now you should be able to get the change in momentum.

Since you have the change in momentum and you have the change in time, divide the change in momentum by the change in time to get the average force. Note that I wrote average force. Thats because we are doing simplifications by ignoring how the momentum changes over that .001s interval. Does most of it change in the beginning? Does it change at the end? We might need a high speed camera to figure that out. But without that info we can just pretend that it changes uniformly during the entire .001s interval. (In reality the force will be higher at some times and lower at other times over that interval).

We have the average force. But you want the pressure (thats what force per square inch or area is). To figure this out we need to know how much the ball deforms upon impact. Does it squeeze a very tiny bit so that only a square millimeter touches the wall? Does it squeeze a lot so that multiple square centimeters touch the wall? The reality is that during the impact the area of the ball that contacts the wall is constantly changing. This means that the pressure is also constantly changing. This means that, again, we need a high speed camera to figure this out. We can do what we did before though and speak of an average pressure by making an educated guess to the average contact area the ball makes with the wall during the impact interval.

You can see from this why physics can be so hard. Even a seemingly simple question has many subtleties involved. To start we usually have to make approximations like we did here with average values. Then we have to turn to more data (high speed cameras of the impact in this case) to refine our theoretical average value into an observed experimental value.
 
As it turns out, real research has been done on the physics of baseball -- you should google it and read a bunch.
 
I have found lots of research but very little on a "weight" comparison for what a ball would feel like at 50 mph versus the standard 5 1/4 ounces.
 
Posts on a thread like this should be prefaced with a massive caveat. We get regular questions about Car Accidents and other damage related situations and there is never an easy definitive answer..
Questions about "Force" during a collisions tend to be a bit fruitless unless you know an awful lot about the structures involved. The forces during an impact are not constant and will depend upon the actual distortion of the object during the collision. It would be hard enough if you were dealing with a mass colliding with a wall, with an ideal spring in between - in which case, you could at least have a chance by applying Hooke's Law throughout the collision and coming up with an equation of motion which could be applied. With a 'rubber??' ball, the way the force varies would be much harder to analyse, due to the shape of the ball and the non-ideal behaviour of the material.
You can achieve the same effective 'bounce' by applying a small force for a long time or a large force for a short time (force times time is called Impulse).

Happily, it turns out that many collision problems can be sorted out, fairly accurately by considering change of momentum, using the idea of Impulse and the use of 'coefficient of restitution', which is the ratio of parting speed to approach speed. But the answers don't give you the actual Force.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Back
Top