How Is the Average Force Calculated When a Steel Ball Hits a Wall?

Click For Summary

Homework Help Overview

The problem involves a steel ball striking a wall at an angle and determining the average force exerted by the wall on the ball during the impact. The context includes concepts from dynamics and impulse-momentum theory.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of average force, questioning the use of gravitational force in the context of the problem. There are attempts to apply impulse concepts and evaluate the momentum change during the impact.

Discussion Status

Multiple interpretations of the problem are being explored, with some participants providing guidance on the correct approach to calculating average force using impulse. There is no explicit consensus on the correct method yet, as various attempts and corrections are being discussed.

Contextual Notes

Participants are navigating the implications of vector components and the distinction between gravitational force and the force exerted by the wall. The time of contact and the angle of impact are also relevant factors under consideration.

chocolatelover
Messages
238
Reaction score
0

Homework Statement


A 3.00 kg stell ball strikes a wall with a speed of 9.0m/s at an angle of 60° with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for .200 s, what is the average force exerted by the wall on the ball?


Homework Equations


ΣFavg=1/Δt integral ΣFdt from ti to tf



The Attempt at a Solution



F avg=1/.200(3.00kg)(9.8m/s^2)x|0 to .200
=29.4N

Could someone please tell me if this looks correct and if now could someone please show me where I went wrong?

Thank you very much
 
Physics news on Phys.org
chocolatelover said:

The Attempt at a Solution



F avg=1/.200(3.00kg)(9.8m/s^2)x|0 to .200
=29.4N

The force you use here is the force of gravity, which isn't what the question is asking. The question is asking for the average force of the water on ball. You don't need to worry about gravity for this problem, because it is merely an impulse problem. Remember:
\int F dt=\Delta P
 
Thank you very much

Does this look right?

integral from 1.25 to .800 (.110kg)(9.8m/s^2)dt

=.8624-1.35
=-.4876

Thank you
 
Edit:

Does this look right?

integral 3.00(9.8)dt from 0 to .200=
3.00kg(9.8m/s^2)x=
3.00(9.8)(.200)=
5.88

Thank you very much
 
Hi chocolatelover,

I think you misunderstood Nabeshin's post. We are searching for the force, we don't know what it is. (When you set the force as 3 kg * 9.8 m/s^2, that is the force of gravity on the ball; but you want the force from the wall.)

If they are looking for the average force, the integral form reduces to

<br /> \vec F_{\rm avg}(\Delta t)= \Delta \vec p<br />

How do you find \vec F_{\rm avg}? Can you evaluate the other parts of the equation?
 
Does this look right?

Favg(.200s)=3.0kg(9.0)
Favg=-135N

Thank you
 
chocolatelover said:
Does this look right?

Favg(.200s)=3.0kg(9.0)
Favg=-135N

Thank you

You need to take into account the vector nature of the formula. In the x direction, and for a one object system, the formula is:

<br /> F_{{\rm avg},x} (\Delta t)= m v_{f,x} - m v_{i,x}<br />

and a similar equation for y.

To evaluate this, you'll need to find the x and y components of the initial and final velocities, and plug them into the x equation and the y equation.
 
Thank you very much

Regards
 

Similar threads

Replies
5
Views
6K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 2 ·
Replies
2
Views
7K
Replies
16
Views
3K
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 4 ·
Replies
4
Views
6K
Replies
3
Views
5K
  • · Replies 10 ·
Replies
10
Views
4K