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

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SUMMARY

The average force exerted by a wall on a 3.00 kg steel ball striking at 9.0 m/s at a 60° angle is calculated using impulse principles. The correct approach involves determining the change in momentum, not the gravitational force. The average force is derived from the equation F_avg(Δt) = Δp, leading to a calculated average force of -135 N when considering the vector components of the ball's velocity before and after impact.

PREREQUISITES
  • Understanding of impulse and momentum principles
  • Knowledge of vector decomposition in physics
  • Familiarity with Newton's second law of motion
  • Basic calculus for evaluating integrals
NEXT STEPS
  • Study the concept of impulse and its relation to momentum
  • Learn how to decompose vectors into their x and y components
  • Review Newton's laws of motion, particularly the second law
  • Practice solving problems involving average force and momentum changes
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Students studying physics, particularly those focusing on mechanics and impulse-momentum theory, as well as educators looking for problem-solving strategies in dynamics.

chocolatelover
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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
 
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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
 

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