# Average force exerted by a wall on a ball

1. Jun 24, 2010

### Agent M27

1. The problem statement, all variables and given/known data
A 3.72 kg steel ball strikes a massive wall at 15.1 m/s at an angle of 58.6 with the plane of the wall. It bounces off with the same speed and angle. If the ball is in contact with the wall for .198s what is the average force exerted on the ball by the wall? Answer in units of N.

The picture is depicted with the wall along the y axis and the ball is originating from the 3rd quadrant with an angle of 58.6 between the trajectory and the negative x axis. It then bounce off and continues into the second quadrant making the 58.6 angle with the negative x axis again.

2. Relevant equations

(1) $$\Delta$$Px = -mvfcos$$\theta$$-mvicos$$\theta$$

(2) $$\Delta$$Py = mvfsin$$\theta$$ - mvisin$$\theta$$

(3) Favg=$$\frac{dP}{dt}$$

3. The attempt at a solution

So I began by using EQ 1 and subbing in known values and arrived at an answer of -58.53231 N s. I then use that value in EQ 3 and arrive at a final answer of -295.61771 N. The online module doesn't give me the correct answer but says this is incorrect. Where am I going wrong? Thanks

I also tried using the angle between the y axis and the trajectory which gave me an angle of 31.4. This yielded an answer of -95.89 N s. Subbing that into EQ 3 I get an answer of 484.2995 N, yet again this is wrong...

Joe

2. Jun 24, 2010

### Staff: Mentor

Your numbers seem OK to me. I assume they just want the magnitude of the force. (As to which one is correct, it depends on what that angle is with respect to. "an angle of 58.6 with the plane of the wall" sounds like your second calculation.)

Sometimes those on-line systems can be picky about significant figures. (Some insist on 3 sig figs.)

3. Jun 24, 2010

### Agent M27

Ya this system requires at least 6 sig figs, which is why I gave those long answers. I am going to have to ask my professor because I tried both of those answers with a negative and positive, and still they are registering incorrect. Am I correct in assuming the reason I can disregard change of momentum in the y direction because it is still travelling in the positive why direction, is that a correct assumption? Thanks

Joe

4. Jun 24, 2010

### Staff: Mentor

Sure. You're not really disregarding it. The change in momentum parallel to the wall is zero.

5. Jun 24, 2010

### Agent M27

I can't believe I spelled the word why when refering to the y axis... I had a chukle at that one. Anyway. Thanks for the description, it makes more sense that way.

Joe