Is this problem really this simple?

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Homework Help Overview

The problem involves a 50-gram ball rebounding off a wall, with given velocities before and after impact, and a contact time with the wall. Participants are exploring the calculation of average acceleration during this interaction.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the interpretation of the problem, questioning whether it asks for average velocity or average acceleration. There is confusion regarding the correct calculation of change in velocity and the application of sign conventions.

Discussion Status

Some participants have provided guidance on the correct approach to calculating average acceleration, emphasizing the importance of direction in velocity changes. There is ongoing exploration of different interpretations and calculations, with no explicit consensus reached.

Contextual Notes

Participants are navigating the terminology and concepts in their first physics class, which may contribute to misunderstandings about the problem's requirements and the calculations involved.

wakejosh
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3. A 50-gram ball traveling at 25 m/s is bounced off a brick wall and
rebounds at 22 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 3.5 ms, what is the average acceleration of the ball during this time interval?

so is this as simple as 25+22= 47 then, 47/2= 23.5 m/s/s ?

im confused with the wording of this question. Is it asking what is the average velocity of the ball during the 3.5 seconds while its in contact with the wall? if so, is it just an average of the entry velocity and the exit velocity? (sorry i don't know the real terms). I am also guessing that the answer would be in M/s/s but I am not sure how to do the conversion, this is my first physics class.. any help is appreciated
 
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wakejosh said:
Is it asking what is the average velocity of the ball during the 3.5 seconds while its in contact with the wall?
No, it's asking for the average acceleration. (Note that velocity has units of m/s, but acceleration has units of m/s/s.)

To find the average acceleration, first find the change in velocity. Remember that direction (and thus sign) counts.

[tex]a_{ave} = \Delta v / \Delta t[/tex]
 
ok.. so

25 m/s -22 m/s = 3 m/s is the change in velocity

so my change in time is just 3.5 ms or 0.0035 s ?

so is it just 3 / 0.0035 = 857 m/s/s?
 
wakejosh said:
25 m/s -22 m/s = 3 m/s is the change in velocity
No. Realize that the initial and final velocities are in opposite directions and thus have opposite signs.

Pick a sign convention. Let's call "towards the wall" to be positive; "away from the wall", negative. Now state the initial and final velocities. (The change will be final minus initial.)

so my change in time is just 3.5 ms or 0.0035 s ?
Right.
 
yea.. i understand the negative part. in my work I actually figured it to be -3 m/s but my answers to choose from were all positive.

so is it going to be -857 m/s/s ? or something totally different?
 
How did you get that answer?

(In selecting your answer, I'm sure all they care about is the magnitude of the acceleration, which is always positive.)
 
Doc Al said:
How did you get that answer?
3 / 0.0035 = 857 m/s/s
 
Reread what I wrote in post #4. The change in velocity is not 3 (or -3).
 
Doc Al said:
Reread what I wrote in post #4. The change in velocity is not 3 (or -3).

25 - (-22) = 47

so

47/0.0035 = 13,428 m/s/s

thats got to be it. durrr.. sorry.. ty for the help
 
  • #10
Now you're cooking. Using the sign convention I suggested, I get:

Vi = +25 m/s
Vf = -22 m/s

Change in V = Vf - Vi = -22 - 25 = -47
Thus the average acceleration is -13400 m/s/s.

The minus sign just means that the direction of the acceleration is away from the wall.
 

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