What Is the Average Acceleration of a Super Ball Bouncing Off a Wall?

  • Thread starter Thread starter Rob21
  • Start date Start date
  • Tags Tags
    Ball Motion
AI Thread Summary
The discussion centers on calculating the average acceleration of a Super Ball that rebounds off a wall. The ball, initially traveling at 30.0 m/s and rebounding at 22.0 m/s, is in contact with the wall for 5.00 ms. The average acceleration is calculated using the formula (final velocity - initial velocity) divided by the time interval, resulting in -1600 m/s². It is emphasized that attention to the direction of velocity is crucial, as velocity is a vector quantity. The final answer for the magnitude of the average acceleration is 1600 m/s².
Rob21
Messages
1
Reaction score
0

Homework Statement



A 50.0 g Super Ball traveling at 30.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 5.00 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms = 10-3 s.)


Homework Equations



avg acceleration=(final V-initial V)/change in time

The Attempt at a Solution



(22m/s-30m/s)/0.005s= -1600m/s^2 and since it asks for the magnitude my final answer is
1600m/s^2

This was from my online physics homework. I really don't see what I'm doing wrong.
 
Physics news on Phys.org
Rob21 said:

Homework Statement



A 50.0 g Super Ball traveling at 30.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 5.00 ms, what is the magnitude of the average acceleration of the ball during this time interval? (Note: 1 ms = 10-3 s.)

Homework Equations



avg acceleration=(final V-initial V)/change in time

The Attempt at a Solution



(22m/s-30m/s)/0.005s= -1600m/s^2 and since it asks for the magnitude my final answer is
1600m/s^2

This was from my online physics homework. I really don't see what I'm doing wrong.

I think you need to look at your direction signs more carefully.

Remember velocity is a vector. The sign you assign it needs to be carried through the calculation.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top