# Particle Travels Between Two Walls

1. Sep 5, 2009

### Warmacblu

Hi all, I am a long time reader but first time poster. I figure I would start posting my homework problems because I seem to have a bunch of them and I do not have the ability to see my professor on a daily basis. Thanks for any future help.

1. The problem statement, all variables and given/known data

A particle travels horizontally between two parallel walls separated by 18.4 m. It moves
toward the opposing wall at a constant rate of 8.7 m/s. Also, it has an acceleration in the
direction parallel to the walls of 2.7 m/s^2.

Part 1 - What will be its speed when it hits the opposing wall?
Part 2 - At what angle with the wall will the particle strike?

2. Relevant equations

I first tried the constant acceleration equations but then I re-read the question to see that the particle does not have a constant acceleration in the Y-direction.

Vf = Vi + at

3. The attempt at a solution

After twenty minutes or so I tried this:

v^2 - u^2 = 2aS

where u = 8.7 and S = 18.4

v^2 = u^2 + 2aS
v = (u^2 + 2aS)^(1/2)
v = (8.7^2 + (2)(2.7)(18.4))^(1/2)
v = 13.23 m/s

I found this equation in my notes I took during class but wasn't sure if it was the right one to use or not.

I did not attempt part two.

Thanks for any help,
Ben

Last edited: Sep 5, 2009
2. Sep 5, 2009

### Staff: Mentor

Lets call the horizontal direction the x-axis. The way I understand the problem is that you have a particle that starts at one wall with a velocity in the x-direction. It experiences an acceleration in the y-direction. Hint: Treat the x and y motions separately.

How long does it take to reach the opposite wall?
What will be the y-component of its velocity at that time? The x-component?

3. Sep 5, 2009

### Warmacblu

Thanks for the reply Doc Al.

How long does it take to reach the opposite wall?

(18.4 m) / (8.7 m/s) = 2.11 s

What will be the y-component of its velocity at that time? The x-component?

I am unsure about the components part but here is what I think:

x-component = 18.4
y-component = (2.11) x (2.7) = 5.697

4. Sep 5, 2009

### Chewy0087

hmm personally i would find out the angle by seeing how far it's gone in the x and y direction between the hits, also remember though that it's a constant acceleration, not constant velocity,so i would use this equation

x = ut + 0.5at²

then it's geometry

Edit:also where did you get that 2.7 from? 0.o, you've said earlier the acceleration was 2.3

Last edited: Sep 5, 2009
5. Sep 5, 2009

### Staff: Mentor

Good.

Careful! Only the y-component is accelerated.
Is the acceleration 2.7 or 2.3 m/s^2?

6. Sep 5, 2009

### Staff: Mentor

What determines the angle is its velocity when it hits, not where it hits the wall.

7. Sep 5, 2009

### Warmacblu

Whoops, right.

In that case, the x-component would be 8.7 m/s.

I typed the problem wrong, the actual acceleration is 2.7 m/s^2; I will fix the original post.

Just to be clear I believe the y-component is (2.11) x (2.7) = 5.697

8. Sep 5, 2009

Looks good.

9. Sep 6, 2009

### Warmacblu

I solved both parts last night, thanks for all the help.