Physics Problem

[PCMarine]

A brick weighing 24.5N is released from rest on a 1.00m frictionless plane, inclinded at an angle of 30.0 degrees. The brick slides down the incline and strikes a second brick weighing 36.8 N.

a) calculate the speed of the first brick at the bottom of the incline.

b) if the force of friction acting on the two bricks is 5N, how much time will elapse before the bricks come to rest.


Since my physics class has been doing electrical circuits for the past few months, it's been awhile since I've solved problems involving velocity and time. For part a, I could find velocity, but only if I knew the time interval... which I am unsure of how to find.

Any help would be appreciated, thanks!

*edit* Here is a diagram

 

[Heisenberg]

For the first part, I'd use both conservation of momentum and conservation of energy. Is the second block initially at rest?

 

[cronos]

m1v1 + m2v2 = m1v1' + m2v2'

 

[PCMarine]

Originally posted by: Heisenberg
For the first part, I'd use both conservation of momentum and conservation of energy. Is the second block initially at rest?

Yes, the second brick is initially at rest.

Hmmm, I'm trying to work with the forumla that Saxguy suggested, as this is the correct formula to use (it was mentioned in the chapter)..

 

[Heisenberg]

Originally posted by: PCMarine

Originally posted by: Heisenberg
For the first part, I'd use both conservation of momentum and conservation of energy. Is the second block initially at rest?

Yes, the second brick is initially at rest.

Hmmm, I'm trying to work with the forumla that Saxguy suggested, as this is the correct formula to use (it was mentioned in the chapter)..

What happens after the collision? Does the first brick stop and the second continue, or do they both slide down together? If they both slide together, you need to use the combined masses in the "after part" of the conservation of momentum equation.

 

[Shivatron]

Originally posted by: Heisenberg

Originally posted by: PCMarine

Originally posted by: Heisenberg
For the first part, I'd use both conservation of momentum and conservation of energy. Is the second block initially at rest?

Yes, the second brick is initially at rest.

Hmmm, I'm trying to work with the forumla that Saxguy suggested, as this is the correct formula to use (it was mentioned in the chapter)..

What happens after the collision? Does the first brick stop and the second continue, or do they both slide down together? If they both slide together, you need to use the combined masses in the "after part" of the conservation of momentum equation.

You have the right idea, but you're wording it wrong. What you should be asking is, "Do the two bricks stick together at the bottom?" If they do, the new formula becomes

(1.1) m1v1 + m2v2 = m12v12'

where m12, v12 are the combined mass and velocity of the stuck together blocks respectively.

So, you are looking at something like this:

For v1:

(1.2) (0.5)*((v1)^2) = gh where h is the height that the block falls (ie. h = 1.00 sin(30))

So now you have a v1 at the bottom of the ramp. Now you can use Conservation of Momentum (either the originally suggested equation or 1.1, as required) to solve for v1' and v2'.

HTH.

 

[PCMarine]

Here is a diagram of the setup

 

[Heisenberg]

Ah ok, there's a flat part after the incline. I was thinking it was just one big incline. So just use conservation of energy (or regular kinematics) to find the velocity of the first block at the bottom, then conservation of momentum to find the velocity of the two block system. For the frictional part, you have a force of 5N acting the opposite way as the velocity. Use F=ma to find the acceleration, then you can find the time.

 

[sash1]

grr nm, that was wrong

 

[MAME]

uh, bricks don't slide!

 

[Yossarian]

I think the 1m distance is along the angle, not horizontal. v1^2-v2^2=2as is the answer to the first part.

 

[MoPHo]

for 1) mgh = 1/2mv².

 

[PCMarine]

Originally posted by: MoPHo
for 1) mgh = 1/2mv².

Excellent, I got 4.43 M/S for the velocity of the first brick at the bottom of the incline.

 

[Evadman]

I figured aout an awesome trick for doing those electrical circuits without the TONS of work. You can get damn close (.01 or so) doing it in your head after a few too.

 

[Howard]

Originally posted by: Evadman
I figured aout an awesome trick for doing those electrical circuits without the TONS of work. You can get damn close (.01 or so) doing it in your head after a few too.

?