Computer Science Canada Grade 12 Physics Momentum |
Author: | Panphobia [ Mon Apr 15, 2013 3:29 pm ] | ||
Post subject: | Grade 12 Physics Momentum | ||
So I have my Unit 2 physics test tomorrow, and I was going through the review and I could not figure out how to do this question
Now for a) I think the normal force would be mg, for b) I think the force of friction would be 0.28 * 18, I am not completely sure about that, and I do not know what to do for c and e, but I do now that kinetic energy is not conserved so we only use momentum. |
Author: | Insectoid [ Mon Apr 15, 2013 4:04 pm ] |
Post subject: | RE:Grade 12 Physics Momentum |
A) That's correct. B) Fk = uk * N. What you have is uk * d, which as far as I know isn't anything useful. C) The only force acting on the block during deceleration is friction, and we know the mass. Since F = ma, we can calculate acceleration. Given acceleration, vf, and distance, we can calculate v0. D) What? There's no D? E)You know the reaction velocity and mass of the block, and the mass of the bullet. Given P=mv and P is conserved, you can calculate the initial velocity of the bullet. Bear in mind I haven't done this in a while (Actually, I'm lying. I'm taking a college physics class right now, but it's way, way easier than my university or even high school physics were) so I might be forgetting something. |
Author: | DemonWasp [ Mon Apr 15, 2013 4:08 pm ] |
Post subject: | RE:Grade 12 Physics Momentum |
a) The normal force is Fn = mg; presumably you would use the mass of the block plus bullet (because that's what's sliding). b) The force of friction is Ff = umg = uFn (which you calculated in part (a)) c) Start by using your kinematics equations to turn the (friction force, mass) into (acceleration); then use (acceleration) and (distance) and (final velocity = 0) to get (initial velocity). e) Use conservation of momentum. |
Author: | Insectoid [ Mon Apr 15, 2013 4:13 pm ] |
Post subject: | RE:Grade 12 Physics Momentum |
Heh, I beat ya, DemonWasp. Quote: presumably you would use the mass of the block plus bullet (because that's what's sliding).
Hmm, I didn't think of that. I assumed the bullet bounced off or for whatever reason was not embedded in the block. |
Author: | Panphobia [ Mon Apr 15, 2013 4:17 pm ] | ||
Post subject: | RE:Grade 12 Physics Momentum | ||
Oh crap on the review sheet there was no d, I just copied the question straight off, sorry. But thank you for your help, it cleared it up. Also, lets say if you had a spring how would you change the conservation of momentum m1v1 + m2v2 =..... to accommodate for the springs elastic potential energy 1/2kx^2, one example is
So would v1 and v2 be 0 and then would the equation be 0 = m1v1' + m2v2' + (would there be something here?), and to find the energy that was stored in the spring I do not know in the slightest. |
Author: | crossley7 [ Tue Apr 16, 2013 12:26 am ] |
Post subject: | RE:Grade 12 Physics Momentum |
keep in mind it is conservation of momentum. Energy does not affect momentum so the elastic potential energy has no affect. you end up with 0=m1v1+m2v2 and solve for the 1 unknown. the elastic potential would be used if you were doing this using conservation of energy but clearly that is not the case here. |