Recent content by Probie1

I was just re reading your posts... ...when I re read this post the light started to flicker. Thanks again Steely Dan

Thanks Steely Dan... that wasn't as hard as I thought it would be... mind you that was an easy momentum problem for you, but difficult for me. I imagine it would be harder if you had 1 vehcile t-bone another and they both have pre and post impact velocities.

So because they have no momentum they cancel out which means you rewrite the equation to solve for velocity1 (1000*2.2+1100*0)=(1000*0+1100*2) (1000)=(1100*2) (m1)=(m2v4) (m2V4)/m1 (1100*2)/1000 2.2ms2= 2200/1000 Is that how it is done?

(1000*2.2+1100*0)=(1000*0+1100*2) V2 and v3 have no velocity so they have no momentum...correct?

Homework Statement Car 1 weighing 1000 kg crashes into the rear of parked car 2 weighing 1100 kg and stops. Car2 moves ahead 2 ms2. What is the velocity of car1 at impact? Homework Equations (m1v1+m2v2)=(m1v3+m2v4) The Attempt at a Solution (m2V4)/m1 (1100*2)/1000 2.2ms2=...

D= at + vi2 Alright... stop laughing. P = momentum m=mass v=velocity g = gravity μ = friction w= weight I thought that if P=mv then v = a = μg but then I remembered where I left my brain because a = change in velocity over a change in time. So it can't possibly be the way I was...

Does product not mean... umm the make up... it is part of or makes up? I guess the context of all this is I am trying to undertand how formula's come about. P = mv so do this mean that the product of v is μg and the product of m is weight? So it could be written P = wμg This is...

P = mv so do this mean that the product of v is μg and the product of m is weight? So it could be written P = wμg How is this formula derived Vf = √(Vi^2 + (2ad))

Ah, never mind this question really makes no sense anyway. Obviously we are dealing with forces which can not be changed to velocity. ( ya... I loose arguments with myself all the time). Thanks

Wait

Homework Statement I'm trying to figure out how much velocity is required to move 1000 kg car up a hill that has a .02 percent grade and a friction value 0.7. Homework Equations F= μmg cosθ The Attempt at a Solution F = 6866.99 N If this is correct how do you get this...

Yes... you are correct about the velocities and angles being rounded, and I will never again say that (P)i doesn't = (P)o. Thanks a Million.

That results are very close... probably be right on if I did not shorten my after decimal digits. Thank you for all the patients you have shown me through this... you should get a medal for all you do. Now if I can copy this stuff down properly I will never forget how to do it on my own...

p1i = 323.687 i + 0.000 j p2i = -7434.830 i - 1310.961 j p1i = -7111.143 i + -1310.961 j p1o = -6446.483 i + 112.523 j p2o = -663.790 i - 1423.501 j p2o = -7110.273 i + -1310.978 j

p1i = 323.687 i + 0.000 j p2i = -7434.830 i - 1310.961 j p1i = -7111.143 i + 1310.961 j p1o = -6446.483 i + 112.523 j p2o = -663.790 i - 1423.501 j p2o = -7110.273 i + -1310.978 j