Recent content by Striders

Thanks, that's pretty clear-cut. A few followup questions: 1) I had been under the impression that the sole criterion for a collision to be considered inelastic is that there is some permanent deformation of the objects involved. Is that incorrect? Or is it correct and the sticking together of...

Homework Statement A ball with mass 2kg is traveling 2m/s. It impacts and sticks to a hanging, stationary 4kg pendulum. How high will the pendulum + ball go? Homework Equations KE = 1/2mv2 PEg = mgh mava = m(a+b)v(a+b) The Attempt at a Solution I tried to solve this question assuming...

"and wrong." Truth hurts! Better to learn here than on a test. I understand the issue now- basically, as you said earlier, use consistent units and it will work out. "In post 3 you did not calculate the potential energy. That was the total energy in an orbit, otherwise the factor 2 would be...

"m has units of mass, so your result is "mass*energy", that is not an energy. You get the right units naturally if you do not remove them everywhere." Perhaps my notation was confusing (as you noted, lacking units makes it difficult to work with). What I meant was the 31 300 000 value is not in...

"And I think there is a wrong number in the third line for the geostationary orbit." Yes, you're right. I edited it to reflect the proper value. "And don't forget the units." Etotal = 31 303 248m J Etotal = 31.3m MJ

D'oh good point! In orbit, it's energy is comprised of KE and PEg. PEg has been calculated, so to that must be added the KE. KE = mv2/2 I don't have v but can re-write it as V = 2rpi / T KE = m(2rpi/T)2/2 KE = 4mr2pi2/2T2 KE = 4m(42 250 474)2pi2/2(86400)2 KE = 4 720 254m, answer in joules So...

"With that approach, putting it in an orbit one meter above the surface (ignoring atmosphere, mountains and so on) would need an insane amount of energy, and launching it to outer space would just need a very gentle push. That is unrealistic." Yep, that's a good way to think about it. I had...

Hi all, This question is for an assignment due Friday. My friend and I spent the better part of an hour on it and are thoroughly stumped. Any help is much appreciated! 1. Homework Statement Calculate the energy required to put a satellite into geostationary orbit. (note that no mass is...

50m/1.812615574v = 0.422618261v*s2/-9.81m 50m/1.812615574v * -9.81m = 0.422618261v*s2 (50m * -9.81m) / 1.812615574v = 0.422618261v*s2 50m * -9.81m = 1.812615574v * 0.422618261v*s2 -490.5m2 = 0.766044441v2 * s2 -490.5m2/s2 = 0.766044441v2 I guess I'll change the acceleration due to gravity...

So 50m/vcos25 = 2vsin25/a 50m/vcos25 = 2vsin25/-9.81m/s2 50m/vcos25/2 = vsin25/-9.81m/s2 50m/2vcos25 = vsin25*s2/-9.81m

No- I take it that it is indeed possible to state the two different expressions in terms of v, without the use of vx or vy, but I don't see how it can be done. Also, I wrote an addendum to a post above but I believe it got buried because I wrote it as an edit and we had moved on to successive...

vx = vcos25 v = vx/cos25 v = vy/sin25 vy/sin25 = vx/cos25

The expression for the time taken to travel to the target in the horizontal frame is 50 / vxcos25 As for 2vf/a = t vertical final velocty = vsin25 So 2vsin25/a = t I don't really follow though, I'm not sure why my previously calculated velocities are incorrect. Nor do I understand why the...

vi = 15.66045976m/s vx = 15.66045976m/s * cos25 vx = 14.193197m/s "So what is t in terms of v? (not vy or vf or vi)" I'm a tad confused here because we said earlier that v is equal to vi. From post #6, "We assumed an initial speed of v, and you found the components." So assuming v = vi t =...

This elicits the answer that t=3.19s, which matches what I got in my post above, but differs from the textbook's listed answer.