Recent content by Gill987

Whoops, thanks for catching that. Whoops again, v3=(2v1-v2)/3. Thanks to both of you for helping me out on all of this!

Alright, so the velocity of the left bob just before collision is v1=√(2gL(1-cosθ1)) and the right is v2=√(2gL(1-cosθ2)). Velocity of the combined bobs is v3=(2v2-v1)/3. Using cons. of e. I get (1/2)3mv32=3mgL(1-cosθ3). Simplifying and plugging in for the values of v, I get...

Sorry about the confusion. I solved for the final angle symbolically, but the answer didn't look right. When I tried to plug in an angle that the bobs started at, such as 0, I got some final angle that didn't make sense for the starting angle, such as 60. I used cons. of e. to get the...

I attempted this using cons. of e. to find each bob's velocity when it meets in the middle and used that velocity in my equations for conservation of angular momentum. I used cons. of e. again from when they were now combined and moving to their final swing angle, got a height, and tried solving...

Whoops, sorry, I was trying to find a thread or something with it but I guess I didn't look hard enough.

My teacher originally drew the diagram with large angles, and labeled the left θo and the right θu. I was thinking maybe he just didn't close the top of his u (although it has a big gap). If the angle was the same for both this would be much simpler as they collide in the middle, right?

Yeah, I was looking at this problem and was kinda baffled, because what I thought might work didn't and I had no other ideas. I made some force diagrams though, got T-2mg=2m(v^2/L) and a = 2gsinθ for the left side and T-mg=m(V^2/L) and A = gsinΦ for the right side. I tried solving for time from...

Yeah, I wasn't sure about finding one angle in terms of the other. I tired saying P before was 0, P after (right before they collide) was 2mv-mV, ∴ 2v=V, and substituting the v's for what i got using cons. of e, but that would only work assuming the collided in the middle. And I didn't think...

I've recently been posting new threads on the homework forum, and when I preview my post it seems that a second template appears below everything I typed. Not sure why it's happening, but I thought I'd let someone know.

Homework Statement Two pendulums collide and stick together after being released from rest at angles given. (See diagram) What is the final swing angle of the stuck-together masses ? (I'm assuming my teacher means the angle when they get to the highest point in their swing) In what direction...

Great, thanks for all your help!

Awesome! For the second part, I know that the energy when it was initially compressed was (1/2)kD^2, and when it compressed again energy was mV^2 + (1/2)ks^2, so can I say total energy came from D initially, and then came from both V and s, so s has a smaller value? Or how could I word that...

I found velocity of center of mass when both blocks are moving at the same speed to get V, set that equal to (D/2)√(k/m), solved for V, plugged in, and got D/√2 = s. Heh, I actually started typing this and then saw your reply. I hope this is right.

Thanks for responding! At the moment of max compression, the velocities of both blocks are equal. I tried using conservation of energy, and got (1/2)mv^2 = (1/2)mV^2 + (1/2)mV^2 + (1/2)ks^2, where V is the after velocity of the blocks and v is the before velocity of the right hand block...

Homework Statement A system consists of two blocks each of mass M, connected by a spring of force constant k. The system is initially shoved against a wall so that the spring is compressed a distance D from its original uncompressed length. Floor is frictionless. The system is now released...