Recent content by bulldog23

I tried using 1/2(m1+m2)v2=(m1+m2)gy. I tried solving for y, but I am not getting the right answer. Can you explain where I am going wrong?

alright so I did do it right. The initial height of ball 1 equals the final height of ball 2 because the collision is elastic. So how do I approach part B?

so I set 1/2mv^2 equal to mgh, and solved for the velocity. I got 1.95 m/s. Am I doing this right?

How do I find the height of the first ball?

can you explain part A more. How do I apply conservation of energy and conservation of momentum?

[SOLVED] collision problem Homework Statement Two identical steel balls, each of mass 3.4 kg, are suspended from strings of length 27 cm so that they touch when in their equilibrium position. We pull one of the balls back until its string makes an angle theta = 74o with the vertical and let it...

[SOLVED] Conservation of Momentum Homework Statement Two blocks with masses m1 = 2.6 kg and m2 = 3.6 kg are at rest on a frictionless surface with a compressed spring between them. The spring is initially compressed by 60.0 cm and has negligible mass. When both blocks are released...

never mind I got it, thanks

What do I put in for m1 and m2?

so then I just take 125 and divide by -9.8?

can you explain to me how to do that?

So the initial velocity is 125 m/s. The accel is -9.8 m/s^2. I still need the distance right? And then I solve for the final velocity?

I am still confused, I don't understand how to figure out the height.

I'm guessing that you are to assume that all of the KE goes into the second half. Can you help me out, I'm lost?

[SOLVED] Horizontal Speed of bullet fragment Homework Statement A 14-kg shell is fired from a gun with a muzzle velocity 125 m/s at 33o above the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the...