Recent content by berrytea

Blocks? There is only one block and one stone though. So, in your equation what would m2 be?

Yes, it is during the collision, so from the start to the end when the spring is fully compressed.

This is what I have gotten so far: ΔE = ΔK + ΔU I previously calculated the ΔU to be 30.25 J ΔK = Ki- Kf = 1/2m1v12- 1/2m1v12+1/2m2v22 = 1/2(15.0)(8.00)^2 - 1/2(15)(8.00)^2+1/2(15.0)(2)^2 = 480 - 510 = -30 ΔE = ΔK + ΔU ΔE = -30 + 30.25 = 0.25 ?? I am super confused at this point and don't know...

I think ΔU = 30.35 J

The stone is traveling to the right and it (the stone) rebounds at 2.00 m/s to the left. There are other parts to the question that I have answered which are: the maximum compression of the spring = 0.11m the speed of the block after the collision being 2 m/s the work done by the spring during...

I am unsure, but I assume it is the block and the ball together as the question does not specify.

Hello, and thanks! My attempt was to find the change in kinetic and potential energies and add them. What is confusing to me, is a friend of mine got -52 J for this solution. I calculated ΔPE = -30.25 J and I got simply ΔE = 30 J, which gives me -0.25 as ΔKE. So I am not sure what I'm doing...

Homework Statement A 15 kg block is attached to a very light horizontal spring of force constant 5000.0 N/m and is resting on a frictionless table. It is struck by a 3.00 kg stone at 8.00 m/s to the right, then rebounds at 2.00 m/s to the left. Homework Equations ΔE = ΔK + ΔU The Attempt at...