Recent content by NikolasLund

And I will attempt to translate it more clearly. Sorry. :)

No, I thought so myself as well, but the mass is 22,4 g, which yields a gravitational force smaller than the buoyancy, so the ball, i guess, will reach some point where it's donward velocity is 0 and will from there ascend to the surface.

I can't help it but feel stupid now, having overlooked this very basic assumption. But still, thank you, I will be more mindful next time.

I first attempted to solve the problem by ##A_{ext} = \Delta E_{mech} = F_{ext}s##. Here, ##F_{ext} = F_{friction}## and ##\Delta E_{mech} = E_{k2} + E_{p2} - E_{k1}##. We obtain then the following equation: $$F_{friction} = (m((v_2)^2 - (v_1)^2))/2h + mg,$$where ##v_1## is the velocity at time...

I managed the optimization with little problem, and got the correct answer there. As for the hint you gave regarding the geometric approach, I think I understand why the angle is ##arctan 3/4## (##|v_2|/|v_1|## ?), but I am not sure how I should correctly use the relative velocity, or the given...

I attempted it, but could not figure out why the angle formed with the horizontal line is approx. 37 degrees. The resulting vector of the blue plane, in the frame of the red plane, I found to have a length of ##a\sqrt 2##,being constructed by the already given vectors with lengths a. This...

This problem is from Prof. Jaan Kalda's study guide to the IPHO. The problem can be solved by optimization, but there is apparently also a geometric approach, which is the one Kalda suggests. Initially I, being naive, tried to solve the problem by calculating the resulting distance the red plane...