PHYS 1050 Final: PHYS 1050 Final Exam 2007 Answers
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3 (a) write newton"s second law at top of loop: -n mg = -mv2/r. Since n = 0 (given) then m cancels and we have g = v2/r. 3 (b) ei = mgh and ef = mv2 + mg(2r) (height at the top of the loop is the diameter or twice the radius). Equate the energies to find mgh = mv2 + mg(2r). Cancel mass to find gh = v2 + g(2r). Substitute in v2 = rg (from (a)) to find gh = rg + g(2r). Cancel g to find h = r + (2r) = 2. 5 r. 3 (d) write newton"s second law at bottom of loop to obtain n mg = mv2/r. Substitute in v = (5gr) to find n mg = m(5gr)/r = 5mg. 4 (c) a = 3. 92 m/s2 (up the hill for the 50 kg mass and down for the 100 kg mass)
