Gayle runs at a speed of 4.00 m/s and dives on a sled, which is initially at res
ID: 2191136 • Letter: G
Question
Gayle runs at a speed of 4.00 m/s and dives on a sled, which is initially at rest on the top of a frictionless snow-covered hill. After she has descended a vertical distance of 5.00 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 17.0 m? Gayle's mass is 50.0 kg, the sled has a mass of 5.00 kg, and her brother has a mass of 30.0 kg. Answer is not 17.4 m/s. Please explain and have the correct answer.Explanation / Answer
Do this one in steps. Starting at the top of the hill, find the speed of the sled plus Gayle at the instant she lands on it. From the law of conservation of momentum (perfectly inelastic collision), the velocity is: p(i) = p(f) m1v1(i) + m2v2(i) = (m1 + m2)v(f) v(f) = [m1v1(i) + m2v2(i)] / (m1 + m2) = [50.0kg)(3.8m/s) + 0] / (50.0kg + 5.00kg) = 3.45m/s Her brother jumps on when Gayle and the sled have descended 5.00m. The velocity of Gayle plus sled at the instant her brother jumps on is found from the law of conservation of energy: E(i) = E(f) KE(i) + PE(i) = KE(f) + PE(f) 0.5mv²(i) + mgh(i) = 0.5mv²(f) + mgh(f) v(f) = v[v²(i) + 2g{h(i) - h(f)}] Initial velocity here, is final velocity from the first stage. Therefore: v(f) = v[(3.45m/s)² + 2(9.80m/s²)(5.00m - 0)] = 10.48m/s At this instant. her brother jumps on, so we have another perfectly inelastic collision as in the first stage. Initial velocity here is 10.48m/s, m1is 55.0kg, and m2of course is 30.0kg : v(f) = [m1v1(i) + m2v2(i)] / (m1 + m2) = [(55.0kg)(10.48m/s) + 0] / (55.0kg + 30.0kg) = 6.78m/s Conservation of energy again gets you the final velocity at the bottom of the hill: v(f) = v[v²(i) + 2g{h(i) - h(f)}] = v[(6.78m/s)² + 2(9.80m/s²)(12m - 0)] = 16.76m/s
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