Two blocks of masses m1 = 2.00 kg and m2 = 4.65 kg are each released from rest a

Question

Two blocks of masses m1 = 2.00 kg and m2 = 4.65 kg are each released from rest at a height of y = 4.10 m on a frictionless track, as shown in Figure P6.56, and undergo an elastic head-on collision. (a) Determine the velocity of each block just before the collision. Let the positive direction point to the right. v1i .8.97 m/s---correct v2i ?m/s (b) Determine the velocity of each block immediately after the collision. v1f -1.9857----wrong...i know this needs to be a negative number v2f 15.95---wrong Your response differs from the correct answer by more than 100%. m/s (c) Determine the maximum heights to which m1 and m2 rise after collision. y1f 13m-correct y2f .20 -wrong

Explanation / Answer

Ef = Ei Ef = (1/2) m v2 = m g h = Ei v2 = 2 g h v2 = 2 (9.8 m/s2) ( 4.1 m) v2 = 80.36 m2/s2 v = 8.96 m/s That is, the 2.0-kg block, m1, moves to the right with v1 = + 8.96 m/s while the 4.65-kg block, m2, moves to the left with v2 = - 8.96 m/s The total momentum of the two-block system before the collision is PTot,i = pi1 + p2i PTot,i = (2.0 kg) (8.96 m/s) + (4.65 kg) ( - 8.96 m/s) PTot,i = - 23.74 kg m/s The masses are not the same so the momentum carried by each block is not the same, even tho' their speeds are the same. So the total momentum is not zero! The two blocks undergo an elastic collision so their final velocities are given by Equations 9.20 and 9.21, v1f = [(m1 - m2)/(m1 + m2)] v1i + [ 2 m2/(m1 + m2) ] v2i v1f = [(2 kg - 4.65 kg)/(2 kg + 4.65 kg)](8.96 m/s) + [ 2(.65 kg)/(2 kg + 4.65 kg)]( - 8.96 m/s) v1f = [(2 - 4.65)/(2 + 4.65)](8.96 m/s) + [ 2(4.65)/(2 + 4.65)]( - 8.96 m/s) v1f = [(-2.65 / 6.65)](8.96m/s) + [ 9.3 / 6.65]( - 8.96 m/s) v1f = -3.57 - 12.53 v1f = - 16.1 m/s Mass m1 moves to the left after the collision. v2f = [2 m1 / (m1 + m2)] v1i + [ (m2 - m1) / (m1 + m2) ] v2i v2f = [2 (2 kg) / (2 kg + 4.65 kg)] (8.96 m/s) + [ (4.65kg - 2 kg) / (2 kg + 4.65 kg)] ( - 8.96 m/s) v2f = [2 (2) / (2+ 4.65)] (8.96 m/s) + [ (4.65 - 2) / (2 + 4.65)] ( - 8.96 m/s) v2f = [4 / 6.65] (8.96 m/s) + [2.65 / 6.65] ( - 8.96 m/s) = 5.38 - 3.57 v2f = 1.81 m/s Mass m2 moves to the right after the collision. With these velocities, how high does each block rise? This is another conservation of energy question: For block 1, Eend = Eslide Eend = Uend = m g h1 = (1/2) m v1f2 = Kslide = Eslide m g h1 = (1/2) m v1f2 h1 = [(1/2) v1f2] / g h1 = [0.5(16.1 m/s)^2] / (9.8 m/s2) h1 = 13.22 m For block 2, Eend = Eslide Eend = Uend = m g h2 = (1/2) m v2f2 = Kslide = Eslide m g h2 = (1/2) m v2f2 h2 = [(1/2) v2f2] / g h2 = [0.5 (1.81 m/s)^2] / (9.8 m/s2) h2 = 0.167 m As a check, we can calculate the final total energy of the system Ef = U1 + U2 = m1 g h1 + m2 g h2 Ef = (2.0 kg) (9.8 m/s2) (13.22 m) + (4.65 kg) (9.8 m/s2) (0.167 m) Ef = 259.1 J + 7.61 J Ef = 266.7 J What was the initial total energy of the sysetm? Ef = U1 + U2 = m1 g h1 + m2 g h2 Ef = (2.0 kg) (9.8 m/s2) (4.1 m) + (4.65 kg) (9.8 m/s2) (4.1 m) Ef = 80.36 J + 186.8 J Ef = 267.2 J

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