Two blocks of masses m1 = 2.00 kg and m2 = 4.40 kg are each released from rest a

Question

Two blocks of masses m1 = 2.00 kg and m2 = 4.40 kg are each released from rest at a height of h = 5.40 m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.) Determine the velocity of each block just before the collision. v1i = m/s v2j = m/s Determine the velocity of each block immediately after the collision. v1f = m/s v2f = m/s Determine the maximum heights to which m1 and rise after the collision. Y1f = m Y2f = m

Explanation / Answer

1) V = (2*9.8*5.4)^0.5 = 10.28 m/s.......each have velocity = 10.28 m/s before collision...answer.... 2) apply conservation of momentum:...m1u1 + m2u2 =m1v1 +m2v2......=> 2*10.28 - 4*10.28 = 2v1+4v2.......and as elastic collision=> v2 - v1 = u1 -u2 = 10.28+10.28=>v2 = 20.56 +v1.........use these 2 equations to get the answer.......

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