Two blocks of masses m1 = 2.00 kg and m2 = 4.55 kg are each released from rest a

Question

Two blocks of masses m1 = 2.00 kg and m2 = 4.55 kg are each released from rest at a height of y = 5.40 m on a frictionless track, as shown in Figure P6.56, and undergo an elastic head-on collision. Figure P6.56 (a) Determine the velocity of each block just before the collision. Let the positive direction point to the right. v1i m/s v2i m/s (b) Determine the velocity of each block immediately after the collision. v1f m/s v2f m/s (c) Determine the maximum heights to which m1 and m2 rise after collision. y1f m y2f m

Explanation / Answer

a)Velocity of each block before collision = v =>0.5mv^2 = mgy =>v = sqrt(2gy) = sqrt(2*9.81*5.4) = 10.29 m/s Velocity of m1 = 10.29 m/s = v1i Velocity of m2 = -10.29 m/s = v2i b)Elastic collision =>v1i-v2i = v2f-v1f =>v2f = v1f+10.29+10.29 = v1f+20.59 Initial momentum = final momentum =>(2*10.29)-(4.55*10.29) = (2*v1f)+(4.55*v2f) =>-26.25 = 2v1f+4.55v1f+(4.55*20.59) =>v1f = -18.31 m/s v2f = -18.31+20.59 = 2.28 m/s c)m1: y1f = (v1f^2)/(2*9.81) = 17.083 m m2: y2f = (v2f^2)/(2*9.81) = 0.26 m

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