Two blocks of masses m1 = 2.00 kg and m2 = 4.75 kg are each released from rest a

Question

Two blocks of masses m1 = 2.00 kg and m2 = 4.75 kg are each released from rest at a height of y = 5.10 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

Since there is no friction, velocity is independent of the mass of the object. Both has a velocity v = v2gh = v (19.6*5.1) = 9.99 m/s ------------------------ Since there is no friction, the collision is elastic. Choosing a reference frame riding with the right side mass, Velocity of approach = 9.99+9.99 = 19.99 v1’ = (m1-m2) u1/ (m1+m2) = (2-4.75)* 19.99/ (6.75) = -8.14 m/s v2’ = 19.99-8.14 =11.84m/s ---------------------------------- In the ground reference frame, The final velocity of 2kg is -8.14-9.99 = -18.13m/s The final velocity of 4.5kg is 11.84-9.99 = 1.85m/2 --------------------------------- Height is found using h =v^2/ 2g The height of 2kg is 18.13^2/(19.6) = 16.77 m The height of 4.5 kg is 1.85^2/(19.6) = 0.1746m ============================

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