Two blocks of masses m 1 = 2.00 kg and m 2 = 4.10 kg are each released from rest

Question

Two blocks of masses m1 = 2.00 kg and m2 = 4.10 kg are each released from rest at a height of h = 4.30 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.)

(a) Determine the velocity of each block just before the collision.

(b) Determine the velocity of each block immediately after the collision.

(c) Determine the maximum heights to which m1 and m2 rise after the collision.

Explanation / Answer

Here is the answer Since there is no friction, velocity is independent of the mass of the object. Both has a velocity v = v2gh = v (19.6*4.8 = 9.7 m/s ------------------------ Since there is no friction, the collision is elastic. Choosing a reference frame riding with the right side mass, Velocity of approach = 9.7+9.7 = 19.4 v1’ = (m1-m2) u1/ (m1+m2) = (2-4.5)* 19.4/ (6.5) = -7.46 m/s v2’ = 19.4-7.46 =11.94m/s ---------------------------------- In the ground reference frame, The final velocity of 2kg is -7.46-9.7 = -17.16m/s The final velocity of 4.5kg is 11.94-9.7 = 2.24m/2 --------------------------------- Height is found using h =v^2/ 2g The height of 2kg is 17.16^2/(19.6) = 15.0 m The height of 4.5 kg is 2.24^2/(19.6) = 0.256m ============================

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