Two blocks of masses m 1 = 1.81 kg and m 2 = 4.16 kg are each released from rest

Question

Two blocks of masses m1 = 1.81 kg and m2 = 4.16 kg are each released from rest at a height of h = 4.83 m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision.

Determine the magnitude of the velocity of the m2 = 4.16 kg block just after the collision.
(in m/s)

Determine the magnitude of the velocity of the m1 = 1.81 kg block just after the collision.
(in m/s)

Determine the maximum heights to which m2 rises after collision.
(in m)

Determine the maximum heights to which m1 rises after collision.
(in m)

Explanation / Answer

v = sqrt(2 g h) = sqrt(2 x 9.8 x 4.83) = 9.73 m/s

(A) for elatic collision,

velocity of approach = velocity of separation

2(9.73) = v2 + v1

v1 = 19.6 - v2

applying momentum conservation,

(1.81 x 9.73) - (4.16 x 9.73) = - 1.81 v1 + 4.16 v2

- 22.87 = - 35.48 + 1.81v2 + 4.16v2

v2 = 2.11 m/s

(B) v1 = 19.6 2.11 = 17.5 m/s

(C) h2 = v2^2 / 2 g = 0.227 m

(D) h1 = v1^2 /2 g = 15.6 m

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