Two objects of masses m 1 = 0.54 kg and m 2 = 0.90 kg are placed on a horizontal

Question

Two objects of masses m1 = 0.54 kg and m2 = 0.90 kg are placed on a horizontal frictionless surface and a compressed spring of force constant k = 290 N/m is placed between them as in figure (a) shown below. Neglect the mass of the spring. The spring is not attached to either object and is compressed a distance of 9.4 cm. If the objects are released from rest, find the final velocity of each object as shown in figure (b). (Let the positive direction be to the right. Indicate the direction with the sign of your answer.)

v1= m/s

v2= m/s

Explanation / Answer

givens: m1 = 0.54 kg m2 = 0.90 kg k = 290 N/m A = 0.094 m conservation of energy: stored PE(spring) converts to KE (0.5)kA^2 = (0.5)(m1)(v1)^2 + (0.5)(m2)(v2)^2 kA^2 = (m1)(v1)^2 + (m2)(v2)^2 290(.094)^2 = .54v1^2 + .9v2^2 2.56244 = .54v1^2 + .9v2^2 conservation of momentum: blocks start at rest and have no momentum; therefore final momentum must be zero as well m1v1 + m2v2 = 0 .54v1 + .9v2 = 0 .9v2 = -.54v1 v2 = -.6v1 2.56244 = .54v1^2 + .9(-.6v1)^2 .864v1^2 = 2.56244 v1 = 1.722 m/s v2 = 1.033 m/s v1 is -1.722 m/s (moves to the left) v2 is 1.033 m/s (moves to the right) BOL

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