Question

A particle of mass 5.00 kg is attached to a spring with a force constant of 210 N/m. It is oscillating on a frictionless, horizontal surface with an amplitude of 4.00 m. A 8.00-kg object is dropped vertically on top of the 5.00-kg object as it passes through its equilibrium point. The two objects stick together.
(a) What is the new amplitude of the vibrating system after the collision?_______________

(b) By what factor has the period of the system changed?______________
(c) By how much does the energy of the system change as a result of the collision?____________J

Answer 1

since no external force is acting on the masses spring system hence the energy of the system remains conserved i.e

initial energy = final energy

TO Skg a) Anitial time period = 21 m m=5kg k=210 N/m 210 = 0.97 8econds. Now since when object is dropped momentum remains conserved so if initicul velocity at equilibrium = ve .: 56= (5+8)x - fincel velocity

Balancing energy we get KAS me for second code & KA,? = 4 m, 12 for initial case 13 I. Az = 1145. 10.2.4m hence final amplitude = 1024 b) Second time period 20 m2 = 27113 V 210 = 1.56 seconds fraction change = 1.56 -0.97 - 0.97 0.61 factor = 1.56 -1.61 0.97