A particle of mass 5.00 kg is attached to a spring with a force constant of 300
ID: 1402665 • Letter: A
Question
A particle of mass 5.00 kg is attached to a spring with a force constant of 300 N/m. It is oscillating on a horizontal frictionless surface with an amplitude of 3.00 m. A 6.00 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) Does the amplitude of the vibrating system increase or decrease as a result of the collision?
increases
decreases
no change
By how much does the amplitude of the vibrating system change as a result of collision?
m
(b) How does the period change?
decreases
increases
no change
By how much does the period change?
s
(c) How does the mechanical energy of the system change?
decreases
no change
increases
By how much does the energy change?
J
(d) Account for the change in energy.
Explanation / Answer
Vibration w = sqrt(k/m) = sqrt(300/5) = 7.75 rad/sec
V at neutral point = Vo = Aw = 3*7.75 = 23.25 m/sec
x-direction momentum remain conserve
This momentum will be the same before and after dropping the 6 on the 4
6*0 + 23.25*5 = (6+5)*v
v = 10.57 m/s
The new KE = .5mv^2= .5*11*10.57^2 = 614.49 J
This will equal the new spring energy:
614.49 J = .5kA^2
= A = sqrt(614.49/150) = 2.02 m
part a ) amplitude decrease
3 - 2.02 = 0.98 m
b) New period: T = 2*pi*sqrt(m/k) = 2pi*sqrt(11/300) = 1.203 sec
Old period: To = 2pi*sqrt(5/300) = 0.8112 sec
period increase
increase by T- To = 0.3918 sec
c) KE1 = 1/2*5*23.25^2 = 1351.4 J
From (above) KE2 = 614.49 J
energy is decreasing
energy change is -736.91 J
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