A particle of mass 5.00 kg is attached toa spring with a force constant of 100 N

Question

A particle of mass 5.00 kg is attached toa spring with a force constant of 100 N/m.It is oscillating on a horizontal frictionless surface with anamplitude of 3.00 m. A 9.00 kg object is dropped vertically on top of the5.00 kg object as it passes up through itsequilibrium point. The two objects stick together. (a) Does theamplitude of the vibrating system increase or decrease as a resultof the collision? 1 decreases
increases    
no change

By how much does the amplitude of the vibrating system change as aresult of collision?
2 m

(b) How does the period change? 3 increases
decreases    
no change

By how much does the period change?
4 s

(c) How does the mechanical energy of the system change? 5 no change
decreases    
increases

By how much does the energy change?
6 J

(d) Account for the change in energy. 1 decreases
increases    
no change
1 decreases
increases    
no change
1 3 increases
decreases    
no change
3 increases
decreases    
no change
3 5 no change
decreases    
increases
5 no change
decreases    
increases
5

Explanation / Answer

the kinetic energy of mass m1 = 5.00 kg is K = (1/2)k * A2 when the 9.00 kg object falls and sticks with the 5.00 kgobject then (1/2)k * A2 = (1/2) * (m1 +m2) * V2 or V = (k/(m1 + m2))1/2 * A----------(1) the amplitude of vibration is A1 = (V/w) w = (k/m1)1/2 or A1 = ((k/(m1 +m2))1/2 * A/(k/m1)1/2)= (m1/(m1 + m2))1/2 * A---------(2) m1 = 5.00 kg,m2 = 9.00 kg and A = 3.00m the amplitude of the vibrating system decreases as a result ofthe collision. the change in the amplitude of vibration as a result ofcollision is obtained from equation (2) therefore we get A1 = (5.00/5.00 + 9.00)1/2 * 3.00 =1.8 m or A = A - A1 = 3.00 - 1.8 = 1.2 m (b)from equation (1) we get V = (k/(m1 + m2))1/2 *A or (2A/T1) = (k/(m1 +m2))1/2 * A or T1 = 2 * ((m1 +m2)/k)1/2 or T1 = 2 * 3.14 * ((5.00 +9.00)/100)1/2 = 2.35 s the period is given by T = 2 * (m1/k)1/2 or T = 2 * 3.14 * (5.00/100)1/2 or T = 1.40 s the change in period is T = T1 - T = 2.35 - 1.40 = 0.95 s (c)the mechanical energy of the system decreases. the change in the mechanical energy is E = (1/2)k * A2 - (1/2)m2 *(k/(m1 + m2)) * A2 or E = ((1/2)k * A2) * [1 -(m2/(m1 + m2)] = ((1/2)k *A2) * [(m1 + m2 -m2/(m1 + m2)] = ((1/2)k * A2) * (m1/(m1 +m2)) or E = (1/2) * 100 * (3.00)2 * (5.00/5.00 +9.00) = 160.7 J (d)the change in energy is because of the increase in mass ofthe vibrating system as a result of collision.After the collisionthe total mass of the vibrating system increases and hence thespeed of the system decreases.As speed of the system decreases thenthe kinetic energy of the system decreases. or E = ((1/2)k * A2) * [1 -(m2/(m1 + m2)] = ((1/2)k *A2) * [(m1 + m2 -m2/(m1 + m2)] = ((1/2)k * A2) * (m1/(m1 +m2)) or E = (1/2) * 100 * (3.00)2 * (5.00/5.00 +9.00) = 160.7 J (d)the change in energy is because of the increase in mass ofthe vibrating system as a result of collision.After the collisionthe total mass of the vibrating system increases and hence thespeed of the system decreases.As speed of the system decreases thenthe kinetic energy of the system decreases.

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