Question

A particle of mass 4.00 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 5.00 m. A 9.00 kg object is dropped vertically on top of the 4.00 kg object as it passes through its equilibrium point. The two objects stick together.

a) By how much does the amplitude of the vibrating system change as a result of collision?

b) By how much does the period change?

c) By how much does the energy change?

Answer 1

m = mass of particle = 4 kg

k = force constant = 300 N/m

A = initial amplitude = 5 m

A_new = Amplitude after collision

V= maximum speed after collision

V_max = maximum speed before collision

M = mass of object = 9 kg

(a)

Due to collision , the Total mass will increase and hence the speed will decreases to keep the momentum same.

As the speed decrease , the maximum kinetic energy decrease . From conservation of energy , Maxiumum kinetic energy must be equal to spring potential energy. So since Kinetic energy decreases, spring potential energy decrease. we know that spring constant does not change , so due to decrease in Spring potential energy , Amplitude decreases.

Angular frequency is given as

w = sqrt(k/m) = sqrt (300/4) = 8.66 rad/s

At the equilibrium Position , Speed of Block is maximum and is given as

V_max = A*w = 5*8.66 = 43.30m/s

After Collision , the objects stick together , hence the final velocity after the collision is given as

V = m V_max / (m + M) = 4 * 43.30 / (4 + 9) = 13.32 m/s

Using Conservation of energy :

New Kinetic energy = New Spring Potential energy

(0.5) (m + M) V² = (0.5) k A²_new

(4 + 9) (8.66)² = 300 A²_new

A_new = 1.80 m

change in Amplitude = A - A_new = 5 - 1.80 = 3.2 m

(b)

period of oscillation is given as

T = (2*pi) sqrt(m/k)

K remains constant

as mass has increased after the two objects combine, hence the period increases.

T_initial = (2*pi) sqrt(m/k) = (2 * 3.14) sqrt(4 /300) = 0.725 sec
T_final = (2*pi) sqrt((m+M)/k) = (2 *3.14) sqrt(13 /300) = 1.30 sec

Change = T_final - T_initial = 1.30 - 0.725 = 0.575 sec

(c)

As the maximum speed decreases , the total mechanical energy decreases

KE_initial = (0.5)*m*V_max² = (0.5)*(4) *(43.30)² = 3749.78 J

KE_final = (0.5) (m + M) V² = (0.5) (4 + 9) (8.66)² = 487.47 J

Change = 487.47 - 3749.78 = -3262.30 J