Question

A 20.0-kg mass is traveling to the right with a speed of 3.00 m/s on a smooth horizontal surface when it collides with and sticks to a second 20.0-kg mass that is initially at rest but is attached to a light spring with force constant 170 N/m . Find the frequency of the subsequent oscillations.
Express your answer with the appropriate units. Find the amplitude of the subsequent oscillations.

Express your answer with the appropriate units.

Answer 1

Answer:

a) 0.328 Hz b) 73 cm (0.73 m)

Explanation:

Using law of conservation of momentum:

M1U1 + M2U2 = V(M1+M2) since the mass stuck together. where M1 is the mass 20 kg travelling with the U1  speed 3m/s and M2 is the 20kg at rest with U2 = 0 and V is the final speed of the mass after collision.

substitute the values into the equation

20 × 3 + 20 ×0 = V (20+20)

V = 60 / 40 = 1.5 m/s

The period of the oscillation can be calculated with the formula:

T = 2π√ ( m/k)  where T is the period in seconds and F is the frequency of the oscillation

substitute the values into the formula

T = 2×3.142×√ (40/170) = 3.05s

T = 1/F

F = 1/T = 1/ 3.05 = 0.328Hz

b) The amplitude of the oscillation can calculated using the formula

V = x√(K/m) where x is the amplitude in cm and V is the maximum speed that caused the displacement (amplitude) x

substitute the values into the formula

1.5 = x × √ (170/40) = x × 2.062

divide both side by 2.062

x = 1.5 / 2.062 = 0.73m = 0.73 cm