A spring-mass system is initially at rest is shown above. A mass m = 1.0 kg trav

Question

A spring-mass system is initially at rest is shown above. A mass m = 1.0 kg traveling at 12ms collides and
sticks to the mass that is attached to the spring. Suppose k = 1000N
m and M = 3.0 kg.
1. Calculate the speed of the two masses immediately after the collision.
2. Calculate the maximum distance the spring will be compressed.
3. Is the system a simple harmonic oscillator? Prove by stating the appropriate equation describing the
system.
4. If the system is a simple harmonic oscillator calculate the period of oscillation of the system. If not,
qualitatively describe the motion of the system for one oscillation.

Explanation / Answer

given m = 1 kg

M = 3 kg

initial speed of m = v1 = 12 m/s

initial speed of M = v2 = 0

initial momentum Pi = m*v1 + M*v2 = 1*12 = 12 kg m/s

after collision the two massses stick with each other

final momentum Pf = (m+M)*V = (1+3)*V

from momentum conservation

Pf = PI

V = 12/4 = 3 m/s <----answer

============

2)

after the spring is compressed by x PE = 0.5*k*x^2

after collision KE of the spring masses system = 0.5*(M+m)*V^2

from energy conservation

PE = KE

0.5*K*x^2 = 0.5*(M+m)*V^2

1000*x^2 = (1+3)*3^2

x = 0.189 m <<-----answer

3)

elastic force acting Fe = -k*dx

from newtons law

net force Fnet = (M+m)*a

therefore

(M+m)*a = -k*dx

a = -(k/(M+m)*dx

a is proportional to -dx

acceleartioan is proportiona to displacement and is directed towards the equilibrium

so the motion is SHM

4)

time period T = 2*pi/w

w = sqrt(K/(M+m)) = sqrt(1000/4) = 15.811

T = 2*3.14/15.811 = 0.397 s <<<---answer

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