Question

A 4.0kg block is suspended from a spring with force constant

Answer 1

Answer:

(a) Amplitude of the motion is 0.184 m.

(b) 0.285 of the original kinetic energy appears as the mechanical energy in the harmonic oscillation.

Explanation:

(a) First we will use the conservation of linear momentum to find the velocity of the combined objects.

$P_1 = P_2\\mv_1 = (M + m) v_2\\0.05*150 = (4 + 0.05)v_2\\v_2 = 1.85~m/s$

Before the collision, the spring is in equilibrium, so it is stretched by the amount of x.

$kx = Mg\\x = \frac{Mg}{k} = \frac{4*9.8}{500} = 0.0784~m$

Just after the collision, the motion starts from this position.

The simple harmonic motion equation is

$x(t) = A\cos(\omega t + \phi)\\v(t) = \frac{dx(t)}{dt} = -\omega A\sin(\omega t + \phi)$

ω is the angular frequency, and given as

$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{500}{4.05}} = 11.11$

Φ is the phase angle and can be found by the initial conditions.

At t = 0 (Just after the collision);

$x(t=0) = 0.0784 = A\cos(\omega * (0) + \phi) = A\cos(\phi)\\v(t=0) = 1.85 = -(11.11)A\sin(\phi)\\\\\cos(\phi) = 0.0784/A\\A = -0.1665/\sin(\phi)\\\\\cos(\phi) = \frac{0.0784}{-\frac{0.1665}{\sin(\phi)}} = 0.4709\sin(\phi)\\\cot(\phi) = 0.4709\\\phi = 64.7^\circ\\A = \frac{0.1665}{\sin(\phi)} = \frac{0.1665}{0.9044} = 0.184~m$

(b) The original kinetic energy is

$K_0 = \frac{1}{2}mv^2 = 562.5~J$

The mechanical energy of the simple harmonic motion is

$K + U = \frac{1}{2}(M+m)v_2^2 + \frac{1}{2}k_x^2 = \frac{1}{2}(4.05)(1.85)^2 + \frac{1}{2}500(0.0784)^2\\K+ U = 160.59~J$

So 0.285 of the original kinetic energy appears as the mechanical energy in the harmonic oscillation. The rest of the original energy is converted to heat during the collision.