A mass of 3.7 kg is released down a frictionless slope from a height of h = 2.3

Question

A mass of 3.7 kg is released down a frictionless slope from a height of h = 2.3 m. When it reaches the bottom of the slope, it undergoes a completely inelastic collision with a mass of 1.0 kg attached to a spring of spring constant k = 5.4 kN/m as illustrated in the figure below. The combined mass then undergoes periodic motion. Calculate the following. (Show step by step)

(a). Maximum velocity of two- mass system

(b) frequency

(c) maximum amplitude

(d) period

(e) Determine expressions for position as a function of time and velocity as a function of time for the combined mass of this oscillating system. (Use the following as necessary: t.)

x(t)=

v(t)=

Explanation / Answer

The kinetic energy at the bottom of the slope is equal to the potential energy at the top of the slope
mgh = 3.7*9.8*2.3 = 83.398 J
KE = 83.398 J = 0.5*3.7 *V^2 => V = 6.7m/s
Use conservaton of momentum to determine the velocity of the combined mass
3.7*6.7= [3.7+1]V => V = 5.27 m/s <------------ a)
The kinetic energy of the combined mass is then 0.5*4.7*5.27^2 = 65.27 J
That is the energy that is stored in the compressed spring 1/2 *k*x^2 = 65.27J
65.27 * 2/5400 = x^2 => x = 0.155m = 15.5cm = A <----------- c)
? = 2*pi*f = ?k/m) = 33.89 rad/s => f = ?/2pi = 5.39 Hz <----------- b)
T = 1/f = 0.185s <------------------- d)
(e)
x(t) = 0.155*cos(33.89*t)
v(t) = x'(t) = -33.89.*0.155*sin(33.89*t) = --5.25*sin(33.89*t)

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