A 2.0 kg box is traveling at 5.0 m/s on a smooth horizontal surface when it coll

Question

A 2.0 kg box is traveling at 5.0 m/s on a smooth horizontal surface when it collides with and sticks to a stationary 6.0 kg box. The larger box is attached to an ideal spring of force constant (spring constant) 150 N/m. Write down the equation for the horizontal position x of the boxes as a function of time t, using the cosine function. Note:  you must show all of your calculation for each  numbers in the position equation. Additionally, please take the origin at the equilibrium point of the boxes, with with rightward being the positive x-direction .

Explanation / Answer

Here,

mass , m1 = 2 Kg

u1 = 5 m/s

m2 = 6 Kg

k = 150 N/m

Here , for the initial speed of the blocks moving after collision

using conservation of momentum

(2 + 6) * v = 5 * 2

v = 1.25 m/s

Now, for the amplitude of the motion

0.5 * (2 + 6) * 1.25^2 = 0.5 * 150 * A^2

solving for x

A = 0.408 m

angular frequency , w = sqrt(k/m)

w = 4.33 rad/s

Now , for the function of position

as initial blocks are at the equilibrium position

x = A * cos(wt - pi/2)

x = 0.408 * cos(4.33 * t - pi/2)

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