In the figure below m1 = 1 kg and m2 = 2 kg. These two masses are pushed togethe

Question

In the figure below m1 = 1 kg and m2 = 2 kg. These two masses are pushed together on a horizontal surface to an initial separation of xi = 5 cm. The spring that is attached to the mass m1 has a spring constant of k = 3750 N/m and an equilibrium length of 7 cm. Thus, the spring is initially compressed between the two masses. When the masses are released, they move apart from each other, but eventually come to a rest because of the friction between the masses and the horizontal surface. The coefficient of kinetic friction between the masses and the surface is Uk = 0.051. What is the final separation distance xf (in m)?

Explanation / Answer

blocks have the same initial momentum, so m2 moves with speed v and m1 moves with speed 2v.

total initial KE = PE stored in spring

(1/2) m1 (2v)^2 + (1/2) m2 v^2 = (1/2) k s^2

solve for v:

1 * 4 v^2 + 2 v^2 = 3750 * 0.02^2

v^2 = 0.3

and so (2v)^2 = 1.2

Now... the distance each slides is given by

speed^2 = 2ugd

or: d = speed^2/2ug

For each block then:

block m1: d = 1.2 / 2 * 0.051*9.80 = 4.802 meters

block m2: d = 0.3 / 2 * 0.051 * 9.80 = 1.2005 meters

So their final separation = 4.802 + 1.2005 + 0.050 = 6.052 meters

(Note: the 0.050 is the initial separation... have to add that to the distances each block traveled.)

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