I dropped a 5kg plate straight down onto the ground and it shattered into three

Question

I dropped a 5kg plate straight down onto the ground and it shattered into three pieces. The pieces shot off into three different directions and slid across the ground until they came to a stop. The pieces of plate experience friction against the ground with coefficient of kinetic friction, µk = .4. Call the point where I dropped the plate the origin, (x, y) = (0, 0). I found a 2kg piece at (x, y) = (3, 7), and a 1.2kg piece at (x, y) = (9, 4). Where did the third piece come a stop? Note all (x,y) positions are in meters

Explanation / Answer

We know, v2 = u2 + 2as

so, initial speed, |u| = (2as)1/2

For 2 kg piece:

|u2| = (2µgs2)1/2 = [2 * 0.4 * 9.81 * (32 + 72)1/2]1/2 = 7.73 m/s

So, u2 = 7.73[cos(tan-1(7/3))x + sin(tan-1(7/3))y] = 3.04x + 7.10y

For 1.2 kg piece:

|u1.2| = (2µgs1.2)1/2 = [2 * 0.4 * 9.81 * ((-9)2 + (-4)2)1/2]1/2 = 8.79 m/s

So, u1.2 = 8.79[cos( + tan-1({-4}/{-9}))x + sin( + tan-1({-4}/{-9}))y] = -8.03x - 3.57y

Let the initial velocity of the third part be u1.8

Using conservation of linear momentum,

m2u2 + m1.2u1.2 + m1.8u1.8 = 0

=> 2(3.04x + 7.10y) + 1.2(-8.03x - 3.57y) + 1.8u1.8 = 0

=> u1.8 = 1.98x - 5.51y

So,

sx,1.8 = ux,1.82/2µg = 1.982/(0.4 * 9.81) = 1 m

sy,1.8 = uy,1.82/2µg = (-5.51)2/(0.4 * 9.81) = 7.7 m

So the third piece came to stop at (1, -7.7)

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