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

Answer 1

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:

|u_1.2| = (2µgs_1.2)^1/2 = [2 * 0.4 * 9.81 * ((-9)² + (-4)²)^1/2]^1/2 = 8.79 m/s

So, u_1.2 = 8.79[cos(π + tan^-1({-4}/{-9}))x + sin(π + tan^-1({-4}/{-9}))y] = -8.03x - 3.57y

Let the initial velocity of the third part be u_1.8

Using conservation of linear momentum,

m₂u₂ + m_1.2u_1.2 + m_1.8u_1.8 = 0

=> 2(3.04x + 7.10y) + 1.2(-8.03x - 3.57y) + 1.8u_1.8 = 0

=> u_1.8 = 1.98x - 5.51y

So,

s_x,1.8 = u_x,1.8²/2µg = 1.98²/(0.4 * 9.81) = 1 m

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

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