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. Thw pieces shot off into three different directions and slid across the ground untill they came to a stop. The pieces of plate experience friction against the ground with coefficient of kinetic friction = 0.4. 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 1.2kg piece at (x,y) = (-9, -4). Where did the third piece come to a stop? Note: all (x,y) positions are in meters, also my teacher said that first it conservation of momentum and then kinematics. Thanks!

Explanation / Answer

Let the coordnate of third piece by (x,y)

After pieces are shattered, all the pieces are exerted a force umg or a deceleration of a= ug = 3.92 m/s^2. Just after the pieces are shattered, their sum of momentum is conserved = zero

By third equation of motion, v^2 = 2as

v = sqrt(2as) = sqrt(7.84s)

v1 = sqrt(7.84*sqrt(9+49)) = 7.727 m/s

v1x = 7.727*3/sqrt(9+49) = 3.044 m/s

v1y = 7.727*7/sqrt(9+49)= 7.102m/s

v2= sqrt(7.84*sqrt(81+16)) = 8.787 m/s

v2x =  8.787*-9/sqrt(81+16) = -8.030 m/s

v2y=   8.787*-4/sqrt(81+16) = -3.569 m/s

momentum is conserved just plate strike

so v3x = -(m1v1x +m2v2x) / m3 = -(2* 3.044 +1.2*-8.030)/1.8 = 1.971 m/s

v3y =  -(m1v1y +m2v2y) / m3 =-(2* 7.102+1.2*-3.569)/1.8 = -5.512 m/s

v3 = sqrt(1.971^2 +5.512^2)= 5.854 m/s

s= v^2/2a = 5.854^2/7.84 = 4.371 m

x = 4.371* 1.971/5.854 = 1.471

y = 4.371*  -5.512/5.854= -4.116

Hence coardinate is ( 1.471, -4.116)

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