Three iceboats are moving as shown in the figure. Boat 1 has a mass of 150 kg an

Question

Three iceboats are moving as shown in the figure. Boat 1 has a mass of 150 kg and a velocity of 10.39 m/s due south. Boat 2 has a mass of 225 kg, a velocity 8.0 m/s due east, and is located 10 m south and 20 m east of Boat 1. Boat 3 has a mass of 100 kg, is moving at 18.0 m/s at an angle of theta = 60 degree as shown, and is located 8 m south and 8 m west of Boat 2. Find the center of mass of the system of iceboats at the moment indicated in the description above. Use the general directions indicated, but be sure to clearly define the remainder of your coordinate system. Find the velocity (magnitude and direction) of the system taken as a whole (i.e. of the center of mass). A strong wind picks up, blowing due west (i.e. left in the figure) with a constant and uniform force of 300 N for exactly 3.0 seconds. What is the velocity of the system (magnitude and direction) after this force has had a chance to act? Think about it: Was momentum conserved throughout this problem? Did it need to be? Can you similarly determine the final velocities of each of the boats after the wind has blown?

Explanation / Answer

Assume boat 1 is at origin

m1 = 150 kg, x1 = 0 , y1 = 0

v1x = 0 , v1y = -10.39 m/s

m2 = 225 kg, x2 = 20 m, y2 = -10 m

v2x = 8 m/s, v2y = 0

m3 = 100 kg, x3 = 20 - 8 = 12 m, y3 = -10 - 8 = -18 m

v3x = 18*cos(60) = 9 m/s
v3y = 18*sin(60) = 15.6 m/s

a) Xcm = (m1*x1 + m2*x2 + m3*x3)/(m1+m2+m3)

= (150*0 + 225*20 + 100*12)/(150 + 225 + 100)

= 12 m

Ycm = (m1*y1 + m2*y2 + m3*y3)/(m1+m2+m3)

= (150*0 + 225*(-10) + 100*(-18))/(150 + 225 + 100)

= -8.53 m

location of center of mass is 12 m East and 8.53 m South of the first boat.

b)

Vcm,x = (m1*v1x + m2*v2x + m3*v3x)/(m1+m2+m3)

= (150*0 + 225*8 + 100*9)/(150 + 225 + 100)

= 5.68 m/s

Vcm,y = (m1*v1y + m2*v2y + m3*v3y)/(m1+m2+m3)

= (150*0 + 225*0 + 100*15.6)/(150 + 225 + 100)

= 3.28 m/s

direction : theta = tan^-1(3.28/5.68)

= 30 degrees North of east

C)

ax = -300/(m1+m2+m3)

= -300/(150 + 225 + 100)

= -0.6316 m/s^2

after t = 3s,

Vcm,xf = Vcm,x + ax*t

= 5.68 - 0.6316*3

= 3.78 m/s

Vcm,yf = Vcm,y

= 3.28 m/s

direction : theta = tan^-1(3.28/3.78)

= 41 degrees North of east

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