Question

# Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a...

Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a 2000-kg truck traveling north at 25 m/s (55.9 mph). The cars lock together in such a way as to prevent them from separating or rotating significantly. One second before the collision, the car is located at position (x,y) = (-20,0) m and the truck is located at position (x,y) = (0,-25) m.

Part a) Calculate the x-coordinate of the center of mass of the two-automobile system, xCM​​, one second before the collision.

________ m

Part b) Calculate the y-coordinate of the center of mass of the two-automobile system, xCM​​, one second before the collision.

________ m

Part c) Calculate the magnitude of the velocity of the center of mass of the two-automobile system, vCM​​, one second before the collision.

________ m/s

Part d) Calculate the direction of the velocity of the center of mass of the two-automobile system, vCM​​ one second before the collision. (Give you answer in degrees relative to the positive x-axis which points to the east.)

________ degrees

Part e) Calculate the magnitude of the velocity of the center of mass of the two-automobile system, vCM, immediately after the collision.

________ m/s

Part f) Calculate the direction of the velocity of the center of mass of the two-automobile system, vCM immediately after the collision. (Give you answer in degrees relative to the positive x-axis which points to the east.)

________ degrees

let

m1 = 1500 kg, u1 = 20 m/s

x1 = -20 m, y1 = 0

m2 = 2000 kg, u2 = 25 m/s

x2 = 0 , y2 = -25 m

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

= (1500*(-20) + 2000*0)/(1500 + 2000)

= -8.57 m

b) Ycm = (m1*y1 + m2*y2)/(m1 + m2)

= (1500*0 + 2000*(-25))/(1500 + 2000)

= -14.3 m

c) Vcmx = (m1*u1x + m2*u2x)/(m1+ m2)

= (1500*20 + 2000*0)/(1500 + 2000)

= 8.57 m/s

Vcmy = (m1*u1y + m2*u2y)/(m1+ m2)

= (1500*0 + 2000*25)/(1500 + 2000)

= 14.3 m/s

Vcm = sqrt(Vcmx^2 + Vcmy^2)

= sqrt(8.57^2 + 14.3^2)

= 16.7 m/s

d) direction : theta = tan^-1(Vcmy/Vcmx)

= tan^-1(14.3/8.57)

= 59.1 degrees

e) Vcm = 16.7 m/s

f) direction : theta = 59.1 degrees

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