Question

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, *x _{CM}*, one second

________ m

Part b) Calculate the y-coordinate of the center of mass of the
two-automobile system, *x _{CM}*, one second

________ m

Part c) Calculate the magnitude of the velocity of the center of
mass of the two-automobile system, *v _{CM}*, one
second

________ m/s

Part d) Calculate the direction of the velocity of the center of
mass of the two-automobile system, *v _{CM}* one
second

________ degrees

Part e) Calculate the magnitude of the velocity of the center of
mass of the two-automobile system, *v _{CM}*,
immediately

________ m/s

Part f) Calculate the direction of the velocity of the center of
mass of the two-automobile system, *v _{CM}*
immediately

________ degrees

Answer #1

**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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