Question

a 900kg car traveling 30.0 degree south of east at 12.0m/s collides with a 750kg car traveling North at 17.0m/s . the cars stick together. what is the speed of the wrekage just after the collision

Answer 1

Using Momentum conservation:

In x-direction

Pix = Pfx

m1v1x + m2v2x = (m1 + m2)*Vx

m1 = 900 kg car

m2 = 750 kg car

v1x = 12*cos 30 deg = 10.4 m/sec

v2x = 0

m1 + m2 = 1650 kg

Vx = (900*10.4 + 750*0)/1650

Vx = 5.67 m/sec

Momentum conservation in y-direction

Piy = Pfy

m1v1y + m2v2y = (m1 + m2)*Vy

m1 = 900 kg car

m2 = 750 kg car

v1y = -12*sin 30 deg = -6 m/sec

v2x = +17 m/sec

m1 + m2 = 1650 kg

Vy = (900*(-6) + 750*17)/1650

Vy = 4.45 m/sec

Vx = 5.67 m/sec

|V| = sqrt (Vx^2 + Vy^2) = final speed of both cars

|V| = sqrt (5.67^2 + 4.45^2) = 7.21 m/sec

direction = arctan (4.45/5.67) = 38.12 deg

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