Two cars approach an ice-covered intersection. One car, of mass 1.26 103 kg, is initially traveling north at 10.6 m/s. The other car, of mass 1.66 103 kg, is initially traveling east at 10.6 m/s. The cars reach the intersection at the same instant, collide, and move off coupled together. Find the velocity of the center of mass of the two-car system just after the collision.
magnitude
direction
north of east
To find the velocity of the center of mass of the two-car system just after the collision, we need to apply the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In this case, since there are no external forces acting on the system, the total momentum before the collision should be equal to the total momentum after the collision.
Before the collision, the momentum of the first car (mass = 1.26 * 10^3 kg) heading north at 10.6 m/s can be calculated as:
Momentum1 = mass1 * velocity1 = (1.26 * 10^3 kg) * (10.6 m/s) = 1.3356 * 10^4 kg·m/s (north)
Before the collision, the momentum of the second car (mass = 1.66 * 10^3 kg) heading east at 10.6 m/s can be calculated as:
Momentum2 = mass2 * velocity2 = (1.66 * 10^3 kg) * (10.6 m/s) = 1.7576 * 10^4 kg·m/s (east)
To find the total momentum before the collision, we need to break down both momenta into x and y components. Since one car is heading north and the other is heading east, we can use trigonometry to find the components.
Momentum1 = Momentum1x + Momentum1y
Momentum1x = 0 (since the car is heading north)
Momentum1y = Momentum1 = 1.3356 * 10^4 kg·m/s (north)
Momentum2 = Momentum2x + Momentum2y
Momentum2x = Momentum2 = 1.7576 * 10^4 kg·m/s (east)
Momentum2y = 0 (since the car is heading east)
Now, since momentum is a vector quantity, we can add the x and y components separately to find the total momentum in the x and y directions:
Total momentum in the x-direction = Momentum1x + Momentum2x = 1.7576 * 10^4 kg·m/s (east)
Total momentum in the y-direction = Momentum1y + Momentum2y = 1.3356 * 10^4 kg·m/s (north)
To find the magnitude of the total momentum after the collision, we can use the Pythagorean theorem since the x and y components form a right-angled triangle:
Magnitude of total momentum = sqrt((Total momentum in the x-direction)^2 + (Total momentum in the y-direction)^2)
Plugging in the values:
Magnitude of total momentum = sqrt((1.7576 * 10^4 kg·m/s)^2 + (1.3356 * 10^4 kg·m/s)^2) = sqrt(4.938816 * 10^8 + 1.78213616 * 10^8) = sqrt(6.72095216 * 10^8) = 8.204 * 10^3 kg·m/s
The magnitude of the velocity of the center of mass of the two-car system just after the collision is 8.204 * 10^3 kg·m/s.
To find the direction, we can use the inverse tangent formula:
Direction = atan((Total momentum in the y-direction) / (Total momentum in the x-direction))
Plugging in the values:
Direction = atan((1.3356 * 10^4 kg·m/s) / (1.7576 * 10^4 kg·m/s)) = atan(0.759) = 37.2°
Therefore, the velocity of the center of mass of the two-car system just after the collision has a magnitude of 8.204 * 10^3 kg·m/s and is directed north of east at an angle of 37.2°.