A 1500-kg car moving north at 27.0 m/s is struck by a 2165-kg car moving east at 13.0 m/s. The cars are stuck together. How fast and in what direction do they move immediately after the collision?
1. what is the degree north of east?
2. how was do they move after the collision (M/S)
The momentum in each direction is conserved.
angle=arctan(momentum N/momentum E)
velocity= (momnetum N + Momentum E)/total mass
To find the speed and direction of the cars immediately after the collision, we can use the principles of conservation of momentum and the concepts of vector addition.
1. To determine the direction of the cars after the collision, we need to find the resultant direction of the two component velocities. This can be done using trigonometry and the angle between the north and east directions.
Let's call the angle between the north direction and the resultant direction θ.
tan(θ) = (opposite/adjacent) = (north velocity/east velocity)
tan(θ) = (27.0 m/s) / (13.0 m/s)
θ = arctan (27.0/13.0)
θ ≈ 63.77 degrees
So the resultant direction is 63.77 degrees north of east.
2. To find the speed at which the cars move after the collision, we can use the principles of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
Momentum before collision = Momentum after collision
(mass1 x velocity1) + (mass2 x velocity2) = (mass1 + mass2) x (final velocity)
Substituting the given information:
(1500 kg x 27.0 m/s) + (2165 kg x 13.0 m/s) = (1500 kg + 2165 kg) x (final velocity)
Simplifying the equation:
(40,500 kg·m/s) + (28,145 kg·m/s) = (3665 kg) x (final velocity)
68,645 kg·m/s = 3665 kg x (final velocity)
Dividing both sides by 3665 kg:
final velocity = (68,645 kg·m/s) / (3665 kg)
final velocity ≈ 18.74 m/s
So, immediately after the collision, the cars move at a speed of approximately 18.74 m/s in a direction that is 63.77 degrees north of east.