A 1140 kg car skidding due north on a level frictionless icy road at 178.56 km/h collides with a 1653 kg car skidding due east at 124 km/h in such a way that the two cars stick together.
At what angle (−180� � � � +180�) East
of North do the two coupled cars skid off at?
Answer in units of �.
Add the two momentum vectors.
The resultant vector will have the direction of the coupled cars.
To find the angle at which the two cars skid off, we can use the principles of conservation of momentum and conservation of kinetic energy.
First, let's identify the given information:
- Mass of the first car (car 1): m1 = 1140 kg
- Velocity of car 1: v1 = 178.56 km/h
- Mass of the second car (car 2): m2 = 1653 kg
- Velocity of car 2: v2 = 124 km/h
Now, let's convert the velocities to m/s:
v1 = 178.56 km/h * (1000 m/km) / (3600 s/h) = 49.60 m/s
v2 = 124 km/h * (1000 m/km) / (3600 s/h) = 34.44 m/s
Next, let's find the final velocity (vf) of the coupled cars using the concepts of conservation of momentum and conservation of kinetic energy.
Conservation of Momentum:
The total momentum before the collision is equal to the total momentum after the collision.
m1 * v1 + m2 * v2 = (m1 + m2) * vf
Substituting the given values:
(1140 kg * 49.60 m/s) + (1653 kg * 34.44 m/s) = (1140 kg + 1653 kg) * vf
56184 + 57013 = 2793 * vf
113197 = 2793 * vf
Thus, the final velocity (vf) of the coupled cars is equal to 113197 / 2793 = 40.53 m/s
Conservation of Kinetic Energy:
The total kinetic energy before the collision is equal to the total kinetic energy after the collision.
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * (m1 + m2) * vf^2
Substituting the given values:
(1/2) * 1140 kg * (49.60 m/s)^2 + (1/2) * 1653 kg * (34.44 m/s)^2 = (1/2) * (1140 kg + 1653 kg) * (40.53 m/s)^2
1405469.28 + 1962977.36 = 2793 * vf^2
3368446.64 = 2793 * vf^2
Now, solve for vf^2:
vf^2 = 3368446.64 / 2793
vf^2 = 1204.068542
Taking the square root of both sides:
vf = √1204.068542
vf = 34.68 m/s
The final velocity of the coupled cars is approximately 34.68 m/s.
Finally, we can find the angle of the velocity vector of the coupled cars relative to the north direction using trigonometry. The angle can be found using the inverse tangent function:
θ = arctan(vf_x / vf_y)
In this case, the x-direction is east and the y-direction is north.
vf_x = 34.68 m/s (eastward direction)
vf_y = 0 m/s (northward direction)
θ = arctan(34.68 / 0)
Note that dividing by 0 is undefined, indicating that the final velocity vector of the coupled cars is purely along the x-axis (east direction). Therefore, the angle relative to the north direction is 90 degrees or π/2 radians.
In terms of the given range of angles, the angle can be expressed as +90 degrees or +π/2 radians.