two cars, one with mass m1=1000kg and the other with mass m2= 4000kg crash in an intersection. Before the crash, car m1 was headed east with a speed of v1i= 25 m/s and the car m2 was headed north with a speed of v2i= 10m/s. After the crash, the cars stick together. what is the speed of the 2-car wreck after the collision and with what angle does it leave the crash point?
The vector sum of the momenta of the two cars before collision equals the momentum of the combined cars after collision.
By solving the momentum conservation equations for two perpendicular axes, you can solve for both final speed and angle.
We will happily critique your work, if needed.
To solve this problem, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy.
1. First, let's calculate the initial momentum of each car:
- Car m1: momentum1 = m1 * v1i = 1000 kg * 25 m/s = 25,000 kg·m/s (east)
- Car m2: momentum2 = m2 * v2i = 4000 kg * 10 m/s = 40,000 kg·m/s (north)
2. Next, let's calculate the total momentum before the collision:
- Total momentum before = momentum1 + momentum2
3. Since the cars stick together after the collision, the total mass of the wreck is the sum of the individual masses:
- Total mass = m1 + m2 = 1000 kg + 4000 kg = 5000 kg
4. Now, let's use the principle of conservation of momentum to find the velocity of the two-car wreck after the collision:
- Total momentum after = Total momentum before
- Total momentum after = Total mass * v_after
Rearranging the equation, we find: v_after = (momentum1 + momentum2) / Total mass
5. Finally, we can calculate the angle at which the two-car wreck leaves the crash point using trigonometry:
- tan(angle) = velocity NORTH / velocity EAST
- angle = arctan(velocity NORTH / velocity EAST)
Let's plug in the given values and calculate the results:
Using the above formula, we find:
- Total momentum before = 25,000 kg·m/s (east) + 40,000 kg·m/s (north) = 65,000 kg·m/s
- v_after = 65,000 kg·m/s / 5000 kg = 13 m/s
Now, we can calculate the angle:
- tan(angle) = velocity NORTH / velocity EAST
- angle = arctan(10 m/s / 25 m/s) = arctan(0.4)
- angle ≈ 21.8 degrees
Therefore, the speed of the two-car wreck after the collision is 13 m/s, and it leaves the crash point at an angle of approximately 21.8 degrees north of east.