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?
Use the law of conservation of momentum.
This will require you to add the momentum vectors of the colliding cars.
To find the speed and angle at which the two cars move after the collision, we can use the principle of conservation of momentum.
Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision, provided no external forces are acting on the system.
The momentum of an object is given by the product of its mass and velocity.
Before the collision:
- Car m1 has a mass m1 = 1000 kg and a velocity v1i = 25 m/s to the east.
- Car m2 has a mass m2 = 4000 kg and a velocity v2i = 10 m/s to the north.
To simplify calculations, we can decompose the velocity of car m2 into its east and north components. Since the car is heading north, the east component will be zero.
Using the concept of vector addition, we can find the east and north components of the combined velocity of the two cars after the collision.
East component of the combined velocity = mass of m1 * velocity of m1 (east component)
= m1 * v1i
North component of the combined velocity = mass of m2 * velocity of m2 (north component)
= m2 * v2i
Since the two cars stick together after the collision, the mass of the combined system is equal to the sum of the masses of the individual cars.
Total mass after the collision = m1 + m2
The combined velocity after the collision is given by the vector sum of the east and north components:
Combined velocity after the collision = √[(East component)^2 + (North component)^2]
Finally, we can determine the angle at which the two-car wreck moves using the inverse tangent function.
Angle = arctan(North component / East component)
Now, let's calculate the values:
East component of combined velocity = (1000 kg) * (25 m/s) = 25000 kg·m/s
North component of combined velocity = (4000 kg) * (10 m/s) = 40000 kg·m/s
Total mass after the collision = 1000 kg + 4000 kg = 5000 kg
Combined velocity after the collision = √[(25000 kg·m/s)^2 + (40000 kg·m/s)^2]
= √(625000000 + 1600000000) = √2225000000 ≈ 47140 m/s (rounded to 2 decimal places)
Angle = arctan(40000 kg·m/s / 25000 kg·m/s) ≈ arctan(1.6) ≈ 57.99 degrees (rounded to 2 decimal places)
Therefore, the speed of the two-car wreck after the collision is approximately 47140 m/s, and it leaves the crash point at an angle of approximately 57.99 degrees.