Answer:
See below.
Step-by-step explanation:
When the 1500 kg car collides with the wall and rebounds at a speed of 5.00 m/s, we can calculate the change in the car's velocity using the following formula:
Δv = v2 - v1
Where Δv is the change in velocity, v2 is the final velocity, and v1 is the initial velocity. Substituting the given values, we get:
Δv = 5.00 m/s - 20.0 m/s
Δv = -15.0 m/s
The negative sign indicates that the direction of the car's velocity has reversed, or that the car is now moving to the left. To calculate the magnitude of the change in velocity, we take the absolute value:
|Δv| = |-15.0 m/s|
|Δv| = 15.0 m/s
Therefore, the magnitude of the change in velocity is 15.0 m/s.
Now,
To find the magnitude of the average force acting on the car during the collision, we can use the impulse-momentum theorem, which states that:
Impulse = change in momentum
Average force = Impulse / time
The change in momentum of the car is given by:
Δp = mΔv
where Δv is the change in velocity calculated in the previous answer and m is the mass of the car.
Δp = 1500 kg × (-15.0 m/s)
Δp = -22500 kg·m/s
The impulse acting on the car during the collision is equal to the change in momentum:
Impulse = Δp = -22500 kg·m/s
To find the magnitude of the average force acting on the car during the 250 ms collision, we divide the impulse by the duration of the collision:
Average force = Impulse / time
Average force = -22500 kg·m/s / 0.250 s
Average force ≈ -90,000 N
The negative sign indicates that the force is in the opposite direction of the car's motion, or to the left. Therefore, the magnitude of the average force acting on the car during the collision is approximately 90,000 N.