In a particular crash test, an automobile of mass 1295 kg collides with a wall and bounces back off the wall. The x components of the initial and final speeds of the automobile are 25 m/s and 1.5 m/s, respectively.

If the collision lasts for 0.11 s, find the magnitude of the impulse due to the collision.
Answer in units of kg · m/s.

I'm having a lot of trouble with this. Where do I start? What equation?

Impulse = M*(V-Vo) = 1295*(-1.5-25) = 34,318 kg*m/s

To find the magnitude of the impulse due to the collision, we can use the Impulse-Momentum theorem, which states that the change in momentum of an object is equal to the impulse applied to it. The formula for impulse is:

Impulse = Δp = m(Vf - Vi)

Where:
Δp = change in momentum
m = mass of the object
Vf = final velocity
Vi = initial velocity

In this case, we are given the initial and final velocities, as well as the mass of the automobile. So we can plug in the values and calculate the magnitude of the impulse. Let's follow the steps:

Step 1: Identify the given information:
- Mass of the automobile (m) = 1295 kg
- Initial velocity (Vi) = 25 m/s
- Final velocity (Vf) = 1.5 m/s

Step 2: Calculate the change in velocity:
Δv = Vf - Vi
Δv = 1.5 m/s - 25 m/s
Δv = -23.5 m/s

Step 3: Calculate the impulse:
Impulse = Δp = m Δv
Impulse = 1295 kg × (-23.5 m/s)
Use a calculator to find the product:

Impulse ≈ -30408.3 kg·m/s

Step 4: Take the magnitude of the impulse:
The magnitude of a negative value is its positive equivalent. So, we take the absolute value:

Impulse (magnitude) ≈ |-30408.3 kg·m/s| ≈ 30408.3 kg·m/s

Therefore, the magnitude of the impulse due to the collision is approximately 30408.3 kg·m/s.