A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. At the bottom of the incline the cart collides with a solid object and comes to a stop. The distance between the cart and start position is 20 cm.
Velocity Before Velocity After Impulse
0.66 m/s -0.59 m/s 0.287 N*s
1. Find the momentum before and after the collision
2. Next, find the change in momentum
3. Calculate the percent different between the change and momentum and the impulse

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

it about 600 mph from what i calculated

c because the action was the roller coaster reaches bottom of hill (action) reaction would be blood tends to drain away from head

hope this !