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3 years ago

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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. A

t 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 answer:

Julli [10]3 years ago

3 0

Answer:

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%

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