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3 years ago

a 1000kg car uses a breaking force of 10,000N to stop in two second. What is the change in momentum of the car?

Physics

1 answer:

gayaneshka [121]3 years ago

8 0

Answer:

ΔP = 20000 N s

Explanation:

To solve this problem we use the relation between momentum and moment

I = Δp

let's calculate the momentum

I = ∫F dt

if we use the average force

I = F t

I = 10000 2

I = 20000 N s

therefore with the first equation

ΔP = I = 20000 N s

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You want to close an open door by throwing either a 400-g lump of clay or a 400-g rubber ball toward it. you can throw either ob

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Refer to the diagram shown below.

Let m = the mass (g) of the door.
Let v = the launch velocity
Let u = the velocity of the door after impact.

Elastic impact (rubber ball):
The rubber ball bounces off the door with presumably elastic impact, which means that both momentum and kinetic energy are conserved.
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400v = -400v + mu
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Inelastic impact (clay):
The clay sticks to the door after impact.
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Therefore
$u=( \frac{400}{m+400} )v$

When we compare magnitudes of u for the door, we find that
$u_{1}=( \frac{400}{m} )(2v), \,\, elastic \\\\ u_{2}=( \frac{400}{m+400} )v , \,\, inelastic$
Clearly, the elastic impact creates a greater value of u for the door.

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