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A 2 kg ball with an initial velocity of 10 m/s moves at an angle 60º above the +x-direction. The ball hits a vertical wall and bounces off so that it is moving 60º above the −x-direction with the same speed.

What is the impulse delivered by the wall?

Respuesta :

Answer:

I = 20 i ^ N s

Explanation:

For this problem let's use the Impulse equation

       I = Δp = m [tex]v_{f}[/tex]- v₀

The impulse and the velocity are vector quantities, let's calculate on each axis, let's decompose the velocity

     cos 60 = vₓ / v

     vₓ = v cos 60

     sin60 = [tex]v_{y}[/tex] / v

     [tex]v_{y}[/tex] = v sin60

     vₓ = 10 cos 60

     [tex]v_{y}[/tex] = 10 sin60

    vₓ = 5.0 m / s

    [tex]v_{y}[/tex] = 8.66 m / s

Let's calculate the impulse on each axis

X axis

     Iₓ = m [tex]v_{xf}[/tex] - m vₓ₀

How the ball bounces

    [tex]v_{xf}[/tex] = - vₓ₀ = vₓ

    Iₓ = 2 m vₓ

    Iₓ = 2 2 5

    Iₓ = 20 N s

Y axis

   [tex]I_{y}[/tex] = m [tex]v_{yf}[/tex] - m vyo

On the axis and the ball does not change direction so

   [tex]v_{yf}[/tex] = vyo

  [tex]I_{y}[/tex] = 0

The total momentum is

   I = Iₓ i ^ + [tex]I_{y}[/tex] j ^

   I = 20 i ^ N s

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