Question

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?

Answer 1

Answer:

I = 20 i ^ N s

Explanation:

For this problem let's use the Impulse equation

       I = Δp = m $v_{f}$- 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 = $v_{y}$ / v

     $v_{y}$ = v sin60

     vₓ = 10 cos 60

     $v_{y}$ = 10 sin60

    vₓ = 5.0 m / s

    $v_{y}$ = 8.66 m / s

Let's calculate the impulse on each axis

X axis

     Iₓ = m $v_{xf}$ - m vₓ₀

How the ball bounces

    $v_{xf}$ = - vₓ₀ = vₓ

    Iₓ = 2 m vₓ

    Iₓ = 2 2 5

    Iₓ = 20 N s

Y axis

   $I_{y}$ = m $v_{yf}$ - m vyo

On the axis and the ball does not change direction so

   $v_{yf}$ = vyo

  $I_{y}$ = 0

The total momentum is

   I = Iₓ i ^ + $I_{y}$ j ^

   I = 20 i ^ N s