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marusya05 [52]
3 years ago
11

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 b

ounces off so that it is moving 60º above the −x-direction with the same speed.
What is the impulse delivered by the wall?
Physics
1 answer:
kow [346]3 years ago
7 0

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

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In our problem we have q=-2.5 nC=-2.5\cdot 10^{-9} C and F=18 nN = 18 \cdot 10^{-9} N, so we can find the magnitude of the electric field:

E= \frac{F}{q}= \frac{18\cdot 10^{-9}N}{2.5\cdot 10^{-9}C}=7.2 V/m

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Answer:

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Explanation:

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