Question

A ball of mass m= 450.0 g traveling at a speed of 8.00 m/s impacts a vertical wall at an angle of θi =45.00 below the horizontal (x axis) and bounces away at an angle of θf= 45.00 above the horizontal. What is the average force exerted by the wall on the ball if the ball is in contact with the wall for 250.0 ms?

Answer 1

Answer:

   F = 20.4 i ^

Explanation:

This exercise can be solved using the ratio of momentum and amount of movement.

     I = F t = Dp

Since force and amount of movement are vector quantities, each axis must be worked separately.

X axis

Let's look for speed

      cos 45 = vₓ / v

      vₓ = v cos 45

      vₓ = 8 cos 45

      vₓ = 5,657 m / s

We write the moment

Before the crash                          p₀ = m vₓ

After the shock                            $p_{f}$ = -m vₓ

The variation of the moment      Δp = mvₓ - (-mvₓ) = 2 m vₓ

The impulse on the x axis           Fₓ t = Δp

       

        Fₓ = 2 m vₓ / t

        Fx = 2 0.450 5.657 / 0.250

        Fx = 20.4 N

We perform the same calculation on the y axis

       sin  45 = vy / v

       vy = v sin 45

       vy = 8 sin 45

       vy = 5,657 m / s

We calculate the initial momentum   po = m $v_{y}$

Final moment                                      $p_{f}$ = m $v_{y}$

Variations moment                             Δp = m$v_{y}$ - m$v_{y}$ = 0

Force in the Y-axis                             $F_{y}$ = 0

Therefore the total force is

       F = fx i ^ + Fyj ^

       F = Fx i ^

       F = 20.4 i ^