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 ^

Answer 2

The average force exerted is 20.34 N.

Impact and Collision:

According to the question, the ball bounces back away after hitting the wall. So splitting the velocity of the ball in horizontal and vertical components we get:

(i) before collision:

vₓ = vcos = 8cos45

vₓ = 5.65 m/s

and

$v_y=vsin\theta=8sin45\\\\v_y=5.65\;m/s$

(ii) after collision:

v'ₓ = -vcos = -8cos(-45)

v'ₓ = -5.65 m/s

since the direction of velocity is opposite to the initial direction and we have considered the angle below horizontal as positive and above it as negative.

and

$v'_y=-vsin\theta=-8sin(-45)\\\\v'_y=5.65\;m/s$

the figure attached below clarifies why the horizontal velocity changes and the vertical velocity remains the same.

The change in momentum in x and y direction is:

Pₓ = mvₓ - mv'ₓ = m(vₓ - v'ₓ)

Pₓ = 0.45 (5.65-(-5.65)) kgm/s

Pₓ = 5.085 kgm/s

and

$P_y=mv_y-mv'_y=m(v_y-v'_y)\\\\P_y=0.45(5.65-(5.65))\;kgm/s\\\\P_y= 0$

Now, the force is given by the change in momentum:

$F_x=\frac{P_x}{t}=\frac{5.085\;kgm/s}{0.25\;s}\\\\F_x=20.34\;N$

and

$F_y=\frac{P_y}{t}=0$

Net force:

F = Fₓ

F = 20.34 N

Learn more about laws of collision:

https://brainly.com/question/2356330?referrer=searchResults