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3 years ago

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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?

1 answer:

Mkey [24]3 years ago

3 0

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 ^

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