Question

A 1.00 kg ball traveling towards a soccer player at a velocity of 5.00 m/s rebounds off the soccer

Answer 1

Answer:

A)   F = - 8.5 10² N,  B)   I = 21 N s

Explanation:

A) We can solve this problem using the relationship of momentum and momentum

          I = Δp

in this case they indicate that the body rebounds, therefore the exit speed is the same in modulus, but with the opposite direction

         v₀ = 8.50 m / s

         v_f = -8.50 m / s

         F t = m v_f -m v₀

         F = $m \frac{(v_f - v_o)}{t}$

let's calculate

         F = $1.00 \ \frac{(-8.5-8.5)}{2 \ 10^{-2}}$

         F = - 8.5 10² N

B) let's start by calculating the speed with which the ball reaches the ground, let's use the kinematic relations

         v² = v₀² - 2g (y- y₀)

as the ball falls its initial velocity is zero (vo = 0) and the height upon reaching the ground is y = 0

         v = $\sqrt{2g y_o}$

calculate  

         v = $\sqrt{2 \ 9.8 \ 10}$

         v = 14 m / s

to calculate the momentum we use

         I = Δp

         I = m v_f - mv₀

when it hits the ground its speed drops to zero

we substitute

         I = 1.50 (0-14)

         I = -21 N s

the negative sign is for the momentum that the ground on the ball, the momentum of the ball on the ground is

        I = 21 N s