Question

A 1.00 kg ball traveling towards a soccer player at a velocity of 5.00 m/s rebounds off the soccer
player's foot at a velocity of 8.50 m/s. If the time of contact between the ball and the player's foot
was 2.00 x 10-2 seconds, what was the force that the foot applied on the ball?

this is the second question

A 1.50 kg rock falls from the top of a 10.0 m high building and strikes the ground below. Calculate
the impulse experienced by the rock during its fall.​

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

Answer 2

(a)The force that the foot applied on the ball will be 175 N.

(b)The impulse experienced by the rock during its fall will be 21 Ns.

What is impulse?

The change in momentum of an item when it is operated upon by a force for a period of time is known as an impulse.Impulse is given by the change in momentum,

I=ΔP

The given data in the question will be

m is the mass of soccer ball = 1.00 kg

u is the velocity by which ball hits =5.00 m/sec

v is the velocity by which ball rebounds=8.50 m/sec.

(a) The force that the foot applied on the ball will be 175 N.

If the ball rebounds  with the the velocity of 8.50m/sec then

$\rm F= \frac{m(v-u)}{t} \\\\\rm F= \frac{1(8.50-5.00)}{2.00\times10^{-2}}\\\\\rm F=175 N$

Hence the force that the foot applied on the ball will be 175 N.

(b)The impulse experienced by the rock during its fall will be 21 Ns.

According to Newton's third equaton of motion.

$\rm v^2=u^2+2gh\\\\\rm v^2=2\times9.81\times10.0\\\\\rm v=\sqrt{2\times9.81\times10.0} \\\\\rm v= 14.0071 m/sec$

Due to the free fall condition the initial velocity is zero,

Impulse is given by the change in momentum,

I=ΔP

I=m(v-u)

$\rm I=m(v-u)\\\\\rm I=1.50(14.0-0)\\\\\rm I=21 Ns$

Hence the impulse experienced by the rock during its fall will be 21 Ns.

To learn more about the impulse refer to the link;

https://brainly.com/question/904448