Question

A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mass of 0.425 kg and the player's foot is in contact with it for 5.10 ✕ 10−2 s, find the x- and y-components of the soccer ball's change in momentum and the magnitude of the average force exerted by the player's foot on the ball.

PART A: the x- and y-components of the soccer ball's change in momentum (in kg · m/s)

Δpx ?????????? = kg · m/s

Δpy ?????????? = kg · m/s

Answer 1

The soccer ball's change in momentum in the x- and y-directions can be found by calculating the velocity components using the given launch angle and then multiplying those components by the mass of the ball. The average force exerted by the player's foot is found by dividing the change in momentum by the time interval of contact.

Step-by-step explanation:

When a soccer player kicks a stationary ball at an angle above the horizon, we can calculate the change in momentum (Δp) and the average force exerted by using principles from physics. Considering that the initial velocity of the ball is zero and that after the kick the velocity vector points at an angle of 33.0° above the horizontal with a magnitude of 15.0 m/s, we can decompose the final velocity into its x- and y-components:

• Vx = V × cos(θ)
• Vy = V × sin(θ)

To find the momentum change, we multiply the components of the velocity by the mass of the ball (m).

• Δpx = m × Vx = 0.425 kg × (15.0 m/s × cos(33.0°))
• Δpy = m × Vy = 0.425 kg × (15.0 m/s × sin(33.0°))

We then calculate the average force (№) exerted on the ball by the player's foot using the relationship between impulse (№ × Δt), where Δt is the contact time, and change in momentum (Δp).

• №x = Δpx / Δt
• №y = Δpy / Δt

Therefore, the x- and y-components of the soccer ball's change in momentum are:

Δp_x = 5.31 kg · m/s (positive x-direction)

Δp_y = 3.40 kg · m/s (positive y-direction)

Answer 2

Step-by-step explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal,
`\theta=33^(\circ)`

Time for which the player's foot is in contact with it,
`\Delta t = 5.1* 10^(-2)\ s`

Part A,

The x component of the soccer ball's change in momentum is given by :

`\Delta p_x=mv\ cos\theta`

`\Delta p_x=0.425* 15\ cos(33)`

`p_x=5.34\ kg-m/s`

The y component of the soccer ball's change in momentum is given by :

`\Delta p_y=mv\ sin\theta`

`\Delta p_y=0.425* 15\ sin(33)`

`p_y=3.47\ kg-m/s`

Hence, this is the required solution.