Answer to Question #245083 in Molecular Physics | Thermodynamics for Kelani

Question

A soccer player takes a corner kick, lofting a stationary ball 30.0° above the horizon at 23.0 m/s. If the soccer ball has a mass of 0.425 kg and the player's foot is in contact with it for 3.90 ✕ 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.

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

Δp_x= ___kg · m/s

Δp_y= ___kg · m/s

(b) the magnitude of the average force exerted by the player's foot on the ball (in N)

 __N

Answer 1

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

The initial velocity of the ball will be V₁ = 0 m/s because it is at rest. Then, to find the final velocity for each component we use the trigonometric relations:

$v_1=v_{1,y}=v_{1,x}=0\,m/s \\ v_2=23.0\, m/s \\ v_{2,x}=v_2 \cos{\theta} \\ v_{2,x}=(23.0\, m/s)\cos{(30°)}=19.9\, m/s \\ v_{2,y}=v_2 \sin{\theta} \\ v_{2,x}=(23.0\, m/s)\sin{(30°)}=11.5\, m/s$

Then we can calculate the change of momentum for the x-axis:

$\Delta p_{x}=m(v_{2,x}-v_{1,x}) \\ \Delta p_{x}=(0.425\,kg)(19.9-0)\, m/s \\ \Delta p_{x}=8.4575 \,{kg}\cdot {m/s}$

Following for the y-axis we find:

$\Delta p_{y}=m(v_{2,y}-v_{1,y}) \\ \Delta p_{y}=(0.425\,kg)(11.5-0)\, m/s \\ \Delta p_{y}=4.8875 \,{kg}\cdot {m/s}$

In conclusion, we have Δp_x= 8.4575 kg · m/s and Δp_y= 4.8875 kg · m/s.

(b) Then, for the second part the magnitude of the average force exerted by the player's foot on the ball (in N) is found with the impulse-force relation:

$\vec{J}=\sum{\vec{F}}\cdot \Delta t=\vec{p_2}-\vec{p_1} \\ \sum{\vec{F}}\cdot \Delta t=m(\vec{v_2}-\vec{v_1}) \\ \implies \sum{{\vec{\mid F \mid}}}=\cfrac{m(\vec{\mid v_2 }-\vec{v_1\mid})}{\Delta t}$

We use the definitions for v1 and v2 to find the vector that describes the change in velocity and then we can find the magnitude of such change:

$\mid({v_{2,x}},{v_{2,y}}) - ({v_{1,x}},{v_{1,y}}) \mid= \mid(19.9,11.5) -(0,0)\mid\,m/s \\ \implies \mid(19.9,11.5)\mid\,m/s=(\sqrt{(19.9)^2+(11.5)})\, m/s \approxeq 23\,m/s \\ \sum{{\vec{\mid F \mid}}}=\cfrac{(0.425\,kg)(23\,m/s)}{3.90\times 10^{-2}s} \\ \implies \sum{{\vec{\mid F \mid}}}=250.6\,N$

In conclusion, the average force exerted by the player's foot is 250.6 N.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.