Question

Mike kicks soccer ball at a speed of 2.5m/s his friend Jake kicks it back at a speed of 5.5 m/s (Mass of ball is 0.5kg) a.What was the change in momentum? b. If the ball was in contact with Jakes foot for 0.07 seconds, what force did Jake apply?

Answer 1

The information given is

• The initial velocity is $\vec{V_{i}}=+2.5\;\text{m}/\text{s}$
• The final velocity $\vec{V_{f}}=-5.5\;\text{m}/\text{s}$ city is
• The mass is $m=0.5\;\text{kg}$
• The interaction time is $t=0.07\;\text{s}$

Part A

The change in momentum is given by

$\Delta \vec{P}=\vec{V}_{f}\;m-\vec{V}_{i}\;m$

Where.

$V_{i}$ is the initial velocity

$V_{f}$ is the final velocity.

$m$ is the mass.

Evaluating numerically.

$\Delta \vec{P}=\vec{V}_{f}\;m-\vec{V}_{i}\;m\\$

$\Delta \vec{P}=-5.5\;\text{m}/\text{s}_{f}\times 0.5\;\text{Kg}-2.5\;\text{m}/\text{s}_{f}\times 0.5\;\text{Kg}\;m\\$

$\Delta \vec{P}=-4.0\;\text{Kg}\;\text{m}/\text{s}$

Answer A

The change in momentum is $\displaystyle \color{red}{\boxed{\Delta P=-4.0\;\text{Kg}\;\text{m}/\text{s}}}$

Part B

The average force is

$\vec{F}_{avg}=\dfrac{\Delta \vec{P}}{t}$

Where.

$\Delta \vec{P}$ is the change of momentum.

$t$ is the interaction time.

Evaluating numerically.

$\vec{F}_{avg}=\dfrac{\Delta \vec{P}}{t}\\ \vec{F}_{avg}=\dfrac{-4.0\;\text{Kg}\;\text{m}/\text{s} }{0.07\;\text{s}}\\ \vec{F}_{avg}=-57\;\text{N}$

Answer B

The average force is $\displaystyle \color{red}{\boxed{\vec{F}_{avg}=-57\;\text{N}}}$

The negative sign indicates that the force and the change in momentum (impulse) points in the opposite direction to the initial movement, that is, when it goes back towards Mike