Question

A soccer player kicks a soccer ball initially at rest setting it in motion at a velocity of 30 m/s. If the ball has a mass of 0.50 kg and the time of contact is 0.025 s, what is the force exerted on the player’s foot?

Answer 1

Answer:

 F= 600 N

Explanation:

Given that

Initial velocity ,u= 0 m/s

Final velocity ,v= 30 m/s

mass ,m = 0.5 kg

time ,t= 0.025 s

The change in the linear momentum is given as

ΔP= m (v - u)

ΔP= 0.5 ( 30 - 0 ) kg.m/s

ΔP= 15 kg.m/s

We know that from second law of Newtons

$F=\dfrac{dP}{dt}$

$F=\dfrac{\Delta P}{t}$

Now by putting the values

$F=\dfrac{15}{0.025}\ N$

F= 600 N