Question

A player kicks a football of mass \[0.5kg\] and the football begins to move with velocity of \[10m/s\]. If the contact between the leg and the football lasts for \[\dfrac{1}{{50}}\sec \], then the force acted on the football should be
A. \[2500N\]
B. \[1250N\]
C. \[250N\]
D. \[625N\]

Answer 1

Hint: In question it is given that the mass of the ball is \[0.5kg\]. The final velocity is also given as \[10m/s\]. Since the ball would have been in the rest state before the player kicked the ball, we can tell what the initial velocity of the ball is\[0m/s\]. We can find the force acting upon the ball by the given formula below.

Formula Used:
\[F = \dfrac{{\Delta p}}{{\Delta t}}\]
Here, \[F\] is the force acting upon the ball
\[\Delta p\] is the change in the momentum.
\[\Delta t\] is the change in time.


Complete step by step solution:
By applying all the known values in the above-mentioned formula we would arrive at the answer to the question.
Change in the momentum formula is given as
\[\Delta p = m{v_2} - m{v_1}\]
Here \[m\]is the mass of the football
\[{v_1}\]and\[{v_2}\]are the initial and final velocities respectively.
It is given that the ball started moving with a velocity of \[10m/s\] after the point of contact of the player. So we can say that \[{v_2}\]=\[10m/s\]
Since the ball was on rest before the force given to the ball we can say that the initial velocity
\[{v_1}\]=\[0m/s\].
We know that =\[m = 0.5kg\]
Substituting all these in the change of momentum formula we get,
\[\Delta p = \left( {0.5} \right)\left( {10} \right) - \left( {0.5} \right)\left( 0 \right)\]
\[\Delta p = 5kgm/s\]
Now we know the point of contact duration. So we can substitute that value for \[\Delta t\]
\[\Delta t = \dfrac{1}{{50}}\sec \]
\[ \Rightarrow \Delta t = 0.02\sec \]
Substituting\[\Delta p\]and\[\Delta t\]values in the force formula we will get
\[F = \dfrac{{5kgm/s}}{{0.02\sec }}\]
\[ \Rightarrow F = 250N\]
Therefore the correct option is C.

Note:
Note that the effect of the force exerted by the player over the ball for only a short duration. And this force changes the momentum of the ball. This effect is called the impulse. The force in the above problem used is called impulse and it is a vector quantity.