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

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.