Question

In a football match, a ball is kicked by a player with a force of $25N$ for $0.2$ sec and then by another player with a force of $70N$ for $0.1$ sec in the same direction. If the football gains a velocity of $24m{s^{ - 1}}$ after two kicks, the mass of the football is nearly
A. $5$ kg
B. $10$ kg
C. $2.5$ kg
D. $0.5$ kg

Answer 1

Hint: We can find the mass of the football by using the concept of impulse which is when a large force acts on a body for a very small time, the product of the average of total force for that small time period is impulse. Also, we use the concept of change in momentum of the body and the impulse-momentum theorem.

Complete step by step answer:
Consider the football moving along +X direction. Football is kicked by a player with a force of $25N$ for $0.2$ sec and then by another player with a force of $70N$ for $0.1$ sec in the same direction.
So, ${F_1} = 25N,{F_2} = 70N$ and $\Delta {t_1} = 0.2$ sec , $\Delta {t_2} = 0.1$ sec
Using the formula of impulse for each kick of the players, we get
${I_1} = {F_1} \times \Delta {t_1} \\
\Rightarrow {I_1} = 25 \times 0.2 \\
\Rightarrow {I_1} = 5\,Ns$
$\Rightarrow {I_2} = {F_2} \times \Delta {t_2} \\
\Rightarrow {I_2} = 70 \times 0.1 \\
\Rightarrow {I_2} = 7\,Ns$

The net impulse acting on the football is $I = {I_1} + {I_2} = 5 + 7 = 12\,Ns$. Now, we are given that football gains a velocity of $24\,m{s^{ - 1}}$ , thus $v = 24\,m{s^{ - 1}}$.We know that, The change in momentum of the body is equal to the impulse of the body.
$I = m \times v$
where, $m$ - mass of the football
$m = \dfrac{I}{v} \\
\Rightarrow m = \dfrac{{12}}{{24}} \\
\therefore m = 0.5$ kg
The mass of the football is $5$ kg.

Hence, option D is correct.

Note: The impulse-momentum theorem states that impulse is equal to the total change in momentum of the body. The impulse after two kicks of players is added and net impulse is calculated because the motion is in the same direction. We always use the concept of impulse for a small period of time. The unit of impulse is the same as that of momentum.