Question

When a ball of mass $5$$kg$ hits a bat with a velocity of $3$$m{s^{ - 1}}$, in a positive direction and it moves back with a velocity of $4$$m{s^{ - 1}}$, find the impulse in SI units.
(A) $5$
(B) $15$
(C) $25$
(D) $35$

Answer 1

Hint: Impulse is used to describe the force acting over a period of time to bring a change in the momentum. As the ball hits the bat, the momentum of the ball will change. Think of what can be used to find the impulse as defined above, what are the quantities that can describe impulse.

Complete step by step answer:
The impulse is defined as the integral of the force applied on the body over the time for which it is applied. Impulse is denoted by $J$. Mathematically,
$J = \int {Fdt} $
But, as force is a vector quantity, impulse is also a vector quantity.
Therefore, impulse generated in a time interval from ${t_1}$ to ${t_2}$ in vector form is given by
${\mathbf{J}} = \int\limits_{{t_1}}^{{t_2}} {{\mathbf{F}}dt} $

Now, the second law of motion is mathematically given as \[{\mathbf{F}} = m{\mathbf{a}}\], since, acceleration is the rate of change of velocity with respect to time, that is \[{\mathbf{a}}\] is derivative of velocity with respect to time.
${\mathbf{a}} = \dfrac{{d{\mathbf{v}}}}{{dt}}$
Substituting value of acceleration in the force equation, we have
${\mathbf{F}} = m\dfrac{{d{\mathbf{v}}}}{{dt}}$
Usually we deal with cases where mass is constant and therefore, we can take the mass $m$ inside of the derivative,
$
{\mathbf{F}} = \dfrac{{d\left( {m{\mathbf{v}}} \right)}}{{dt}} \\
\Rightarrow{\mathbf{F}} = \dfrac{{d{\mathbf{p}}}}{{dt}} \\
 $
$d{\mathbf{p}} = {\mathbf{F}}dt$, here, $d{\mathbf{p}}$ is the change in momentum due to the force. Substituting this value in the equation of impulse, we get,
${\mathbf{J}} = \int\limits_{{{\mathbf{p}}_1}}^{{{\mathbf{p}}_2}} {d{\mathbf{p}}} $
$\Rightarrow {\mathbf{J}} = {{\mathbf{p}}_2} - {{\mathbf{p}}_1}$
Therefore, impulse is the change brought in the momentum when a force acts on a body over an interval of time.
Since, our problem is one dimensional, we can write the impulse simply as $J = {p_2} - {p_1}$
In our case,
${p_2} = m{v_f} \\
\Rightarrow{p_2} = \left( 5 \right)\left( { - 4} \right) \\
\Rightarrow {p_2} = - 20\,Ns$
$\Rightarrow {p_1} = m{v_i}
\Rightarrow {p_1} = \left( 5 \right)\left( 3 \right)
\Rightarrow {p_1}= 15\,Ns$
Therefore, $J = \left( { - 20} \right) - 15 = - 35\,Ns$. The magnitude of the impulse will be $35Ns$.Hence, when a ball of mass $5\,kg$ hits a bat with a velocity of $3m{s^{ - 1}}$, in a positive direction and moves back with a velocity of $4m{s^{ - 1}}$, the impulse will be $35\,Ns$ in the negative direction.

Hence,Option D is correct.

Note: Here, as the ball is moving with velocity of $3$$m{s^{ - 1}}$ to hit the bat, the direction of the velocity is positive. After hitting the bat, the ball travels in the opposite direction or the negative direction with a velocity of $4$$m{s^{ - 1}}$. Hence keep in mind the sign of the velocity both before and after the impact.