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**Hint:**Impulse can be defined mathematically as the product of force and time. The Impulse momentum theorem can be gotten from Newton’s second law.

Formula used: In this solution we will be using the following formulae;

\[F = \dfrac{{dp}}{{dt}}\] where \[F\] is the force acting on a body, \[p\] is the momentum of a body, and \[t\] is time, and \[\dfrac{{dp}}{{dt}}\] signifies instantaneous rate of change of momentum.

**Complete Step-by-Step solution:**

Generally, impulse is defined as the product of and time. It is generally used to quantify how long a force acts on a particular body. Its unit in Ns. 1 Ns is defined as the amount of impulse when 1 N of force acts on a body for one second. However, by relation, it is equal to the change in momentum of the body

The impulse – momentum theorem generally states that the impulse applied to a body is equal to the change in momentum of that body. This theorem can be proven from Newton’s law.

According to Newton’s second law, we have that

\[F = \dfrac{{dp}}{{dt}}\] where \[F\] is the force acting on a body, \[p\] is the momentum of a body, and \[t\] is time, and \[\dfrac{{dp}}{{dt}}\] signifies instantaneous rate of change of momentum.

Hence, by cross multiplying, we have

\[Fdt = dp\]

Then, integrating both sides from initial point to final point for both momentum and time, we have

\[\int_0^1 {Fdt} = \int_{{p_0}}^{{P_f}} {dp} \]

Hence, by integrating, we have that

\[Ft = {p_f} - {p_o}\]

\[ \Rightarrow Ft = \Delta p\]

**Hence, the impulse is equal to change in momentum.**

**Note:**Alternately, we can use the constant form of Newton's second law equation. Which can be given as,

\[F = \dfrac{{mv - mu}}{t}\]

Hence, simply by cross multiplying we have

\[Ft = mv - mu\]

Hence, we have that

\[I = mv - mu\] which is the impulse-momentum theorem.

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