Calculating impulse of a constant force

Impulse is the change in momentum of an object. Impulse is interchangeably symbolized as change in momentum, \(\Delta p\), or simply \(\vec{J}\). This falls out of the mathematical definition of impulse:

\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vec{J}=\Delta \vec{p}.\)

In investigations of the processes that cause an object's momentum to change, such as cars colliding, or a ball bouncing, impulse is the path to studying the forces involved in the change.

As it is an expression of momentum, an acceptable unit for impulse is the \(\text{Kg m s}^{-1}\), however common units for impulse include the Newton-second (\(\text{Ns}\)).

When a force is applied to an object, no matter what the state of that object's motion is, acceleration will occur. This is in fact Newton's second law, \(F = ma\). The equation for impulse, \(\Delta p,\) can in fact be derived from this law:

\[F=ma, ~a=\frac{\Delta v}{\Delta t} \Rightarrow F=m \frac{\Delta v}{\Delta t}=\frac{m\Delta v}{\Delta t}. \]

Since \(m \Delta v = \Delta p\) and \(F=\frac{\Delta p}{\Delta t}\) see below, we have the following theorem:

\[\Delta p = F \Delta t,\]

where \(\Delta p\) represents impulse, \(F\) represents applied force, and \(\Delta t\) represents the time over which the force is applied.

A force of \(10N\) was applied to an object for \(5s\). Calculate the impulse experienced by the object.

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Let \(\Delta p\) represent impulse, \(F\) force, and \(\Delta t\) the time that the force acted. Then

\[\Delta p = F \Delta t=10\times 5= 50 \text{ (Ns)}. \]

Newton's second law states that the acceleration of an object due to a net force is in the same direction as that force and inversely proportional to the mass of the object, or, mathematically, \(\vec{F}=m\vec{a}\).

Since an acceleration causes a change in velocity, a reformulation of Newton's second law is possible in terms of velocity.

\[F=ma=m \frac{\Delta v}{\Delta t} = \frac{m\Delta v}{\Delta t} = \frac{\Delta p}{\Delta t}\]

\[F=\frac{\Delta p}{\Delta t}\]

The equality of the two formulations of Newton's second law is best demonstrated through a simple example.

A 2 \(\text{kg}\) ball is dropped near the surface of a planet. If \(v(t=5\text{s})=10 \frac{\text(m)}{\text{s}},\) find the average force on the ball throughout its fall.

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Method 1: \(F=ma\)

The problem provides \(m=2\text{kg}\). The acceleration is not given, but since the motion takes place "near the surface of a planet", it is assumed to be constant. Hence, find it using one dimensional kinematics:

\[a=\dfrac{\Delta v}{\Delta t} = \dfrac{v_f-v_i}{\Delta t}=\dfrac{10\dfrac{\text{m}}{\text{s}} - 0\dfrac{\text{m}}{\text{s}}} {5 \text{s}}=2\dfrac{\text{m}}{\text{s}^2}. \]

Finally, evaluate Newton's second law:

\[F=ma=(2\text{kg})(2\dfrac{\text{m}}{\text{s}^2})=4\text{N}.\]

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Method 2: \(F=\frac{\Delta p}{\Delta t}\)

\[F=\frac{\Delta p}{\Delta t}=\dfrac{p_f-p_i}{\Delta t}.\]

Since \(p=mv\),

\[F=\dfrac{mv_f-mv_i}{\Delta t}=m\dfrac{v_f-v_i}{\Delta t}=(2\text{kg})(\dfrac{10\frac{\text{m}}{\text{s}} - 0\frac{\text{m}}{\text{s}}} {5 \text{s}}) =4\text{N}. \]

An object which was initially at rest has a momentum of \(10\text{ Kg ms}^{-1}\) after being influenced by a force for \(2s\). Calculate the magnitude of the force that acted upon the object.

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The \(\text{ Kg ms}^{-1}\) is equivalent to the \(Ns\), so substitute that value, along with the provided time, into \(F=\frac{\Delta p}{\Delta t}\) to obtain:

\[F=\dfrac{\Delta p}{\Delta t}=\dfrac{p_f-p_i}{\Delta t}=\dfrac{10-0}{2}=5\text{N}. \]