Impulse
Impulse
When a net force acts on a body it will result in an acceleration which alters the motion of the body. A large net force will cause a larger acceleration than a small net force. The total change in motion of the object can be the same if the large and small forces act for different time intervals. The combination of the force and time that it acts is a useful quantity which leads us to define impulse.
Definition: Impulse
Impulse is the product of the net force and the time interval for which the force acts.
\[\text{Impulse}=\vec{F}_{net}·\Delta t\]However, from Newton's Second Law, we know that
\begin{align*} \vec{F}_{net}& = \frac{\Delta \vec{p}}{\Delta t} \\ \therefore \vec{F}_{net}·\Delta t& = \Delta \vec{p} \\& = \text{Impulse}\end{align*}
Therefore we can define the impulse-momentum theorem:
\[\text{Impulse}=\Delta \vec{p}\]
Impulse is equal to the change in momentum of an object. From this equation we see, that for a given change in momentum, \(\vec{F}_{net}\Delta t\) is fixed. Thus, if \(\vec{F}_{net}\) is reduced, \(\Delta t\) must be increased (i.e. a smaller resultant force must be applied for longer to bring about the same change in momentum). Alternatively if \(\Delta t\) is reduced (i.e. the resultant force is applied for a shorter period) then the resultant force must be increased to bring about the same change in momentum.
The graphs below show how the force acting on a body changes with time.
The area under the graph, shaded in, represents the impulse of the body.
Example: Impulse and Change in Momentum
Question
A \(\text{150}\) \(\text{N}\) resultant force acts on a \(\text{300}\) \(\text{kg}\) trailer. Calculate how long it takes this force to change the trailer's velocity from \(\text{2}\) \(\text{m·s$^{-1}$}\) to \(\text{6}\) \(\text{m·s$^{-1}$}\) in the same direction. Assume that the forces acts to the right which is the direction of motion of the trailer.
Step 1: Identify what information is given and what is asked for
The question explicitly gives
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the trailer's mass as \(\text{300}\) \(\text{kg}\),
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the trailer's initial velocity as \(\text{2}\) \(\text{m·s$^{-1}$}\) to the right,
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the trailer's final velocity as \(\text{6}\) \(\text{m·s$^{-1}$}\) to the right, and
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the resultant force acting on the object.
We are asked to calculate the time taken \(\Delta t\) to accelerate the trailer from the \(\text{2}\) to \(\text{6}\) \(\text{m·s$^{-1}$}\). From the Newton's second law,
\begin{align*}\vec{F}_{net}\Delta t& = \Delta \vec{p} \\& = m\vec{v}_{f}-m\vec{v}_{i} \\\Delta t& = \frac{m}{\vec{F}_{net}}\left(\vec{v}_{f}-\vec{v}_{i}\right).\end{align*}Thus we have everything we need to find \(\Delta t\)!
Step 2: Choose a frame of reference
Choose right as the positive direction.
Step 3: Do the calculation and quote the final answer
\begin{align*} \Delta t& = \frac{m}{\vec{F}_{net}}\left(\vec{v}_{f}-\vec{v}_{i}\right) \\ \Delta \text{t}& = \left(\frac{300}{+150}\right)\left(\left(+6 \right)-\left(+2 \right)\right) \\ \Delta \text{t}& = \left(\frac{300}{150}\right)\left(4 \right) \\ \Delta t& = \frac{\left(300\right)\left(+4 \right)}{150} \\ \Delta t& = \text{8}\text{ s}\end{align*}It takes \(\text{8}\) \(\text{s}\) for the force to change the object's velocity from \(\text{2}\) \(\text{m·s$^{-1}$}\) to the right to \(\text{6}\) \(\text{m·s$^{-1}$}\) to the right.
This lesson is part of:
Momentum and Impulse