3.5.1(d)-(e) - Impulse and force-time graphs

Impulse As Force Acting For Time

Impulse is a way of measuring the effect of a force during a time interval. A large force for a short time can have the same impulse as a smaller force for a longer time, provided the product of force and time is the same.

For a constant force, or for an average force over a stated time interval:

Here $J$ is impulse in $\mathrm{N} \mathrm{s}$, $F$ is force in N, and $\Delta t$ is the time interval in s. Some books use $I$ instead of $J$ for impulse; the physics is the same. Be careful not to confuse the symbol $J$ with the unit joule, because impulse is not energy.

The unit checks the idea:

So impulse has the same base units as momentum. That is a clue that impulse is linked to changing momentum, not to doing work.

Worked example: constant force impulse

A bat exerts an average force of $320\,\mathrm{N}$ on a ball for $0.045\,\mathrm{s}$. Calculate the impulse on the ball.

To two significant figures, the impulse is $14\,\mathrm{N}\,\mathrm{s}$ in the direction of the average force exerted by the bat on the ball.

Keep that product in mind when you read a force-time graph: the same multiplication becomes a physical area.

Impulse And Momentum Change

The previous lesson introduced the momentum form of Newton's second law:

For a finite interval, take $F$ to be the average resultant force and rearrange:

This means the impulse of the resultant force on an object is equal to the object's change in momentum.

Here the mass $m$ is constant and measured in kg, $u$ is initial velocity, and $v$ is final velocity. Use velocities, not just speeds, because direction matters. If right is chosen as positive, then a leftward velocity is negative.

Worked example: signed impulse

A $0.080\,\mathrm{kg}$ ball moves to the right at $12\,\mathrm{m}\,\mathrm{s}^{-1}$. After a short contact, it moves to the left at $8.0\,\mathrm{m}\,\mathrm{s}^{-1}$. Take right as positive. Determine the impulse on the ball.

Initial momentum:

Final momentum:

Change in momentum:

The impulse is $-1.60\,\mathrm{N}\,\mathrm{s}$, or $1.60\,\mathrm{N}\,\mathrm{s}$ to the left. The negative sign is not a problem; it says the impulse is opposite to the positive direction.

If the contact time were $0.020\,\mathrm{s}$, the average force on the ball would be:

The average force is $80\,\mathrm{N}$ to the left.

The safest calculation habit is to write the momentum change before deciding the force direction.

Area Under Force-time Graphs

When a force varies during an interaction, $F\Delta t$ is still useful if $F$ means the average force. A force-time graph gives a more direct method: the impulse is the area between the graph and the time axis.

[DIAGRAM: asset_name: 3.5.1(d)-(e) - Impulse and force-time graphs - diagram 1; asset_slug: 3_05_1b_force_impulse_and_force_time_graphs__diagram_01; recommended_method: drawn_physics; caption: The force-time rectangle has an impulse of 400 N s.; description: A force-time rectangle of height 800 N and width 0.50 s has an impulse of 400 N s.]

The units of graph area confirm the result:

For straight-line sections, split the area into simple shapes:

• rectangle: $\text{area}=\text{base}\times\text{height}$
• triangle: $\text{area}=\frac12\times\text{base}\times\text{height}$
• trapezium: $\text{area}=\frac12\times\text{sum of parallel sides}\times\text{width}$

Worked example: triangular force pulse

A force-time graph for a short impact is triangular. The force rises from zero to a maximum of $650\,\mathrm{N}$ and returns to zero. The whole contact lasts $0.040\,\mathrm{s}$.

The impulse is the area of the triangle:

This is also a momentum change of $13.0\,\mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-1}$ in the direction of the force.

Signed area

If a force-time graph is below the time axis, the force is in the negative direction. The area then represents a negative impulse. If a graph has positive and negative sections, add the signed areas to find the net impulse.

For example, a force of $+5.0\,\mathrm{N}$ for $0.20\,\mathrm{s}$, followed by $-2.0\,\mathrm{N}$ for $0.10\,\mathrm{s}$, gives $J=1.0-0.20=+0.80\,\mathrm{N\,s}$. Adding the magnitudes would wrongly give 1.2 N s.