How to Calculate Impulse from a Triangular Force-Time Graph

A trolley rolls into a force sensor mounted at the end of a horizontal track. The sensor records a force that rises from \(0\,\text{N}\) to \(600\,\text{N}\), then falls back to zero over \(8.0\,\text{ms}\).

A student multiplies the peak force by the collision time:

\[
J=600\times0.008=4.8\,\text{N s}
\]

There is a problem with that calculation. The trolley did not experience \(600\,\text{N}\) for the entire collision. The force only reached that value at its peak.

The force-time graph is triangular, so you need half the area of the corresponding rectangle:

\[
\boxed{|J|=\frac{1}{2}F_{\text{peak}}\Delta t}
\]

For this collision:

\[
|J|=\frac{1}{2}(600)(0.008)
=2.4\,\text{N s}
\]

The impulse magnitude is \(2.4\,\text{N s}\), not \(4.8\,\text{N s}\). The average force is \(300\,\text{N}\), exactly half the peak force.

This is the method you need for a triangular force-time graph. The important part is recognising which force value belongs in the calculation, using the full collision time, and understanding when the half-peak shortcut is valid.

All numerical data in the examples below are illustrative.

01Why a triangular graph gives half the peak force

A triangular force-time graph has three important features:

• The force begins at zero.
• It increases to a maximum value, called the peak force.
• It decreases back to zero.

The area under the graph is a triangle. Its area is calculated using:

\[
A=\frac{1}{2}\times\text{base}\times\text{height}
\]

For a force-time graph:

• The base is the total time interval, \(\Delta t\), measured in seconds.
• The height is the peak force, \(F_{\text{peak}}\), measured in newtons.

Therefore:

\[
J=\frac{1}{2}F_{\text{peak}}\Delta t
\]

Here, \(J\) represents impulse, measured in newton seconds (\(\text{N s}\)).

This formula gives the magnitude of the impulse. If the graph is below the time axis, the impulse is negative when upward force is defined as positive.

Does the peak have to occur halfway through?

No. This is a useful detail that students sometimes overlook.

Imagine a force that reaches its maximum after \(3.0\,\text{ms}\), then takes another \(5.0\,\text{ms}\) to return to zero.

The total duration is \(8.0\,\text{ms}\), not \(3.0\,\text{ms}\) and not \(5.0\,\text{ms}\).

The area of a triangle depends on its total base and perpendicular height. It does not depend on where the peak sits along the base.

02Peak force versus mean force

The peak force and mean force are not the same quantity.

The peak force is the greatest instantaneous force during the collision. The mean force is the constant force that would produce the same impulse over the same time interval.

To calculate mean force, use:

\[
F_{\text{mean}}=\frac{J}{\Delta t}
\]

For a triangular force-time graph that begins and ends at zero:

\[
F_{\text{mean}}
=\frac{\frac{1}{2}F_{\text{peak}}\Delta t}{\Delta t}
\]

Therefore:

\[
\boxed{F_{\text{mean}}=\frac{F_{\text{peak}}}{2}}
\]

This relationship works even when the peak occurs away from the midpoint of the collision.

For example, if the peak force is \(600\,\text{N}\), the mean force is \(300\,\text{N}\).

You can picture this by flattening the triangular area into a rectangle with the same width. The rectangle must have half the triangle’s height to preserve its area.

However, the trolley does not actually experience a constant \(300\,\text{N}\) force. Its force still rises and falls. The rectangle is simply a useful way to represent the same total impulse.

Why multiplying peak force by time gives the wrong answer

Consider two possible calculations for a triangular pulse with a peak of \(600\,\text{N}\) and a duration of \(0.008\,\text{s}\).

The first calculation assumes the force remains at its maximum value throughout the collision. That would produce a rectangular graph, not a triangular one.

For a zero-to-zero triangular pulse, multiplying peak force by the full duration always gives twice the correct impulse.

03How the force sensor produces the graph

It helps to understand where the triangular shape comes from.

Consider the experimental arrangement:

1. A trolley sits on a horizontal track.
2. A force sensor is securely attached to a fixed support at one end of the track.
3. The trolley moves towards the sensor, with a bumper making contact.
4. The sensor converts the contact force into an electrical signal.
5. A data-logging interface sends the measurements to a computer, which plots force against time.

The sensor measures the force during contact. As the bumper compresses, the force increases. As the bumper unloads, the force decreases.

What should the graph look like?

If the contact force were constant throughout the collision, the graph would be rectangular.

Instead, the force changes as the bumper compresses and releases. An idealised collision may be represented by a triangular pulse.

The straight sides of the triangle are a model of the changing force. Real experimental traces may have rounded peaks, small fluctuations, or noise.

A reliable experiment also needs a correctly zeroed sensor and consistent alignment. If the graph begins above zero before contact, check for a background force or sensor offset before treating the entire trace as a zero-to-zero triangle.

There is another experimental detail worth checking: what force is actually being recorded?

A sensor may report the force exerted by the trolley on the sensor, rather than the force exerted by the sensor on the trolley. These forces are equal in magnitude and opposite in direction, according to Newton’s third law.

Also, the area under a contact-force graph gives the impulse from that contact force. It equals the total momentum change only when other impulses are negligible or have been included.

04Worked method: finding impulse from a triangular graph

Use this sequence in an exam:

1. Identify the beginning and end of the triangular pulse.
2. Calculate the full time interval.
3. Read the peak force from the vertical axis.
4. Convert milliseconds into seconds.
5. Calculate the triangular area.
6. Apply the correct sign if the graph is below the time axis.

Then, if the question asks for mean force, divide the impulse by the total duration.

05When the half-peak formula does not work

There is one important condition behind the shortcut:

The triangular pulse must begin and end at zero force.

Suppose a force-time graph starts at \(40\,\text{N}\), rises to \(280\,\text{N}\), and returns to \(40\,\text{N}\) over \(12\,\text{ms}\).

The changing part is triangular, but the entire region under the graph is actually a trapezium.

You cannot simply use half the peak force.

Worked contrast: A triangular increase above a non-zero baseline

Calculate the impulse and mean force for this graph.

Step 1: Identify the rectangular and triangular areas.

The graph has a constant baseline force of \(40\,\text{N}\).

The additional force rises from \(0\,\text{N}\) to:

\[
280-40=240\,\text{N}
\]

So the total area consists of:

• A rectangle with height \(40\,\text{N}\).
• A triangle with height \(240\,\text{N}\).

Step 2: Convert the duration.

\[
12\,\text{ms}=0.012\,\text{s}
\]

Step 3: Calculate both areas.

The rectangle contributes:

\[
J_1=40(0.012)=0.48\,\text{N s}
\]

The triangle contributes:

\[
J_2=\frac{1}{2}(240)(0.012)
=1.44\,\text{N s}
\]

Step 4: Add the areas.

\[
\begin{aligned}
J&=J_1+J_2\\
&=0.48+1.44\\
&=1.92\,\text{N s}
\end{aligned}
\]

Step 5: Calculate the mean force.

\[
\begin{aligned}
F_{\text{mean}}&=\frac{1.92}{0.012}\\
&=160\,\text{N}
\end{aligned}
\]

The impulse is \(1.92\,\text{N s}\), and the mean force is \(160\,\text{N}\).

Using half the peak force would give \(140\,\text{N}\) and an impulse of \(1.68\,\text{N s}\). Both answers would be incorrect because the calculation would ignore the rectangular baseline.

07Questions and solutions

The following questions progress from a straightforward triangular-area calculation to situations where you need to combine impulse with other ideas.

Question 1

A force sensor records the net force on a small block.

The force-time graph is a triangle that:

• Begins at \(t=1.0\,\text{ms}\) with a force of \(0\,\text{N}\).
• Reaches a peak of \(+500\,\text{N}\) at \(t=3.0\,\text{ms}\).
• Returns to \(0\,\text{N}\) at \(t=9.0\,\text{ms}\).

Calculate:

a) The impulse delivered to the block.

b) The mean force during the interaction.

Take the plotted force direction as positive.

Solution 1

The impulse is \(+2.0\,\text{N s}\), and the mean force is \(+250\,\text{N}\).

Step 1: Find the full duration.

The graph starts at \(1.0\,\text{ms}\) and finishes at \(9.0\,\text{ms}\).

\[
\Delta t=9.0-1.0=8.0\,\text{ms}
\]

Convert to seconds:

\[
\Delta t=0.0080\,\text{s}
\]

Step 2: Calculate the impulse.

\[
\begin{aligned}
J&=\frac{1}{2}F_{\text{peak}}\Delta t\\
&=\frac{1}{2}(500)(0.0080)\\
&=2.0\,\text{N s}
\end{aligned}
\]

The impulse is positive because the graph is above the time axis.

Step 3: Calculate the mean force.

\[
\begin{aligned}
F_{\text{mean}}&=\frac{J}{\Delta t}\\
&=\frac{2.0}{0.0080}\\
&=250\,\text{N}
\end{aligned}
\]

The mean force is half the peak force.

Notice that the peak occurs at \(3.0\,\text{ms}\), even though the total interval is \(8.0\,\text{ms}\). You must use the full base of the triangle.

Question 2

A \(0.30\,\text{kg}\) trolley initially moves to the right at \(4.0\,\text{m s}^{-1}\).

It experiences two successive net-force pulses:

• Pulse A is directed left, lasts \(12\,\text{ms}\), and has a triangular shape with a peak magnitude of \(200\,\text{N}\).
• Pulse B immediately follows pulse A, is directed right, lasts \(8.0\,\text{ms}\), and has a triangular shape with a peak magnitude of \(100\,\text{N}\).

Both pulses begin and end at zero force.

Calculate:

a) The impulse from each pulse.

b) The trolley’s momentum immediately after pulse A.

c) Its final velocity after pulse B.

Explain what happens to the trolley at the transition between the two pulses.

Solution 2

The trolley momentarily comes to rest at the end of pulse A, then finishes moving right at approximately \(1.33\,\text{m s}^{-1}\).

The two pulses produce impulses in opposite directions, so their effects must be combined algebraically.

Step 1: Calculate the initial momentum.

Take right as positive.

\[
\begin{aligned}
p_i&=mv_i\\
&=0.30(4.0)\\
&=1.2\,\text{kg m s}^{-1}
\end{aligned}
\]

Step 2: Calculate the impulse from pulse A.

Convert its duration:

\[
12\,\text{ms}=0.012\,\text{s}
\]

Since pulse A acts left:

\[
\begin{aligned}
J_A&=-\frac{1}{2}(200)(0.012)\\
&=-1.2\,\text{N s}
\end{aligned}
\]

Step 3: Find the momentum after pulse A.

\[
\begin{aligned}
p_A&=p_i+J_A\\
&=1.2-1.2\\
&=0\,\text{kg m s}^{-1}
\end{aligned}
\]

The trolley has zero momentum at the end of the first pulse. Since its mass is non-zero, its velocity is also zero at that instant.

Step 4: Calculate the impulse from pulse B.

Convert its duration:

\[
8.0\,\text{ms}=0.0080\,\text{s}
\]

Pulse B acts right, so:

\[
\begin{aligned}
J_B&=\frac{1}{2}(100)(0.0080)\\
&=+0.40\,\text{N s}
\end{aligned}
\]

Step 5: Calculate the final momentum.

\[
\begin{aligned}
p_f&=p_A+J_B\\
&=0+0.40\\
&=+0.40\,\text{kg m s}^{-1}
\end{aligned}
\]

Step 6: Calculate the final velocity.

\[
\begin{aligned}
v_f&=\frac{p_f}{m}\\
&=\frac{0.40}{0.30}\\
&=1.33\,\text{m s}^{-1}
\end{aligned}
\]

The final velocity is \(1.33\,\text{m s}^{-1}\) to the right.

The important detail is that pulse A exactly removes the trolley’s initial momentum. Pulse B then gives it new rightward momentum.

Comparing the peak forces alone would not reveal this sequence. The two triangular areas determine the actual momentum changes.

Question 3

A force sensor records the net force on a block over a \(12\,\text{ms}\) interval.

The force begins at \(+40\,\text{N}\), rises linearly to \(+280\,\text{N}\) halfway through the interval, then decreases linearly back to \(+40\,\text{N}\).

Calculate:

a) The total impulse.

b) The mean force.

c) Explain why using half the peak force multiplied by the full duration gives an incorrect answer.

Solution 3

The total impulse is \(+1.92\,\text{N s}\), and the mean force is \(+160\,\text{N}\).

The graph is a trapezium, not a zero-to-zero triangle. Its rectangular baseline must be included.

Step 1: Convert the time interval.

\[
\Delta t=12\times10^{-3}
=0.012\,\text{s}
\]

Step 2: Calculate the rectangular contribution.

The baseline force is \(40\,\text{N}\).

\[
\begin{aligned}
J_1&=F\Delta t\\
&=40(0.012)\\
&=0.48\,\text{N s}
\end{aligned}
\]

Step 3: Calculate the additional triangular contribution.

The additional height is:

\[
280-40=240\,\text{N}
\]

Therefore:

\[
\begin{aligned}
J_2&=\frac{1}{2}(240)(0.012)\\
&=1.44\,\text{N s}
\end{aligned}
\]

Step 4: Calculate the total impulse.

\[
\begin{aligned}
J&=J_1+J_2\\
&=0.48+1.44\\
&=1.92\,\text{N s}
\end{aligned}
\]

Step 5: Calculate the mean force.

\[
\begin{aligned}
F_{\text{mean}}&=\frac{J}{\Delta t}\\
&=\frac{1.92}{0.012}\\
&=160\,\text{N}
\end{aligned}
\]

The tempting mistake is to treat the entire graph as a triangle with a height of \(280\,\text{N}\).

That would give:

\[
J=\frac{1}{2}(280)(0.012)
=1.68\,\text{N s}
\]

This misses the area beneath the \(40\,\text{N}\) baseline.

Always check where the graph meets the time axis before using the triangular shortcut.

Question 4

A \(0.80\,\text{kg}\) trolley changes velocity from \(+2.5\,\text{m s}^{-1}\) to \(-1.0\,\text{m s}^{-1}\) over \(14\,\text{ms}\).

During this interval, the net force on the trolley is represented by a single triangular force-time pulse. The force is zero at the beginning and end of the interval and acts entirely to the left.

Calculate:

a) The impulse.

b) The mean net force.

c) The peak force magnitude.

A student claims that the peak must occur at \(7.0\,\text{ms}\) because the collision lasts \(14\,\text{ms}\). Is this conclusion justified?

Solution 4

The impulse is \(-2.8\,\text{N s}\), the mean net force is \(-200\,\text{N}\), and the peak force magnitude is \(400\,\text{N}\). The peak time cannot be determined from the information given.

Step 1: Calculate the initial and final momentum.

Using:

\[
p=mv
\]

The initial momentum is:

\[
p_i=0.80(2.5)
=+2.0\,\text{kg m s}^{-1}
\]

The final momentum is:

\[
p_f=0.80(-1.0)
=-0.80\,\text{kg m s}^{-1}
\]

Step 2: Calculate the impulse.

Impulse is the change in momentum:

\[
J=p_f-p_i
\]

Substitute:

\[
\begin{aligned}
J&=-0.80-2.0\\
&=-2.8\,\text{N s}
\end{aligned}
\]

The negative sign confirms that the momentum change is towards the left.

Step 3: Calculate the mean force.

The duration is:

\[
14\,\text{ms}=0.014\,\text{s}
\]

Therefore:

\[
\begin{aligned}
F_{\text{mean}}&=\frac{J}{\Delta t}\\
&=\frac{-2.8}{0.014}\\
&=-200\,\text{N}
\end{aligned}
\]

The mean net force is \(200\,\text{N}\) to the left.

Step 4: Calculate the peak force.

For a triangular graph:

\[
|J|=\frac{1}{2}F_{\text{peak}}\Delta t
\]

Rearrange:

\[
F_{\text{peak}}=\frac{2|J|}{\Delta t}
\]

Substitute:

\[
\begin{aligned}
F_{\text{peak}}&=\frac{2(2.8)}{0.014}\\
&=400\,\text{N}
\end{aligned}
\]

The peak force magnitude is \(400\,\text{N}\), directed left.

Step 5: Evaluate the student’s claim.

The claim is not justified.

A triangular graph does not have to reach its peak halfway through the interval. The peak could occur at \(4.0\,\text{ms}\), \(7.0\,\text{ms}\), or \(11\,\text{ms}\), provided the graph remains triangular and has the same base and height.

All these possibilities produce the same impulse.

The total area determines the peak height when the base is known. It does not determine the exact time at which the peak occurs.

Question 5

A \(0.50\,\text{kg}\) trolley initially moves to the right at \(2.0\,\text{m s}^{-1}\) along a horizontal track.

It collides with a bumper attached to a force sensor. The sensor records the bumper’s force on the trolley over a \(20\,\text{ms}\) interval.

The measured force is a triangular pulse:

• At \(t=0\), the force is zero.
• At \(t=10\,\text{ms}\), the force reaches \(300\,\text{N}\) to the left.
• At \(t=20\,\text{ms}\), the force returns to zero.

A constant rolling resistance of magnitude \(5.0\,\text{N}\) also acts throughout the collision. It always acts opposite to the trolley’s instantaneous direction of motion.

For this question, assume the resistance reverses direction immediately when the trolley reverses direction. Ignore other horizontal forces.

Calculate:

a) The time at which the trolley first reaches zero velocity.

b) Its final velocity at the end of the \(20\,\text{ms}\) interval.

c) The mean bumper force over the collision.

d) Explain why the area under the sensor’s triangular graph is not, by itself, the total momentum change.

Solution 5

The trolley first reaches zero velocity at \(8.0\,\text{ms}\), then reverses direction. Its final velocity is approximately \(3.96\,\text{m s}^{-1}\) to the left, and the mean bumper force is \(150\,\text{N}\) to the left.

This question has an important complication: the force sensor records the bumper force, but rolling resistance also contributes to the trolley’s momentum change.

More importantly, rolling resistance changes direction when the trolley reverses.

Step 1: Choose a positive direction and calculate initial momentum.

Take right as positive.

\[
\begin{aligned}
p_i&=mv_i\\
&=0.50(2.0)\\
&=+1.0\,\text{kg m s}^{-1}
\end{aligned}
\]

Step 2: Calculate the total bumper impulse.

The bumper force forms a triangle with:

\[
F_{\text{peak}}=-300\,\text{N}
\]

and:

\[
\Delta t=0.020\,\text{s}
\]

Therefore:

\[
\begin{aligned}
J_b&=-\frac{1}{2}(300)(0.020)\\
&=-3.0\,\text{N s}
\end{aligned}
\]

The bumper delivers an impulse of \(3.0\,\text{N s}\) to the left.

Step 3: Find when the trolley first stops.

This is the most important part of the problem.

Initially, the trolley moves right, so rolling resistance acts left.

The bumper force also acts left. Both forces reduce the trolley’s rightward momentum.

During the first \(10\,\text{ms}\), the bumper force increases linearly from zero to \(300\,\text{N}\).

If \(t\) is measured in seconds, the magnitude of the bumper force during this rising section is:

\[
F_b(t)=\frac{300}{0.010}t
\]

Therefore:

\[
F_b(t)=30000t
\]

The impulse from the bumper during the first \(t\) seconds is the area of a smaller triangle:

\[
\begin{aligned}
J_b(t)&=\frac{1}{2}t(30000t)\\
&=15000t^2
\end{aligned}
\]

This impulse acts left.

Rolling resistance also acts left during this initial motion:

\[
J_r(t)=-5.0t
\]

The trolley reaches zero velocity when its momentum becomes zero.

Therefore:

\[
p_i+J_b(t)+J_r(t)=0
\]

Substituting:

\[
1.0-15000t^2-5.0t=0
\]

Rearrange:

\[
15000t^2+5.0t-1.0=0
\]

Using the positive solution of the quadratic:

\[
t=\frac{-5+\sqrt{5^2+4(15000)(1)}}{30000}
\]

\[
t=\frac{240}{30000}
\]

\[
t=0.0080\,\text{s}
\]

Therefore:

\[
\boxed{t=8.0\,\text{ms}}
\]

The trolley first stops at \(8.0\,\text{ms}\), before the bumper force reaches its peak at \(10\,\text{ms}\).

Step 4: Calculate the total rolling-resistance impulse.

The trolley moves right for the first \(8.0\,\text{ms}\), so rolling resistance acts left.

\[
J_{r,1}=-5.0(0.0080)
\]

\[
J_{r,1}=-0.040\,\text{N s}
\]

After \(8.0\,\text{ms}\), the trolley moves left. Rolling resistance now acts right.

The remaining time is:

\[
20-8.0=12\,\text{ms}
\]

Therefore:

\[
J_{r,2}=+5.0(0.012)
\]

\[
J_{r,2}=+0.060\,\text{N s}
\]

The total rolling-resistance impulse is:

\[
\begin{aligned}
J_r&=J_{r,1}+J_{r,2}\\
&=-0.040+0.060\\
&=+0.020\,\text{N s}
\end{aligned}
\]

Notice that the net rolling-resistance impulse is positive, even though the trolley initially moves right.

This happens because it spends longer moving left than right during the collision.

Step 5: Calculate the total impulse and final momentum.

The total impulse includes both the bumper and rolling resistance:

\[
J_{\text{net}}=J_b+J_r
\]

\[
\begin{aligned}
J_{\text{net}}&=-3.0+0.020\\
&=-2.98\,\text{N s}
\end{aligned}
\]

Now use:

\[
p_f=p_i+J_{\text{net}}
\]

\[
\begin{aligned}
p_f&=1.0-2.98\\
&=-1.98\,\text{kg m s}^{-1}
\end{aligned}
\]

Step 6: Calculate the final velocity.

\[
\begin{aligned}
v_f&=\frac{p_f}{m}\\
&=\frac{-1.98}{0.50}\\
&=-3.96\,\text{m s}^{-1}
\end{aligned}
\]

The trolley finishes moving left at \(3.96\,\text{m s}^{-1}\).

Step 7: Calculate the mean bumper force.

The question asks for the mean bumper force, not the mean net force.

\[
\begin{aligned}
F_{\text{mean}}&=\frac{J_b}{\Delta t}\\
&=\frac{-3.0}{0.020}\\
&=-150\,\text{N}
\end{aligned}
\]

The mean bumper force is \(150\,\text{N}\) to the left.

The main trap is treating the bumper’s impulse as the total momentum change. The sensor only records the bumper interaction. Rolling resistance contributes another impulse, and its direction changes when the trolley reverses.

This is why identifying the actual force being graphed matters just as much as calculating the area correctly.

08What to remember in an HSC exam

For a triangular force-time graph that begins and ends at zero:

\[
\boxed{J=\frac{1}{2}F_{\text{peak}}\Delta t}
\]

For the corresponding mean force:

\[
\boxed{F_{\text{mean}}=\frac{F_{\text{peak}}}{2}}
\]

Use the following checks before submitting your answer:

• Full duration: Use the time from the beginning to the end of the pulse, not just the time taken to reach the peak.
• Peak versus mean: The peak is the maximum instantaneous force. The mean is half the peak for a zero-to-zero triangular pulse.
• Time conversion: Convert milliseconds into seconds before calculating.
• Direction: A graph below the time axis gives a negative impulse if upward force is positive.
• Graph shape: Check that the force really starts and ends at zero before using the triangular shortcut.
• Net force: Check whether the graph represents the net force or just one interaction force.

For the broader treatment of signed areas, momentum changes, and other force-time graph shapes, read Force-Time Graphs: Impulse and Momentum Change in HSC Physics.

The next useful step is connecting impulse to conservation of momentum. That lets you track how two objects exchange momentum during a collision, rather than looking at just one object’s force-time graph.