Why Impulse Cannot Tell You the Peak Force in a Collision

A force sensor records an impulse of \(6.0\text{ N s}\) during a collision lasting \(0.040\text{ s}\). Can you tell whether the peak force was \(200\text{ N}\), \(300\text{ N}\), or something else?

No. Impulse tells you the area under a force-time graph, not its maximum height. Even when you know the collision duration, you can calculate the mean force, but you cannot determine the exact peak force without additional information about how the force changes during the collision.

This is an important distinction in HSC Physics. Two collisions can produce exactly the same change in momentum while exposing an object to very different maximum forces.

01What impulse tells you, and what it doesn’t

The impulse delivered to an object is the signed area under its net force-time graph:

\[ J=\int_{t_1}^{t_2}F_{\text{net}}(t)\,dt =\Delta p \]

Here:

• \(J\) is impulse, measured in newton seconds (\(\text{N s}\)).
• \(F_{\text{net}}\) is the net force, measured in newtons (\(\text{N}\)).
• \(t\) is time, measured in seconds (\(\text{s}\)).
• \(\Delta p\) is the change in momentum, measured in \(\text{kg m s}^{-1}\).

For a collision lasting a total time \(T\), the mean net force is:

\[ F_{\text{avg}}=\frac{J}{T} \]

This is the constant force that would produce the same impulse over the same time interval.

Peak force is different. It is the greatest magnitude of the net force reached at any instant during the collision.

\[ F_{\text{peak}}=\max |F_{\text{net}}(t)| \]

The difference is easier to see when you compare what each quantity represents.

Think about average speed and maximum speed.

If you travel 100 km in two hours, your average speed is \(50\text{ km/h}\). That doesn’t tell someone whether your maximum speed was 60, 80, or \(110\text{ km/h}\).

You need information about how your speed changed throughout the journey.

The same reasoning applies to force. Impulse gives you the overall effect of the force, but it doesn’t preserve the details of its time history.

There is one useful mathematical result. If the force acts in one direction throughout the collision, its peak magnitude must be at least as large as its mean magnitude:

\[ F_{\text{peak}}\geq |F_{\text{avg}}| \]

But this is a lower bound, not an exact answer.

If you are given impulse without the collision duration, you cannot even calculate the mean force. If you are given both impulse and duration, you still need the force-time profile to determine the peak.

02How two collisions can have the same impulse but different peak forces

Let’s return to the original collision.

Both of the following examples have:

• Collision duration of \(0.040\text{ s}\).
• Impulse of \(6.0\text{ N s}\).
• A force that acts in the same direction throughout contact.

However, their force-time profiles are different.

Collision A: A triangular force pulse

Suppose the force rises uniformly from zero to \(300\text{ N}\), then falls uniformly back to zero over \(0.040\text{ s}\).

The impulse is the triangular area:

\[ J=\frac{1}{2}TF_{\text{peak}} \]

Substituting:

\[ J=\frac{1}{2}(0.040)(300) \]

\[ J=6.0\text{ N s} \]

The peak force is \(300\text{ N}\).

Collision B: A trapezoidal force pulse

Now suppose the force:

• Increases uniformly from zero to \(200\text{ N}\) over \(0.010\text{ s}\).
• Remains at \(200\text{ N}\) for \(0.020\text{ s}\).
• Decreases uniformly to zero over \(0.010\text{ s}\).

Calculate the area using the rectangle and two triangles:

\[ \begin{aligned} J={}&\frac12(0.010)(200)\\ &+(0.020)(200)\\ &+\frac12(0.010)(200) \end{aligned} \]

\[ J=1.0+4.0+1.0 \]

\[ J=6.0\text{ N s} \]

The peak force is only \(200\text{ N}\).

Compare the results

The following values are illustrative, not measurements from an actual collision.

Both collisions produce the same impulse and therefore the same change in momentum.

However, Collision A produces a peak force that is 50% greater than Collision B.

The important point is that knowing the area doesn’t tell you the height of the graph. You need to know how that area is distributed across time.

03How would an experiment actually measure peak force?

Imagine you want to measure the peak force experienced by a trolley during a collision.

There are two ways to approach the experiment. The first measures the momentum change. The second measures the force throughout the collision.

First approach: Measure the change in momentum

Start with a trolley of known mass on a low-friction horizontal track.

Use a motion detector or light gates to measure its velocity immediately before and after it hits a bumper.

You can then calculate:

\[ \Delta p=m(v_f-v_i) \]

Since impulse equals the change in momentum:

\[ J=\Delta p \]

This gives you the total impulse delivered during the collision.

But there is a problem.

The velocity measurements tell you the initial and final states. They don’t tell you exactly what happened between those measurements.

The force might have increased gradually, reached one sharp peak, or fluctuated several times.

All of these possibilities could produce the same momentum change.

The original measurement therefore cannot answer the peak-force question.

Revised approach: Measure the force throughout the collision

Now add a force sensor.

Attach the sensor to a rigid support at the end of the track. Connect it to a data logger, then allow the trolley to collide with the sensor.

The sensor records the force at closely spaced time intervals, producing a force-time graph.

The revised experiment provides information that the velocity-only method cannot.

You can now identify:

1. The impulse from the area under the recorded force-time graph.
2. The collision duration from the width of the force pulse.
3. The measured peak force from the highest force recorded.

You can also compare the area under the graph with the momentum change calculated independently from the trolley’s velocities.

If the two impulse measurements agree within experimental uncertainty, this provides a useful check on the results.

For this horizontal arrangement, assume friction is negligible during the short collision. The measured horizontal contact force can then be treated as approximately equal to the net horizontal force.

One experimental limitation: Sampling rate

There is still something to watch.

A data logger does not measure force continuously. It records values at discrete time intervals.

Suppose the actual force briefly reaches \(800\text{ N}\) between two measurements, but the logger records \(500\text{ N}\) immediately before and after that instant.

The recorded maximum could be only \(500\text{ N}\), even though the true peak was higher.

This is why the sampling rate and sensor response matter. The equipment must be fast enough to resolve the changes in force during the collision.

A force-time graph with too few measurements can underestimate the actual peak.

The experiment has therefore revealed an important distinction:

• Measuring momentum change determines impulse.
• Measuring collision duration allows you to determine mean force.
• Recording the force throughout the collision allows you to estimate peak force, provided the measurement system has sufficient time resolution.

04When can you calculate peak force from impulse?

You can calculate peak force if you are given additional information about the shape of the force-time graph.

For example, consider three idealised force profiles.

These relationships come from the geometry of each graph.

For a triangular pulse:

\[ J=\frac12 TF_{\text{peak}} \]

Since \(F_{\text{avg}}=J/T\):

\[ F_{\text{peak}}=2F_{\text{avg}} \]

But be careful. You cannot automatically assume a collision produces a triangular force-time graph.

A triangular model is useful when the question specifies that shape or provides enough information to justify it. It is not a universal property of collisions.

A useful HSC decision process

When a question asks you to calculate force from impulse, work through these decisions.

One final detail: if the force changes direction, impulse is the signed area, not the sum of the magnitudes of all areas. The mean force is also a signed time average. Peak force, however, refers to the greatest force magnitude.

For a more detailed explanation of signed areas and force-time graph calculations, see Force-Time Graphs: Impulse and Momentum Change in HSC Physics.

05Worked examples

06Questions and solutions

Question 1

A net force acts in one direction on an object for \(0.030\text{ s}\). The impulse delivered is \(3.0\text{ N s}\).

a) Calculate the mean net force.

b) Can the exact peak force be determined from this information? Explain.

Solution 1

The mean net force is \(100\text{ N}\), but the exact peak force cannot be determined.

Part a

Use:

\[ F_{\text{avg}}=\frac{J}{T} \]

Substitute:

\[ F_{\text{avg}}=\frac{3.0}{0.030} \]

\[ F_{\text{avg}}=100\text{ N} \]

Part b

The mean force is the height of a rectangle with the same area as the actual force-time graph.

The actual graph could have a relatively broad peak or a much narrower, taller peak.

For a force that remains in one direction, the peak must be at least \(100\text{ N}\), but its exact value cannot be established without knowing the force-time profile.

The key distinction is that impulse and duration determine mean force, not peak force.

Question 2

Two objects undergo collisions lasting \(0.050\text{ s}\). Both experience an impulse of \(7.5\text{ N s}\).

The force-time profiles are:

• Collision A: A triangular pulse, starting and ending at zero.
• Collision B: A trapezoidal pulse, rising uniformly from zero over \(0.010\text{ s}\), remaining constant for \(0.030\text{ s}\), and falling uniformly to zero over the final \(0.010\text{ s}\).

Calculate the mean force and peak force for each collision.

Explain why the two collisions produce different peak forces despite having identical impulses and durations.

Solution 2

Collision A has a peak force of \(300\text{ N}\), while Collision B has a peak force of \(187.5\text{ N}\). Both have a mean force of \(150\text{ N}\).

Step 1: Calculate the mean force for both collisions.

They have the same impulse and duration:

\[ F_{\text{avg}}=\frac{7.5}{0.050} \]

\[ F_{\text{avg}}=150\text{ N} \]

Step 2: Calculate the peak force for Collision A.

For a triangular graph:

\[ J=\frac12 TF_{\text{peak}} \]

Rearrange:

\[ F_{\text{peak}}=\frac{2J}{T} \]

Substitute:

\[ F_{\text{peak}}= \frac{2(7.5)}{0.050} \]

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

Step 3: Calculate the peak force for Collision B.

Let its peak force be \(P\).

The trapezoidal area consists of two triangles and one rectangle:

\[ J=\frac12(0.010)P +(0.030)P +\frac12(0.010)P \]

Collecting terms:

\[ 7.5=0.040P \]

Therefore:

\[ P=\frac{7.5}{0.040} \]

\[ P=187.5\text{ N} \]

Step 4: Compare the results.

The tempting assumption is that identical impulses and durations must produce identical peak forces.

However, those two quantities only determine the area and mean height. The different shapes distribute the area differently, producing different maximum heights.

Question 3

A \(0.25\text{ kg}\) trolley moves to the right at \(8.0\text{ m s}^{-1}\). It collides with a bumper and rebounds to the left at \(4.0\text{ m s}^{-1}\).

The contact lasts \(0.020\text{ s}\).

Take right as positive. Assume the bumper provides the only significant horizontal force during the collision and that this force acts to the left throughout contact.

a) Calculate the impulse delivered to the trolley.

b) Calculate the mean net force.

c) Explain why the peak force cannot be determined from your answers to parts a) and b).

Solution 3

The impulse is \(-3.0\text{ N s}\), and the mean net force is \(-150\text{ N}\). The exact peak force cannot be determined from these values.

Step 1: Establish the sign convention.

Right is positive, so:

\[ v_i=+8.0\text{ m s}^{-1} \]

\[ v_f=-4.0\text{ m s}^{-1} \]

The negative final velocity represents motion to the left after the rebound.

Step 2: Calculate the change in momentum.

Use:

\[ \Delta p=m(v_f-v_i) \]

Substitute:

\[ \Delta p=0.25(-4.0-8.0) \]

\[ \Delta p=0.25(-12) \]

\[ \Delta p=-3.0\text{ kg m s}^{-1} \]

Since impulse equals change in momentum:

\[ J=-3.0\text{ N s} \]

The negative sign indicates that the impulse acts to the left.

Step 3: Calculate the mean net force.

\[ F_{\text{avg}}=\frac{J}{T} \]

\[ F_{\text{avg}}=\frac{-3.0}{0.020} \]

\[ F_{\text{avg}}=-150\text{ N} \]

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

Step 4: Explain why the peak is unknown.

The calculation gives the average force throughout the collision.

The actual force could increase well above \(150\text{ N}\) for a short period, then decrease again. Alternatively, it could remain relatively close to its mean value for much of the collision.

Both possibilities can produce the same impulse.

Since the force-time profile has not been provided, the exact peak force cannot be established.

Question 4

A force sensor records the following readings during a collision.

Assume the force changes linearly between consecutive readings.

a) Calculate the impulse.

b) Calculate the mean force.

c) Determine the peak force of the reconstructed force-time graph.

d) Explain why the largest recorded value might not equal the true peak force in an actual experiment.

Solution 4

The impulse is \(4.0\text{ N s}\), the mean force is \(250\text{ N}\), and the reconstructed peak is \(500\text{ N}\). However, the actual experimental peak may be higher.

Step 1: Convert the time intervals into seconds.

The collision lasts:

\[ T=16\text{ ms}=0.016\text{ s} \]

Each interval is:

\[ \Delta t=0.004\text{ s} \]

Step 2: Calculate the four trapezium areas.

Use:

\[ A=\frac{F_1+F_2}{2}\Delta t \]

First interval:

\[ A_1=\frac{0+300}{2}(0.004) =0.60\text{ N s} \]

Second interval:

\[ A_2=\frac{300+500}{2}(0.004) =1.60\text{ N s} \]

Third interval:

\[ A_3=\frac{500+200}{2}(0.004) =1.40\text{ N s} \]

Fourth interval:

\[ A_4=\frac{200+0}{2}(0.004) =0.40\text{ N s} \]

Step 3: Calculate total impulse.

\[ \begin{aligned} J&=0.60+1.60+1.40+0.40\\ &=4.0\text{ N s} \end{aligned} \]

Step 4: Calculate mean force.

\[ F_{\text{avg}}=\frac{4.0}{0.016} \]

\[ F_{\text{avg}}=250\text{ N} \]

Step 5: Determine the peak of the reconstructed graph.

Under the piecewise-linear assumption, the highest force is:

\[ F_{\text{peak}}=500\text{ N} \]

Step 6: Consider the experimental limitation.

The sensor only records values at specific times.

The actual force might briefly rise above \(500\text{ N}\) between two measurements before falling again.

Therefore, \(500\text{ N}\) is the peak of the reconstructed graph, but it is not necessarily the true peak experienced during the collision.

A sufficiently high sampling rate and suitable sensor response are needed to measure the actual peak reliably.

Question 5

A student is investigating a collision lasting \(0.040\text{ s}\).

An independent momentum measurement establishes that the impulse delivered to the object is exactly \(6.0\text{ N s}\).

A data logger records the following force measurements:

The student joins consecutive readings with straight lines and concludes that the peak force was \(200\text{ N}\).

The actual force is known to be continuous and nonnegative, but its behaviour between measurements has not been established.

a) Calculate the mean force.

b) Explain why the student’s conclusion about the peak force is not necessarily correct.

c) Construct an alternative continuous force-time profile that passes through every recorded point, produces the same impulse, but has a peak force of \(300\text{ N}\).

d) Explain what additional experimental information would be needed to distinguish between the two profiles.

Solution 5

The mean force is \(150\text{ N}\), but the student’s conclusion that the peak must be \(200\text{ N}\) is not justified. A different continuous force-time profile can satisfy every measurement and have a peak of \(300\text{ N}\).

This question tests a subtle distinction: even knowing the exact impulse and several force measurements does not necessarily determine what happened between those measurements.

Step 1: Calculate the mean force.

The impulse is:

\[ J=6.0\text{ N s} \]

The duration is:

\[ T=0.040\text{ s} \]

Therefore:

\[ F_{\text{avg}}=\frac{6.0}{0.040} \]

\[ F_{\text{avg}}=150\text{ N} \]

The mean force is \(150\text{ N}\).

Step 2: Calculate the area of the student’s reconstructed graph.

The first section is a triangle:

\[ A_1=\frac12(0.010)(200) =1.0\text{ N s} \]

The middle section is a rectangle lasting \(0.020\text{ s}\):

\[ A_2=(0.020)(200) =4.0\text{ N s} \]

The final section is another triangle:

\[ A_3=\frac12(0.010)(200) =1.0\text{ N s} \]

Therefore:

\[ J=1.0+4.0+1.0 \]

\[ J=6.0\text{ N s} \]

The straight-line reconstruction is consistent with the measured impulse.

Its peak is \(200\text{ N}\).

However, this only proves what the peak would be if the force changed linearly between the recorded points.

The experiment has not established that assumption.

Step 3: Construct a different force-time profile.

Start with the student’s reconstructed graph.

Now make two changes between the recorded measurement times.

First, between \(10\text{ ms}\) and \(20\text{ ms}\), introduce a narrow triangular peak.

• The peak occurs at \(15\text{ ms}\).
• The force increases from \(200\text{ N}\) to \(300\text{ N}\).
• It then decreases back to \(200\text{ N}\).
• The triangular increase has a base of \(4\text{ ms}\).

The additional area is:

\[ \Delta A_1= \frac12(0.004)(100) \]

\[ \Delta A_1=0.20\text{ N s} \]

The force now reaches \(300\text{ N}\), but the total impulse has increased by \(0.20\text{ N s}\).

We need to compensate for that increase.

Second, between \(20\text{ ms}\) and \(30\text{ ms}\), introduce a triangular dip.

• The force starts at \(200\text{ N}\).
• It decreases to \(160\text{ N}\) at \(25\text{ ms}\).
• It returns to \(200\text{ N}\) at \(30\text{ ms}\).

The area removed is:

\[ \Delta A_2= \frac12(0.010)(40) \]

\[ \Delta A_2=0.20\text{ N s} \]

The additional area from the peak is exactly cancelled by the area removed by the dip.

Therefore:

\[ J_{\text{new}} =6.0+0.20-0.20 \]

\[ J_{\text{new}}=6.0\text{ N s} \]

Step 4: Check the measurement conditions.

The alternative profile:

• Passes through every recorded measurement.
• Remains continuous.
• Never produces a negative force.
• Has exactly the same total impulse.
• Reaches a peak of \(300\text{ N}\).

It therefore satisfies every condition in the question.

The original profile has a peak of \(200\text{ N}\), while the alternative has a peak of \(300\text{ N}\).

Both are consistent with the available evidence.

Step 5: Identify the missing experimental information.

To distinguish between these profiles, the experiment needs more information about how the force behaves between the existing measurements.

A faster data logger with an appropriate sensor response could record additional points during the collision, revealing the narrow peak and compensating dip.

Alternatively, a justified physical model could constrain how quickly the force is able to change.

Without such information, the peak cannot be uniquely determined.

This is the deeper reason impulse alone cannot reveal peak force: calculating an area does not reconstruct the unique shape of the curve that produced it.

07What to remember for HSC Physics

When a question involves impulse and peak force, keep these three quantities separate:

\[ J=\Delta p \]

\[ F_{\text{avg}}=\frac{J}{T} \]

\[ F_{\text{peak}}=\text{maximum force magnitude} \]

The first relationship gives the change in momentum. The second gives the mean force over the collision. The third requires information about the actual force-time profile.

Never assume that mean force equals peak force, or that a triangular graph can be assumed without justification.

The next useful application is understanding how cushioning, crumple zones, and airbags change the force-time profile. They can produce the same required momentum change over a longer collision time, reducing mean force and often reducing peak force as well.

For that application, see How to Compare Airbag Designs Using Force-Time Graphs.