How to Calculate Average Collision Force from Velocity Change

Two identical trolleys approach a bumper at \(3.0\text{ m s}^{-1}\). Both are brought to rest. The first collision lasts \(0.10\text{ s}\), while the second lasts just \(0.020\text{ s}\).

Which trolley experiences the larger average collision force?

The second one. Both trolleys experience the same change in momentum, but the second experiences it in one-fifth of the time. Its average force is therefore five times greater.

This is the key to finding average collision force from a change in velocity: calculate the change in momentum, then divide by the collision time.

01The calculation: from velocity change to average force

The relationship comes from the impulse-momentum theorem:

\[
F_{\mathrm{net,avg}}=\frac{\Delta p}{\Delta t}
\]

Momentum is mass multiplied by velocity, so:

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

Substituting gives the equation you need:

\[
\boxed{
F_{\mathrm{net,avg}}
=\frac{m(v_f-v_i)}{\Delta t}
}
\]

Here is what each symbol represents:

There are two important conditions to understand.

First, you need the collision duration. Knowing the mass and change in velocity tells you the impulse, but not the average force.

Second, this equation calculates the average resultant force. It gives the average contact force from the bumper when other forces are negligible during the collision. If other forces are significant, you must account for them separately.

Why the direction of velocity matters

A common mistake is to calculate the change in speed rather than the change in velocity.

Suppose a ball travels towards a wall at \(+8.0\text{ m s}^{-1}\), then rebounds at \(-5.0\text{ m s}^{-1}\).

Taking right as positive:

\[
\Delta v=v_f-v_i
\]

\[
\Delta v=-5.0-(+8.0)
\]

\[
\Delta v=-13.0\text{ m s}^{-1}
\]

The change in velocity is \(-13.0\text{ m s}^{-1}\), not \(3.0\text{ m s}^{-1}\).

Why? The ball does not simply slow down by \(3.0\text{ m s}^{-1}\). It reverses direction.

The negative sign tells us that the change in momentum, and therefore the average resultant force, acts to the left.

If signed velocity is still confusing, revisit Average and Instantaneous Velocity in HSC Physics.

02How the force-time graph gives the same answer

There is another way to find average collision force.

Instead of calculating the change in momentum from the initial and final velocities, you can measure the force throughout the collision.

The important relationship is:

\[
\boxed{
\text{Impulse}=\text{signed area under the force-time graph}
}
\]

For the resultant force:

\[
\Delta p=\int_{t_i}^{t_f}F_{\mathrm{net}}(t)\,dt
\]

This means that the area under a force-time graph represents the impulse delivered to the object.

Average force is therefore:

\[
\boxed{
F_{\mathrm{net,avg}}
=\frac{\text{signed area}}{\text{collision duration}}
}
\]

The two methods are connected:

• Change in velocity gives you the change in momentum.
• Change in momentum gives you the impulse.
• The force-time graph gives you the same impulse through its area.
• Dividing impulse by collision duration gives average force.

Average force versus peak force

These are not interchangeable.

Average force is the constant force that would produce the same impulse over the entire collision duration.

Peak force is the largest instantaneous force reached during the collision.

Imagine a force-time pulse that starts at zero, rises to \(240\text{ N}\), and then falls back to zero. Its peak force is \(240\text{ N}\).

However, its average force depends on the area underneath the curve.

For a triangular pulse:

\[
F_{\mathrm{avg}}=\frac{1}{2}F_{\mathrm{peak}}
\]

But this relationship only applies to a triangular pulse acting in one direction. A rectangular or trapezoidal pulse gives a different relationship.

Never assume that the average force is half the peak force unless the graph actually supports that assumption.

Finding the area from experimental data

Sometimes an HSC question provides a force-time graph with numerical readings rather than a simple triangle or rectangle.

You can estimate the area using the trapezoidal rule:

\[
A\approx\sum
\frac{F_i+F_{i+1}}{2}\Delta t
\]

Each pair of adjacent readings forms a trapezoid.

Remember to convert milliseconds into seconds before calculating the area. The resulting impulse must be in \(\text{N s}\).

For a force that remains in one direction, the magnitude of the average force cannot exceed the peak magnitude.

03How an experiment measures average collision force

Consider a trolley travelling along a horizontal, low-friction track towards a padded force sensor mounted at the end.

The force sensor is connected to a data logger, which records force against time. A motion sensor at the opposite end of the track measures the trolley’s position and allows its velocities before and after the collision to be determined.

What should the experiment show?

If the contact force is the only significant horizontal force, the area under the force-time graph should equal the change in momentum calculated from the velocity measurements.

The graph should show a force pulse: the force rises as the bumper compresses, reaches a maximum, and falls as the trolley separates from the bumper.

The highest point is the peak force. The area divided by the contact duration gives the average force.

Why measuring just the maximum force is insufficient

An initial experimental design might use a spring balance to record the largest force during impact.

This gives a peak reading, but it does not tell us the average force. It also does not show the complete force-time relationship.

A better design connects the force sensor to a data logger so the entire collision is recorded. Combining this record with the motion sensor measurements allows the results to be checked independently.

There are a few experimental limitations to consider:

• The force sensor must be calibrated and zeroed.
• The data logger must sample quickly enough to capture the collision.
• The contact duration must be measured, not confused with the total recording time.
• Friction and other external forces should be negligible or accounted for.
• If the sensor records the force exerted by the trolley on the sensor, Newton’s Third Law means its direction is opposite to the force exerted on the trolley.

That final point is particularly important when interpreting signed force-time graphs.

04Worked examples

05A reliable method for HSC questions

When you encounter an average collision force question, use this decision process.

Common mistakes to avoid

• Using speed instead of velocity: Always calculate \(v_f-v_i\), including the signs.
• Forgetting the collision duration: A change in momentum alone gives impulse, not average force.
• Using milliseconds directly: Convert the time into seconds.
• Confusing peak and average force: Peak force is the highest reading; average force comes from impulse divided by time.
• Ignoring other forces: The momentum change gives the impulse from the resultant force, not automatically the impulse from the bumper alone.
• Using the wrong force direction: Check whether the sensor measures force on the trolley or force on the sensor.

06Questions and solutions

Question 1

A \(0.75\text{ kg}\) trolley moves right at \(2.0\text{ m s}^{-1}\). It collides with a bumper and rebounds left at \(1.0\text{ m s}^{-1}\).

The collision lasts \(0.030\text{ s}\).

Calculate the average horizontal resultant force acting on the trolley. Assume other horizontal forces are negligible.

Solution 1

The average resultant force is \(-75\text{ N}\), meaning \(75\text{ N}\) to the left.

Take right as positive.

First, calculate the change in velocity:

\[
\begin{aligned}
\Delta v&=v_f-v_i\\
&=-1.0-2.0\\
&=-3.0\text{ m s}^{-1}
\end{aligned}
\]

Calculate the change in momentum:

\[
\begin{aligned}
\Delta p&=m\Delta v\\
&=0.75(-3.0)\\
&=-2.25\text{ N s}
\end{aligned}
\]

Now divide by the collision duration:

\[
\begin{aligned}
F_{\mathrm{avg}}&=\frac{\Delta p}{\Delta t}\\
&=\frac{-2.25}{0.030}\\
&=-75\text{ N}
\end{aligned}
\]

The negative sign indicates that the force acts opposite to the trolley’s initial motion.

The important detail is that the trolley reverses direction. Its velocity change is \(3.0\text{ m s}^{-1}\) to the left, not \(1.0\text{ m s}^{-1}\).

Question 2

A \(0.20\text{ kg}\) ball travels towards a vertical wall at \(15\text{ m s}^{-1}\) to the right.

Two different collisions are investigated:

• Trial A: The ball hits the wall and comes to rest.
• Trial B: The ball rebounds to the left at \(9.0\text{ m s}^{-1}\).

Both collisions last \(0.012\text{ s}\).

Calculate the average horizontal resultant force in each trial. Which collision produces the greater average force magnitude?

Assume the wall provides the only significant horizontal force.

Solution 2

Trial A produces an average force of \(-250\text{ N}\), while Trial B produces an average force of \(-400\text{ N}\). Trial B has the greater average force magnitude.

Take right as positive.

Trial A: Ball comes to rest

The initial and final velocities are:

\[
v_i=+15\text{ m s}^{-1}
\]

\[
v_f=0
\]

Calculate the change in momentum:

\[
\begin{aligned}
\Delta p&=0.20(0-15)\\
&=-3.0\text{ N s}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
F_{\mathrm{avg}}&=\frac{-3.0}{0.012}\\
&=-250\text{ N}
\end{aligned}
\]

Trial B: Ball rebounds

The final velocity is now:

\[
v_f=-9.0\text{ m s}^{-1}
\]

The change in momentum is:

\[
\begin{aligned}
\Delta p&=0.20(-9.0-15)\\
&=0.20(-24)\\
&=-4.8\text{ N s}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
F_{\mathrm{avg}}&=\frac{-4.8}{0.012}\\
&=-400\text{ N}
\end{aligned}
\]

The rebound produces the greater average force because the ball undergoes a larger change in velocity.

A tempting mistake is to subtract the speeds in Trial B and use \(15-9=6\). However, the ball reverses direction. Its velocity changes by \(24\text{ m s}^{-1}\), not \(6\text{ m s}^{-1}\).

Question 3

A \(1.0\text{ kg}\) trolley moves right at \(4.0\text{ m s}^{-1}\) and rebounds left at \(2.0\text{ m s}^{-1}\).

Two bumper designs are tested:

• Bumper A produces a contact time of \(0.015\text{ s}\).
• Bumper B produces a contact time of \(0.060\text{ s}\).

Both collisions produce the same initial and final velocities.

a) Calculate the average collision force for each bumper.

b) How many times greater is the average force for Bumper A?

c) A student claims that Bumper B must also produce a smaller peak force. Is this necessarily true? Explain.

Assume other horizontal forces are negligible.

Solution 3

Bumper A produces an average force of \(-400\text{ N}\), while Bumper B produces an average force of \(-100\text{ N}\). Bumper A’s average force is four times greater. However, the information does not establish which bumper has the larger peak force.

Step 1: Calculate the change in momentum.

Taking right as positive:

\[
\begin{aligned}
\Delta p&=m(v_f-v_i)\\
&=1.0(-2.0-4.0)\\
&=-6.0\text{ N s}
\end{aligned}
\]

Both bumpers produce the same momentum change.

Step 2: Calculate the average force for Bumper A.

\[
\begin{aligned}
F_{\mathrm{avg,A}}&=\frac{-6.0}{0.015}\\
&=-400\text{ N}
\end{aligned}
\]

Step 3: Calculate the average force for Bumper B.

\[
\begin{aligned}
F_{\mathrm{avg,B}}&=\frac{-6.0}{0.060}\\
&=-100\text{ N}
\end{aligned}
\]

Step 4: Compare the magnitudes.

\[
\frac{400}{100}=4
\]

Bumper A produces four times the average force magnitude.

Step 5: Consider peak force.

The student’s claim is not necessarily correct.

The average force depends on the total area under the force-time curve and the collision duration. Peak force depends on the highest point of that curve.

For example, Bumper B could produce a relatively small force for most of the collision but still have a brief, sharp peak. Its average force could be lower while its peak force is higher.

The velocity changes and contact durations determine the averages, not the exact shapes of the force-time pulses.

Question 4

A \(0.60\text{ kg}\) trolley moves right at \(3.0\text{ m s}^{-1}\) and rebounds left at \(1.0\text{ m s}^{-1}\).

It collides with a force sensor mounted on a wall to the right of the trolley. The collision lasts \(0.020\text{ s}\).

The sensor records the force exerted by the trolley on the sensor. Its positive direction is to the right.

The recorded force-time graph has:

• A signed area of \(+2.4\text{ N s}\).
• A peak reading of \(+180\text{ N}\).

a) Are the force-time data consistent with the trolley’s velocity measurements?

b) Calculate the average contact force exerted on the trolley.

c) State the peak contact force acting on the trolley, including its direction.

Solution 4

Yes, the data are consistent. The average contact force on the trolley is \(-120\text{ N}\), and its peak contact force is \(-180\text{ N}\).

Step 1: Calculate the trolley’s change in momentum.

Taking right as positive:

\[
\begin{aligned}
\Delta p&=m(v_f-v_i)\\
&=0.60(-1.0-3.0)\\
&=-2.4\text{ N s}
\end{aligned}
\]

The impulse on the trolley must therefore be \(-2.4\text{ N s}\).

Step 2: Compare the sensor reading with the momentum change.

The sensor records \(+2.4\text{ N s}\).

However, this is the impulse delivered by the trolley to the sensor, not the impulse delivered by the sensor to the trolley.

Newton’s Third Law states that these forces are equal in magnitude and opposite in direction.

Therefore, the impulse on the trolley is:

\[
J_{\mathrm{trolley}}=-2.4\text{ N s}
\]

This agrees with the momentum change, so the measurements are consistent.

Step 3: Calculate the average contact force on the trolley.

\[
\begin{aligned}
F_{\mathrm{avg}}&=\frac{-2.4}{0.020}\\
&=-120\text{ N}
\end{aligned}
\]

Step 4: Identify the peak contact force.

The sensor records a peak of \(+180\text{ N}\) for the force exerted by the trolley on the sensor.

The force exerted by the sensor on the trolley is equal and opposite:

\[
F_{\mathrm{peak}}=-180\text{ N}
\]

The peak magnitude is \(180\text{ N}\), while the average magnitude is \(120\text{ N}\).

This question tests two separate ideas: Newton’s Third Law determines the direction of the contact force, while the force-time area determines the impulse.

Question 5

A \(0.80\text{ kg}\) trolley travels down a smooth \(30^\circ\) incline towards a very soft bumper at the bottom.

Take down the incline as the positive direction.

The trolley approaches the bumper at \(2.5\text{ m s}^{-1}\), then rebounds up the incline at \(0.50\text{ m s}^{-1}\).

Contact with the bumper lasts \(0.40\text{ s}\).

During the collision, the only significant forces along the incline are:

• The contact force from the bumper.
• The component of gravitational force parallel to the incline.

Friction is negligible.

A calibrated force sensor reports a peak bumper force magnitude of \(12\text{ N}\). No information about the shape of the force-time pulse is provided.

a) Calculate the average resultant force along the incline.

b) Calculate the average contact force exerted by the bumper on the trolley.

c) Is the measured peak force of \(12\text{ N}\) consistent with your result? Explain why assuming a triangular force-time pulse would change your conclusion.

Use \(g=9.8\text{ m s}^{-2}\).

Solution 5

The average contact force is \(-9.92\text{ N}\), acting up the incline. The measured peak of \(12\text{ N}\) is possible, but the available information does not establish the actual force-time profile.

This question has an important extra complication: gravity also acts along the incline during the collision.

Step 1: Calculate the change in momentum.

Taking down the incline as positive:

\[
v_i=+2.5\text{ m s}^{-1}
\]

\[
v_f=-0.50\text{ m s}^{-1}
\]

Therefore:

\[
\begin{aligned}
\Delta p&=m(v_f-v_i)\\
&=0.80(-0.50-2.5)\\
&=-2.4\text{ N s}
\end{aligned}
\]

Step 2: Calculate the average resultant force.

The collision lasts \(0.40\text{ s}\).

\[
\begin{aligned}
F_{\mathrm{net,avg}}&=\frac{\Delta p}{\Delta t}\\
&=\frac{-2.4}{0.40}\\
&=-6.0\text{ N}
\end{aligned}
\]

The average resultant force is \(6.0\text{ N}\) up the incline.

However, this is not the average bumper force.

Step 3: Calculate the gravitational force along the incline.

The component of gravitational force parallel to the slope is:

\[
F_{g,\parallel}=mg\sin\theta
\]

Substituting:

\[
\begin{aligned}
F_{g,\parallel}
&=0.80(9.8)\sin30^\circ\\
&=0.80(9.8)(0.5)\\
&=3.92\text{ N}
\end{aligned}
\]

Gravity acts down the incline, so it is positive.

Step 4: Calculate the average bumper force.

The resultant force is the sum of the contact force and the gravitational component:

\[
F_{\mathrm{net,avg}}
=
F_{\mathrm{bumper,avg}}+F_{g,\parallel}
\]

Rearranging:

\[
F_{\mathrm{bumper,avg}}
=
F_{\mathrm{net,avg}}-F_{g,\parallel}
\]

Substitute:

\[
\begin{aligned}
F_{\mathrm{bumper,avg}}
&=-6.0-3.92\\
&=-9.92\text{ N}
\end{aligned}
\]

The average bumper force is therefore \(9.92\text{ N}\) up the incline.

The tempting approach was to use the change in momentum divided by time and call the result the bumper force. That gives the resultant force, which includes gravity. Because gravity acts throughout the collision, its contribution must be removed.

Step 5: Check whether the peak force is consistent.

The average bumper force magnitude is:

\[
|F_{\mathrm{avg}}|=9.92\text{ N}
\]

The measured peak is:

\[
F_{\mathrm{peak}}=12\text{ N}
\]

A peak greater than the average is physically possible.

We can check this using the required contact impulse:

\[
J_{\mathrm{bumper}}
=(-9.92)(0.40)
\]

\[
J_{\mathrm{bumper}}=-3.968\text{ N s}
\]

If the force remained at its maximum magnitude of \(12\text{ N}\) for the entire \(0.40\text{ s}\), the impulse magnitude would be:

\[
J_{\max}=12(0.40)=4.8\text{ N s}
\]

The required impulse magnitude, \(3.968\text{ N s}\), is less than this maximum. Therefore, a suitable force-time curve with a peak of \(12\text{ N}\) is possible.

However, suppose you assumed the pulse was triangular.

Its impulse magnitude would be:

\[
J=\frac12 F_{\mathrm{peak}}\Delta t
\]

\[
J=\frac12(12)(0.40)
\]

\[
J=2.4\text{ N s}
\]

That is smaller than the required contact impulse of \(3.968\text{ N s}\).

A triangular pulse would therefore be inconsistent with the velocity measurements.

The important conclusion is that the peak force alone does not determine the area under the graph. A relatively flat-topped pulse could produce the required impulse, while a triangular pulse with the same peak and duration could not.

07Where to go next

Once you can calculate average collision force, the next step is understanding how the forces on different objects relate during an interaction.

In particular, Can a Collision Have Zero Net Impulse but a Large Force? explores how large contact forces can produce cancelling impulses, and why the system you choose to analyse matters.

The key distinction to carry into your next collision question is simple: velocity change determines impulse, collision duration determines average force, and the force-time curve reveals the peak.