How to Calculate Impact Duration from Momentum and Average Force

A 0.20 kg ball travels towards a wall at 10 m/s and rebounds at 6 m/s in the opposite direction. A force sensor records a peak impact force of 1000 N. Can you use that reading to calculate how long the ball was in contact with the wall?

Not directly. You first need the ball’s change in momentum, then the average net force during the collision. The peak force is not necessarily the average force.

Take the ball’s initial direction as positive. Its change in momentum is:

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

\[ =0.20(-6.0-10.0) =-3.2\ \mathrm{N\,s} \]

Suppose the average net force during contact is 640 N to the left, or \(-640\) N. The impact duration is:

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

\[ =\frac{-3.2}{-640} =0.0050\ \mathrm{s} \]

Therefore, the impact lasts 5.0 ms.

Notice that using the peak force of 1000 N instead would give 3.2 ms. That is too short.

The key is to use the average force over the entire collision, not the largest force recorded at one instant.

01The relationship between momentum, force, and impact duration

The method comes from the impulse-momentum relationship:

\[ J=\Delta p \]

Impulse is the change in an object’s momentum.

For a constant net force, impulse is also:

\[ J=F_{\mathrm{net}}\Delta t \]

Combining these gives:

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

During a collision, however, the force usually changes throughout the impact. It might rise rapidly, reach a maximum, and then decrease as the objects separate.

This means we use the average net force over the collision interval:

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

Rearranging gives the equation you need:

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

For a one-dimensional collision, the momentum change is:

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

Therefore:

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

Here:

• \(\Delta t\) = impact duration, in seconds (s).
• \(m\) = mass of the object, in kilograms (kg).
• \(v_i\) = initial velocity, in metres per second (m/s).
• \(v_f\) = final velocity, in metres per second (m/s).
• \(F_{\mathrm{avg,net}}\) = average net force during impact, in newtons (N).
• \(\Delta p\) = change in momentum, in newton-seconds (N s).

The average force in this equation is the average net force. In many short-collision HSC questions, other forces are assumed negligible, so the average collision force is approximately equal to the average net force. We’ll examine when that assumption needs checking later.

Get the direction right before calculating

A common mistake is to use the difference between the initial and final speeds instead of the difference between the initial and final velocities.

Speed has no direction. Velocity does.

Suppose a ball travels right at 15 m/s and rebounds left at 5 m/s.

Taking right as positive:

\[ v_i=+15\ \mathrm{m/s} \]

\[ v_f=-5\ \mathrm{m/s} \]

The change in velocity is:

\[ \Delta v=-5-15=-20\ \mathrm{m/s} \]

It is not \(5-15=-10\ \mathrm{m/s}\).

The ball has changed direction, so its velocity change is the sum of the two speed magnitudes, with the appropriate sign.

For impact duration, the final answer must be positive. You can use the magnitudes of momentum change and average force once you have established their directions:

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

This magnitude form is particularly useful when a question gives force and momentum change as positive magnitudes.

02Why average force matters more than peak force

Imagine pushing a shopping trolley into a padded barrier.

The force doesn’t necessarily remain constant. It builds as the barrier compresses, reaches a maximum, and decreases as the trolley slows.

A force-time graph captures this changing force.

The important relationship is:

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

The area represents the momentum change.

The average force is the constant force that would produce exactly the same impulse over the same time interval.

Think of replacing a complicated, uneven force pulse with a rectangle. The rectangle must have the same area as the original pulse.

Its height is the average force.

Mathematically:

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

Rearranging:

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

This explains why the peak force cannot automatically be used to calculate impact duration.

Three common force-time pulse shapes

The shape of the force-time graph determines the relationship between peak and average force.

For a triangular pulse, the impulse is:

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

Therefore:

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

This shortcut only works when the force-time pulse is triangular.

For an irregular pulse, you need to determine its area or average force from the actual graph.

A useful comparison: same peak, different impact durations

Consider two collisions that each produce an impulse of 3.0 N s and reach a peak force of 600 N.

In the first collision, the force is constant at 600 N.

\[ \Delta t=\frac{3.0}{600} =0.0050\ \mathrm{s} \]

The impact lasts 5.0 ms.

In the second collision, the force-time graph is triangular, with a peak of 600 N. Its average force is only 300 N.

\[ \Delta t=\frac{3.0}{300} =0.010\ \mathrm{s} \]

The impact lasts 10.0 ms.

Both collisions have the same peak force and the same impulse, but their durations are different.

This is why a peak-force reading alone is insufficient to determine impact duration. You need the average force or enough information about the force-time graph to calculate it.

03How to find impact duration in an experiment

The relationship between momentum and average force can also be tested using a simple collision experiment.

Imagine a low-friction trolley travelling along a horizontal track towards a bumper attached to a force sensor.

The force sensor is connected to a data logger, which records the force on the trolley throughout contact. Two photogates positioned along the track measure the trolley’s speed before and after the collision.

The arrangement works as follows:

1. The trolley moves towards the bumper.
2. It contacts the force sensor, which records a changing force.
3. The trolley slows down and may rebound.
4. The photogates provide the incoming and outgoing speeds. The order in which the trolley passes through the gates identifies its direction of travel.
5. The trolley’s mass and signed velocities give its momentum change.

What should the experiment show?

The impulse-momentum model predicts that the signed area under the force-time graph should equal the trolley’s change in momentum:

\[ J=\Delta p \]

If the trolley experiences a larger impulse, its momentum changes by a larger amount.

If two collisions produce the same momentum change but have different contact durations, the longer collision should have the smaller average net force.

The actual force trace will generally have a peak rather than remain perfectly flat. This is precisely why the complete force-time record is more useful than a single maximum-force reading.

What can go wrong with the measurement?

Suppose the force sensor displays a maximum reading of 500 N. Dividing the momentum change by 500 N assumes that the force remained at 500 N throughout the collision.

That assumption is usually incorrect.

A better design records the complete force-time trace. The average force can then be estimated from the area under the curve divided by the contact duration.

There is another possible problem: a photogate measures speed, but the calculation requires velocity. If the trolley rebounds, its final velocity has the opposite sign to its initial velocity.

A reliable experiment must therefore establish direction as well as speed.

In practice, the sensor’s baseline, sampling rate, and uncertainty in identifying the start and end of contact also affect the result. These limitations matter when comparing experimental impulse with the calculated momentum change.

04A reliable method for exam questions

When a question asks you to calculate collision time, use this sequence.

The main decision is whether the question actually gives you the average force.

• If it gives average net force, use it directly.
• If it gives a peak force and specifies the pulse shape, calculate the average force first.
• If it gives a force-time graph, calculate its signed area or average force.
• If it gives only peak force without the pulse shape, you cannot uniquely determine the impact duration.

Remember that:

\[ 1\ \mathrm{N\,s}=1\ \mathrm{kg\,m\,s^{-1}} \]

And:

\[ 1\ \mathrm{ms}=0.001\ \mathrm{s} \]

A useful extension is to consider what happens when forces act in opposite directions during a collision. Their impulses must be added with signs, rather than treating every force as a positive contribution. This is especially important in situations involving changing force directions and zero net impulse. See Can a Collision Have Zero Net Impulse but a Large Force?.

05Worked examples

06Questions and solutions

The following questions progress from a straightforward application of the impulse-momentum relationship to more challenging situations involving force-time graphs, experimental data, and forces that act throughout a collision.

All numerical values are illustrative.

Question 1

A 0.30 kg handball travels horizontally at 6.0 m/s towards a rigid wall. It comes to rest during the collision.

The average net force acting on the ball during contact is 180 N, directed opposite to its initial motion.

Calculate the impact duration.

Solution 1

The impact duration is 10 ms.

The ball comes to rest, so its final velocity is zero. We can calculate its momentum change and divide by the average net force.

Step 1: Choose a positive direction.

Take the initial direction of motion as positive.

\[ v_i=+6.0\ \mathrm{m/s} \]

\[ v_f=0\ \mathrm{m/s} \]

The average net force is:

\[ F_{\mathrm{avg,net}}=-180\ \mathrm{N} \]

Step 2: Calculate the momentum change.

\[ \begin{aligned} \Delta p&=m(v_f-v_i)\\ &=0.30(0-6.0)\\ &=-1.8\ \mathrm{N\,s} \end{aligned} \]

Step 3: Calculate the duration.

\[ \begin{aligned} \Delta t&=\frac{\Delta p}{F_{\mathrm{avg,net}}}\\ &=\frac{-1.8}{-180}\\ &=0.010\ \mathrm{s} \end{aligned} \]

Therefore:

\[ \boxed{\Delta t=10\ \mathrm{ms}} \]

The negative signs cancel because the average force and momentum change act in the same direction.

Question 2

A 0.16 kg cricket ball approaches a bat horizontally at 25 m/s. After being struck, the ball travels in the opposite direction at 35 m/s.

The average net force on the ball during contact is 960 N, directed opposite to its initial motion.

(a) Calculate the change in momentum of the ball.

(b) Calculate the impact duration.

(c) Explain why calculating the change in velocity as \(35-25=10\ \mathrm{m/s}\) produces an incorrect result.

Solution 2

(a) The change in momentum is \(-9.6\ \mathrm{N\,s}\).

The ball reverses direction, so its final velocity must be negative if the initial direction is positive.

Take the initial direction as positive:

\[ v_i=+25\ \mathrm{m/s} \]

\[ v_f=-35\ \mathrm{m/s} \]

Calculate the momentum change:

\[ \begin{aligned} \Delta p&=m(v_f-v_i)\\ &=0.16(-35-25)\\ &=0.16(-60)\\ &=-9.6\ \mathrm{N\,s} \end{aligned} \]

(b) The impact duration is 10 ms.

The average net force is:

\[ F_{\mathrm{avg,net}}=-960\ \mathrm{N} \]

Using the impulse-momentum relationship:

\[ \begin{aligned} \Delta t&=\frac{\Delta p}{F_{\mathrm{avg,net}}}\\ &=\frac{-9.6}{-960}\\ &=0.010\ \mathrm{s} \end{aligned} \]

Therefore:

\[ \boxed{\Delta t=10\ \mathrm{ms}} \]

(c) The velocity change is 60 m/s in magnitude, not 10 m/s.

The tempting calculation \(35-25=10\) compares the two speed magnitudes. It ignores the fact that the ball travels in opposite directions before and after impact.

The correct calculation is:

\[ \Delta v=-35-(+25)=-60\ \mathrm{m/s} \]

This distinction is essential whenever an object rebounds.

Question 3

A 0.20 kg hockey puck travels horizontally at 8.0 m/s towards a padded barrier and comes to rest.

A force sensor records a triangular force-time pulse during the collision. The force magnitude increases linearly from zero to a peak of 800 N, then decreases linearly back to zero.

Assume the measured force is the net horizontal force on the puck.

(a) Determine the average net force during the collision.

(b) Calculate the impact duration.

(c) Explain what error would result from using the peak force directly in the impact-duration equation.

Solution 3

(a) The average net force magnitude is 400 N.

The force-time pulse is triangular.

The average force magnitude is half the peak force because the area of the triangle is half the area of a rectangle with the same base and height.

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

Substitute:

\[ F_{\mathrm{avg,net}}= \frac{1}{2}(800) \]

\[ \boxed{F_{\mathrm{avg,net}}=400\ \mathrm{N}} \]

The force acts opposite to the puck’s initial motion.

(b) The impact duration is 4.0 ms.

First, calculate the momentum change.

Take the initial direction as positive:

\[ v_i=+8.0\ \mathrm{m/s} \]

\[ v_f=0 \]

Therefore:

\[ \begin{aligned} \Delta p&=m(v_f-v_i)\\ &=0.20(0-8.0)\\ &=-1.6\ \mathrm{N\,s} \end{aligned} \]

The average net force is \(-400\) N.

Now calculate the duration:

\[ \begin{aligned} \Delta t&=\frac{-1.6}{-400}\\ &=0.0040\ \mathrm{s} \end{aligned} \]

Therefore:

\[ \boxed{\Delta t=4.0\ \mathrm{ms}} \]

(c) Using the peak force would underestimate the impact duration.

If you incorrectly used 800 N:

\[ \Delta t=\frac{1.6}{800} =0.0020\ \mathrm{s} \]

This gives 2.0 ms, only half the correct duration.

The mistake is treating the maximum force as though it acted throughout the collision. The force actually varies continuously, so its average is lower than its peak.

Question 4

A 0.40 kg trolley travels along a horizontal track at 3.0 m/s towards a bumper. It rebounds at 1.0 m/s in the opposite direction.

A force sensor records the force exerted by the bumper on the trolley. The following table gives five force readings at equally spaced fractions of the unknown contact duration, \(T\).

Assume:

• The initial direction of motion is positive.
• The force varies linearly between successive readings.
• Friction and other horizontal forces are negligible.
• The force readings are representative of the complete contact interval.

(a) Calculate the change in momentum of the trolley.

(b) Use the trapezoidal rule to estimate the average net force during contact.

(c) Estimate the impact duration.

(d) Explain why using the maximum force reading would give an incorrect duration.

Solution 4

(a) The change in momentum is \(-1.6\ \mathrm{N\,s}\).

Take the initial direction as positive.

The initial velocity is:

\[ v_i=+3.0\ \mathrm{m/s} \]

The final velocity is:

\[ v_f=-1.0\ \mathrm{m/s} \]

Calculate:

\[ \begin{aligned} \Delta p&=m(v_f-v_i)\\ &=0.40(-1.0-3.0)\\ &=0.40(-4.0)\\ &=-1.6\ \mathrm{N\,s} \end{aligned} \]

The negative sign indicates that the momentum change is directed towards the left.

(b) The estimated average net force is \(-275\) N.

The force readings are equally spaced, and the force varies linearly between them.

We can estimate the area under the force-time graph by dividing it into four trapezoids.

The area of a trapezoid is:

\[ A=\frac{1}{2}(F_1+F_2)\Delta t \]

Each time interval has a width of \(0.25T\).

For the first interval:

\[ A_1=\frac{0+(-300)}{2}(0.25T) \]

\[ A_1=-37.5T \]

For the second interval:

\[ A_2=\frac{-300+(-500)}{2}(0.25T) \]

\[ A_2=-100T \]

The third interval has the same area:

\[ A_3=-100T \]

The fourth interval also has:

\[ A_4=-37.5T \]

Add the four areas:

\[ \begin{aligned} J&=A_1+A_2+A_3+A_4\\ &=-37.5T-100T\\ &\quad-100T-37.5T\\ &=-275T \end{aligned} \]

Since the area under the graph represents impulse:

\[ J=F_{\mathrm{avg,net}}T \]

Therefore:

\[ \boxed{F_{\mathrm{avg,net}}=-275\ \mathrm{N}} \]

(c) The estimated impact duration is 5.8 ms.

Use the impulse-momentum relationship:

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

Substitute:

\[ \begin{aligned} \Delta t&=\frac{-1.6}{-275}\\ &=0.00582\ \mathrm{s} \end{aligned} \]

Therefore:

\[ \boxed{\Delta t\approx5.8\ \mathrm{ms}} \]

(d) The maximum force of 500 N is not the average force.

The force reaches a peak magnitude of 500 N, but spends much of the collision at lower values.

Using 500 N directly would give:

\[ \Delta t=\frac{1.6}{500} =0.0032\ \mathrm{s} \]

This predicts 3.2 ms instead of approximately 5.8 ms.

The trapezoidal method gives a better estimate because it accounts for the changing force throughout the collision.

This is also a useful experimental technique when a force sensor records discrete data points rather than a perfectly smooth graph.

Question 5

A 0.50 kg ball falls vertically onto a force plate and rebounds.

Immediately before contact, the ball travels downwards at 3.0 m/s. Immediately after contact, it travels upwards at 1.0 m/s.

A force sensor in the plate records the upward contact force acting on the ball. During the collision, this force:

• Starts at zero.
• Increases linearly to a peak of 100 N.
• Decreases linearly back to zero.
• Has its peak at the midpoint of the contact interval.

Assume air resistance is negligible and gravitational acceleration is \(9.8\ \mathrm{m/s^2}\) downwards.

(a) Calculate the ball’s change in momentum.

(b) Determine the average upward contact force during the collision.

(c) Calculate the impact duration, accounting for all forces acting on the ball.

(d) A student calculates the duration by dividing the momentum change by the average upward contact force. Explain precisely why this approach is incorrect, and calculate the percentage difference between their answer and the correct duration.

Solution 5

(a) The ball’s change in momentum is \(+2.0\ \mathrm{N\,s}\), directed upwards.

Take upwards as positive.

The initial velocity is:

\[ v_i=-3.0\ \mathrm{m/s} \]

The final velocity is:

\[ v_f=+1.0\ \mathrm{m/s} \]

Calculate:

\[ \begin{aligned} \Delta p&=m(v_f-v_i)\\ &=0.50(1.0-(-3.0))\\ &=0.50(4.0)\\ &=+2.0\ \mathrm{N\,s} \end{aligned} \]

The positive result means the ball’s momentum change is upwards.

(b) The average upward contact force is 50 N.

The contact-force graph is triangular.

The average force is half the peak force:

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

Substitute:

\[ F_{\mathrm{avg,contact}} =\frac{1}{2}(100) \]

Therefore:

\[ \boxed{F_{\mathrm{avg,contact}}=50\ \mathrm{N}} \]

This is the average force exerted by the plate on the ball.

However, it is not the average net force.

(c) The impact duration is approximately 44.3 ms.

This is the important complication.

The ball experiences two forces during contact:

• The upward contact force from the plate.
• The downward gravitational force (weight).

The impulse-momentum relationship uses the net force, not just one of the forces acting on the object.

First, calculate the ball’s weight:

\[ W=mg \]

\[ W=0.50(9.8)=4.9\ \mathrm{N} \]

The weight acts downwards, so it is negative under our sign convention.

The average net force is therefore:

\[ F_{\mathrm{avg,net}} =F_{\mathrm{avg,contact}}-W \]

\[ =50-4.9 \]

\[ F_{\mathrm{avg,net}}=45.1\ \mathrm{N} \]

Now use the impulse-momentum relationship:

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

Substitute:

\[ \begin{aligned} \Delta t&=\frac{2.0}{45.1}\\ &=0.04435\ \mathrm{s} \end{aligned} \]

Therefore:

\[ \boxed{\Delta t\approx44.3\ \mathrm{ms}} \]

The impact lasts approximately 44.3 milliseconds.

(d) The student’s calculation ignores the impulse due to gravity.

The student’s approach gives:

\[ \Delta t_{\mathrm{student}} =\frac{2.0}{50} \]

\[ \Delta t_{\mathrm{student}}=0.0400\ \mathrm{s} \]

This is 40.0 ms, which is shorter than the correct duration.

The problem is that the student has treated the average contact force as the average net force.

The plate pushes upwards with an average force of 50 N, but gravity simultaneously pulls downwards with a force of 4.9 N.

Consequently, the average net force is only 45.1 N.

Because the net force is smaller, a longer time is required to produce the same momentum change.

The percentage difference, relative to the correct duration, is:

\[ \text{Percentage difference} = \frac{44.3-40.0}{44.3}\times100 \]

\[ \boxed{\text{Percentage difference}\approx9.7\%} \]

The student’s answer underestimates the impact duration by approximately 9.7%.

This is a useful distinction in experimental questions: a force sensor measures the force exerted by the sensor or contact surface, not automatically the net force on the object.

For a very short collision, gravity may be negligible compared with the contact force. But that is an assumption to justify, not something to apply automatically.

07Where to go next

Finding impact duration is one application of the impulse-momentum relationship. The same reasoning explains why increasing the time over which an object stops can reduce the average force experienced during a collision, provided the momentum change remains the same.

To strengthen the underlying method, work through Force-Time Graphs: Impulse and Momentum Change in HSC Physics. Pay particular attention to calculating signed areas, interpreting non-constant forces, and distinguishing impulse from instantaneous force.

Those skills will help when HSC questions give you a force-time graph rather than directly supplying the average force.