Why Rebounding Produces a Larger Impulse Than Sticking

A rubber ball hits a wall. In one collision, it stops. In another, it bounces backwards. Both balls started with exactly the same velocity.

Which collision produces the larger impulse?

The rebound.

The reason is that stopping a ball removes its original momentum. Rebounding must do something extra: it must remove the original momentum and give the ball momentum in the opposite direction.

This means the change in momentum, and therefore the magnitude of the impulse, is greater when an object rebounds than when it simply stops.

But there is a second question worth investigating. Does a larger impulse always mean a larger force? And what happens if the rebound lasts longer than the stopping collision?

To answer that, we need to look at the area, width, and height of a force-time graph.

01Why rebounding produces a larger impulse

Let’s start with the momentum of the ball.

Momentum is calculated using

\[ p=mv \]

where:

• \(p\) is momentum, measured in \(\text{kg m s}^{-1}\)
• \(m\) is mass, measured in kilograms (\(\text{kg}\))
• \(v\) is velocity, measured in metres per second (\(\text{m s}^{-1}\))

Remember that velocity includes direction. This becomes particularly important when an object reverses direction during a collision.

Impulse is equal to the change in momentum:

\[ J=\Delta p=p_f-p_i \]

Here, \(J\) is impulse, measured in newton seconds (\(\text{N s}\)), while \(p_i\) and \(p_f\) are the initial and final momenta.

Let’s compare two collisions involving the same ball.

Take motion towards the wall as the positive direction.

Case 1: The ball sticks to the wall

The ball initially travels towards the wall at \(6.0\text{ m s}^{-1}\). It finishes at rest.

Its velocity changes from \(+6.0\text{ m s}^{-1}\) to zero.

The magnitude of its momentum change is

\[ |\Delta p|=m(6.0) \]

Case 2: The ball rebounds

The ball initially travels towards the wall at \(6.0\text{ m s}^{-1}\), but rebounds at \(4.0\text{ m s}^{-1}\).

Its velocity changes from \(+6.0\text{ m s}^{-1}\) to \(-4.0\text{ m s}^{-1}\).

The magnitude of its momentum change is

\[ |\Delta p|=m(6.0+4.0) \]

Notice the addition.

The final velocity is negative because the ball travels in the opposite direction. When calculating the change in velocity, we subtract the initial velocity from the final velocity:

\[ \Delta v=-4.0-6.0=-10.0\text{ m s}^{-1} \]

The rebound therefore produces a larger momentum change.

The general relationship

We can describe both situations mathematically.

Let:

• \(m\) be the mass of the ball
• \(u\) be its initial speed towards the wall
• \(w\) be its rebound speed away from the wall

Take towards the wall as positive.

For a ball that sticks:

\[ J_{\text{stick}}=m(0-u) \]

\[ J_{\text{stick}}=-mu \]

For a ball that rebounds:

\[ J_{\text{rebound}}=m(-w-u) \]

\[ J_{\text{rebound}}=-m(u+w) \]

Comparing the magnitudes gives

\[ \boxed{|J_{\text{rebound}}|=m(u+w)} \]

and

\[ \boxed{|J_{\text{stick}}|=mu} \]

Therefore, the ratio is

\[ \boxed{ \frac{|J_{\text{rebound}}|} {|J_{\text{stick}}|} =1+\frac{w}{u} } \]

This gives us an important result.

• If the ball sticks, its momentum change has magnitude \(mu\).
• If the ball rebounds at any non-zero speed, its momentum change is larger.
• If the ball rebounds at its original speed, the impulse magnitude doubles.

A rebound does not automatically double the impulse. It doubles only when the rebound speed equals the initial speed.

02How the force-time graph explains the difference

We know that impulse is the change in momentum. But how does this appear on a force-time graph?

The important relationship is

\[ \boxed{J=\text{signed area under a force-time graph}} \]

The area accounts for both the size of the force and how long it acts.

For a changing force, the impulse is

\[ J=\int F_{\text{net}}\,dt \]

You do not need calculus to interpret the ordinary HSC force-time graphs. You can calculate their areas using familiar shapes such as triangles, rectangles, and trapeziums.

There is one detail to watch.

If we take towards the wall as positive, the force exerted by the wall on the ball points in the negative direction. Its graph therefore lies below the time axis.

The area is negative, representing a negative impulse. When comparing the sizes of two impulses, we compare the magnitudes of their areas.

If a graph instead displays force magnitude, both collision pulses can be drawn above the time axis, and their areas represent impulse magnitudes.

Peak force versus average force

This is where students often make a mistake.

A larger impulse means a larger area under the force-time graph. It does not automatically mean a larger peak force.

There are three quantities to distinguish.

The average force is calculated using

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

where:

• \(F_{\text{avg}}\) is average net force, in newtons (\(\text{N}\))
• \(J\) is signed impulse, in newton seconds (\(\text{N s}\))
• \(\Delta t\) is collision duration, in seconds (\(\text{s}\))

When comparing magnitudes, use

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

The average force can be pictured as the height of a rectangle with the same area and duration as the actual force-time graph.

The peak force is different. It is simply the maximum magnitude reached during the collision.

A useful way to remember the distinction:

• Area tells you how much momentum changes.
• Average force tells you how quickly that change occurs over the whole collision.
• Peak force tells you the greatest force reached at any instant.

A collision can therefore have a larger impulse but a smaller average force if it lasts sufficiently longer.

Its peak force can also be smaller, depending on the shape of the force-time curve.

For the same collision duration, a larger impulse must produce a larger average force magnitude. However, the peak still depends on the shape of the force-time pulse.

03How an experiment could test this

Imagine setting up a controlled experiment to compare a ball sticking to a target with a ball rebounding from it.

The aim is to measure two things independently:

1. The change in momentum of the ball.
2. The area under the force-time graph during contact.

If both methods agree, we have a useful check that our model describes the collision.

Setting up the apparatus

A suitable arrangement uses:

• A ball moving along a horizontal, low-friction track.
• A force sensor mounted behind a target at the end of the track.
• A data logger connected to the force sensor.
• A high-speed camera positioned beside the track, with a scale for measuring the ball’s velocity.

The target has interchangeable surfaces.

One is a putty pad, designed to make the ball stick. The other is a springy pad, designed to make the ball rebound.

The experiment follows a simple sequence.

First, release the same ball towards the putty pad at a controlled initial speed.

Next, record the force-time trace during contact. Use the camera to determine the ball’s velocity immediately before and after the collision.

Finally, replace the putty with the springy pad and repeat the collision using the same ball and initial speed.

The prediction is that the rebound will produce a greater impulse magnitude because the ball finishes with momentum in the opposite direction.

But we cannot assume that it will produce a higher peak force.

Why measuring only peak force is not enough

A first attempt might simply compare the maximum readings from the force sensor.

That would miss an important part of the experiment.

Imagine the springy pad produces a lower peak force, but the force acts for twice as long. Its force-time graph could still have a larger area.

The improved method is to record the entire force-time trace and calculate its area. We then compare this with the momentum change calculated independently from the camera measurements.

There are also important experimental controls:

• Use the same ball to keep mass constant.
• Maintain the same incoming velocity.
• Keep the target position and sensor arrangement unchanged.
• Record the complete collision, not just the maximum force.
• Sample the force quickly enough to resolve the short collision.

One further detail: the sensor measures the force exerted by the ball on the target. By Newton’s third law, the target exerts an equal-magnitude, opposite-direction force on the ball. The impulse magnitudes are therefore equal, but their signs are opposite.

The following example uses illustrative, idealised data rather than results from an actual experiment.

04A useful decision rule for HSC questions

When comparing a sticking collision with a rebound, work through these decisions in order.

1. Check the final velocity. Does the object stop, or does it reverse direction?
2. Calculate the change in momentum. Use \(J=m(v_f-v_i)\), keeping the signs.
3. Compare impulse magnitudes. A rebound from a fixed target produces a greater impulse magnitude than stopping, provided the rebound speed is non-zero.
4. Check the collision duration. A larger impulse does not automatically mean a larger average force.
5. Inspect the peak of the graph. Do not confuse the highest force with the total area.

For more practice with the underlying momentum relationship, see Impulse and Change in Momentum for HSC Physics.

For further practice interpreting areas and shapes, see Force-Time Graphs: Impulse and Momentum Change in HSC Physics.

05Questions and solutions

Question 1

A \(0.12\text{ kg}\) ball travels towards a fixed wall at \(5.0\text{ m s}^{-1}\). It hits the wall and sticks.

Take motion towards the wall as positive.

Calculate the signed impulse experienced by the ball.

Solution 1

The signed impulse is \(-0.60\text{ N s}\), meaning an impulse of magnitude \(0.60\text{ N s}\) directed away from the wall.

The impulse is equal to the change in momentum:

\[ J=m(v_f-v_i) \]

Take towards the wall as positive.

The initial velocity is

\[ v_i=+5.0\text{ m s}^{-1} \]

The final velocity is

\[ v_f=0 \]

Substituting:

\[ \begin{aligned} J&=0.12(0-5.0)\\ &=-0.60\text{ N s} \end{aligned} \]

Therefore,

\[ \boxed{J=-0.60\text{ N s}} \]

The negative sign indicates that the impulse acts opposite to the ball’s initial direction of motion.

Because the ball stops rather than rebounds, the wall only needs to remove its original momentum.

Question 2

A \(0.20\text{ kg}\) ball travels towards a fixed wall at \(8.0\text{ m s}^{-1}\). It rebounds directly backwards at \(3.0\text{ m s}^{-1}\).

The collision lasts \(0.025\text{ s}\).

Take motion towards the wall as positive.

Calculate:

a) The signed impulse on the ball.

b) The magnitude of the average net force on the ball.

c) The impulse magnitude and average force magnitude if the ball had instead stuck to the wall over the same contact time.

Explain why the rebound produces a different result.

Solution 2

The rebound produces an impulse of \(-2.20\text{ N s}\) and an average net force magnitude of \(88\text{ N}\). If the ball sticks, the corresponding values are \(1.60\text{ N s}\) and \(64\text{ N}\).

The key is that the rebound velocity has the opposite sign to the initial velocity.

Part a: Calculate the impulse.

Take towards the wall as positive.

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

\[ v_f=-3.0\text{ m s}^{-1} \]

Apply the impulse-momentum relationship:

\[ \begin{aligned} J&=m(v_f-v_i)\\ &=0.20(-3.0-8.0)\\ &=0.20(-11.0)\\ &=-2.20\text{ N s} \end{aligned} \]

Therefore,

\[ \boxed{J=-2.20\text{ N s}} \]

Part b: Calculate the average net force.

Use

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

Substitute:

\[ \begin{aligned} F_{\text{avg}} &=\frac{-2.20}{0.025}\\ &=-88\text{ N} \end{aligned} \]

Therefore, the average force magnitude is

\[ \boxed{|F_{\text{avg}}|=88\text{ N}} \]

Part c: Compare with the sticking collision.

If the ball sticks, its final velocity is zero.

\[ \begin{aligned} J_{\text{stick}} &=0.20(0-8.0)\\ &=-1.60\text{ N s} \end{aligned} \]

The average force magnitude is

\[ \begin{aligned} |F_{\text{avg}}| &=\frac{1.60}{0.025}\\ &=64\text{ N} \end{aligned} \]

Comparing the two results:

The contact durations are identical, so the larger impulse necessarily produces the larger average force magnitude.

The tempting mistake is to calculate the velocity change as \(8.0-3.0=5.0\text{ m s}^{-1}\). This compares speeds rather than velocities.

The correct velocity change is

\[ \Delta v=-3.0-8.0=-11.0\text{ m s}^{-1} \]

The reversal is exactly why the rebound produces the larger impulse.

Question 3

Two identical \(0.20\text{ kg}\) balls approach fixed targets at \(6.0\text{ m s}^{-1}\) to the right.

The contact force on each ball acts to the left throughout the collision.

The force-magnitude graphs have the following shapes.

Collision A: A triangular pulse that rises from zero to \(300\text{ N}\) over \(4.0\text{ ms}\), then falls to zero over the next \(4.0\text{ ms}\).

Collision B: A trapezoidal pulse that:

• rises from zero to \(100\text{ N}\) over \(4.0\text{ ms}\);
• remains at \(100\text{ N}\) for \(12.0\text{ ms}\);
• falls to zero over the final \(4.0\text{ ms}\).

For each collision, calculate:

a) The magnitude of the impulse.

b) The magnitude of the average net force.

c) The final velocity of the ball.

d) The peak force.

Explain why the collision with the lower peak force can produce the larger impulse.

Solution 3

Collision B produces the larger impulse, \(1.60\text{ N s}\), despite having a peak force of only \(100\text{ N}\). It causes the ball to rebound at \(2.0\text{ m s}^{-1}\), whereas Collision A brings the ball to rest.

The important difference is the total area under each force-time graph.

Because the contact force acts to the left, the signed impulses will be negative. We calculate the positive areas first and then assign the negative sign.

Collision A

Part a: Calculate the impulse.

The graph is triangular.

Convert the duration into seconds:

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

The area is

\[ \begin{aligned} |J_A| &=\frac{1}{2}(0.0080)(300)\\ &=1.20\text{ N s} \end{aligned} \]

Therefore,

\[ \boxed{J_A=-1.20\text{ N s}} \]

Part b: Calculate the average force magnitude.

\[ \begin{aligned} |F_{\text{avg,A}}| &=\frac{1.20}{0.0080}\\ &=150\text{ N} \end{aligned} \]

Part c: Calculate the final velocity.

Take right as positive.

The initial momentum is

\[ p_i=0.20(6.0)=1.20\text{ kg m s}^{-1} \]

The final momentum is

\[ \begin{aligned} p_f&=p_i+J_A\\ &=1.20-1.20\\ &=0 \end{aligned} \]

Therefore,

\[ \boxed{v_f=0\text{ m s}^{-1}} \]

The ball stops.

Part d: Identify the peak force.

The peak is given directly by the graph:

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

Collision B

Part a: Calculate the impulse.

The graph consists of two triangles and one rectangle.

The two triangular regions each have a base of \(0.0040\text{ s}\).

The rectangular region has a width of \(0.012\text{ s}\).

Calculate the total area:

\[ \begin{aligned} |J_B| ={}&\frac{1}{2}(0.0040)(100)\\ &+(0.012)(100)\\ &+\frac{1}{2}(0.0040)(100) \end{aligned} \]

\[ |J_B|=0.20+1.20+0.20 \]

\[ |J_B|=1.60\text{ N s} \]

Therefore,

\[ \boxed{J_B=-1.60\text{ N s}} \]

Part b: Calculate the average force magnitude.

The total duration is

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

Therefore,

\[ \begin{aligned} |F_{\text{avg,B}}| &=\frac{1.60}{0.020}\\ &=80\text{ N} \end{aligned} \]

Part c: Calculate the final velocity.

The initial momentum is again

\[ p_i=1.20\text{ kg m s}^{-1} \]

The final momentum is

\[ \begin{aligned} p_f&=p_i+J_B\\ &=1.20-1.60\\ &=-0.40\text{ kg m s}^{-1} \end{aligned} \]

Therefore,

\[ \begin{aligned} v_f&=\frac{p_f}{m}\\ &=\frac{-0.40}{0.20}\\ &=-2.0\text{ m s}^{-1} \end{aligned} \]

Hence,

\[ \boxed{v_f=-2.0\text{ m s}^{-1}} \]

The ball rebounds to the left.

Part d: Identify the peak force.

\[ \boxed{F_{\text{peak,B}}=100\text{ N}} \]

Final comparison

Collision B has the larger impulse because its force acts over a much wider time interval.

Its peak and average forces are both smaller because the force is distributed over a longer collision.

This demonstrates why looking only at the tallest point on a force-time graph can lead to the wrong conclusion about momentum change.

Question 4

A force sensor is attached to a fixed vertical wall.

A \(0.30\text{ kg}\) ball approaches the wall from the left at \(5.0\text{ m s}^{-1}\).

The force sensor records the force exerted by the ball on the wall. Taking right as positive, the signed area under its force-time graph is \(+1.8\text{ N s}\).

The collision lasts \(0.040\text{ s}\), and the peak force magnitude is \(90\text{ N}\).

Neglect other horizontal impulses.

Calculate:

a) The impulse experienced by the ball.

b) The ball’s final velocity.

c) The magnitude and direction of the average force on the ball.

d) Whether the recorded force-time pulse could be triangular.

Solution 4

The ball experiences an impulse of \(-1.8\text{ N s}\), finishes moving left at \(1.0\text{ m s}^{-1}\), and experiences an average force of \(45\text{ N}\) to the left. The measured pulse is consistent with a triangular force-time graph.

The important detail is that the sensor measures the force exerted by the ball on the wall, not the force exerted by the wall on the ball.

Part a: Determine the impulse on the ball.

By Newton’s third law, the force exerted by the wall on the ball is equal in magnitude and opposite in direction to the force exerted by the ball on the wall.

Therefore, their impulses are also equal in magnitude and opposite in direction.

The sensor records

\[ J_{\text{wall}}=+1.8\text{ N s} \]

So the impulse on the ball is

\[ \boxed{J_{\text{ball}}=-1.8\text{ N s}} \]

Part b: Calculate the final velocity.

The initial velocity is

\[ v_i=+5.0\text{ m s}^{-1} \]

The initial momentum is

\[ p_i=mv_i=(0.30)(5.0)=1.50\text{ kg m s}^{-1} \]

Using

\[ J=p_f-p_i \]

we obtain

\[ \begin{aligned} -1.8&=p_f-1.50\\ p_f&=-0.30\text{ kg m s}^{-1} \end{aligned} \]

Now calculate the final velocity:

\[ \begin{aligned} v_f&=\frac{p_f}{m}\\ &=\frac{-0.30}{0.30}\\ &=-1.0\text{ m s}^{-1} \end{aligned} \]

Therefore,

\[ \boxed{v_f=-1.0\text{ m s}^{-1}} \]

The ball rebounds to the left.

Part c: Calculate the average force.

Use the impulse relationship:

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

\[ \begin{aligned} F_{\text{avg}} &=\frac{-1.8}{0.040}\\ &=-45\text{ N} \end{aligned} \]

Therefore,

\[ \boxed{45\text{ N to the left}} \]

Part d: Check whether the pulse could be triangular.

For a triangular force-time graph,

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

Substitute the supplied values:

\[ \begin{aligned} |J| &=\frac{1}{2}(90)(0.040)\\ &=1.8\text{ N s} \end{aligned} \]

This matches the recorded impulse magnitude.

Therefore, the pulse is consistent with a triangular shape.

The useful lesson here is that a force sensor’s reading must be interpreted carefully. Always identify which object experiences the force being measured before assigning the direction of the impulse.

Question 5

A \(0.25\text{ kg}\) ball travels to the right at \(4.0\text{ m s}^{-1}\) towards a stationary trolley of mass \(0.75\text{ kg}\).

The trolley is on a horizontal, low-friction track and is free to move.

Two collision trials are conducted.

Trial S: Sticking

The ball sticks to the trolley, and they move together after the collision.

Trial R: Rebounding

The ball rebounds to the left at \(2.0\text{ m s}^{-1}\), while the trolley moves to the right at \(2.0\text{ m s}^{-1}\).

The contact times are:

• Trial S: \(0.030\text{ s}\)
• Trial R: \(0.060\text{ s}\)

Assume negligible external horizontal impulse. Both force-magnitude pulses are idealised as triangles.

Answer the following:

a) Calculate the common final velocity in Trial S using conservation of momentum.

b) Calculate the impulse on the ball in each trial.

c) Calculate the impulse on the trolley in each trial.

d) Compare the average contact force magnitudes and peak contact forces in the two trials.

e) Explain why assuming that the ball stops in Trial S would produce an incorrect answer. Also explain why the larger rebound impulse does not produce a larger peak force in this example.

Solution 5

The rebound trial produces twice the impulse magnitude of the sticking trial: \(1.50\text{ N s}\) compared with \(0.75\text{ N s}\). However, both trials have the same average contact force magnitude of \(25\text{ N}\) and the same peak force of \(50\text{ N}\).

The important twist is that the target is a free trolley, not a fixed wall. When the ball sticks, it continues moving with the trolley. Its final velocity is therefore not zero.

Part a: Calculate the common final velocity in Trial S.

Because there is negligible external horizontal impulse, total horizontal momentum is conserved.

Initially, only the ball is moving.

The initial momentum is

\[ \begin{aligned} p_i&=m_bv_b+m_tv_t\\ &=(0.25)(4.0)+(0.75)(0)\\ &=1.0\text{ kg m s}^{-1} \end{aligned} \]

where \(m_b\) is the ball’s mass and \(m_t\) is the trolley’s mass.

After sticking, both objects move together.

Their combined mass is

\[ m_{\text{total}}=0.25+0.75=1.00\text{ kg} \]

Let their common velocity be \(v\).

Conservation of momentum gives

\[ \begin{aligned} 1.0&=(1.00)v\\ v&=1.0\text{ m s}^{-1} \end{aligned} \]

Therefore,

\[ \boxed{v=+1.0\text{ m s}^{-1}} \]

The ball does not stop. It continues moving right with the trolley.

Part b: Calculate the impulse on the ball.

For Trial S:

\[ \begin{aligned} J_{\text{ball,S}} &=m_b(v_f-v_i)\\ &=0.25(1.0-4.0)\\ &=-0.75\text{ N s} \end{aligned} \]

Therefore,

\[ \boxed{J_{\text{ball,S}}=-0.75\text{ N s}} \]

For Trial R:

The ball rebounds at \(-2.0\text{ m s}^{-1}\).

\[ \begin{aligned} J_{\text{ball,R}} &=0.25(-2.0-4.0)\\ &=0.25(-6.0)\\ &=-1.50\text{ N s} \end{aligned} \]

Therefore,

\[ \boxed{J_{\text{ball,R}}=-1.50\text{ N s}} \]

The rebound impulse magnitude is twice the sticking impulse magnitude.

Part c: Calculate the impulse on the trolley.

The trolley initially has zero momentum.

For Trial S:

\[ \begin{aligned} J_{\text{trolley,S}} &=0.75(1.0-0)\\ &=+0.75\text{ N s} \end{aligned} \]

For Trial R:

\[ \begin{aligned} J_{\text{trolley,R}} &=0.75(2.0-0)\\ &=+1.50\text{ N s} \end{aligned} \]

Therefore,

\[ \boxed{J_{\text{trolley,S}}=+0.75\text{ N s}} \]

and

\[ \boxed{J_{\text{trolley,R}}=+1.50\text{ N s}} \]

These results agree with Newton’s third law.

The ball and trolley experience equal-magnitude, opposite-direction impulses during each collision.

We can also check momentum conservation in Trial R:

\[ \begin{aligned} p_f &=(0.25)(-2.0)+(0.75)(2.0)\\ &=-0.50+1.50\\ &=+1.00\text{ kg m s}^{-1} \end{aligned} \]

This matches the initial momentum of \(+1.00\text{ kg m s}^{-1}\).

Part d: Compare average and peak forces.

The magnitude of the average contact force is

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

For Trial S:

\[ \begin{aligned} |F_{\text{avg,S}}| &=\frac{0.75}{0.030}\\ &=25\text{ N} \end{aligned} \]

For Trial R:

\[ \begin{aligned} |F_{\text{avg,R}}| &=\frac{1.50}{0.060}\\ &=25\text{ N} \end{aligned} \]

Both trials have the same average contact force magnitude.

Now use the triangular force-time relationship:

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

Rearrange:

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

For Trial S:

\[ F_{\text{peak,S}} =\frac{2(0.75)}{0.030} =50\text{ N} \]

For Trial R:

\[ F_{\text{peak,R}} =\frac{2(1.50)}{0.060} =50\text{ N} \]

Therefore,

\[ \boxed{F_{\text{peak,S}}=F_{\text{peak,R}}=50\text{ N}} \]

The comparison is:

Part e: Explain the result.

There are two important misconceptions to avoid.

First, sticking does not always mean an object finishes at rest.

That assumption works for a ball sticking to a fixed wall, but not for a ball sticking to a free trolley. Here, the ball and trolley continue moving together at \(1.0\text{ m s}^{-1}\).

Second, a larger impulse does not automatically produce a larger average or peak force.

In Trial R, the rebound produces twice the impulse, but the contact time is also twice as long.

The increased area is spread over twice the duration, giving the same average force. Because both force-time pulses are triangular, they also have the same peak.

This is the deeper distinction: momentum change determines impulse, while collision duration and force-time profile determine how that impulse is delivered.

06What to remember

When a ball rebounds from a fixed target, it experiences a greater impulse magnitude than if it simply stops, because its velocity reverses direction.

However, three separate ideas must stay separate:

• Impulse: determined by the change in momentum and represented by the signed area under the force-time graph.
• Average force: determined by dividing impulse by collision duration.
• Peak force: determined by the highest point of the force-time graph.

A rebound can produce a larger impulse without producing a larger peak force. A longer collision can spread that impulse over more time.

The next useful step is to apply this reasoning to conservation of momentum in collisions involving two moving objects. That is where you need to track not just what happens to the ball, but where its lost or gained momentum goes.