How to Calculate Momentum Change from Multiple Force Pulses

A force sensor records two distinct peaks during a collision. The first reaches \(300\text{ N}\), and the second reaches \(200\text{ N}\). Has the object experienced a momentum change of \(500\text{ N s}\)?

No. Adding the peak forces tells you nothing useful about the total momentum change.

To read momentum change from a force-time graph with multiple pulses, calculate the area of every pulse, assign its sign, and add the areas together. The result is the net impulse, which equals the object’s change in momentum.

There is another detail worth watching. The peak force is the highest instantaneous force, while the mean force depends on the total area and the time interval you choose. If two pulses are separated by a gap, that gap contributes zero impulse but still counts towards the mean force over the entire event.

Let’s work through how to handle both quantities without mixing them up.

01What do multiple force pulses tell you?

Imagine a trolley receives two separate pushes in the same direction.

The first push is relatively short and strong. The second is weaker but lasts longer. Between the pushes, the trolley experiences approximately zero horizontal force.

You might predict that the first push produces the greater momentum change because its peak force is higher.

But that prediction isn’t necessarily correct.

A force of \(300\text{ N}\) acting for \(2\text{ ms}\) produces a different impulse from a force of \(150\text{ N}\) acting for \(8\text{ ms}\).

The first produces:

\[ J_1=300(0.002)=0.60\text{ N s} \]

The second produces:

\[ J_2=150(0.008)=1.20\text{ N s} \]

The weaker force produces twice the momentum change because it acts for longer.

This is why the area under a force-time graph matters more than the height of an individual peak.

The central equation

For a one-dimensional force-time graph:

\[ \boxed{\Delta p=J_{\text{total}}=\sum_i J_i} \]

Here:

• \(\Delta p\) is the total change in momentum, measured in \(\text{kg m s}^{-1}\).
• \(J_{\text{total}}\) is the net impulse, measured in newton seconds (\(\text{N s}\)).
• \(J_i\) is the impulse contributed by an individual pulse.

For pulses entirely above the time axis, simply add their areas:

\[ J_{\text{total}}=A_1+A_2+A_3+\cdots \]

If a pulse falls below the time axis, its area is negative. You must subtract its magnitude rather than add it.

This is the same area principle used in Force-Time Graphs: Impulse and Momentum Change in HSC Physics, but now we’re applying it to several separate interactions within one time interval.

02How to calculate the area of each pulse

A force-time graph can contain different shapes. You need to identify each shape before calculating its area.

For HSC questions, the most common shapes are rectangles, triangles, and trapeziums.

Remember that the horizontal axis represents time in seconds, while the vertical axis represents force in newtons.

Therefore:

\[ \text{Area}=\text{N}\times\text{s}=\text{N s} \]

And because impulse equals momentum change:

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

What if the graph contains several different shapes?

Split the graph into manageable sections.

For example, imagine a force-time graph containing:

• A triangular pulse.
• A rectangular pulse.
• A trapezium-shaped pulse.

Calculate each area separately, then combine the results.

You do not need a special formula for the entire graph. You just need the correct area formula for each section.

A useful decision process

When you encounter a multiple-pulse question, follow this sequence:

1. Choose a positive direction. For example, define right as positive.
2. Identify every pulse. Separate the graph into rectangles, triangles, and trapeziums.
3. Convert time into seconds. Remember that \(1\text{ ms}=0.001\text{ s}\).
4. Calculate each area. Keep the areas separate until the end.
5. Assign signs. Areas above the time axis are positive; areas below are negative.
6. Add the signed areas. This gives the total impulse.
7. Calculate the mean force if required. Divide the net impulse by the specified time interval.

The important part is step six. Do not add the peak forces, and do not calculate the mean of the peak values.

03Peak force versus mean force

These two quantities are easy to confuse when a graph contains multiple pulses.

They answer different questions.

Peak force

The peak force is the greatest instantaneous force reached anywhere on the graph.

For a graph containing several positive and negative pulses, use the greatest force magnitude:

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

For example, suppose three pulses reach:

• \(+250\text{ N}\)
• \(+400\text{ N}\)
• \(-350\text{ N}\)

The peak force magnitude is:

\[ F_{\text{peak}}=400\text{ N} \]

You do not add the three peaks.

Mean force

The mean net force is the constant force that would produce the same net impulse over the same total time interval.

Its equation is:

\[ \boxed{F_{\text{mean}}=\frac{J_{\text{total}}}{\Delta t}} \]

Here, \(\Delta t\) is the time between the beginning of the first included pulse and the end of the last included pulse.

This means that zero-force gaps must be included when calculating the mean over the entire event.

Consider two positive pulses separated by a \(2\text{ ms}\) gap.

Suppose their combined area is \(1.4\text{ N s}\), and the entire event lasts \(12\text{ ms}\).

The mean force is:

\[ F_{\text{mean}}=\frac{1.4}{0.012} \]

\[ \boxed{F_{\text{mean}}=116.7\text{ N}} \]

The object does not actually experience a constant force of \(116.7\text{ N}\). This is an equivalent average over the complete interval.

Why the time interval matters

Suppose the two pulses in the previous example are active for a combined \(10\text{ ms}\), with a \(2\text{ ms}\) gap.

There are now two possible averages:

Both calculations are mathematically valid, but they answer different questions.

Unless the question specifically asks for the mean force while the force is active, use the entire stated event interval.

Exam tip: Always identify the time interval before calculating mean force. A correct impulse divided by the wrong time interval gives an incorrect answer.

04What happens when pulses act in opposite directions?

So far, we’ve considered pulses acting in the same direction.

Now imagine a trolley receives a rightward push, followed by a leftward push.

The second pulse reduces some of the momentum gained during the first.

If right is positive:

• Rightward pulses have positive areas.
• Leftward pulses have negative areas.

The total impulse becomes:

\[ \boxed{J_{\text{total}}=A_1-A_2+A_3-\cdots} \]

You must preserve the signs throughout the calculation.

For example, suppose an object experiences:

\[ J_1=+2.0\text{ N s} \]

followed by:

\[ J_2=-1.5\text{ N s} \]

Its total impulse is:

\[ J_{\text{total}}=2.0-1.5 \]

\[ \boxed{J_{\text{total}}=+0.50\text{ N s}} \]

The object gains \(0.50\text{ kg m s}^{-1}\) of momentum in the positive direction.

Notice that the second pulse does not necessarily reverse the object’s motion. It changes the momentum in the negative direction, but the resulting momentum may still be positive.

For a more detailed treatment of this situation, see How to Calculate Impulse When Force Changes Sign.

07One experimental detail that can change your answer

There is a limitation to the method we’ve used so far.

The area under a force-time graph gives the change in momentum only when the graph represents the net force acting on the object.

A force sensor often measures just one force, such as the contact force from an actuator. Other forces may still be acting.

Consider a trolley being pushed along a horizontal track. The sensor measures two rightward pushes, but friction acts to the left throughout the experiment.

The impulse from the sensor alone is not the total change in momentum.

Instead:

\[ J_{\text{net}}=J_{\text{contact}}+J_{\text{friction}} \]

You must account for all relevant forces.

This is why a good experimental design uses a level, low-friction track, calibrates the force sensor, and checks for background forces.

A simple hand-pushed trolley might produce several peaks, but it is difficult to control the timing and strength of each push. Sensor vibration could also produce an apparent extra peak.

A better design uses a programmable actuator to deliver two controlled pushes separated by a known interval. A calibrated force sensor records the interaction, while photogates or a motion sensor independently measure the trolley’s velocity before and after the pushes.

The prediction is straightforward:

• After the first positive pulse, the trolley gains momentum.
• During the gap, its momentum should remain approximately constant if other horizontal forces are negligible.
• After the second positive pulse, it gains further momentum.

The measured change in momentum should agree with the sum of the two force-time areas.

If it doesn’t, check the sensor’s zero offset, friction, and whether the graph actually represents the net force.

This distinction becomes particularly important when interpreting real experimental data rather than an idealised HSC graph.

08Questions and solutions

The following questions build from straightforward area calculations to situations where multiple pulses interact with the object’s motion.

All numerical data are illustrative.

Question 1

A trolley experiences two rightward force pulses.

• Pulse A is rectangular, with a force of \(120\text{ N}\) acting for \(5.0\text{ ms}\).
• Pulse B is triangular, with a peak force of \(240\text{ N}\) and a duration of \(4.0\text{ ms}\).
• There is a \(1.0\text{ ms}\) gap between the pulses.

The graph represents the net horizontal force on the trolley.

Calculate:

a) The total impulse.

b) The peak force.

c) The mean net force over the entire event.

Solution 1

The total impulse is \(1.08\text{ N s}\), the peak force is \(240\text{ N}\), and the mean net force is \(108\text{ N}\) to the right.

Part a: Total impulse

Calculate the area of the rectangle:

\[ J_1=(120)(0.0050) \]

\[ J_1=0.60\text{ N s} \]

Calculate the area of the triangle:

\[ J_2=\frac12(240)(0.0040) \]

\[ J_2=0.48\text{ N s} \]

Both pulses are positive, so:

\[ \begin{aligned} J_{\text{total}}&=0.60+0.48\\ &=\boxed{1.08\text{ N s}} \end{aligned} \]

Part b: Peak force

The largest instantaneous force is the peak of the triangular pulse:

\[ \boxed{F_{\text{peak}}=240\text{ N}} \]

Part c: Mean net force

The complete event lasts:

\[ \Delta t=5.0+1.0+4.0=10.0\text{ ms} \]

\[ \Delta t=0.010\text{ s} \]

Therefore:

\[ F_{\text{mean}}=\frac{1.08}{0.010} \]

\[ \boxed{F_{\text{mean}}=108\text{ N}} \]

The gap contributes zero impulse, but it is included in the \(10.0\text{ ms}\) interval.

Question 2

A \(0.25\text{ kg}\) trolley initially moves right at \(4.0\text{ m s}^{-1}\).

It experiences two force pulses:

• Pulse A is a positive triangular pulse lasting \(5.0\text{ ms}\), with a peak force of \(400\text{ N}\).
• Pulse B is a negative triangular pulse lasting \(4.0\text{ ms}\), with an unknown peak force magnitude \(F\).

There is a \(3.0\text{ ms}\) gap between the pulses.

After both pulses, the trolley is measured moving right at \(2.0\text{ m s}^{-1}\).

Assume the graph represents the net horizontal force.

Calculate:

a) The impulse from each pulse.

b) The unknown peak force \(F\).

c) The mean net force over the entire event.

d) The peak force magnitude across both pulses.

Solution 2

The second pulse has a peak force magnitude of \(750\text{ N}\), the net impulse is \(-0.50\text{ N s}\), and the mean net force is approximately \(41.7\text{ N}\) to the left.

Part a: Calculate the impulse from each pulse

For the first triangular pulse:

\[ \begin{aligned} J_1&=\frac12(400)(0.0050)\\ &=+1.00\text{ N s} \end{aligned} \]

Next, use the initial and final velocities to determine the total impulse.

\[ J_{\text{total}}=m(v-u) \]

Substitute:

\[ \begin{aligned} J_{\text{total}} &=0.25(2.0-4.0)\\ &=-0.50\text{ N s} \end{aligned} \]

Therefore, the impulse from the second pulse must be:

\[ J_2=J_{\text{total}}-J_1 \]

\[ J_2=-0.50-1.00 \]

\[ \boxed{J_2=-1.50\text{ N s}} \]

Part b: Find the unknown peak force

The second pulse is triangular and lies below the time axis.

Use the magnitude of its impulse:

\[ |J_2|=\frac12 F\Delta t \]

Substitute:

\[ 1.50=\frac12 F(0.0040) \]

Rearrange:

\[ F=\frac{2(1.50)}{0.0040} \]

\[ \boxed{F=750\text{ N}} \]

The second pulse has a peak force of \(750\text{ N}\) to the left.

Part c: Calculate the mean net force

The total event lasts:

\[ \Delta t=5.0+3.0+4.0=12.0\text{ ms} \]

\[ \Delta t=0.0120\text{ s} \]

Therefore:

\[ F_{\text{mean}}=\frac{-0.50}{0.0120} \]

\[ \boxed{F_{\text{mean}}=-41.7\text{ N}} \]

The negative sign indicates a leftward mean net force.

Part d: Identify the peak force

The first pulse reaches \(400\text{ N}\), while the second reaches \(750\text{ N}\) in magnitude.

Therefore:

\[ \boxed{F_{\text{peak}}=750\text{ N}} \]

The second pulse has the greater peak, even though the trolley still finishes moving right.

This is because the leftward impulse reduces the trolley’s rightward momentum without eliminating it.

Question 3

A \(0.50\text{ kg}\) trolley initially moves right at \(0.40\text{ m s}^{-1}\).

A force sensor measures the contact force from a motorised actuator. Its graph contains two rightward pulses:

• Pulse A is triangular, lasting \(5.0\text{ ms}\), with a peak of \(300\text{ N}\).
• Pulse B is rectangular, lasting \(3.0\text{ ms}\), with a constant force of \(150\text{ N}\).

The entire interval from the beginning of the first pulse to the end of the second pulse is \(0.050\text{ s}\).

However, the trolley also experiences a constant frictional force of \(5.0\text{ N}\) to the left throughout the entire interval, including the gap.

Assume these are the only horizontal forces.

Calculate:

a) The impulse from the actuator.

b) The impulse from friction.

c) The trolley’s final velocity.

d) Explain why using only the area under the sensor’s graph gives an incorrect final velocity.

Solution 3

The trolley’s final velocity is \(2.30\text{ m s}^{-1}\) to the right. The frictional impulse must be included because the sensor measures the actuator force, not the total net force.

Part a: Calculate the actuator impulse

The first pulse is triangular:

\[ \begin{aligned} J_1&=\frac12(300)(0.0050)\\ &=0.75\text{ N s} \end{aligned} \]

The second pulse is rectangular:

\[ \begin{aligned} J_2&=(150)(0.0030)\\ &=0.45\text{ N s} \end{aligned} \]

Therefore:

\[ \begin{aligned} J_{\text{actuator}}&=0.75+0.45\\ &=1.20\text{ N s} \end{aligned} \]

Part b: Calculate the frictional impulse

Friction acts throughout the full \(0.050\text{ s}\) interval.

It acts to the left, so its impulse is negative:

\[ \begin{aligned} J_{\text{friction}} &=(-5.0)(0.050)\\ &=-0.25\text{ N s} \end{aligned} \]

Part c: Calculate the final velocity

The net impulse is:

\[ \begin{aligned} J_{\text{net}}&=J_{\text{actuator}}+J_{\text{friction}}\\ &=1.20-0.25\\ &=0.95\text{ N s} \end{aligned} \]

The initial momentum is:

\[ p_i=mu \]

\[ p_i=0.50(0.40)=0.20\text{ kg m s}^{-1} \]

The final momentum is:

\[ \begin{aligned} p_f&=p_i+J_{\text{net}}\\ &=0.20+0.95\\ &=1.15\text{ kg m s}^{-1} \end{aligned} \]

Therefore:

\[ v_f=\frac{p_f}{m} \]

\[ v_f=\frac{1.15}{0.50} \]

\[ \boxed{v_f=2.30\text{ m s}^{-1}\text{ right}} \]

Part d: Explain the experimental limitation

The tempting approach is to calculate the area under the sensor graph and treat it as the net impulse.

That would give:

\[ J=1.20\text{ N s} \]

However, the sensor only measures the actuator’s contact force. It does not include the frictional force.

Ignoring friction would incorrectly give:

\[ v_f=\frac{0.20+1.20}{0.50} =2.80\text{ m s}^{-1} \]

The correct answer is \(2.30\text{ m s}^{-1}\).

The distinction is between the impulse from one force and the net impulse from all forces acting on the object.

Question 4

The same \(0.50\text{ kg}\) trolley starts from rest in three separate experiments. Each experiment produces a different force-time graph.

All forces act to the right.

The three profiles are:

Profile A

Two triangular pulses, each with a peak force of \(300\text{ N}\) and a duration of \(4.0\text{ ms}\). The pulses are separated by an \(8.0\text{ ms}\) gap.

The total event lasts \(16.0\text{ ms}\).

Profile B

A constant rectangular force of \(75\text{ N}\) acting for \(16.0\text{ ms}\).

Profile C

A single triangular pulse with a peak force of \(300\text{ N}\), lasting \(8.0\text{ ms}\).

For each profile, calculate:

a) The total impulse.

b) The final velocity.

c) The peak force.

d) The mean net force over the complete event.

Then explain why the three experiments can produce the same final velocity despite having different peak forces and mean forces.

Solution 4

All three experiments produce the same impulse of \(1.20\text{ N s}\) and the same final velocity of \(2.40\text{ m s}^{-1}\). However, their peak and mean forces are different.

Part a: Calculate the impulse for each profile

For Profile A, each triangle has an area of:

\[ J_1=\frac12(300)(0.0040) =0.60\text{ N s} \]

There are two pulses:

\[ J_A=0.60+0.60 \]

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

For Profile B:

\[ \begin{aligned} J_B&=(75)(0.016)\\ &=\boxed{1.20\text{ N s}} \end{aligned} \]

For Profile C:

\[ \begin{aligned} J_C&=\frac12(300)(0.0080)\\ &=\boxed{1.20\text{ N s}} \end{aligned} \]

All three profiles have equal areas.

Part b: Calculate the final velocity

The trolley starts from rest, so its initial momentum is zero.

For each experiment:

\[ J=m(v-u) \]

Since \(u=0\):

\[ v=\frac{J}{m} \]

\[ v=\frac{1.20}{0.50} \]

\[ \boxed{v=2.40\text{ m s}^{-1}} \]

The final velocities are identical because all three experiments produce the same total momentum change.

Part c: Compare the peak forces

Profile A reaches \(300\text{ N}\).

Profile B reaches \(75\text{ N}\).

Profile C reaches \(300\text{ N}\).

Therefore:

Part d: Calculate the mean forces

For Profile A:

\[ F_{\text{mean,A}}=\frac{1.20}{0.016} \]

\[ \boxed{F_{\text{mean,A}}=75\text{ N}} \]

For Profile B:

\[ F_{\text{mean,B}}=\frac{1.20}{0.016} \]

\[ \boxed{F_{\text{mean,B}}=75\text{ N}} \]

For Profile C:

\[ F_{\text{mean,C}}=\frac{1.20}{0.008} \]

\[ \boxed{F_{\text{mean,C}}=150\text{ N}} \]

The results are:

The key is that impulse determines momentum change, not peak force or mean force individually.

Profiles A and B have the same impulse and the same total duration, so they also have the same mean force. However, their peak forces are very different.

Profile C produces the same impulse in half the time, so its mean force is twice as large.

Equal impulse does not require equal peak force. Equal impulse only guarantees equal mean force when the time intervals are also equal.

Question 5

A \(0.50\text{ kg}\) trolley initially rests on a horizontal, low-friction track.

A motorised actuator delivers two force pulses.

First pulse: A symmetrical triangular force pulse acts to the right for \(4.0\text{ ms}\), reaching a peak of \(500\text{ N}\).

Gap: The horizontal force is zero for \(0.30\text{ s}\).

Second pulse: A constant force of \(10\text{ N}\) acts to the left for \(0.10\text{ s}\).

Assume these are the only horizontal forces acting on the trolley.

Take right as positive.

Calculate:

a) The impulse from each pulse and the total impulse.

b) The final velocity.

c) The peak force and mean net force over the entire event.

d) The total displacement from the beginning of the first pulse to the end of the second pulse.

e) Suppose the zero-force gap is increased from \(0.30\text{ s}\) to \(0.50\text{ s}\), with both pulses unchanged. Which of your previous answers change, and by how much?

Solution 5

The total impulse is zero, and the trolley finishes at rest. However, it travels \(0.704\text{ m}\) during the event. The peak force is \(500\text{ N}\), while the mean net force is zero.

This question combines impulse, momentum, mean force, and displacement. The important distinction is that returning to the original momentum does not mean returning to the original position.

Part a: Calculate the impulse from each pulse

The first pulse is triangular:

\[ \begin{aligned} J_1&=\frac12(500)(0.0040)\\ &=+1.0\text{ N s} \end{aligned} \]

The second pulse is rectangular and acts to the left:

\[ \begin{aligned} J_2&=(-10)(0.10)\\ &=-1.0\text{ N s} \end{aligned} \]

The total impulse is:

\[ \begin{aligned} J_{\text{total}}&=1.0-1.0\\ &=\boxed{0\text{ N s}} \end{aligned} \]

The two pulses have equal and opposite impulses.

Part b: Calculate the final velocity

Use the impulse-momentum theorem:

\[ J_{\text{total}}=m(v-u) \]

The trolley initially rests, so \(u=0\).

\[ 0=0.50(v-0) \]

Therefore:

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

The trolley finishes at rest.

However, it does not remain at rest throughout the experiment.

After the first pulse, its momentum is:

\[ p=+1.0\text{ kg m s}^{-1} \]

Its velocity is:

\[ v=\frac{1.0}{0.50} \]

\[ v=+2.0\text{ m s}^{-1} \]

During the gap, it continues travelling right at approximately \(2.0\text{ m s}^{-1}\).

The second pulse then reduces its rightward momentum to zero.

Part c: Calculate the peak and mean net force

The first pulse reaches \(500\text{ N}\).

The second pulse reaches only \(10\text{ N}\) in magnitude.

Therefore:

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

Now calculate the complete event duration:

\[ \Delta t=0.004+0.30+0.10 \]

\[ \Delta t=0.404\text{ s} \]

The mean net force is:

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

\[ F_{\text{mean}}=\frac{0}{0.404} \]

\[ \boxed{F_{\text{mean}}=0\text{ N}} \]

A zero mean net force does not mean that no force acted. It means the positive and negative impulses cancelled over the complete interval.

Part d: Calculate the total displacement

This is the most important part of the question.

We need to calculate the displacement during all three stages.

Stage 1: First triangular pulse

The trolley starts at rest and finishes the first pulse travelling at \(2.0\text{ m s}^{-1}\).

Because the triangular force profile is symmetrical in time, the acceleration profile is also symmetrical. The average velocity during this pulse is therefore the average of the initial and final velocities:

\[ v_{\text{average}}=\frac{0+2.0}{2} \]

\[ v_{\text{average}}=1.0\text{ m s}^{-1} \]

The displacement during the \(0.0040\text{ s}\) pulse is:

\[ \begin{aligned} s_1&=v_{\text{average}}\Delta t\\ &=(1.0)(0.0040)\\ &=0.0040\text{ m} \end{aligned} \]

Stage 2: The zero-force gap

The trolley moves at a constant velocity of \(2.0\text{ m s}^{-1}\) during the gap.

Therefore:

\[ \begin{aligned} s_2&=v\Delta t\\ &=(2.0)(0.30)\\ &=0.60\text{ m} \end{aligned} \]

Stage 3: Second rectangular pulse

The constant leftward force decelerates the trolley from \(2.0\text{ m s}^{-1}\) to rest.

Because the acceleration is constant, the average velocity is:

\[ v_{\text{average}}=\frac{2.0+0}{2} \]

\[ v_{\text{average}}=1.0\text{ m s}^{-1} \]

The displacement during this pulse is:

\[ \begin{aligned} s_3&=(1.0)(0.10)\\ &=0.10\text{ m} \end{aligned} \]

Total displacement

Add the three displacements:

\[ \begin{aligned} s_{\text{total}}&=s_1+s_2+s_3\\ &=0.0040+0.60+0.10\\ &=\boxed{0.704\text{ m}} \end{aligned} \]

The trolley finishes at rest, but it has travelled \(0.704\text{ m}\) to the right.

The tempting mistake is to assume that zero net impulse means zero displacement. It only means that the final momentum equals the initial momentum.

Momentum describes the object’s state of motion at an instant. Displacement describes how its position changes over an interval.

Part e: Increase the gap to \(0.50\text{ s}\)

The two pulses remain unchanged, so their individual impulses are unchanged.

The total impulse is still:

\[ J_{\text{total}}=0 \]

The final velocity is still zero.

The peak force is still \(500\text{ N}\).

The mean net force is also still zero because the total impulse remains zero:

\[ F_{\text{mean}}=\frac{0}{0.604}=0\text{ N} \]

However, the displacement during the gap increases.

The new gap displacement is:

\[ s_2=(2.0)(0.50) \]

\[ s_2=1.0\text{ m} \]

The new total displacement is:

\[ \begin{aligned} s_{\text{total}}&=0.0040+1.0+0.10\\ &=\boxed{1.104\text{ m}} \end{aligned} \]

The displacement increases by:

\[ 1.104-0.704=0.400\text{ m} \]

This is the subtle result: increasing the gap changes the distance travelled, but it does not change the total impulse, final velocity, peak force, or mean net force.

The two force pulses determine the final momentum. The time between them determines how long the trolley travels with the momentum gained from the first pulse.

08What to remember in an HSC exam

When a force-time graph contains multiple pulses, keep these quantities separate.

Before submitting your answer, check these five things:

1. Have you converted every time measurement from milliseconds into seconds?
2. Have you calculated each pulse’s area rather than adding its peak force?
3. Have you assigned negative signs to pulses below the time axis?
4. Have you included zero-force gaps when calculating mean force over the entire event?
5. Does the graph represent the net force, or are other forces acting on the object?

The next useful step is connecting the momentum change to the object’s motion between pulses. Once you can calculate the momentum after each individual pulse, you can track changes in velocity, identify when an object reverses direction, and determine how far it travels during the gaps.