How to Calculate Impulse When Force Changes Sign

A force sensor records a force that starts at \(-12\text{ N}\), rises to zero, and then reaches \(+8\text{ N}\). You need to calculate the impulse over the entire interval.

Do you add both areas under the force-time graph, or subtract one from the other?

You subtract them. When force changes sign, impulse is the signed area under the force-time graph. Areas above the time axis contribute positive impulse, while areas below it contribute negative impulse.

The calculation is:

\[ \boxed{J=A_+-A_-} \]

where:

• \(J\) is the net impulse, measured in newton seconds (\(\text{N s}\)).
• \(A_+\) is the total area above the time axis.
• \(A_-\) is the magnitude of the total area below the time axis.

You then calculate the mean net force using:

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

This is different from the peak force, which is the greatest instantaneous force magnitude recorded during the interval.

Let’s work through why the signs matter, how to calculate the areas, and what happens when positive and negative impulses almost cancel.

01Why the area changes sign

Imagine pushing a trolley to the right and then pulling it to the left.

During the push, the trolley receives a rightward impulse. During the pull, it receives a leftward impulse.

These impulses act in opposite directions. They cannot both count as positive contributions to the trolley’s net change in momentum.

Take right as the positive direction:

• A force acting to the right is positive.
• A force acting to the left is negative.
• A force of zero produces no impulse at that instant.

The sign of the force therefore determines the sign of its contribution to impulse.

This is the physical meaning of signed area. The graph is not simply measuring how much force has acted. It records how much force has acted in each direction.

What if the positive and negative areas are equal?

Suppose a trolley receives a positive impulse of \(3.0\text{ N s}\), followed by a negative impulse of \(-3.0\text{ N s}\).

Its net impulse is:

\[ J=3.0-3.0=0 \]

The trolley finishes with the same momentum it had initially.

However, this does not mean that no force acted. The trolley could have experienced substantial forces in both directions and undergone significant changes in momentum during the interval.

The positive and negative contributions simply cancel when calculating the overall momentum change.

02An experiment: measuring a force that changes direction

Consider an experiment designed to investigate this effect.

A trolley sits on a low-friction horizontal track. A bidirectional force sensor connects the trolley to a motorised actuator at the right-hand end of the track.

The force sensor is connected to a computer that records force against time. Its positive direction is defined as right.

The actuator first pulls the trolley towards the right-hand end, exerting a leftward force on the trolley. It then changes direction and pushes the trolley towards the right.

What should the force-time graph look like?

Initially, the measured force is negative because the actuator pulls left. When the actuator changes direction, the force passes through zero and becomes positive.

The resulting graph has a region below the time axis followed by a region above it.

Illustrative experimental data

Suppose the force sensor produces the following readings.

Assume the force changes linearly between consecutive readings.

The graph contains two triangles.

The first triangle is below the time axis, so its contribution is negative.

The second triangle is above the time axis, so its contribution is positive.

Step 1: Calculate the negative area.

The negative triangle has:

• Base = \(0.20\text{ s}\)
• Height = \(12\text{ N}\)

Using the triangle area formula:

\[ \begin{aligned} A_-&=\frac12 bh\\ &=\frac12(0.20)(12)\\ &=1.2\text{ N s} \end{aligned} \]

Its contribution to impulse is \(-1.2\text{ N s}\).

Step 2: Calculate the positive area.

The positive triangle has:

• Base = \(0.20\text{ s}\)
• Height = \(8\text{ N}\)

\[ \begin{aligned} A_+&=\frac12 bh\\ &=\frac12(0.20)(8)\\ &=0.80\text{ N s} \end{aligned} \]

Step 3: Calculate the net impulse.

Subtract the magnitude of the negative area from the positive area:

\[ \begin{aligned} J&=A_+-A_-\\ &=0.80-1.20\\ &=-0.40\text{ N s} \end{aligned} \]

Therefore:

\[ \boxed{J=-0.40\text{ N s}} \]

The net impulse is \(0.40\text{ N s}\) to the left.

Notice that the negative area is larger, even though the graph also contains a positive force region.

Peak force versus mean force

This experiment also demonstrates why peak force and mean force must not be confused.

The peak force magnitude is the largest instantaneous force magnitude recorded:

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

Therefore:

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

The mean net force is calculated using the net impulse divided by the entire time interval.

The total time is:

\[ \Delta t=0.40\text{ s} \]

Therefore:

\[ \begin{aligned} F_{\text{mean}} &=\frac{J}{\Delta t}\\ &=\frac{-0.40}{0.40}\\ &=-1.0\text{ N} \end{aligned} \]

So the mean net force is \(1.0\text{ N}\) to the left.

Compare the results:

The peak force is twelve times the magnitude of the mean force.

There is another useful distinction here. If you average the magnitudes of the forces over time, rather than their signed values, you obtain:

\[ \begin{aligned} F_{\text{mean magnitude}} &=\frac{A_++A_-}{\Delta t}\\ &=\frac{0.80+1.20}{0.40}\\ &=5.0\text{ N} \end{aligned} \]

This is the average force magnitude, not the mean net force.

For impulse and momentum-change questions, use the signed mean net force:

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

An important experimental limitation

There is one more detail to consider.

The force sensor in this arrangement measures the force exerted by the actuator on the trolley. It does not automatically measure the net force if friction is significant.

For example, if friction also acts on the trolley, the net horizontal force is:

\[ F_{\text{net}}=F_{\text{actuator}}+F_{\text{friction}} \]

Both forces must be included with their correct signs.

A revised experiment could use an air track to minimise friction, or measure the remaining friction in a control trial and correct the recorded force values.

The illustrative calculations above assume that friction is negligible or has already been accounted for. Under that assumption, the measured force represents the net horizontal force.

This is important because the impulse-momentum relationship applies to net impulse, not automatically to the impulse from one individual force.

03A reliable method for calculating impulse

When a force changes sign, use the following procedure.

Step 1: Choose a positive direction.

For example, take right as positive. This determines which parts of the graph are positive and negative.

Step 2: Identify every region.

Split the graph wherever the force crosses the time axis.

Step 3: Calculate each area.

Use the appropriate geometric formula:

If a straight-line section crosses the time axis, split it at the crossing before calculating the positive and negative contributions.

Step 4: Add the signed areas.

\[ J=A_+-A_- \]

Do not add the magnitudes of all regions as though they were positive.

Step 5: Calculate the mean net force if required.

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

Use the entire time interval, including the periods when the force acts in the opposite direction.

Step 6: Identify the peak force separately.

Find the largest absolute force value on the graph.

Do not use the peak force in place of the mean force.

A useful way to remember the difference is:

• Area under the graph determines impulse.
• Impulse divided by total time determines mean net force.
• The highest force magnitude determines peak force.

06Questions and solutions

Question 1

A trolley experiences a net force of \(+10\text{ N}\) for \(0.20\text{ s}\), followed by a net force of \(-5\text{ N}\) for \(0.20\text{ s}\).

Take right as positive.

Calculate:

a) The net impulse over the entire interval.

b) The mean net force.

c) The peak force magnitude.

Solution 1

The net impulse is \(+1.0\text{ N s}\), the mean net force is \(+2.5\text{ N}\), and the peak force magnitude is \(10\text{ N}\).

Step 1: Calculate the positive impulse.

The positive region is a rectangle:

\[ \begin{aligned} A_+&=F\Delta t\\ &=(10)(0.20)\\ &=2.0\text{ N s} \end{aligned} \]

Step 2: Calculate the negative impulse.

The negative region is also a rectangle:

\[ \begin{aligned} A_-&=(5)(0.20)\\ &=1.0\text{ N s} \end{aligned} \]

Its contribution is \(-1.0\text{ N s}\).

Step 3: Calculate the net impulse.

\[ \begin{aligned} J&=2.0-1.0\\ &=+1.0\text{ N s} \end{aligned} \]

Step 4: Calculate the mean net force.

The total time is \(0.40\text{ s}\).

\[ \begin{aligned} F_{\text{mean}} &=\frac{1.0}{0.40}\\ &=+2.5\text{ N} \end{aligned} \]

Step 5: Identify the peak force magnitude.

The largest absolute force is \(10\text{ N}\).

\[ \boxed{F_{\text{peak}}=10\text{ N}} \]

The positive impulse is greater than the negative impulse, so the overall momentum change is to the right.

Question 2

A force sensor records the following net force readings for a trolley.

Assume the force changes linearly between consecutive readings.

Calculate:

a) The net impulse.

b) The mean net force.

c) The peak force magnitude.

d) The mean force magnitude.

Solution 2

The net impulse is \(+1.6\text{ N s}\), the mean net force is \(+3.2\text{ N}\), the peak force magnitude is \(12\text{ N}\), and the mean force magnitude is \(6.4\text{ N}\).

Step 1: Calculate the positive areas.

From \(0\) to \(0.10\text{ s}\), there is a triangle:

\[ A_1=\frac12(0.10)(12)=0.60\text{ N s} \]

From \(0.10\) to \(0.20\text{ s}\), there is a rectangle:

\[ A_2=(0.10)(12)=1.20\text{ N s} \]

From \(0.20\) to \(0.30\text{ s}\), there is another triangle:

\[ A_3=\frac12(0.10)(12)=0.60\text{ N s} \]

The total positive area is:

\[ A_+=0.60+1.20+0.60 \]

\[ A_+=2.40\text{ N s} \]

Step 2: Calculate the negative areas.

The negative section contains two triangles.

The first is:

\[ A_4=\frac12(0.10)(8)=0.40\text{ N s} \]

The second is identical:

\[ A_5=\frac12(0.10)(8)=0.40\text{ N s} \]

Therefore:

\[ A_-=0.40+0.40=0.80\text{ N s} \]

Step 3: Calculate the net impulse.

\[ \begin{aligned} J&=A_+-A_-\\ &=2.40-0.80\\ &=+1.60\text{ N s} \end{aligned} \]

Step 4: Calculate the mean net force.

\[ \begin{aligned} F_{\text{mean}} &=\frac{1.60}{0.50}\\ &=+3.2\text{ N} \end{aligned} \]

Step 5: Calculate the peak force magnitude.

The largest force magnitude is:

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

Step 6: Calculate the mean force magnitude.

For this quantity, add the magnitudes of both positive and negative areas.

\[ \begin{aligned} F_{\text{mean magnitude}} &=\frac{2.40+0.80}{0.50}\\ &=6.4\text{ N} \end{aligned} \]

Therefore:

\[ \boxed{F_{\text{mean magnitude}}=6.4\text{ N}} \]

Notice that the mean net force and mean force magnitude are different.

The former accounts for direction and determines the net impulse. The latter measures the average strength of the force without allowing opposite directions to cancel.

Question 3

A \(1.0\text{ kg}\) cart initially moves to the left at \(1.0\text{ m/s}\).

The net force acting on it follows this pattern:

• From \(0\) to \(0.20\text{ s}\), the force forms a positive triangle with a peak of \(+20\text{ N}\) at \(0.10\text{ s}\).
• From \(0.20\) to \(0.40\text{ s}\), the force forms a negative triangle with a minimum of \(-20\text{ N}\) at \(0.30\text{ s}\).

Determine:

a) The net impulse.

b) The final velocity.

c) The mean net force and peak force magnitude.

d) Whether the cart changes direction during the interval.

Solution 3

The net impulse is zero, the final velocity is \(1.0\text{ m/s}\) to the left, the mean net force is zero, and the peak force magnitude is \(20\text{ N}\). However, the cart reverses direction twice during the interval.

Step 1: Calculate the positive impulse.

The positive triangle has a base of \(0.20\text{ s}\) and a height of \(20\text{ N}\).

\[ \begin{aligned} J_+&=\frac12(0.20)(20)\\ &=+2.0\text{ N s} \end{aligned} \]

Step 2: Calculate the negative impulse.

The negative triangle has the same dimensions.

\[ J_-=-\frac12(0.20)(20) \]

\[ J_-=-2.0\text{ N s} \]

Step 3: Calculate the net impulse.

\[ \begin{aligned} J&=J_++J_-\\ &=2.0-2.0\\ &=0 \end{aligned} \]

Therefore:

\[ \boxed{J=0\text{ N s}} \]

Step 4: Calculate the final velocity.

Take right as positive.

The initial velocity is:

\[ v_i=-1.0\text{ m/s} \]

The initial momentum is:

\[ p_i=mv_i=(1.0)(-1.0)=-1.0\text{ kg m s}^{-1} \]

Since the net impulse is zero:

\[ p_f=p_i+J \]

\[ p_f=-1.0+0=-1.0\text{ kg m s}^{-1} \]

Therefore:

\[ v_f=\frac{-1.0}{1.0} \]

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

The cart finishes travelling left at its original speed.

Step 5: Calculate the mean and peak forces.

The mean net force is:

\[ F_{\text{mean}}=\frac{0}{0.40}=0\text{ N} \]

The peak force magnitude is:

\[ \boxed{F_{\text{peak}}=20\text{ N}} \]

Step 6: Determine what happens during the interval.

This is the interesting part.

The cart starts with momentum:

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

The positive triangular pulse delivers \(+2.0\text{ N s}\).

Therefore, at the end of the positive pulse:

\[ p=-1.0+2.0=+1.0\text{ kg m s}^{-1} \]

The cart has reversed direction and is moving right.

The first half of that positive triangle delivers \(+1.0\text{ N s}\), so the cart momentarily reaches zero velocity at \(0.10\text{ s}\) before moving right.

The negative pulse then removes \(2.0\text{ N s}\), returning the cart to its original momentum. It momentarily stops again at \(0.30\text{ s}\), before moving left.

This demonstrates an important distinction:

Zero net impulse does not mean zero force or no change in motion during the interval.

It only means that the final momentum equals the initial momentum.

Question 4

A \(2.0\text{ kg}\) trolley initially moves to the right at \(3.0\text{ m/s}\) along a horizontal track.

A motorised actuator exerts:

• \(+10\text{ N}\) for the first \(0.20\text{ s}\).
• \(-6\text{ N}\) for the next \(0.30\text{ s}\).

A constant friction force of \(2.0\text{ N}\) acts to the left throughout the interval.

Assume the trolley continues moving to the right.

Calculate:

a) The impulse delivered by the actuator.

b) The impulse delivered by friction.

c) The net impulse on the trolley.

d) The final velocity.

e) The mean net force and peak net-force magnitude.

Solution 4

The actuator delivers \(+0.20\text{ N s}\), friction delivers \(-1.0\text{ N s}\), and the net impulse is \(-0.80\text{ N s}\). The final velocity is \(2.6\text{ m/s}\) to the right.

The important point is that the actuator force is not the same as the net force.

Step 1: Calculate the actuator impulse.

During the first interval:

\[ J_1=(10)(0.20)=+2.0\text{ N s} \]

During the second interval:

\[ J_2=(-6)(0.30)=-1.8\text{ N s} \]

Therefore:

\[ \begin{aligned} J_{\text{actuator}}&=2.0-1.8\\ &=+0.20\text{ N s} \end{aligned} \]

Step 2: Calculate the friction impulse.

Friction acts to the left throughout the entire \(0.50\text{ s}\).

\[ \begin{aligned} J_{\text{friction}}&=(-2.0)(0.50)\\ &=-1.0\text{ N s} \end{aligned} \]

Step 3: Calculate the net impulse.

Add the impulses from both forces:

\[ \begin{aligned} J_{\text{net}}&=J_{\text{actuator}}+J_{\text{friction}}\\ &=0.20-1.0\\ &=-0.80\text{ N s} \end{aligned} \]

Therefore:

\[ \boxed{J_{\text{net}}=-0.80\text{ N s}} \]

The net impulse acts to the left.

Step 4: Calculate the final velocity.

The initial momentum is:

\[ p_i=mv_i \]

\[ p_i=(2.0)(3.0)=6.0\text{ kg m s}^{-1} \]

The final momentum is:

\[ \begin{aligned} p_f&=p_i+J_{\text{net}}\\ &=6.0-0.80\\ &=5.2\text{ kg m s}^{-1} \end{aligned} \]

Therefore:

\[ \begin{aligned} v_f&=\frac{p_f}{m}\\ &=\frac{5.2}{2.0}\\ &=2.6\text{ m/s} \end{aligned} \]

So:

\[ \boxed{v_f=2.6\text{ m/s to the right}} \]

The trolley continues moving right, consistent with the stated assumption.

Step 5: Calculate the mean net force.

\[ \begin{aligned} F_{\text{mean}}&=\frac{-0.80}{0.50}\\ &=-1.6\text{ N} \end{aligned} \]

The mean net force is \(1.6\text{ N}\) to the left.

Step 6: Calculate the peak net-force magnitude.

During the first interval:

\[ F_{\text{net}}=10-2=+8\text{ N} \]

During the second interval:

\[ F_{\text{net}}=-6-2=-8\text{ N} \]

Therefore:

\[ \boxed{F_{\text{peak}}=8\text{ N}} \]

Notice that the actuator’s peak force magnitude is \(10\text{ N}\), but the peak net-force magnitude is \(8\text{ N}\).

Always check whether the question asks for an individual force or the net force.

Question 5

A \(2.0\text{ kg}\) trolley starts at \(1.0\text{ m/s}\) to the right on a frictionless horizontal track.

It undergoes two different force histories in separate trials.

Trial A

A constant net force of \(+4.0\text{ N}\) acts for \(0.50\text{ s}\).

Trial B

The net force changes as follows:

• \(+16\text{ N}\) for the first \(0.10\text{ s}\).
• \(-8\text{ N}\) for the next \(0.10\text{ s}\).
• \(+4.0\text{ N}\) for the final \(0.30\text{ s}\).

Assume each force is constant over its stated interval.

A student argues:

“The two trials have the same total impulse and duration, so the trolley must travel the same distance in both.”

Test this claim.

For each trial, calculate:

a) The net impulse.

b) The mean net force.

c) The peak force magnitude.

d) The final velocity.

e) The total displacement.

Finally, determine whether the net work done on the trolley is the same in both trials. Explain your reasoning.

Solution 5

The student is incorrect about the displacement.

Both trials have the same net impulse, mean net force, and final velocity. However, Trial A produces a displacement of \(0.75\text{ m}\), while Trial B produces a displacement of \(0.81\text{ m}\).

The reason is that equal impulse fixes the overall change in momentum, not the exact velocity at every instant.

Step 1: Calculate the impulse in Trial A.

The force is constant:

\[ \begin{aligned} J_A&=F\Delta t\\ &=(4.0)(0.50)\\ &=2.0\text{ N s} \end{aligned} \]

Step 2: Calculate the impulse in Trial B.

The first interval gives:

\[ J_1=(16)(0.10)=+1.6\text{ N s} \]

The second interval gives:

\[ J_2=(-8)(0.10)=-0.8\text{ N s} \]

The third interval gives:

\[ J_3=(4.0)(0.30)=+1.2\text{ N s} \]

Therefore:

\[ \begin{aligned} J_B&=1.6-0.8+1.2\\ &=2.0\text{ N s} \end{aligned} \]

Both trials have the same impulse:

\[ \boxed{J_A=J_B=2.0\text{ N s}} \]

Step 3: Compare the mean net forces.

Both trials last \(0.50\text{ s}\).

For Trial A:

\[ F_{\text{mean,A}}=\frac{2.0}{0.50}=4.0\text{ N} \]

For Trial B:

\[ F_{\text{mean,B}}=\frac{2.0}{0.50}=4.0\text{ N} \]

Therefore, both have the same mean net force.

Step 4: Compare the peak force magnitudes.

For Trial A:

\[ F_{\text{peak,A}}=4.0\text{ N} \]

For Trial B:

\[ F_{\text{peak,B}}=16\text{ N} \]

Trial B has four times the peak force magnitude, despite having the same mean force.

Step 5: Calculate the final velocity in both trials.

The initial momentum is:

\[ p_i=mv_i \]

\[ p_i=(2.0)(1.0)=2.0\text{ kg m s}^{-1} \]

Both trials deliver \(2.0\text{ N s}\) of impulse.

Therefore:

\[ p_f=2.0+2.0=4.0\text{ kg m s}^{-1} \]

The final velocity is:

\[ \begin{aligned} v_f&=\frac{4.0}{2.0}\\ &=2.0\text{ m/s} \end{aligned} \]

Thus:

\[ \boxed{v_{f,A}=v_{f,B}=2.0\text{ m/s}} \]

The final velocities must match because both trolleys have the same initial momentum and receive the same net impulse.

Step 6: Calculate the displacement in Trial A.

The acceleration is constant:

\[ a=\frac{F}{m} \]

\[ a=\frac{4.0}{2.0}=2.0\text{ m/s}^2 \]

Using the constant-acceleration displacement equation:

\[ s=ut+\frac12at^2 \]

Substitute:

\[ \begin{aligned} s_A&=(1.0)(0.50)\\ &\quad+\frac12(2.0)(0.50)^2\\ &=0.50+0.25\\ &=0.75\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{s_A=0.75\text{ m}} \]

Step 7: Calculate the displacement in Trial B.

This is where the problem becomes more interesting.

The acceleration changes between intervals, so we cannot use the constant-acceleration equation for the entire \(0.50\text{ s}\).

Instead, calculate the velocity and displacement during each interval.

First interval: \(0\) to \(0.10\text{ s}\).

The acceleration is:

\[ a_1=\frac{16}{2.0}=8.0\text{ m/s}^2 \]

The initial velocity is \(1.0\text{ m/s}\).

The final velocity of this interval is:

\[ \begin{aligned} v_1&=u+a_1t\\ &=1.0+(8.0)(0.10)\\ &=1.8\text{ m/s} \end{aligned} \]

The displacement is:

\[ \begin{aligned} s_1&=ut+\frac12a_1t^2\\ &=(1.0)(0.10)\\ &\quad+\frac12(8.0)(0.10)^2\\ &=0.14\text{ m} \end{aligned} \]

Second interval: \(0.10\) to \(0.20\text{ s}\).

The force is now negative.

The acceleration becomes:

\[ a_2=\frac{-8}{2.0}=-4.0\text{ m/s}^2 \]

The cart starts this interval at \(1.8\text{ m/s}\).

Its velocity after another \(0.10\text{ s}\) is:

\[ \begin{aligned} v_2&=1.8+(-4.0)(0.10)\\ &=1.4\text{ m/s} \end{aligned} \]

The displacement is:

\[ \begin{aligned} s_2&=(1.8)(0.10)\\ &\quad+\frac12(-4.0)(0.10)^2\\ &=0.16\text{ m} \end{aligned} \]

The trolley slows down but continues moving right.

Third interval: \(0.20\) to \(0.50\text{ s}\).

The acceleration is:

\[ a_3=\frac{4.0}{2.0}=2.0\text{ m/s}^2 \]

The initial velocity for this interval is \(1.4\text{ m/s}\).

The final velocity is:

\[ \begin{aligned} v_3&=1.4+(2.0)(0.30)\\ &=2.0\text{ m/s} \end{aligned} \]

The displacement is:

\[ \begin{aligned} s_3&=(1.4)(0.30)\\ &\quad+\frac12(2.0)(0.30)^2\\ &=0.51\text{ m} \end{aligned} \]

Now add the three displacements:

\[ \begin{aligned} s_B&=s_1+s_2+s_3\\ &=0.14+0.16+0.51\\ &=0.81\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{s_B=0.81\text{ m}} \]

Step 8: Explain why the displacements differ.

The two trials have identical:

• Initial velocities.
• Final velocities.
• Total time intervals.
• Net impulses.
• Mean net forces.

But their velocity-time graphs have different shapes.

Displacement is the area under a velocity-time graph. Therefore, different velocity histories can produce different displacements, even when the initial and final velocities match.

The tempting mistake is to use the average acceleration as though it were constant throughout the interval.

Both trials have an average acceleration of:

\[ a_{\text{mean}}=\frac{2.0-1.0}{0.50} =2.0\text{ m/s}^2 \]

However, Trial B does not have a constant acceleration. The constant-acceleration displacement equation cannot be applied to the entire interval using this average value.

Step 9: Compare the net work.

There is another useful result.

Both trials have the same initial and final speeds. Since the mass is also the same, their changes in kinetic energy must be equal.

Using the work-energy theorem:

\[ W_{\text{net}}=\Delta E_k \]

\[ \begin{aligned} W_{\text{net}} &=\frac12m(v_f^2-v_i^2)\\ &=\frac12(2.0)(2.0^2-1.0^2)\\ &=4.0-1.0\\ &=3.0\text{ J} \end{aligned} \]

Therefore:

\[ \boxed{W_A=W_B=3.0\text{ J}} \]

The net work is the same, even though the displacements differ.

This is consistent with the work-energy theorem: the total work depends on the initial and final kinetic energies, not on whether the force was constant.

The deeper lesson is that impulse and work describe different aspects of motion:

• Impulse relates force over time to change in momentum.
• Work relates force over displacement to change in kinetic energy.

Equal impulse does not guarantee equal displacement.

07What to remember when force changes sign

When a force-time graph crosses the horizontal axis, do not treat every region as positive.

Use signed areas:

\[ \boxed{J=A_+-A_-} \]

Then distinguish the three quantities:

Always check whether the graph represents the net force or an individual force, and make your positive direction clear before calculating.

For broader practice with force-time graphs, see Force-Time Graphs: Impulse and Momentum Change in HSC Physics.

The next useful connection is conservation of momentum. Once you can calculate the signed impulse on an individual object, you can investigate how interacting objects exchange momentum during collisions. See Impulse and Change in Momentum for HSC Physics.