How to Choose Between Force and Momentum Equations in HSC Physics

A 0.50 kg trolley hits a bumper and rebounds. Its velocity changes from \(+2.0\ \mathrm{m\,s^{-1}}\) to \(-1.0\ \mathrm{m\,s^{-1}}\) in just 0.10 seconds. A force sensor records a peak force of 25 N.

You might be tempted to multiply 25 N by 0.10 s to calculate the impulse. But that gives 2.5 N s, which doesn’t match the trolley’s actual change in momentum of 1.5 N s.

The problem? The peak force isn’t the average force.

This is exactly why choosing the right physics equation matters. Sometimes you need Newton’s second law. Sometimes you need the impulse-momentum relationship. And when multiple objects collide, conservation of momentum may be the simplest approach.

Here’s the distinction:

• Use Newton’s second law, \(F_{\text{net}}=ma\), when you know the net force on an object and want to calculate its acceleration, or vice versa.
• Use the impulse-momentum equation, \(\Delta p=F_{\text{net,avg}}\Delta t\), when you want to connect a force acting over a time interval with an object’s change in momentum.
• Use conservation of momentum, \(\sum p_{\text{before}}=\sum p_{\text{after}}\), when analysing an interaction between objects in an isolated system.

These equations are connected, not competing. The trick is recognising which physical situation each one describes.

01The three equations: what are they actually telling you?

Before choosing an equation, identify what you are trying to calculate. The three relationships describe different aspects of motion.

Let’s build up each one.

01When to use Newton’s second law: \(F_{\text{net}}=ma\)

Imagine pushing a shopping trolley along a supermarket aisle.

If you push harder, the trolley accelerates more. If the trolley is heavier, the same push produces less acceleration.

That’s Newton’s second law.

The important word is net. The equation uses the resultant force on an object, not necessarily the force you apply.

\[
\vec F_{\text{net}}=m\vec a
\]

Where:

• \(F_{\text{net}}\) = net force, measured in newtons (N).
• \(m\) = mass, measured in kilograms (kg).
• \(a\) = acceleration, measured in metres per second squared (\(\mathrm{m\,s^{-2}}\)).

For example, suppose you push a trolley to the right with 20 N, while friction acts to the left with 5 N.

The net force is:

\[
F_{\text{net}}=20-5=15\ \mathrm N
\]

You must use 15 N in Newton’s second law, not 20 N.

When should you reach for \(F_{\text{net}}=ma\)?

Use it when:

• The question asks for an object’s acceleration.
• The mass and individual forces acting on an object are known.
• You need to determine the net force required to produce a particular acceleration.
• You are analysing an object at rest or moving with constant velocity and need to establish equilibrium.

A particularly important point: an object can be moving without accelerating.

If its velocity is constant, its acceleration is zero, so:

\[
F_{\text{net}}=0
\]

This does not mean there are no forces. It means the forces balance.

02When to use the impulse-momentum equation

Now consider a different situation.

A tennis ball travels towards a player, hits their racquet, and flies backwards. During the collision, the racquet exerts a large force on the ball for a very short time.

You want to calculate the average force exerted by the racquet.

Newton’s second law could work if you knew the acceleration. But acceleration during a collision can change extremely rapidly.

Instead, you can compare the ball’s momentum immediately before and after contact.

Momentum is the product of mass and velocity:

\[
\vec p=m\vec v
\]

Where:

• \(p\) = momentum, measured in \(\mathrm{kg\,m\,s^{-1}}\).
• \(m\) = mass, measured in kg.
• \(v\) = velocity, measured in \(\mathrm{m\,s^{-1}}\).

The change in momentum is:

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

For an object with constant mass:

\[
\Delta\vec p=m(\vec v-\vec u)
\]

Here, \(u\) is the initial velocity and \(v\) is the final velocity.

Impulse describes the effect of a force acting over a period of time.

For a constant net force, or using the average net force over the interval:

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

Rearranging gives:

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

This is the equation to use when a question gives you a time interval and asks about the change in momentum or average force.

Impulse is measured in newton-seconds (N s).

Importantly:

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

Both units describe the same physical quantity.

Why does the time interval matter?

Imagine catching a cricket ball.

You can catch it by stopping your hands almost immediately, or by moving your hands backwards as the ball enters your gloves.

In both cases, the ball starts with the same momentum and finishes at rest. Therefore, the change in momentum is the same.

However, moving your hands backwards increases the time taken to stop the ball.

Since:

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

Increasing the stopping time decreases the average net force.

This is the physical principle behind cushioning and many protective designs.

An experiment that shows why impulse is useful

Consider an illustrative investigation using a trolley and a force sensor.

A trolley moves along a low-friction horizontal track towards a fixed bumper. A force sensor is attached between the bumper and the end stop. A light gate measures the trolley’s speed before impact and again after it rebounds.

The force sensor connects to a data logger, which records the force throughout the collision.

What would you expect the force sensor to record?

If the force remained constant throughout the collision, the force-time graph would be horizontal. But a real collision usually produces a changing force: it increases rapidly as the trolley compresses the bumper, reaches a maximum, and then decreases as contact ends.

This creates a problem for someone relying on a single force reading.

The maximum force alone cannot tell you the total impulse.

A better experimental design records the entire force-time relationship. The area under the force-time graph gives the impulse, which can then be compared with the change in momentum calculated from the measured velocities.

Let’s use some illustrative measurements.

Take right as positive. Assume friction is negligible, so the horizontal contact force is approximately the net horizontal force on the trolley.

Step 1: Calculate the change in momentum.

\[
\Delta p=m(v-u)
\]

\[
=0.50(-1.0-2.0)
\]

\[
=-1.5\ \mathrm{N\,s}
\]

Step 2: Calculate the average net force.

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

\[
=\frac{-1.5}{0.10}
\]

\[
=-15\ \mathrm N
\]

The average force on the trolley is 15 N to the left.

The sensor’s peak reading was 25 N, but the average force was only 15 N. Using the peak reading as though it acted throughout the collision would overestimate the impulse.

The investigation demonstrates why impulse is useful when the force changes during an interaction.

For a more detailed treatment of the graph and its area, see Force-Time Graphs: Impulse and Momentum Change in HSC Physics.

03When to use conservation of momentum

So far, we’ve considered the forces acting on individual objects.

But what happens when two objects collide?

Imagine two trolleys on a smooth track. One moves towards the other, they collide, and they stick together.

You know their masses and initial velocities. You want to calculate their shared final velocity.

Do you need to know the force between the trolleys?

No.

If the system experiences negligible net external impulse, its total momentum remains constant.

This is the law of conservation of momentum:

\[
\boxed{
\sum\vec p_{\text{before}}
=
\sum\vec p_{\text{after}}
}
\]

For two objects:

\[
m_1\vec u_1+m_2\vec u_2
=
m_1\vec v_1+m_2\vec v_2
\]

Where:

• \(m_1,m_2\) are the masses of the objects.
• \(u_1,u_2\) are their initial velocities.
• \(v_1,v_2\) are their final velocities.

The equation applies when the net external impulse on the chosen system is zero or negligible.

Why does momentum stay constant?

During a collision, each object exerts a force on the other.

Newton’s third law tells us these forces are equal in magnitude and opposite in direction.

Because the forces act over the same contact time, the impulses they produce are also equal and opposite.

One object gains the momentum that the other loses.

For the combined system, these internal impulses cancel.

If no significant external impulse acts on the system, its total momentum cannot change.

Does momentum conservation require an elastic collision?

No. This is a common misconception.

Momentum is conserved in both elastic and inelastic collisions, provided the system experiences negligible net external impulse.

Kinetic energy behaves differently:

In an inelastic collision, some initial kinetic energy is transformed into other forms, such as sound, heat, and deformation.

The momentum equation still works.

This distinction is particularly useful when a question tells you that objects stick together. You can immediately set their final velocities equal, without assuming that kinetic energy is conserved.

04How to choose the right equation in an HSC exam

When you encounter a mechanics question, don’t immediately start substituting numbers.

First, identify the object or system being analysed and what information the question gives you.

Use this decision table.

A reliable four-step decision process

Step 1: Identify the system.

Are you analysing one object, or multiple interacting objects?

This is probably the most important decision.

Step 2: Identify what you need to calculate.

Is it acceleration, average force, change in momentum, or final velocity?

Step 3: Check the assumptions.

• For Newton’s second law, calculate the net force on the object.
• For impulse, consider the net force acting over the specified time interval.
• For momentum conservation, check that the net external impulse on the system is negligible.

Step 4: Check your directions and units.

Momentum, force, acceleration, and impulse are vector quantities.

Choose a positive direction and maintain it throughout the calculation.

A result of \(-3.0\ \mathrm{m\,s^{-1}}\) means the object moves at \(3.0\ \mathrm{m\,s^{-1}}\) in the negative direction.

Can you use more than one equation in the same question?

Absolutely.

In fact, some of the better HSC questions require you to connect the equations.

For example, you might first use conservation of momentum to determine a trolley’s final velocity. Then you might use the impulse-momentum equation to calculate the average force during the collision.

The equations are different ways of describing related physical behaviour.

For constant mass, Newton’s second law and the impulse-momentum relationship are connected through:

\[
F_{\text{net,avg}}
=
m\frac{\Delta v}{\Delta t}
=
\frac{\Delta p}{\Delta t}
\]

The difference is the information you have and the time interval you are considering.

Three common mistakes to avoid

Mistake 1: Using the applied force instead of the net force.

If a trolley experiences a 30 N push and 10 N of friction, the net force is 20 N, not 30 N.

Mistake 2: Assuming momentum is always conserved.

Momentum is conserved for an isolated system. If a significant external impulse acts during the interaction, you must account for it.

Mistake 3: Assuming kinetic energy is always conserved during a collision.

Momentum conservation applies to elastic and inelastic collisions. Kinetic energy conservation is a special condition of elastic collisions.

For help locating the relevant relationships and checking their units, see How to Use the HSC Physics Data and Formulae Sheets.

05Syllabus note: 2017 versus 2025 Physics

There is an important distinction for NSW students studying during the syllabus transition.

This article follows the Physics Stage 6 Syllabus (2017), which is the syllabus being taught to Year 11 and Year 12 students in 2026.

Under this syllabus, Year 11 Module 2: Dynamics explicitly includes:

• Newton’s laws of motion, including Newton’s second law.
• The law of conservation of momentum.
• Impulse and its relationship to change in momentum.
• Force-time graphs.
• One-dimensional and two-dimensional interactions.
• Elastic and inelastic collisions.

The new Physics 11-12 Syllabus (2025) has a different implementation schedule:

The 2025 syllabus therefore does not replace the 2017 syllabus for every student at once.

If you are studying under the new syllabus, check the exact content for your focus area rather than assuming that every 2017 Module 2 dot point transfers unchanged.

Official references: NESA Physics Stage 6 Syllabus (2017) and NESA Physics 11-12 Syllabus (2025).

07Questions and solutions

The following questions progress from standard force calculations to more demanding collision problems. Pay particular attention to the system you choose and the assumptions that allow you to use each equation.

Question 1

A 5.0 kg block rests on a horizontal surface. A horizontal force of 28 N pushes the block to the right, while friction acts to the left with a magnitude of 8.0 N.

Calculate the block’s horizontal acceleration.

Solution 1

The block accelerates to the right at \(4.0\ \mathrm{m\,s^{-2}}\).

The question provides the mass and the horizontal forces acting on one object. It asks for acceleration, so Newton’s second law is the most direct choice.

Step 1: Calculate the net horizontal force.

Taking right as positive:

\[
F_{\text{net}}=28-8
\]

\[
F_{\text{net}}=20\ \mathrm N
\]

Step 2: Apply Newton’s second law.

\[
F_{\text{net}}=ma
\]

Rearrange:

\[
a=\frac{F_{\text{net}}}{m}
\]

Substitute:

\[
a=\frac{20}{5.0}
\]

\[
\boxed{a=4.0\ \mathrm{m\,s^{-2}}}
\]

The block accelerates to the right.

The tempting mistake is to use the full 28 N applied force. However, friction opposes the motion, so the block experiences a net force of only 20 N.

Question 2

A 0.16 kg ball travels horizontally to the right at \(18\ \mathrm{m\,s^{-1}}\). A bat exerts an average horizontal force of 480 N to the left on the ball for 0.010 s.

Assume other horizontal forces are negligible during contact.

Calculate:

a) The impulse delivered to the ball.

b) The ball’s final velocity, including its direction.

Solution 2

The impulse is \(4.8\ \mathrm{N\,s}\) to the left, and the ball rebounds at \(12\ \mathrm{m\,s^{-1}}\) to the left.

This question gives the force and contact time, so the impulse-momentum equation is the appropriate starting point.

Step 1: Choose a positive direction.

Take right as positive.

The force acting on the ball is therefore:

\[
F_{\text{net,avg}}=-480\ \mathrm N
\]

Step 2: Calculate the impulse.

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

\[
J=(-480)(0.010)
\]

\[
\boxed{J=-4.8\ \mathrm{N\,s}}
\]

The impulse has a magnitude of \(4.8\ \mathrm{N\,s}\), directed left.

Step 3: Calculate the initial momentum.

\[
p_i=mu
\]

\[
p_i=0.16(18)
\]

\[
p_i=2.88\ \mathrm{kg\,m\,s^{-1}}
\]

Step 4: Calculate the final momentum.

Using:

\[
\Delta p=p_f-p_i
\]

Rearrange:

\[
p_f=p_i+J
\]

\[
p_f=2.88-4.8
\]

\[
p_f=-1.92\ \mathrm{kg\,m\,s^{-1}}
\]

Step 5: Calculate the final velocity.

\[
p_f=mv
\]

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

\[
v=\frac{-1.92}{0.16}
\]

\[
\boxed{v=-12\ \mathrm{m\,s^{-1}}}
\]

The ball finishes moving at \(12\ \mathrm{m\,s^{-1}}\) to the left.

Notice that the final momentum is negative. The impulse was large enough to reverse the ball’s direction, rather than simply slow it down.

A common mistake is to assume that impulse equals final momentum. It actually equals the change in momentum.

Question 3

Two trolleys collide on a horizontal, low-friction track.

• Trolley A has a mass of 2.0 kg and initially travels right at \(5.0\ \mathrm{m\,s^{-1}}\).
• Trolley B has a mass of 1.0 kg and initially travels left at \(1.0\ \mathrm{m\,s^{-1}}\).
• After the collision, trolley A travels right at \(1.0\ \mathrm{m\,s^{-1}}\).
• The collision lasts 0.15 s.

Assume negligible external horizontal impulse during the collision.

Calculate:

a) The final velocity of trolley B.

b) The average contact force acting on trolley A.

c) The average contact force acting on trolley B.

Solution 3

Trolley B finishes moving right at \(7.0\ \mathrm{m\,s^{-1}}\). The average contact forces are 53.3 N in opposite directions.

This question combines conservation of momentum with the impulse-momentum equation.

First, find the unknown final velocity using the whole two-trolley system. Then analyse each trolley separately to calculate the contact forces.

Step 1: Establish a sign convention.

Take right as positive.

The initial velocities are:

\[
u_A=+5.0\ \mathrm{m\,s^{-1}}
\]

\[
u_B=-1.0\ \mathrm{m\,s^{-1}}
\]

Step 2: Calculate the initial total momentum.

\[
p_i=m_Au_A+m_Bu_B
\]

\[
=2.0(5.0)+1.0(-1.0)
\]

\[
p_i=9.0\ \mathrm{kg\,m\,s^{-1}}
\]

Step 3: Apply conservation of momentum.

\[
p_i=p_f
\]

\[
9.0=2.0(1.0)+1.0v_B
\]

\[
9.0=2.0+v_B
\]

\[
\boxed{v_B=+7.0\ \mathrm{m\,s^{-1}}}
\]

Trolley B moves to the right after the collision.

Step 4: Calculate the impulse on trolley A.

Now consider trolley A by itself.

\[
\Delta p_A=m_A(v_A-u_A)
\]

\[
=2.0(1.0-5.0)
\]

\[
\Delta p_A=-8.0\ \mathrm{N\,s}
\]

The average contact force on A is:

\[
F_{A,\text{avg}}=\frac{\Delta p_A}{\Delta t}
\]

\[
=\frac{-8.0}{0.15}
\]

\[
\boxed{F_{A,\text{avg}}=-53.3\ \mathrm N}
\]

The force acts to the left.

Step 5: Calculate the impulse on trolley B.

\[
\Delta p_B=m_B(v_B-u_B)
\]

\[
=1.0(7.0-(-1.0))
\]

\[
\Delta p_B=+8.0\ \mathrm{N\,s}
\]

Therefore:

\[
F_{B,\text{avg}}=
\frac{8.0}{0.15}
\]

\[
\boxed{F_{B,\text{avg}}=+53.3\ \mathrm N}
\]

The force acts to the right.

The forces are equal in magnitude and opposite in direction, as required by Newton’s third law.

The important reasoning step is that conservation of momentum gives the final velocity of B, while the impulse-momentum equation gives the force on each individual trolley.

Question 4

Two pucks collide on a smooth horizontal air-hockey table.

Puck A has a mass of 0.20 kg and initially travels east at \(6.0\ \mathrm{m\,s^{-1}}\). Puck B has a mass of 0.30 kg and is initially stationary.

After the collision, puck A travels north at \(2.0\ \mathrm{m\,s^{-1}}\).

Assuming negligible external horizontal impulse, calculate the speed and direction of puck B immediately after the collision.

Solution 4

Puck B moves at approximately \(4.2\ \mathrm{m\,s^{-1}}\), directed \(18.4^\circ\) south of east.

The key is that momentum is a vector. You cannot simply add or subtract the speeds as though the pucks move along the same straight line.

We must conserve momentum separately in the east-west and north-south directions.

Step 1: Choose coordinate directions.

Take:

• East as positive \(x\).
• North as positive \(y\).

Step 2: Calculate the initial total momentum.

Initially, puck B is stationary, so it has zero momentum.

For the east-west direction:

\[
p_{x,i}=0.20(6.0)
\]

\[
p_{x,i}=1.2\ \mathrm{kg\,m\,s^{-1}}
\]

For the north-south direction:

\[
p_{y,i}=0
\]

Step 3: Calculate puck A’s final momentum.

Puck A moves north at \(2.0\ \mathrm{m\,s^{-1}}\).

Its final momentum components are:

\[
p_{Ax,f}=0
\]

\[
p_{Ay,f}=0.20(2.0)=0.40
\]

Step 4: Determine puck B’s momentum components.

Conservation of momentum in the east-west direction gives:

\[
1.2=0+p_{Bx,f}
\]

Therefore:

\[
p_{Bx,f}=1.2\ \mathrm{kg\,m\,s^{-1}}
\]

In the north-south direction:

\[
0=0.40+p_{By,f}
\]

Therefore:

\[
p_{By,f}=-0.40\ \mathrm{kg\,m\,s^{-1}}
\]

The negative sign means puck B moves south.

Step 5: Calculate puck B’s velocity components.

Using \(p=mv\):

\[
v_{Bx}=\frac{1.2}{0.30}=4.0\ \mathrm{m\,s^{-1}}
\]

\[
v_{By}=\frac{-0.40}{0.30}
\]

\[
v_{By}=-1.33\ \mathrm{m\,s^{-1}}
\]

Step 6: Calculate the speed.

Use Pythagoras because the velocity components are perpendicular.

\[
v_B=\sqrt{v_{Bx}^2+v_{By}^2}
\]

\[
=\sqrt{4.0^2+1.33^2}
\]

\[
\boxed{v_B\approx4.2\ \mathrm{m\,s^{-1}}}
\]

Step 7: Calculate the direction.

\[
\tan\theta=\frac{1.33}{4.0}
\]

\[
\theta=\tan^{-1}(0.3325)
\]

\[
\boxed{\theta\approx18.4^\circ}
\]

Therefore, puck B travels at \(4.2\ \mathrm{m\,s^{-1}}\), \(18.4^\circ\) south of east.

The tempting mistake is to treat momentum as a scalar and calculate using the speeds alone. That loses the directional information.

Whenever an interaction involves two-dimensional motion, resolve momentum into perpendicular components before applying conservation.

Question 5

Two trolleys collide on a horizontal track.

Immediately before the collision:

• Trolley A has a mass of 1.0 kg and moves right at \(4.0\ \mathrm{m\,s^{-1}}\).
• Trolley B has a mass of 2.0 kg and moves left at \(1.0\ \mathrm{m\,s^{-1}}\).

Immediately after the collision:

• Trolley A moves left at \(2.0\ \mathrm{m\,s^{-1}}\).
• Trolley B moves right at \(1.0\ \mathrm{m\,s^{-1}}\).

The collision lasts 0.20 s.

During the entire collision, a brake pad attached to the track applies a constant 10 N horizontal force to trolley B, directed to the left. Assume there are no other significant external horizontal forces.

An excellent student argues:

“The forces between the two trolleys must be equal and opposite because of Newton’s third law. Therefore, their changes in momentum must also be equal and opposite. But the given velocities don’t satisfy that condition, so the collision data must be wrong.”

Is the student’s conclusion correct?

Determine:

a) The contact impulse acting on each trolley.

b) The average contact force acting on each trolley.

c) The total external impulse on the two-trolley system.

d) Whether the given collision data are physically consistent, explaining the student’s error.

Solution 5

The student’s conclusion is incorrect. The collision data are physically consistent because the brake applies an external impulse to trolley B. The contact impulses are equal and opposite, but the total momentum changes of the individual trolleys are not.

This is a system-boundary problem.

Newton’s third law describes the interaction forces between A and B. However, trolley B also experiences a separate external force from the brake.

We must account for both interactions.

Step 1: Establish the sign convention.

Take right as positive.

The given velocities are:

\[
u_A=+4.0,\qquad v_A=-2.0
\]

\[
u_B=-1.0,\qquad v_B=+1.0
\]

All velocities are in \(\mathrm{m\,s^{-1}}\).

Step 2: Calculate the change in momentum of trolley A.

\[
\Delta p_A=m_A(v_A-u_A)
\]

\[
=1.0(-2.0-4.0)
\]

\[
\boxed{\Delta p_A=-6.0\ \mathrm{N\,s}}
\]

The only significant horizontal force acting on A during the collision is the contact force from B.

Therefore, the contact impulse on A is:

\[
J_{B\rightarrow A}=-6.0\ \mathrm{N\,s}
\]

Step 3: Apply Newton’s third law.

The contact force exerted by A on B must be equal in magnitude and opposite in direction to the force exerted by B on A.

Because both forces act over the same contact time, their impulses are also equal and opposite.

Therefore:

\[
\boxed{J_{A\rightarrow B}=+6.0\ \mathrm{N\,s}}
\]

The contact impulse on B is 6.0 N s to the right.

Step 4: Calculate the change in momentum of trolley B.

\[
\Delta p_B=m_B(v_B-u_B)
\]

\[
=2.0(1.0-(-1.0))
\]

\[
\boxed{\Delta p_B=+4.0\ \mathrm{N\,s}}
\]

Notice that B’s total momentum change is only \(+4.0\ \mathrm{N\,s}\), even though its contact impulse is \(+6.0\ \mathrm{N\,s}\).

Why?

Because the brake also applies an impulse to B.

Step 5: Calculate the external impulse from the brake.

\[
J_{\text{brake}}=F_{\text{brake}}\Delta t
\]

\[
=(-10)(0.20)
\]

\[
\boxed{J_{\text{brake}}=-2.0\ \mathrm{N\,s}}
\]

The total impulse on trolley B is therefore:

\[
\Delta p_B=J_{A\rightarrow B}+J_{\text{brake}}
\]

\[
=6.0-2.0
\]

\[
=+4.0\ \mathrm{N\,s}
\]

This matches the momentum change calculated directly from B’s initial and final velocities.

The two calculations agree.

Step 6: Calculate the average contact forces.

For trolley A:

\[
F_{A,\text{contact}}=
\frac{-6.0}{0.20}
\]

\[
\boxed{F_{A,\text{contact}}=-30\ \mathrm N}
\]

For trolley B:

\[
F_{B,\text{contact}}=
\frac{+6.0}{0.20}
\]

\[
\boxed{F_{B,\text{contact}}=+30\ \mathrm N}
\]

The contact forces are equal in magnitude and opposite in direction, as Newton’s third law requires.

Trolley B also experiences the separate 10 N brake force to the left. Its average net force is therefore:

\[
F_{\text{net},B}=30-10
\]

\[
F_{\text{net},B}=20\ \mathrm N
\]

This also agrees with:

\[
F_{\text{net},B}=\frac{\Delta p_B}{\Delta t}
=\frac{4.0}{0.20}=20\ \mathrm N
\]

Step 7: Check the momentum change of the complete system.

Now consider both trolleys together as one system.

Initial total momentum:

\[
p_i=1.0(4.0)+2.0(-1.0)
\]

\[
p_i=+2.0\ \mathrm{kg\,m\,s^{-1}}
\]

Final total momentum:

\[
p_f=1.0(-2.0)+2.0(1.0)
\]

\[
p_f=0
\]

Therefore:

\[
\Delta p_{\text{system}}=0-2.0
\]

\[
\boxed{\Delta p_{\text{system}}=-2.0\ \mathrm{N\,s}}
\]

This equals the external impulse from the brake:

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

The system’s momentum has changed by exactly the amount predicted by the external impulse.

The student’s mistake was assuming that equal and opposite contact impulses must produce equal and opposite total momentum changes for the two objects.

Newton’s third law only guarantees that the interaction forces are equal and opposite. It does not eliminate other forces acting on either object.

For the combined system, the internal contact impulses cancel, but the brake’s external impulse remains.

This is why choosing the correct system boundary is just as important as choosing the correct equation.

The most useful habit to develop is to identify the object or system before looking for a formula. Newton’s second law describes how net force produces acceleration, impulse connects force and momentum change over time, and conservation of momentum describes the behaviour of an isolated system.

Once you can distinguish those three situations, the next step is becoming comfortable with force-time graphs, particularly when a collision involves a changing force rather than a constant one.