Why Colliding Objects Receive Equal and Opposite Impulses

Put a 0.50 kg cart and a 1.50 kg cart on a track and let them collide. The lighter cart might shoot backwards while the heavier cart barely changes speed. Does that mean the lighter cart receives a larger impulse?

No. During the collision, the force exerted by the first cart on the second is always equal in magnitude and opposite in direction to the force exerted by the second cart on the first. Because both forces act over the same time interval, the impulses they produce are also equal in magnitude and opposite in direction.

This is a consequence of Newton’s third law. It remains true regardless of the objects’ masses, initial velocities, or final velocities.

The important distinction is between the impulse from the collision interaction and the total impulse on each object. We’ll use a force-time experiment to see why that distinction matters.

01The experiment: measuring both sides of a collision

Imagine testing the collision using two low-friction carts on a horizontal track.

Cart A is on the left, and cart B is on the right. Each cart has a force sensor mounted on its front, with the sensors facing each other. When the carts collide, the sensors compress and measure the contact force.

Both sensors connect to the same data logger, which records force against time.

Take right as the positive direction. The logger records:

• The force exerted on A by B, which acts to the left.
• The force exerted on B by A, which acts to the right.

What should the experiment show?

Suppose cart A is much heavier than cart B.

You might predict that A exerts a greater force because it is harder to accelerate. Alternatively, you might think the lighter cart experiences a greater force because its velocity changes more dramatically.

Neither prediction is correct.

Newton’s third law states that when two objects interact, the force exerted by the first object on the second is equal in magnitude and opposite in direction to the force exerted by the second object on the first.

The two forces are a pair. They act simultaneously, on different objects, and have the same magnitude.

A first experimental design might place a force sensor on only cart A. This would measure the force experienced by A, but it would not independently test what happens to B.

Adding a second sensor allows us to compare both forces directly. For a reliable comparison, the sensors must be calibrated, their time measurements synchronised, and their readings compared over the same collision interval.

What the force-time measurements tell us

Consider the following illustrative sensor readings from a collision lasting 20 ms.

Take right as positive. Assume the force changes linearly between consecutive measurements.

The readings show that whenever the force on A increases in magnitude, the force on B increases by exactly the same amount in the opposite direction.

But there’s another important observation: the maximum force is not the average force.

The peak force has a magnitude of 600 N. However, the average force over the collision is only 300 N.

Why?

Because impulse depends on the entire force-time graph, not just its highest point.

02Why the impulses are equal and opposite

Impulse is the area under a force-time graph.

For a changing force, the impulse is calculated by integrating the force over the time interval:

\[ J=\int_{t_0}^{t_1}F(t)\,dt \]

Here:

• \(J\) is impulse, measured in newton-seconds (N s).
• \(F(t)\) is the force at a particular time, measured in newtons (N).
• \(t\) is time, measured in seconds (s).
• \(t_0\) and \(t_1\) are the beginning and end of the collision.

The area is signed. A force below the time axis produces negative impulse, while a force above the time axis produces positive impulse.

Let’s apply this to the two carts.

Step 1: Calculate the impulse on cart A.

Using the trapezoidal rule, the magnitude of the area is:

\[ \begin{aligned} |J_A|={}&\frac{0+300}{2}(0.005)\\ &+\frac{300+600}{2}(0.005)\\ &+\frac{600+300}{2}(0.005)\\ &+\frac{300+0}{2}(0.005) \end{aligned} \]

Therefore,

\[ |J_A|=0.75+2.25+2.25+0.75 \]

\[ |J_A|=6.0\text{ N s} \]

Since the force on A acts to the left:

\[ J_A=-6.0\text{ N s} \]

Step 2: Calculate the impulse on cart B.

The force on B has the same magnitude at every instant, but acts to the right.

Therefore,

\[ J_B=+6.0\text{ N s} \]

Step 3: Compare the results.

\[ \boxed{J_A=-J_B} \]

The impulses are equal in magnitude and opposite in direction.

This is not a coincidence caused by the particular shape of the graph. It follows directly from Newton’s third law.

Mathematically, at every instant during the collision:

\[ \vec F_{A\leftarrow B}(t) = -\vec F_{B\leftarrow A}(t) \]

Integrating both sides over the same collision interval gives:

\[ \boxed{ \vec J_{A\leftarrow B} = -\vec J_{B\leftarrow A} } \]

The same force acts over the same time interval, so the signed areas must be equal and opposite.

Notice that the masses do not appear anywhere in this proof. The equality comes from the interaction between the objects, not their individual properties.

03Peak force versus mean force

This is an important distinction in HSC Physics.

Two collisions can produce the same impulse but have very different peak forces.

The mean force over a collision is:

\[ \boxed{\bar F=\frac{J}{\Delta t}} \]

where:

• \(\bar F\) is the mean force in newtons.
• \(J\) is the signed impulse in newton-seconds.
• \(\Delta t\) is the collision duration in seconds.

The peak force, on the other hand, is the greatest instantaneous force recorded during the collision.

Consider two different collisions with triangular force-time profiles.

Both collisions produce the same impulse magnitude:

\[ J=\frac12 F_{\text{peak}}\Delta t \]

For collision A:

\[ J=\frac12(1000)(0.008)=4.0\text{ N s} \]

For collision B:

\[ J=\frac12(400)(0.020)=4.0\text{ N s} \]

The longer collision produces the same impulse with a lower peak force.

For a triangular force-time graph, the mean force magnitude is half the peak force. This is a property of a triangular pulse, not a universal rule for every collision.

There is another subtle point.

Within one collision, the two objects experience equal and opposite peak forces as well as equal and opposite impulses. Newton’s third law applies at every instant, so their force-time curves are mirror images.

However, two different collisions can have the same impulse but different peak forces.

Don’t confuse these two comparisons.

04When the net impulses are not equal and opposite

So far, we’ve focused on the forces the two colliding objects exert on each other.

But what if other forces act during the collision?

Imagine cart A colliding with cart B while B is pressed against a wall.

The force exerted by A on B still has an equal and opposite partner: the force exerted by B on A.

However, B also experiences a force from the wall.

Its total impulse is therefore the sum of two contributions:

\[ J_{\text{net},B} = J_{A\rightarrow B} + J_{\text{wall}\rightarrow B} \]

The impulse from the collision interaction is not necessarily equal to the net impulse on B.

This is the key distinction:

• The interaction impulses between A and B are always equal and opposite.
• The net impulses on A and B are equal and opposite only when other external impulses can be neglected.

If the two-cart system experiences a significant external impulse, its total momentum changes.

This is also why the forces in a Newton’s third law pair do not cancel on either individual cart. They act on different objects. They cancel when considering the combined A-B system.

For the broader relationship between impulse and momentum, see Impulse and Change in Momentum for HSC Physics.

07Questions and solutions

The following questions progress from standard HSC calculations to more challenging applications of Newton’s third law, force-time graphs, and the impulse-momentum theorem.

Question 1

Two carts, A and B, collide on a horizontal track.

The force exerted on A by B has a triangular force-time profile with a peak magnitude of 420 N and a duration of 10 ms.

The force on A acts to the left, and the force on B acts to the right.

Take right as positive.

Calculate:

a) The impulse experienced by each cart.

b) The mean force experienced by each cart during the collision.

Solution 1

The impulse on A is \(-2.1\text{ N s}\), and the impulse on B is \(+2.1\text{ N s}\). Their mean forces are \(-210\text{ N}\) and \(+210\text{ N}\), respectively.

Part a: Calculate the impulses.

Step 1: Convert the collision duration into seconds.

\[ \Delta t=10\text{ ms}=0.010\text{ s} \]

Step 2: Calculate the triangular area.

\[ |J|=\frac12F_{\text{peak}}\Delta t \]

\[ |J|=\frac12(420)(0.010) \]

\[ |J|=2.1\text{ N s} \]

Step 3: Assign the signs.

The force on A acts left:

\[ J_A=-2.1\text{ N s} \]

The force on B acts right:

\[ J_B=+2.1\text{ N s} \]

Part b: Calculate the mean forces.

Use:

\[ \bar F=\frac{J}{\Delta t} \]

For A:

\[ \bar F_A=\frac{-2.1}{0.010} \]

\[ \boxed{\bar F_A=-210\text{ N}} \]

For B:

\[ \bar F_B=\frac{+2.1}{0.010} \]

\[ \boxed{\bar F_B=+210\text{ N}} \]

The mean forces are equal in magnitude and opposite in direction, just like the instantaneous forces throughout the collision.

Question 2

Two force sensors measure the interaction forces during a collision between carts A and B.

The following table shows the magnitude of the contact force at different times.

Assume the force varies linearly between consecutive measurements.

The force on A acts to the left, while the force on B acts to the right.

Calculate:

a) The magnitude of the impulse experienced by each cart.

b) The mean force magnitude over the collision.

c) The ratio of peak force to mean force magnitude.

d) Explain why the peak force cannot be used directly to calculate the impulse.

Solution 2

The impulse magnitudes are \(3.9\text{ N s}\), the mean force magnitude is 260 N, and the peak-to-mean ratio is approximately 1.92.

Part a: Calculate the impulse using the trapezoidal areas.

The area of each trapezium is:

\[ A=\frac12(F_1+F_2)\Delta t \]

Convert the time intervals into seconds before calculating.

From 0 to 3 ms:

\[ A_1=\frac12(0+500)(0.003) \]

\[ A_1=0.75\text{ N s} \]

From 3 to 7 ms:

\[ A_2=\frac12(500+300)(0.004) \]

\[ A_2=1.60\text{ N s} \]

From 7 to 12 ms:

\[ A_3=\frac12(300+200)(0.005) \]

\[ A_3=1.25\text{ N s} \]

From 12 to 15 ms:

\[ A_4=\frac12(200+0)(0.003) \]

\[ A_4=0.30\text{ N s} \]

Add the areas:

\[ |J|=0.75+1.60+1.25+0.30 \]

\[ \boxed{|J|=3.90\text{ N s}} \]

Therefore:

\[ J_A=-3.90\text{ N s} \]

\[ J_B=+3.90\text{ N s} \]

Part b: Calculate the mean force magnitude.

The collision lasts:

\[ \Delta t=15\text{ ms}=0.015\text{ s} \]

Using:

\[ \bar F=\frac{|J|}{\Delta t} \]

\[ \bar F=\frac{3.90}{0.015} \]

\[ \boxed{\bar F=260\text{ N}} \]

Part c: Calculate the peak-to-mean ratio.

The peak force is 500 N.

\[ \text{Ratio}= \frac{F_{\text{peak}}}{\bar F} \]

\[ =\frac{500}{260} \]

\[ \boxed{\text{Ratio}\approx1.92} \]

Part d: Explain the difference between peak force and impulse.

The peak force is the greatest instantaneous force during the collision. It tells us the maximum force experienced at one moment.

Impulse, however, depends on the entire force-time graph.

A peak force of 500 N does not mean that a force of 500 N acts throughout the collision. The force changes over time, so using the peak value multiplied by the full duration would overestimate the impulse.

The correct method is to calculate the area under the force-time graph.

Question 3

Two carts collide on a rough horizontal track.

Cart A has a mass of 0.50 kg and initially moves right at 4.0 m/s. Cart B has a mass of 1.00 kg and is initially stationary.

During the collision:

• The impulse exerted by B on A is \(-2.4\text{ N s}\).
• The impulse exerted by A on B is \(+2.4\text{ N s}\).
• Friction produces an additional impulse of \(-0.10\text{ N s}\) on A.
• Friction produces an additional impulse of \(-0.30\text{ N s}\) on B.

Both friction impulses act to the left.

Calculate:

a) The net impulse on each cart.

b) The final velocity of each cart.

c) The change in total momentum of the two-cart system.

d) Explain whether the unequal net impulses contradict Newton’s third law.

Solution 3

The net impulses are \(-2.50\text{ N s}\) on A and \(+2.10\text{ N s}\) on B. Their final velocities are \(-1.0\text{ m/s}\) and \(+2.1\text{ m/s}\), respectively.

The unequal net impulses do not contradict Newton’s third law because friction produces different additional impulses on the two carts.

Part a: Calculate the net impulse on each cart.

The net impulse includes every force acting on the object.

For cart A:

\[ J_{\text{net},A} = J_{B\rightarrow A}+J_{\text{friction},A} \]

\[ J_{\text{net},A}=-2.4-0.10 \]

\[ \boxed{J_{\text{net},A}=-2.50\text{ N s}} \]

For cart B:

\[ J_{\text{net},B} = J_{A\rightarrow B}+J_{\text{friction},B} \]

\[ J_{\text{net},B}=2.4-0.30 \]

\[ \boxed{J_{\text{net},B}=+2.10\text{ N s}} \]

Notice that the contact impulses are still equal and opposite. It is the additional friction impulses that cause the net impulses to differ.

Part b: Calculate the final velocities.

Use:

\[ J_{\text{net}}=p_f-p_i \]

For cart A, the initial momentum is:

\[ p_{A,i}=0.50(4.0) \]

\[ p_{A,i}=2.0\text{ kg m/s} \]

Its final momentum is:

\[ p_{A,f}=2.0-2.50 \]

\[ p_{A,f}=-0.50\text{ kg m/s} \]

Therefore:

\[ v_{A,f}=\frac{-0.50}{0.50} \]

\[ \boxed{v_{A,f}=-1.0\text{ m/s}} \]

Cart A reverses direction.

For cart B, the initial momentum is zero.

\[ p_{B,f}=0+2.10 \]

\[ p_{B,f}=2.10\text{ kg m/s} \]

Therefore:

\[ v_{B,f}=\frac{2.10}{1.00} \]

\[ \boxed{v_{B,f}=+2.1\text{ m/s}} \]

Part c: Calculate the change in total momentum.

The initial total momentum is:

\[ p_i=2.0+0=2.0\text{ kg m/s} \]

The final total momentum is:

\[ p_f=-0.50+2.10 \]

\[ p_f=1.60\text{ kg m/s} \]

Therefore:

\[ \Delta p_{\text{total}}=1.60-2.0 \]

\[ \boxed{\Delta p_{\text{total}}=-0.40\text{ kg m/s}} \]

Part d: Explain why Newton’s third law is still satisfied.

Newton’s third law applies to the contact forces between A and B.

Their interaction impulses are:

\[ J_{B\rightarrow A}=-2.4\text{ N s} \]

\[ J_{A\rightarrow B}=+2.4\text{ N s} \]

These are equal and opposite.

However, friction is an external force on the two-cart system. Its total impulse is:

\[ J_{\text{friction,total}}=-0.10-0.30 \]

\[ J_{\text{friction,total}}=-0.40\text{ N s} \]

This exactly matches the change in total momentum.

The tempting mistake is to assume that the total momentum changes of the two objects must always be equal and opposite. That is only true when their other external impulses can be neglected.

Question 4

Two students investigate Newton’s third law using force sensors mounted on two colliding carts.

The sensors are calibrated, but each is connected to a separate data logger.

The actual contact force has a triangular time profile:

• The collision begins at \(t=0\).
• The collision ends at \(t=12\text{ ms}\).
• The peak force magnitude is 600 N at \(t=6\text{ ms}\).

Logger A records from 0 to 12 ms.

Logger B records from 2 to 14 ms.

The force on A is negative, while the force on B is positive.

a) Calculate the impulse recorded by logger A.

b) Calculate the impulse recorded by logger B.

c) Explain why the different readings do not contradict Newton’s third law.

d) Describe how the experimental design should be improved.

Solution 4

Logger A records \(-3.6\text{ N s}\), while logger B records \(+3.4\text{ N s}\). The difference is caused by the recording windows, not a failure of Newton’s third law.

Part a: Calculate the impulse recorded by logger A.

Logger A records the complete triangular pulse.

\[ |J_A|=\frac12F_{\text{peak}}\Delta t \]

\[ |J_A|=\frac12(600)(0.012) \]

\[ |J_A|=3.6\text{ N s} \]

The force on A acts left, so:

\[ \boxed{J_A=-3.6\text{ N s}} \]

Part b: Calculate the impulse recorded by logger B.

Logger B starts recording at 2 ms, so it misses the first part of the collision.

During the rising section, the force increases linearly from zero at 0 ms to 600 N at 6 ms.

Therefore, at 2 ms:

\[ F=\frac{2}{6}(600) \]

\[ F=200\text{ N} \]

The missing area is a small triangle with a base of 0.002 s and a height of 200 N.

\[ J_{\text{missing}} = \frac12(200)(0.002) \]

\[ J_{\text{missing}}=0.20\text{ N s} \]

Logger B records the remaining area:

\[ J_B=3.6-0.20 \]

\[ \boxed{J_B=+3.4\text{ N s}} \]

The positive sign is used because the force on B acts to the right.

Part c: Explain the discrepancy.

Newton’s third law has not been violated.

The actual forces on the two carts are equal and opposite at every instant. However, the students have not integrated the forces over the same time interval.

Logger A captures the entire collision, while logger B misses the initial part.

The experiment therefore produces different measured impulses, even though the actual interaction impulses are equal and opposite.

Part d: Improve the experiment.

The students should:

• Connect both force sensors to the same data logger, if possible.
• Otherwise, synchronise the two recording systems using a common trigger.
• Use identical start and end times that capture the complete collision.
• Check the zero readings and calibration of both sensors.
• Use sufficiently rapid sampling to capture the changing force accurately.

With both sensors recording the complete collision, the measured impulses should be approximately \(-3.6\text{ N s}\) and \(+3.6\text{ N s}\), within experimental uncertainty.

Question 5

A 1.50 kg cart A moves right at 4.0 m/s and collides with a 2.00 kg cart B, which is held stationary against a rigid wall.

Take right as positive.

During the collision:

• The A-B contact force increases linearly from zero to a peak magnitude of 1000 N at 6 ms.
• It then decreases linearly to zero at 12 ms.
• Cart B remains stationary throughout the collision because the wall prevents it from moving.
• The wall provides whatever horizontal force is necessary to keep B stationary.
• All other horizontal forces and impulses are negligible.

Answer the following:

a) Calculate the impulse exerted by A on B and the impulse exerted by B on A.

b) Calculate the impulse exerted by the wall on B.

c) Determine the net impulse on each cart and their final velocities.

d) Calculate the mean and peak magnitudes of the A-B contact force.

e) A student argues that because B remains stationary, its impulse must be zero. Therefore, they claim that A cannot have received a non-zero impulse from B. Explain precisely why this reasoning is incorrect.

f) Is the momentum of the two-cart system conserved during the collision? Explain where the transferred momentum goes.

Solution 5

The A-B interaction impulses are \(-6.0\text{ N s}\) on A and \(+6.0\text{ N s}\) on B. The wall supplies an additional impulse of \(-6.0\text{ N s}\) on B, leaving B with zero net impulse.

This is a particularly important distinction: equal and opposite interaction impulses do not guarantee equal and opposite net impulses when a third object also interacts with the system.

Part a: Calculate the A-B interaction impulses.

The contact force-time graph is triangular.

Its base is:

\[ \Delta t=12\text{ ms}=0.012\text{ s} \]

Its height is:

\[ F_{\text{peak}}=1000\text{ N} \]

The impulse magnitude is:

\[ |J|=\frac12F_{\text{peak}}\Delta t \]

Substituting:

\[ |J|=\frac12(1000)(0.012) \]

\[ |J|=6.0\text{ N s} \]

The force on A acts left, while the force on B acts right.

Therefore:

\[ \boxed{J_{B\rightarrow A}=-6.0\text{ N s}} \]

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

The interaction impulses are equal and opposite, exactly as Newton’s third law predicts.

Part b: Calculate the impulse exerted by the wall on B.

Cart B remains stationary throughout the collision.

Therefore, its initial and final momenta are both zero.

Using the impulse-momentum theorem:

\[ J_{\text{net},B}=\Delta p_B \]

\[ J_{\text{net},B}=0 \]

However, B experiences two separate impulses:

\[ J_{\text{net},B} = J_{A\rightarrow B} + J_{\text{wall}\rightarrow B} \]

Substitute the known values:

\[ 0=6.0+J_{\text{wall}\rightarrow B} \]

Therefore:

\[ \boxed{J_{\text{wall}\rightarrow B}=-6.0\text{ N s}} \]

The wall exerts an impulse to the left that exactly balances the impulse from A.

Part c: Calculate the net impulse and final velocity of each cart.

For cart A, there are no other significant horizontal impulses.

Therefore:

\[ J_{\text{net},A}=-6.0\text{ N s} \]

Its initial momentum is:

\[ p_{A,i}=mv \]

\[ p_{A,i}=1.50(4.0) \]

\[ p_{A,i}=6.0\text{ kg m/s} \]

Its final momentum is:

\[ p_{A,f}=p_{A,i}+J_{\text{net},A} \]

\[ p_{A,f}=6.0-6.0 \]

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

Therefore:

\[ \boxed{v_{A,f}=0\text{ m/s}} \]

Cart A comes to rest.

For cart B:

\[ J_{\text{net},B}=+6.0-6.0 \]

\[ J_{\text{net},B}=0 \]

Therefore, its momentum remains zero, and:

\[ \boxed{v_{B,f}=0\text{ m/s}} \]

Both carts finish stationary.

Part d: Calculate the mean and peak contact forces.

The peak contact force magnitude is given:

\[ \boxed{F_{\text{peak}}=1000\text{ N}} \]

The mean contact force magnitude is:

\[ \bar F=\frac{|J|}{\Delta t} \]

\[ \bar F=\frac{6.0}{0.012} \]

\[ \boxed{\bar F=500\text{ N}} \]

The mean is half the peak because the contact force has a triangular time profile.

Part e: Explain why the student’s reasoning is incorrect.

The student’s mistake is treating the net impulse on B as though it were the impulse from A alone.

B receives a rightward impulse of \(+6.0\text{ N s}\) from A.

At the same time, B receives a leftward impulse of \(-6.0\text{ N s}\) from the wall.

These impulses cancel, leaving B with zero net impulse.

That does not mean either individual impulse was zero.

The two Newton’s third law pairs are:

• A exerts a force on B, and B exerts an equal and opposite force on A.
• The wall exerts a force on B, and B exerts an equal and opposite force on the wall.

The wall’s force on B is not the reaction force to A’s force on B. Its reaction force is B’s force on the wall.

This is the crucial distinction between an interaction impulse and a net impulse.

Part f: Is momentum conserved for the two-cart system?

No. The momentum of the two-cart system is not conserved because the wall exerts an external impulse on the system.

Initially, the total momentum is:

\[ p_i=6.0+0=6.0\text{ kg m/s} \]

Finally:

\[ p_f=0+0=0 \]

Therefore:

\[ \Delta p_{\text{system}}=-6.0\text{ kg m/s} \]

This equals the external impulse from the wall:

\[ J_{\text{external}}=-6.0\text{ N s} \]

The momentum has been transferred to the wall and ultimately the Earth. The Earth’s velocity change is negligible because of its enormous mass, but the momentum transfer still occurs.

If we include the Earth and wall in the complete system, momentum is conserved, assuming negligible external impulses on that larger system.

The important lesson is that Newton’s third law is never violated. What changes is the system boundary and which forces contribute to the net impulse.

08The key idea to remember

Whenever you see two objects colliding in an HSC Physics question, separate the interaction forces from the total forces acting on each object.

The interaction forces are always equal and opposite. Integrating them over the same collision interval gives equal and opposite impulses:

\[ \boxed{J_{A\rightarrow B}=-J_{B\rightarrow A}} \]

But if the question asks about the net impulse or change in momentum of either object, you must account for every force acting on it.

Also remember:

• Impulse is the signed area under a force-time graph.
• Mean force is impulse divided by collision duration.
• Peak force is the maximum instantaneous force.
• Equal impulses do not necessarily produce equal changes in velocity.
• Equal and opposite interaction impulses do not necessarily mean equal and opposite net impulses.

Once you can distinguish these ideas, force-time graph questions become much easier to interpret. The next useful step is applying the impulse-momentum theorem to collisions where the force changes over time, particularly when comparing different collision durations and force profiles.