Can a Collision Have Zero Net Impulse but a Large Force?

Two carts collide. For a few milliseconds, each experiences a contact force of hundreds of newtons. Yet if you treat both carts as one system, the collision can produce zero net impulse.

Is that possible? Yes. The large contact forces act in equal and opposite directions on the two carts. When you consider the pair together, those forces cancel. Each cart individually, however, experiences a substantial impulse and changes momentum.

There is another possibility. A single object can experience large forces in opposite directions at different times, with the positive and negative areas under its force-time graph cancelling to give zero net impulse.

The key is to distinguish the force acting at an instant, the impulse accumulated over time, and the system experiencing that impulse.

01How can large forces produce zero net impulse?

Imagine two skaters standing on ice. They push against each other’s hands and move apart.

One skater accelerates to the left. The other accelerates to the right. Both experience a force, and both experience a change in momentum.

However, the momentum gained by one skater is equal and opposite to the momentum gained by the other, assuming external horizontal forces are negligible.

A collision between two carts follows the same principle.

Newton’s Third Law tells us that when cart A exerts a force on cart B, cart B exerts an equal and opposite force on cart A.

At every instant during contact:

\[ F_{A\text{ on }B}=-F_{B\text{ on }A} \]

These forces can both be very large.

The important detail is that they act on different objects.

The system you choose changes the answer

A system is simply the object or group of objects whose motion you are investigating.

Consider a collision between two carts.

There are three different ways to analyse this situation.

This is the central distinction.

For the two-cart system:

\[ J_A+J_B=0 \]

Here, \(J_A\) and \(J_B\) are the contact impulses experienced by carts A and B.

The combined system can therefore have zero net impulse even though each cart experiences a large impulse.

More generally:

\[ J_{\text{system}}=J_{\text{external}} \]

Internal forces cancel when considering the complete system. External forces do not necessarily cancel.

This is why momentum is conserved for an isolated system, even when the objects inside it experience substantial forces.

02Why the area under a force-time graph matters

Now consider a different question.

Suppose a force sensor records a peak force of 800 N during a collision. Does that mean the object experiences an impulse of 800 N s?

No. Force and impulse are different physical quantities.

• Force measures the instantaneous interaction, in newtons (N).
• Impulse measures the accumulated effect of a force over time, in newton-seconds (N s).

The impulse delivered to an object equals its change in momentum:

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

where:

• \(J\) is impulse, measured in N s.
• \(m\) is mass, measured in kg.
• \(u\) is initial velocity, measured in m/s.
• \(v\) is final velocity, measured in m/s.

You can also find impulse from the signed area under a net force-time graph.

\[ \boxed{J=\text{signed area under the }F_{\text{net}}-t\text{ graph}} \]

This is the important connection: the area determines the impulse, not the peak force.

A large peak does not necessarily mean a large impulse

Compare two hypothetical force-time pulses acting in the same direction.

• Pulse A is triangular, with a peak force of 800 N over 2 ms.
• Pulse B is rectangular, with a constant force of 200 N over 4 ms.

Their impulses are:

For pulse A:

\[ J_A=\frac12(800)(0.002) \]

\[ J_A=0.80\text{ N s} \]

For pulse B:

\[ J_B=(200)(0.004) \]

\[ J_B=0.80\text{ N s} \]

Both pulses produce exactly the same impulse.

However, pulse A has four times the peak force.

This comparison is useful in collision problems because two objects can experience the same change in momentum through very different force profiles.

Peak force versus average force

The average force over a time interval is:

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

where \(\Delta t\) is the duration of the interaction.

Peak force is the largest instantaneous force reached during that interval.

These values are generally different.

For a triangular force-time pulse, the area is:

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

Therefore:

\[ F_{\text{peak}}=2|F_{\text{avg}}| \]

This relationship only applies to a triangular pulse that stays in one direction. A rectangular or trapezoidal pulse has a different relationship between its peak and average forces.

Also, remember that average force is a signed quantity. If a force reverses direction, positive and negative areas can cancel, giving a zero average force even though the average magnitude is not zero.

03Can one object experience large forces but zero net impulse?

Yes, but there is an important qualification.

For one object, zero net impulse means:

\[ J_{\text{net}}=\Delta p=0 \]

Therefore:

\[ p_f=p_i \]

The object’s final momentum must equal its initial momentum.

This does not mean that its momentum remained constant throughout the interaction.

Imagine a puck that starts at rest. A striker pushes it to the right, and a second bumper subsequently pushes it back to the left.

Suppose the first contact delivers an impulse of \(+0.60\text{ N s}\), and the second delivers an impulse of \(-0.60\text{ N s}\).

The total impulse is:

\[ J_{\text{net}}=0.60-0.60 \]

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

The puck finishes at rest, just as it started.

However, it was moving between the two collisions.

This reveals an important limitation of using impulse alone.

Impulse tells you the overall change in momentum between two selected times. It does not describe every detail of the motion in between.

There is a useful distinction here:

• A single contact force acting continuously in one direction produces a non-zero impulse.
• Multiple forces or contact events can produce large individual impulses that cancel.
• Opposing forces acting simultaneously can also produce zero resultant force while remaining individually large.

So, if a question states that an object experiences zero net impulse, always check whether it means zero impulse from one particular force or zero impulse from all forces combined.

04How could you measure this in an experiment?

A collision between two carts provides a straightforward way to investigate the relationship between force and impulse.

The experimental question is:

Can two carts experience large contact forces while their combined momentum remains approximately constant?

Setting up the experiment

Place two carts on a horizontal, low-friction track, facing each other.

Fit a force sensor to the front of each cart so that the sensors make contact when the carts collide.

Connect both sensors to the same data logger. Place photogates near the collision region, or use motion tracking, to measure the carts’ velocities before and after the collision.

The apparatus measures two different things:

1. The force sensors record the contact force throughout the collision.
2. The photogates measure the velocities needed to calculate the momentum changes.

What should the experiment show?

Newton’s Third Law predicts that the two force sensors should record equal-sized forces in opposite directions at each instant.

Consequently, the signed areas under their force-time graphs should also be equal and opposite.

If the track is sufficiently level and friction is negligible, the combined momentum of the two carts should remain approximately constant.

A representative illustrative result might look like this:

The combined contact impulse is:

\[ J_{\text{total}}=-1.50+1.50 \]

\[ \boxed{J_{\text{total}}=0\text{ N s}} \]

Yet each cart experiences a peak force of 500 N.

These are illustrative values, not measurements from a particular experiment.

Why the original measurement method is incomplete

Suppose you initially used only photogates.

You could measure the initial and final velocities and calculate the impulse on each cart using its change in momentum.

However, you could not determine the exact peak force or the shape of the force-time graph from those measurements alone.

Adding force sensors solves this problem. A sufficiently fast data logger records the force throughout the collision, allowing you to calculate the impulse from the graph and identify the peak force.

The revised design still has limitations:

• Friction can produce an external impulse.
• A slightly sloping track can change the horizontal momentum.
• Poor sensor calibration can make the two force readings appear unequal.
• A low sampling rate can miss a very short peak force.

The experiment therefore supports the relationship between Newton’s Third Law, impulse, and momentum conservation, but its accuracy depends on controlling external forces and measuring the collision carefully.

05Worked examples

06Questions and solutions

Question 1

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

Cart A has a mass of 0.40 kg and initially moves right at \(3.0\text{ m/s}\). After the collision, it moves right at \(0.50\text{ m/s}\).

Cart B has a mass of 0.60 kg and is initially stationary.

The collision lasts 5.0 ms. Assume negligible external horizontal impulse.

Calculate:

a) The impulse on cart A.

b) The impulse on cart B.

c) The average contact force on cart A.

d) The final velocity of cart B.

e) The net impulse on the two-cart system.

Solution 1

The impulse on cart A is \(-1.0\text{ N s}\), the impulse on cart B is \(+1.0\text{ N s}\), and the combined system has zero net impulse.

Step 1: Calculate the impulse on cart A.

Choose right as positive.

\[ J_A=m_A(v_A-u_A) \]

\[ J_A=0.40(0.50-3.0) \]

\[ \boxed{J_A=-1.0\text{ N s}} \]

Step 2: Calculate the impulse on cart B.

Newton’s Third Law means the contact impulses are equal and opposite.

\[ J_B=-J_A \]

\[ \boxed{J_B=+1.0\text{ N s}} \]

Step 3: Calculate the average force on cart A.

Convert the contact time into seconds:

\[ 5.0\text{ ms}=0.0050\text{ s} \]

\[ F_{\text{avg},A}=\frac{-1.0}{0.0050} \]

\[ \boxed{F_{\text{avg},A}=-200\text{ N}} \]

The average force acts to the left.

Step 4: Calculate the final velocity of cart B.

Cart B starts at rest, so its initial momentum is zero.

\[ J_B=m_B(v_B-u_B) \]

\[ 1.0=0.60(v_B-0) \]

\[ v_B=\frac{1.0}{0.60} \]

\[ \boxed{v_B=1.67\text{ m/s to the right}} \]

Step 5: Calculate the total impulse.

\[ J_{\text{total}}=-1.0+1.0 \]

\[ \boxed{J_{\text{total}}=0\text{ N s}} \]

Both carts change velocity, but their momentum changes cancel. The combined system conserves momentum because its external horizontal impulse is negligible.

Question 2

A 0.40 kg cart moves right at \(3.0\text{ m/s}\) before colliding with a barrier.

Two different bumper designs are tested. Each brings the cart to rest.

The force-time profiles are:

• Design X: A triangular leftward force pulse lasting 4.0 ms, with a peak magnitude of 600 N.
• Design Y: A constant leftward force of 150 N lasting 8.0 ms.

a) Calculate the impulse produced by each design.

b) Verify that both designs bring the cart to rest.

c) Calculate the average force for each design.

d) Explain why the two designs produce the same momentum change despite having different peak forces.

Solution 2

Both designs produce an impulse of \(-1.2\text{ N s}\), so they produce the same momentum change and bring the cart to rest. Design X has a peak force four times greater than Design Y.

Step 1: Calculate the impulse for design X.

The force-time graph is triangular.

\[ J_X=-\frac12(600)(0.0040) \]

\[ \boxed{J_X=-1.2\text{ N s}} \]

Step 2: Calculate the impulse for design Y.

The graph is rectangular.

\[ J_Y=(-150)(0.0080) \]

\[ \boxed{J_Y=-1.2\text{ N s}} \]

The impulses are equal.

Step 3: Verify the final velocity.

Using the impulse-momentum theorem:

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

For either design:

\[ -1.2=0.40(v-3.0) \]

\[ v-3.0=-3.0 \]

\[ \boxed{v=0\text{ m/s}} \]

Both bumper designs bring the cart to rest.

Step 4: Compare average forces.

For design X:

\[ F_{\text{avg},X}=\frac{-1.2}{0.0040} \]

\[ \boxed{F_{\text{avg},X}=-300\text{ N}} \]

For design Y:

\[ F_{\text{avg},Y}=\frac{-1.2}{0.0080} \]

\[ \boxed{F_{\text{avg},Y}=-150\text{ N}} \]

Design X has a peak force of 600 N, whereas design Y has a constant force of 150 N.

Step 5: Explain the difference.

The two force-time graphs have equal signed areas, so their impulses are equal.

Design X delivers the required impulse over a shorter interval, requiring a much larger peak force.

Design Y spreads the same impulse over a longer interval.

The important conclusion is that the change in momentum depends on impulse, not peak force alone.

Question 3

A 0.20 kg puck initially at rest experiences two separate contact events on a horizontal track.

First, a striker on the left exerts a rightward triangular force pulse lasting 3.0 ms, with a peak force of 400 N.

The puck then travels freely for 2.0 ms before contacting a stop on the right. The stop exerts a leftward triangular force pulse lasting 4.0 ms, with a peak force of 300 N.

Assume there are no other significant horizontal forces.

a) Calculate the impulse delivered by each contact.

b) Determine the puck’s velocity immediately after the first contact.

c) Determine its final velocity after the second contact.

d) Calculate the net impulse over the entire 9.0 ms interval.

e) Calculate the average signed force and the average force magnitude over the entire interval.

f) Explain why zero net impulse does not mean the puck remained stationary.

Solution 3

The puck experiences a net impulse of zero and finishes at rest, but it reaches \(3.0\text{ m/s}\) to the right between the two collisions. Its peak force magnitude is 400 N.

Step 1: Calculate the first impulse.

The first force-time graph is triangular and positive.

\[ J_1=\frac12(400)(0.0030) \]

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

Step 2: Calculate the second impulse.

The second force acts to the left, so its impulse is negative.

\[ J_2=-\frac12(300)(0.0040) \]

\[ \boxed{J_2=-0.60\text{ N s}} \]

Step 3: Find the velocity after the first contact.

The puck initially rests.

\[ J_1=m(v-u) \]

\[ 0.60=0.20(v-0) \]

\[ \boxed{v=3.0\text{ m/s to the right}} \]

The puck is now moving, despite having started at rest.

Step 4: Find the final velocity.

The second impulse cancels the first.

\[ J_{\text{net}}=0.60-0.60=0 \]

Therefore:

\[ J_{\text{net}}=m(v_f-u) \]

\[ 0=0.20(v_f-0) \]

\[ \boxed{v_f=0\text{ m/s}} \]

The puck finishes at rest.

Step 5: Calculate the average signed force.

The total interval is:

\[ \Delta t=3.0+2.0+4.0=9.0\text{ ms} \]

\[ \Delta t=0.0090\text{ s} \]

The net impulse is zero:

\[ F_{\text{avg,signed}}=\frac{0}{0.0090} \]

\[ \boxed{F_{\text{avg,signed}}=0\text{ N}} \]

Step 6: Calculate the average force magnitude.

The total unsigned area is the sum of the magnitudes of both impulses:

\[ J_{\text{absolute}}=0.60+0.60 \]

\[ J_{\text{absolute}}=1.20\text{ N s} \]

Therefore:

\[ F_{\text{avg,magnitude}} =\frac{1.20}{0.0090} \]

\[ \boxed{F_{\text{avg,magnitude}}=133\text{ N}} \]

The average signed force is zero, but the average force magnitude is 133 N.

This is an important distinction. Positive and negative forces can cancel algebraically without the object experiencing zero force throughout the interval.

Question 4

Two carts collide on a horizontal track.

During the 5.0 ms collision:

• The contact force on cart A is a triangular leftward pulse with a peak magnitude of 600 N.
• Cart B experiences an equal and opposite contact force.
• The combined rolling resistance acting on both carts is a constant 18 N to the left.

The initial total momentum of the two-cart system is \(+0.80\text{ kg m/s}\).

Assume the rolling resistance acts throughout the collision, and all vertical forces cancel.

a) Calculate the contact impulse on each cart.

b) Calculate the external impulse due to rolling resistance.

c) Determine the net impulse on the two-cart system.

d) Calculate the final total momentum.

e) Explain why the total impulse is not zero, despite the equal and opposite contact forces.

Solution 4

The net impulse on the two-cart system is \(-0.090\text{ N s}\), not zero. The contact impulses cancel, but the external rolling-resistance impulse remains.

Step 1: Calculate the contact impulse on cart A.

The contact pulse is triangular and acts to the left.

\[ J_A=-\frac12(600)(0.0050) \]

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

Step 2: Calculate the contact impulse on cart B.

By Newton’s Third Law:

\[ \boxed{J_B=+1.50\text{ N s}} \]

The combined contact impulse is:

\[ J_{\text{contact,total}}=-1.50+1.50=0 \]

Step 3: Calculate the external impulse.

Rolling resistance is a constant external force acting to the left.

\[ J_{\text{external}}=F\Delta t \]

\[ J_{\text{external}}=(-18)(0.0050) \]

\[ \boxed{J_{\text{external}}=-0.090\text{ N s}} \]

Step 4: Calculate the net impulse on the pair.

The internal contact impulses cancel, leaving the external impulse.

\[ J_{\text{net,system}}=0-0.090 \]

\[ \boxed{J_{\text{net,system}}=-0.090\text{ N s}} \]

Step 5: Calculate the final total momentum.

Using the impulse-momentum theorem:

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

\[ -0.090=p_f-0.80 \]

\[ \boxed{p_f=+0.71\text{ kg m/s}} \]

The total momentum decreases slightly.

The mistake would be to assume that equal and opposite internal forces automatically mean the entire system experiences zero net impulse. They only cancel the internal contribution. External forces must still be considered.

Question 5

Two carts, A and B, each have a mass of 0.50 kg.

They are arranged on a horizontal, low-friction track. Cart A approaches cart B from the left at \(5.0\text{ m/s}\) to the right. Cart B is initially stationary and is attached to a spring anchored to a wall on its right.

During the 4.0 ms collision:

• The contact force on cart A is a triangular pulse acting to the left, with a peak magnitude of 900 N.
• Cart B experiences an equal and opposite contact force.
• The compressed spring exerts an approximately constant force of 30 N to the left on cart B.

Ignore all other external horizontal forces. Assume the spring force remains constant throughout the collision.

a) Calculate the contact impulse on cart A and its final velocity.

b) Calculate the contact impulse on cart B.

c) Calculate the spring impulse on cart B and its total impulse. Determine its final velocity.

d) Calculate the net impulse on the combined two-cart system.

e) Verify the result using the initial and final total momentum.

f) If the system is expanded to include both carts, the spring, the wall, and the Earth, explain why the net external horizontal impulse can be approximately zero.

g) Calculate the average contact force on cart A and explain why the peak force does not contradict the relatively small net impulse on the two-cart system.

Solution 5

Cart A finishes moving right at \(1.4\text{ m/s}\), while cart B moves right at \(3.36\text{ m/s}\). The two-cart system receives a net impulse of \(-0.12\text{ N s}\), because the spring provides an external impulse that does not cancel with the internal contact forces.

Step 1: Establish the sign convention.

Choose right as positive.

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

The spring force on B also acts left.

These directions must be kept separate when calculating the impulses.

Step 2: Calculate the contact impulse on cart A.

The contact force-time graph is triangular.

\[ J_{A,\text{contact}} =-\frac12F_{\text{peak}}\Delta t \]

\[ J_{A,\text{contact}} =-\frac12(900)(0.0040) \]

\[ \boxed{J_{A,\text{contact}}=-1.80\text{ N s}} \]

Step 3: Calculate the final velocity of cart A.

The initial momentum of A is:

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

\[ p_{A,i}=0.50(5.0)=2.50\text{ kg m/s} \]

Its final momentum is:

\[ p_{A,f}=p_{A,i}+J_A \]

\[ p_{A,f}=2.50-1.80 \]

\[ p_{A,f}=0.70\text{ kg m/s} \]

Therefore:

\[ v_{A,f}=\frac{0.70}{0.50} \]

\[ \boxed{v_{A,f}=1.4\text{ m/s to the right}} \]

Step 4: Calculate the contact impulse on cart B.

The contact impulse on B is equal and opposite to that on A.

\[ \boxed{J_{B,\text{contact}}=+1.80\text{ N s}} \]

However, this is not the only force acting on B.

The spring also exerts a leftward force.

Step 5: Calculate the spring impulse on cart B.

The spring force is constant, so:

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

\[ J_{\text{spring}}=(-30)(0.0040) \]

\[ \boxed{J_{\text{spring}}=-0.12\text{ N s}} \]

The total impulse on B is therefore:

\[ J_{B,\text{net}}=1.80-0.12 \]

\[ \boxed{J_{B,\text{net}}=+1.68\text{ N s}} \]

Step 6: Calculate the final velocity of cart B.

Cart B starts at rest.

\[ J_B=m_B(v_B-u_B) \]

\[ 1.68=0.50(v_B-0) \]

\[ v_B=\frac{1.68}{0.50} \]

\[ \boxed{v_{B,f}=3.36\text{ m/s to the right}} \]

Step 7: Calculate the net impulse on the two-cart system.

Add the net impulses on both carts:

\[ J_{\text{system}}=J_A+J_B \]

\[ J_{\text{system}}=-1.80+1.68 \]

\[ \boxed{J_{\text{system}}=-0.12\text{ N s}} \]

Notice what happened.

The contact impulses cancel:

\[ -1.80+1.80=0 \]

But the spring impulse remains:

\[ J_{\text{system}}=-0.12\text{ N s} \]

The spring is outside the two-cart system, so its force is external.

Step 8: Verify the result using momentum.

Initially, the total momentum is:

\[ p_i=0.50(5.0)+0.50(0) \]

\[ p_i=2.50\text{ kg m/s} \]

After the collision:

\[ p_f=0.50(1.4)+0.50(3.36) \]

\[ p_f=0.70+1.68 \]

\[ p_f=2.38\text{ kg m/s} \]

Therefore:

\[ \Delta p=2.38-2.50 \]

\[ \boxed{\Delta p=-0.12\text{ kg m/s}} \]

This matches the calculated external impulse.

The momentum calculation confirms that the spring’s contribution cannot be ignored.

Step 9: Expand the system boundary.

So far, the system included only carts A and B.

Now include the spring, the wall, and the Earth.

The spring’s interaction with cart B is now an internal interaction within the expanded system. The spring and its support receive the corresponding opposite impulse.

Assuming negligible external horizontal impulse on this larger system:

\[ \boxed{J_{\text{external}}\approx0} \]

The total momentum of the expanded system is approximately conserved.

The apparent loss of momentum from the two-cart system has been transferred to the spring, wall, and Earth.

Step 10: Compare average and peak contact force.

The average contact force on cart A is:

\[ F_{\text{avg}}=\frac{-1.80}{0.0040} \]

\[ \boxed{F_{\text{avg}}=-450\text{ N}} \]

The peak force is:

\[ \boxed{F_{\text{peak}}=-900\text{ N}} \]

The peak is twice the average because the contact-force pulse is triangular.

The two carts experience these large, equal and opposite contact forces. Their contributions cancel when considering the combined system.

The spring force does not cancel within that system, leaving a much smaller net impulse of \(-0.12\text{ N s}\).

This is the subtle point: a large internal contact force can coexist with a small net impulse on a system, because the internal forces cancel while external forces determine the system’s overall momentum change.

The next useful step in collision problems is to distinguish momentum conservation from kinetic energy conservation. Momentum depends on the net external impulse, while kinetic energy can change through deformation, heating, sound, and other energy transfers during a collision.