When Do Equal-Speed Objects Have Zero Total Momentum?

Two objects can move quite quickly and still have zero total momentum. Imagine two identical hockey pucks sliding towards each other at exactly the same speed. One moves to the right at 3.0 m/s, and the other moves to the left at 3.0 m/s.

Their individual momenta are non-zero. Yet, when you add their momenta as vectors, the total is zero.

The important conditions are that the objects have equal masses, equal non-zero speeds, and opposite velocities. Equal speeds alone are not enough.

## Why equal speeds can produce zero total momentum

Start with the momentum of a single moving object:

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

Here:
- \(\vec p\) is momentum, measured in kg m/s.
- \(m\) is mass, measured in kg.
- \(\vec v\) is velocity, measured in m/s.

Notice that velocity is a vector. It has both magnitude (speed) and direction.

Momentum is also a vector, so the direction of motion matters when adding the momentum of multiple objects.

Consider two objects with masses of 2.0 kg, travelling at 3.0 m/s in opposite directions.

Choose right as the positive direction. Their momenta are:

\[
p_1 = 2.0 \times (+3.0)
= +6.0\ \mathrm{kg\,m/s}
\]

\[
p_2 = 2.0 \times (-3.0)
= -6.0\ \mathrm{kg\,m/s}
\]

The total momentum is:

\[
\begin{aligned}
p_{\mathrm{total}} &= p_1+p_2\\
&=+6.0-6.0\\
&=0\ \mathrm{kg\,m/s}
\end{aligned}
\]

Neither object has stopped. Their momenta cancel because they are equal in magnitude and opposite in direction.

[IMAGE: Two identical hockey pucks on a horizontal surface, separated by a gap and travelling directly towards each other. Show equal-length velocity arrows, one pointing right and one pointing left, each labelled 3.0 m/s. Include equal mass labels of 2.0 kg and equal, oppositely directed momentum arrows. Clearly distinguish the two objects and show the positive-right sign convention.]

### The exact condition

For two objects, the total momentum is:

\[
\vec p_{\mathrm{total}}
=m_1\vec v_1+m_2\vec v_2
\]

For their total momentum to equal zero:

\[
m_1\vec v_1=-m_2\vec v_2
\]

This tells us two things:

1. Their velocities must point in opposite directions.
2. Their momentum magnitudes must be equal.

If both objects have the same non-zero speed, their masses must also be equal.

There is one trivial exception: if both objects are stationary, their speeds and momenta are zero regardless of whether their masses are equal.

So, for two moving objects with equal speeds, equal masses and opposite directions are the necessary and sufficient conditions for zero total momentum.

## Why direction and mass both matter

It is tempting to see two equal speed readings and immediately conclude that the total momentum is zero. That prediction ignores two important details.

First, equal speeds do not tell you whether the objects are moving towards each other or in the same direction.

Second, even if they move in opposite directions, different masses produce different momentum magnitudes.

Compare these three situations:

| Situation | Object A | Object B | Total momentum |
|---|---|---|---|
| Equal masses, opposite directions | 2 kg at +3 m/s | 2 kg at −3 m/s | 0 kg m/s |
| Unequal masses, opposite directions | 2 kg at +3 m/s | 4 kg at −3 m/s | −6 kg m/s |
| Equal masses, same direction | 2 kg at +3 m/s | 2 kg at +3 m/s | +12 kg m/s |

All three situations have equal speeds. Only the first has zero total momentum.

[IMAGE: Three side-by-side physics comparisons. Panel A shows two equal-mass blocks moving in opposite directions with equal velocity arrows and cancelling momentum arrows. Panel B shows a smaller and larger mass moving oppositely at equal speeds, with the larger momentum arrow visibly longer. Panel C shows two equal-mass blocks moving in the same direction, with momentum arrows adding. Label all masses, velocities, and resulting total momentum.]

### Worked example: Equal speeds, unequal masses

A 3.0 kg trolley travels right at 4.0 m/s. A 5.0 kg trolley travels left at 4.0 m/s. Calculate their total momentum.

**Step 1**: Choose a positive direction.

Take right as positive and left as negative.

**Step 2**: Calculate each trolley's momentum.

\[
p_1=3.0(+4.0)=+12\ \mathrm{kg\,m/s}
\]

\[
p_2=5.0(-4.0)=-20\ \mathrm{kg\,m/s}
\]

**Step 3**: Add the momentum vectors.

\[
\begin{aligned}
p_{\mathrm{total}}&=+12-20\\
&=-8.0\ \mathrm{kg\,m/s}
\end{aligned}
\]

The total momentum is 8.0 kg m/s to the left.

The trolleys have equal speeds and opposite velocities, but their masses are different. The heavier trolley has greater momentum, so the momenta do not cancel.

[IMAGE: Two trolleys on a straight horizontal track, one with a 3.0 kg label moving right and one with a 5.0 kg label moving left. Both velocity arrows have equal lengths and are labelled 4.0 m/s. Show the heavier trolley's momentum arrow as longer, pointing left. Do not show a pre-calculated net momentum.]

## An experiment: Testing momentum cancellation with two carts

How could you test whether two objects with equal speeds really can have zero total momentum?

A useful experiment uses two identical low-friction carts on a level air track. The arrangement allows us to measure their speeds before they collide and examine what happens to the system afterwards.

[IMAGE: A complete horizontal air-track experiment viewed from the side. Show two identical carts approaching each other from opposite ends, photogates positioned along the track near the centre, and hook-and-loop bumpers on the cart fronts so they stick together on collision. Include a ruler or measurement scale and indicate the direction of travel. Show the track, carts, gates, and connections as actual apparatus rather than abstract process boxes.]

### 1. The original prediction

Suppose each cart has a mass of 0.50 kg.

One cart approaches from the left at 0.80 m/s. The other approaches from the right at 0.80 m/s.

Choose right as positive.

The predicted total momentum is:

\[
p_{\mathrm{total}}
=(0.50)(+0.80)+(0.50)(-0.80)
\]

\[
p_{\mathrm{total}}=0\ \mathrm{kg\,m/s}
\]

If the carts stick together when they collide, what should happen afterwards?

The combined mass is 1.00 kg. If the system's momentum remains zero, its final velocity must also be zero.

The prediction is that the carts meet, stick together, and remain stationary.

### 2. What the apparatus measures

Each cart passes through a photogate before reaching the collision point. A small flag attached to the cart interrupts a light beam.

The photogate measures how long the flag blocks the beam. If the flag length is known, the cart's speed can be calculated.

For a flag of length 0.10 m that blocks the beam for 0.125 s:

\[
\begin{aligned}
v&=\frac{\text{distance}}{\text{time}}\\
&=\frac{0.10}{0.125}\\
&=0.80\ \mathrm{m/s}
\end{aligned}
\]

The two readings establish equal speeds. The carts' positions and directions of travel establish that their velocities are opposite.

The distinction is important: a photogate measures speed, not direction. The direction must be established separately from the arrangement or from a signed velocity measurement.

### 3. What the observation means

If the carts stick together and stop, that result is consistent with the predicted zero total momentum.

However, a single demonstration is not enough to establish perfect momentum conservation. Small differences in cart masses, track inclination, friction, and measurement uncertainty can produce a small residual momentum.

A more reliable version measures both cart masses, records their velocities immediately before collision, and measures the combined cart's velocity immediately afterwards. The track must be level, and external horizontal forces should be negligible during the collision.

The revised arrangement tests the momentum of the entire two-cart system rather than relying only on the fact that the carts appear to stop.

[IMAGE: The same two-cart air-track experiment immediately after collision. Show the two carts joined by their hook-and-loop bumpers at the centre of the track, with the combined system stationary. Include the photogates on either side and a signed velocity measurement of zero for the joined carts.]

### 4. Momentum is not the same as kinetic energy

There is a particularly interesting result here.

The carts can have zero total momentum while possessing substantial kinetic energy.

Before collision, each cart has kinetic energy:

\[
E_k=\frac12 mv^2
\]

For each 0.50 kg cart travelling at 0.80 m/s:

\[
E_k=\frac12(0.50)(0.80)^2
=0.16\ \mathrm{J}
\]

The total initial kinetic energy is:

\[
E_{k,\mathrm{initial}}=0.16+0.16
=0.32\ \mathrm{J}
\]

After the carts stick together and stop:

\[
E_{k,\mathrm{final}}=0\ \mathrm{J}
\]

The initial kinetic energy has been transformed into other forms, including thermal energy, sound, and energy associated with deformation.

There is no contradiction. Momentum is a vector quantity, so opposite momenta can cancel. Kinetic energy is a scalar quantity, so the two positive kinetic energies add.

This is why momentum and kinetic energy must be checked separately when analysing a collision.

[IMAGE: A two-stage comparison of the identical carts before and after a perfectly inelastic collision. Before: two equal carts moving towards one another with opposite momentum arrows and positive kinetic-energy labels. After: the carts are joined and stationary, with zero kinetic energy of motion and subtle visual indications of deformation, sound waves, and thermal energy transfer. Make clear that total momentum is conserved while kinetic energy of motion decreases.]

### Choosing the system correctly

There is another detail worth getting right in collision questions: what exactly are you treating as the system?

For the two-cart experiment, choose both carts together as the system.

During their collision:
- The forces between the carts are internal forces. They act in opposite directions and cancel when considering the whole system.
- Friction and other forces from the track are external forces.
- If the net external impulse is negligible, the total momentum of the two-cart system is conserved.

Therefore:

\[
\vec p_{\mathrm{initial}}=\vec p_{\mathrm{final}}
\]

For motion along a straight track, this is a one-dimensional calculation. For motion in two dimensions, the horizontal and vertical components must each be conserved separately.

A zero total momentum vector requires:

\[
\sum p_x=0
\qquad\text{and}\qquad
\sum p_y=0
\]

This is particularly useful when objects approach each other at angles. Equal momentum magnitudes do not guarantee cancellation unless their directions are opposite.

For a fuller explanation of resolving and conserving momentum components, see [Conservation of Momentum in Two Dimensions for HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-conservation-of-momentum-in-two-dimensions/).

### Does the reference frame matter?

Yes. Momentum depends on the reference frame used to measure velocity.

Two objects can have equal speeds and opposite velocities in one frame, giving zero total momentum, but have different velocities and non-zero total momentum in another frame.

This is a classical mechanics result. The final question explores why it matters.

The calculations in this article use Newtonian momentum, \(p=mv\), which is appropriate for ordinary speeds. For objects travelling at speeds approaching the speed of light, relativistic momentum must be used instead. See [Relativistic Momentum: How HSC Physics Explains the Speed Limit](https://tutorgum.com/learn/resources/hsc-physics-relativistic-momentum/).

## Questions and solutions

### Question 1

Two identical pucks, each with a mass of 2.0 kg, slide directly towards each other along a horizontal surface. Each has a speed of 3.0 m/s.

Calculate:

a) The momentum of each puck, taking right as positive.

b) The total momentum of the two-puck system.

c) Whether either puck individually has zero momentum.

### Solution 1

**The total momentum is zero, but neither puck individually has zero momentum.**

**Step 1**: Calculate the momentum of each puck.

Using:

\[
p=mv
\]

For the puck travelling right:

\[
p_1=2.0(+3.0)
=+6.0\ \mathrm{kg\,m/s}
\]

For the puck travelling left:

\[
p_2=2.0(-3.0)
=-6.0\ \mathrm{kg\,m/s}
\]

**Step 2**: Add the momenta.

\[
\begin{aligned}
p_{\mathrm{total}}&=p_1+p_2\\
&=+6.0-6.0\\
&=0\ \mathrm{kg\,m/s}
\end{aligned}
\]

**Step 3**: Interpret the result.

Each puck has a momentum magnitude of 6.0 kg m/s. Their individual momenta are non-zero, but their vector sum is zero.

The important distinction is between the momentum of an individual object and the total momentum of a system.

### Question 2

A 0.60 kg trolley travels right at 2.5 m/s. It collides with a 0.90 kg trolley travelling left at 2.5 m/s.

The trolleys have hook-and-loop bumpers, so they stick together.

Assuming negligible external impulse during the collision, calculate:

a) Their final velocity.

b) Their total kinetic energy before the collision.

c) Their kinetic energy after the collision.

d) The kinetic energy lost during the collision.

[IMAGE: Two trolleys approaching each other along a straight horizontal track. Label the left trolley 0.60 kg with a rightward velocity arrow of 2.5 m/s, and the right trolley 0.90 kg with a leftward velocity arrow of 2.5 m/s. Show hook-and-loop bumpers at the collision ends. Do not show the final velocity or momentum result.]

### Solution 2

**The joined trolleys move left at 0.50 m/s, and 4.50 J of kinetic energy is lost.**

**Step 1**: Choose a sign convention.

Take right as positive and left as negative.

The initial velocities are:

\[
v_1=+2.5\ \mathrm{m/s}
\]

\[
v_2=-2.5\ \mathrm{m/s}
\]

**Step 2**: Calculate the initial total momentum.

\[
\begin{aligned}
p_{\mathrm{initial}}
&=(0.60)(2.5)+(0.90)(-2.5)\\
&=1.50-2.25\\
&=-0.75\ \mathrm{kg\,m/s}
\end{aligned}
\]

**Step 3**: Apply conservation of momentum.

Since the trolleys stick together, their combined mass is:

\[
m_{\mathrm{total}}=0.60+0.90
=1.50\ \mathrm{kg}
\]

Using:

\[
p_{\mathrm{initial}}=p_{\mathrm{final}}
\]

\[
-0.75=(1.50)v_f
\]

Therefore:

\[
v_f=-0.50\ \mathrm{m/s}
\]

The negative sign indicates leftward motion.

**Step 4**: Calculate the initial kinetic energy.

Kinetic energy is a scalar, so both positive contributions are added.

\[
E_{k,i}=\frac12m_1v_1^2+\frac12m_2v_2^2
\]

\[
\begin{aligned}
E_{k,i}
&=\frac12(0.60)(2.5)^2\\
&\quad+\frac12(0.90)(2.5)^2\\
&=1.875+2.8125\\
&=4.6875\ \mathrm{J}
\end{aligned}
\]

**Step 5**: Calculate the final kinetic energy.

The combined trolleys move together at 0.50 m/s.

\[
\begin{aligned}
E_{k,f}
&=\frac12(1.50)(0.50)^2\\
&=0.1875\ \mathrm{J}
\end{aligned}
\]

**Step 6**: Calculate the kinetic energy lost.

\[
\begin{aligned}
\Delta E_k
&=E_{k,i}-E_{k,f}\\
&=4.6875-0.1875\\
&=4.50\ \mathrm{J}
\end{aligned}
\]

The collision is inelastic. Momentum is conserved, but kinetic energy decreases as energy is transferred into deformation, sound, and thermal energy.

Notice that the initial total momentum was not zero. The heavier trolley had the greater momentum magnitude, despite both trolleys having equal speeds.

### Question 3

Two identical pucks, each with a mass of 0.25 kg, move on a smooth horizontal surface.

Puck A travels east at 6.0 m/s. Puck B travels north at 6.0 m/s.

Calculate the magnitude and direction of the total momentum.

Explain why the total momentum is not zero, even though the pucks have equal masses and equal speeds.

[IMAGE: Top-down view of two identical 0.25 kg pucks moving on a horizontal surface. Puck A travels horizontally east at 6.0 m/s, while puck B travels vertically north at 6.0 m/s. Show perpendicular paths and clearly labelled east and north axes. Do not show the resultant momentum vector.]

### Solution 3

**The total momentum is 2.12 kg m/s, directed 45° north of east. It is not zero because the two momentum vectors are perpendicular rather than opposite.**

**Step 1**: Choose perpendicular axes.

Take east as positive \(x\) and north as positive \(y\).

Puck A travels entirely in the \(x\)-direction:

\[
p_{Ax}=(0.25)(6.0)=1.50\ \mathrm{kg\,m/s}
\]

\[
p_{Ay}=0
\]

Puck B travels entirely in the \(y\)-direction:

\[
p_{Bx}=0
\]

\[
p_{By}=(0.25)(6.0)=1.50\ \mathrm{kg\,m/s}
\]

**Step 2**: Add the momentum components.

\[
p_x=1.50+0=1.50\ \mathrm{kg\,m/s}
\]

\[
p_y=0+1.50=1.50\ \mathrm{kg\,m/s}
\]

**Step 3**: Calculate the magnitude of the resultant momentum.

Using Pythagoras:

\[
\begin{aligned}
p_{\mathrm{total}}
&=\sqrt{p_x^2+p_y^2}\\
&=\sqrt{1.50^2+1.50^2}\\
&=2.12\ \mathrm{kg\,m/s}
\end{aligned}
\]

**Step 4**: Calculate the direction.

\[
\tan\theta=\frac{p_y}{p_x}
\]

\[
\theta=\tan^{-1}\left(\frac{1.50}{1.50}\right)
\]

\[
\theta=45^\circ
\]

The resultant momentum points 45° north of east.

The tempting mistake is to assume that equal momentum magnitudes automatically cancel. They only cancel when the vectors point in exactly opposite directions.

### Question 4

Two identical 0.50 kg carts move towards each other along a rough, horizontal track. Each cart initially travels at 1.0 m/s.

Each cart experiences a kinetic friction force of magnitude 0.20 N. Assume the carts have identical friction conditions and that the friction forces remain constant.

The carts are initially 1.2 m apart.

Consider the system consisting of both carts.

a) Calculate the net horizontal external force on the system.

b) Calculate the speed of each cart after 0.50 s.

c) Determine whether the total momentum of the system has changed.

d) Calculate the decrease in the system's kinetic energy.

e) Explain what would happen to the total momentum if friction acted on only one cart.

[IMAGE: Two identical 0.50 kg carts separated by 1.2 m on a rough horizontal track. The left cart moves right at 1.0 m/s, and the right cart moves left at 1.0 m/s. Show the rough track and the velocity arrows, but do not draw friction-force arrows in the question figure.]

### Solution 4

**The total momentum remains zero, even though both carts slow down and lose kinetic energy.**

**Step 1**: Determine the external forces.

The left cart moves right, so friction acts left.

The right cart moves left, so friction acts right.

The two friction forces have equal magnitudes and opposite directions.

Therefore:

\[
F_{\mathrm{net,external}}
=-0.20+0.20=0\ \mathrm{N}
\]

The net horizontal external force on the two-cart system is zero.

This is an important distinction: external forces do act on the system, but their resultant is zero.

**Step 2**: Calculate the acceleration of each cart.

Using Newton's second law:

\[
F=ma
\]

The magnitude of each cart's acceleration is:

\[
a=\frac{0.20}{0.50}
=0.40\ \mathrm{m/s^2}
\]

Each cart accelerates opposite to its direction of motion.

**Step 3**: Calculate the final speeds.

Using:

\[
v=u+at
\]

For the left cart, taking right as positive:

\[
\begin{aligned}
v_1&=1.0-(0.40)(0.50)\\
&=0.80\ \mathrm{m/s}
\end{aligned}
\]

For the right cart, taking left as negative:

\[
\begin{aligned}
v_2&=-1.0+(0.40)(0.50)\\
&=-0.80\ \mathrm{m/s}
\end{aligned}
\]

Both carts still have equal speeds in opposite directions.

They have not collided during this interval. Each travels 0.45 m, leaving a gap of 0.30 m.

**Step 4**: Check the total momentum.

Initially:

\[
\begin{aligned}
p_i&=(0.50)(1.0)+(0.50)(-1.0)\\
&=0\ \mathrm{kg\,m/s}
\end{aligned}
\]

After 0.50 s:

\[
\begin{aligned}
p_f&=(0.50)(0.80)+(0.50)(-0.80)\\
&=0\ \mathrm{kg\,m/s}
\end{aligned}
\]

The total momentum remains zero because the net external impulse is zero.

**Step 5**: Calculate the initial kinetic energy.

\[
\begin{aligned}
E_{k,i}
&=2\left(\frac12(0.50)(1.0)^2\right)\\
&=0.50\ \mathrm{J}
\end{aligned}
\]

**Step 6**: Calculate the final kinetic energy.

\[
\begin{aligned}
E_{k,f}
&=2\left(\frac12(0.50)(0.80)^2\right)\\
&=0.32\ \mathrm{J}
\end{aligned}
\]

The decrease is:

\[
\Delta E_k=0.50-0.32
\]

\[
\Delta E_k=0.18\ \mathrm{J}
\]

This energy is transferred into thermal energy through friction.

**Step 7**: Consider friction acting on only one cart.

Now suppose only the left cart experiences the 0.20 N friction force.

The net external force becomes:

\[
F_{\mathrm{net}}=-0.20\ \mathrm{N}
\]

The external impulse over 0.50 s is:

\[
\begin{aligned}
J&=F_{\mathrm{net}}\Delta t\\
&=(-0.20)(0.50)\\
&=-0.10\ \mathrm{N\,s}
\end{aligned}
\]

Since impulse equals the change in total momentum:

\[
p_f=p_i+J
\]

\[
p_f=0-0.10
\]

\[
p_f=-0.10\ \mathrm{kg\,m/s}
\]

The total momentum is now non-zero.

The key lesson is that momentum conservation depends on the net external impulse on the selected system, not on whether external forces exist individually.

### Question 5

A low-friction laboratory train travels at a constant velocity of 5.0 m/s to the right relative to a stationary platform.

Inside the train are two identical 0.50 kg carts on a horizontal track.

Relative to the train:
- Cart A moves right at 0.80 m/s.
- Cart B moves left at 0.80 m/s.

The carts collide and stick together. Assume negligible external horizontal impulse during the collision.

Two observers analyse the event:

- Observer T is standing on the train.
- Observer P is standing on the stationary platform.

Both observers use classical mechanics.

a) Calculate the total momentum of the two-cart system immediately before and after the collision according to Observer T.

b) Calculate the velocity of each cart immediately before the collision according to Observer P.

c) Calculate the total momentum immediately before the collision according to Observer P.

d) Calculate the final velocity and momentum of the joined carts according to Observer P.

e) Calculate the kinetic energy lost in the collision according to each observer.

f) Explain how both observers can agree that momentum is conserved even though they obtain different values for the total momentum.

[IMAGE: A laboratory train moving horizontally to the right relative to a stationary platform, shown in side view. Inside the train, two identical 0.50 kg carts approach each other on a low-friction track. Relative to the train, label cart A with a rightward velocity of 0.80 m/s and cart B with a leftward velocity of 0.80 m/s. Show the train's rightward velocity of 5.0 m/s relative to the platform. Include a stationary platform and a clearly distinguished train reference frame. Do not show platform-frame cart velocities, resultant momentum, or collision outcomes.]

### Solution 5

**Both observers find that momentum is conserved, but only Observer T measures zero total momentum. Momentum is frame-dependent, while the conservation of momentum applies in both inertial frames.**

This is the most important conceptual distinction in the question.

The carts have equal and opposite velocities relative to the train. That does not mean they have equal and opposite velocities relative to the platform.

**Step 1**: Analyse the collision in the train's reference frame.

Observer T measures:

\[
v_A=+0.80\ \mathrm{m/s}
\]

\[
v_B=-0.80\ \mathrm{m/s}
\]

The initial total momentum is:

\[
\begin{aligned}
p_{i,T}
&=(0.50)(0.80)+(0.50)(-0.80)\\
&=0\ \mathrm{kg\,m/s}
\end{aligned}
\]

After the carts stick together, their combined mass is:

\[
m_{\mathrm{total}}=0.50+0.50
=1.00\ \mathrm{kg}
\]

Momentum is conserved, so:

\[
0=(1.00)v_f
\]

Therefore:

\[
v_{f,T}=0\ \mathrm{m/s}
\]

The final momentum is also zero.

Observer T sees the joined carts stationary inside the train.

**Step 2**: Convert the velocities into the platform's reference frame.

The train moves right at 5.0 m/s relative to the platform.

Using classical velocity addition:

\[
v_{\mathrm{platform}}
=v_{\mathrm{train}}+v_{\mathrm{relative}}
\]

For cart A:

\[
v_A=5.0+0.80
=5.80\ \mathrm{m/s}
\]

For cart B:

\[
v_B=5.0-0.80
=4.20\ \mathrm{m/s}
\]

Both carts now move to the right relative to the platform.

Notice what has happened: the carts no longer have equal speeds in this reference frame.

**Step 3**: Calculate the initial total momentum in the platform frame.

\[
\begin{aligned}
p_{i,P}
&=(0.50)(5.80)+(0.50)(4.20)\\
&=2.90+2.10\\
&=5.00\ \mathrm{kg\,m/s}
\end{aligned}
\]

The total momentum is 5.00 kg m/s to the right.

It is not zero.

**Step 4**: Calculate the final velocity and momentum in the platform frame.

The carts stick together and are stationary relative to the train.

Therefore, they continue moving at the train's velocity relative to the platform:

\[
v_{f,P}=+5.0\ \mathrm{m/s}
\]

Their final momentum is:

\[
\begin{aligned}
p_{f,P}
&=(1.00)(5.0)\\
&=+5.00\ \mathrm{kg\,m/s}
\end{aligned}
\]

Therefore:

\[
p_{i,P}=p_{f,P}
\]

Momentum is conserved in the platform frame too.

**Step 5**: Check the kinetic energy in the train frame.

Initially, each cart has:

\[
E_k=\frac12(0.50)(0.80)^2
=0.16\ \mathrm{J}
\]

The total initial kinetic energy is:

\[
E_{k,i,T}=0.16+0.16
=0.32\ \mathrm{J}
\]

The joined carts are stationary, so:

\[
E_{k,f,T}=0
\]

The kinetic energy lost is:

\[
\Delta E_{k,T}=0.32\ \mathrm{J}
\]

**Step 6**: Check the kinetic energy in the platform frame.

Before collision:

\[
\begin{aligned}
E_{k,i,P}
&=\frac12(0.50)(5.80)^2\\
&\quad+\frac12(0.50)(4.20)^2\\
&=8.41+4.41\\
&=12.82\ \mathrm{J}
\end{aligned}
\]

After collision:

\[
\begin{aligned}
E_{k,f,P}
&=\frac12(1.00)(5.0)^2\\
&=12.50\ \mathrm{J}
\end{aligned}
\]

The kinetic energy lost is:

\[
\begin{aligned}
\Delta E_{k,P}
&=12.82-12.50\\
&=0.32\ \mathrm{J}
\end{aligned}
\]

Both observers calculate the same kinetic energy decrease, even though their initial and final kinetic energies are different.

The missing 0.32 J is transformed into other forms during the inelastic collision.

**Step 7**: Explain the result.

The tempting conclusion is that zero total momentum must be an absolute property of a pair of objects.

It isn't.

Momentum depends on the reference frame because velocity depends on the reference frame.

The train observer sees equal and opposite velocities, giving zero total momentum. The platform observer sees both carts travelling right, giving a total momentum of 5.00 kg m/s.

Neither observer is wrong. Both measure momentum conservation correctly in their own inertial reference frames.

The deeper lesson is that you cannot discuss the total momentum of moving objects without specifying the reference frame.

## The key decision to remember

When a question asks whether two objects have zero total momentum, don't start by comparing their speeds.

Work through these checks:

1. **Choose the system.** Decide which objects are included.
2. **Choose the reference frame.** State where the velocities are measured.
3. **Choose your sign convention.** Identify positive and negative directions.
4. **Calculate individual momenta.** Use mass multiplied by velocity.
5. **Add momentum vectors.** Resolve into \(x\)- and \(y\)-components if necessary.
6. **Check external impulse.** Only conserve the system's momentum when its net external impulse is zero or negligible.
7. **Check energy separately.** Momentum conservation does not automatically mean kinetic energy is conserved.

For two moving objects with equal speeds, the answer is straightforward once the conditions are clear: they have zero total momentum if and only if they have equal masses and opposite velocities in the chosen reference frame.

They can both be moving, both have non-zero momentum, and still have a total momentum of exactly zero.

That distinction is the starting point for analysing more complicated collisions, including two-dimensional interactions and collisions between objects that stick together.