Why Momentum Is Conserved When Kinetic Energy Is Lost

Two air-hockey pucks collide on a smooth table. Before impact, puck A moves east and puck B is stationary. After impact, both pucks move away at different angles.

Add their momentum vectors before and after the collision, and the total can be exactly the same. Add their kinetic energies, however, and the total can be smaller.

There is no contradiction. Momentum is conserved because the net external impulse on the two-puck system is negligible. Kinetic energy can decrease because some of the energy of their motion is transferred into deformation, vibration, sound, and thermal energy.

This is the important distinction: internal collision forces can redistribute momentum between objects without changing their total momentum, while those same forces can do work that reduces the objects’ total kinetic energy.

Before going further, consider one prediction. If puck A changes its momentum during the collision, must puck B keep the same momentum? No. Both pucks change momentum. What stays unchanged is the vector sum of their momenta, provided the net external impulse is negligible.

01Start by choosing the system

Momentum conservation is not a statement about an individual object in every situation. It applies to a chosen system of objects.

For our collision, choose both pucks as the system.

During impact, puck A pushes on puck B, while puck B pushes back on puck A. Newton’s third law tells us that these forces are equal in magnitude and opposite in direction:

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

The forces can be enormous. Neither puck keeps its original momentum.

However, these forces are internal forces because both pucks belong to our system.

Their impulses cancel when we consider the two pucks together. One puck gains momentum while the other experiences an equal and opposite change.

This gives us the momentum-impulse relationship:

\[ \Delta \vec p_{\text{system}} = \vec J_{\text{external}} \]

Here:

• \(\Delta \vec p_{\text{system}}\) is the change in total system momentum, measured in \(\mathrm{kg\,m\,s^{-1}}\).
• \(\vec J_{\text{external}}\) is the net external impulse, measured in \(\mathrm{N\,s}\).

If the net external impulse is zero, total momentum is conserved:

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

If it is small enough to ignore, we use this as an approximation.

Why the size of the collision force isn’t the deciding factor

A common mistake is to assume that momentum cannot be conserved because the collision involves large forces.

That misses the important distinction between internal and external forces.

Consider a pair of trolleys on a low-friction track.

• The force trolley A exerts on trolley B is internal to the two-trolley system.
• The force trolley B exerts on trolley A is also internal.
• Friction from the track is external.
• Any other force from outside the system is external.

The internal forces cancel in the total momentum calculation. External forces can change the total momentum if they produce a significant net impulse.

On a horizontal air table, for example, the vertical normal force and gravitational force approximately balance. Horizontal friction is very small during a brief collision.

This makes conservation of horizontal momentum a good approximation.

What if we choose only trolley A as the system?

Now the force from trolley B is external. Trolley A’s momentum changes during the collision, and there is no reason to conserve its individual momentum.

The system boundary changes which forces count as external. Always decide what belongs to your system before writing a conservation equation.

02Why equal and opposite forces can still reduce kinetic energy

This is the part that makes the question interesting.

If the collision forces are equal and opposite, why don’t they cancel out every effect of the collision?

The answer is that cancelling impulses and cancelling work are two different things.

Impulse changes momentum

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

\[ J=F_{\text{average}}\Delta t \]

For the two-puck system, the internal impulses are equal and opposite.

Their sum is zero, so the total momentum does not change.

Work changes kinetic energy

Work depends on both force and displacement:

\[ W=Fs\cos\theta \]

Here:

• \(W\) is work in joules.
• \(F\) is the force in newtons.
• \(s\) is displacement in metres.
• \(\theta\) is the angle between the force and displacement.

During a collision, the objects can deform. Their contact points move relative to one another as their surfaces compress.

Consequently, equal and opposite forces do not necessarily do equal and opposite amounts of work on the objects’ translational motion.

Imagine a moving trolley catching a slower trolley with soft bumpers.

The force on the faster trolley opposes its motion, reducing its kinetic energy. The force on the slower trolley accelerates it, increasing its kinetic energy.

However, the energy lost by the faster trolley does not all become kinetic energy of the slower trolley. Some is transferred into the deformation of the bumpers and other internal energy stores.

Another way to picture it: centre-of-mass motion

There is an even more useful way to understand the result.

The centre of mass of a system moves according to its total momentum:

\[ v_{\text{cm}}= \frac{p_{\text{total}}}{m_{\text{total}}} \]

Here, \(v_{\text{cm}}\) is the centre-of-mass velocity, and \(m_{\text{total}}\) is the combined mass.

If total momentum is conserved and the total mass stays constant, the centre-of-mass velocity cannot change.

However, the objects can still change their motion relative to each other.

During an inelastic collision, this relative motion decreases. The energy associated with that relative motion can be transferred into deformation, vibration, and internal energy.

The centre of mass keeps moving as before, even though the objects’ total translational kinetic energy decreases.

This is why momentum conservation and a decrease in kinetic energy can happen simultaneously.

03Kinetic energy is not the same as total energy

Kinetic energy is the energy an object has because of its motion:

\[ E_k=\frac12mv^2 \]

For a system containing several objects, calculate the kinetic energy of each object and add the results:

\[ E_{k,\text{total}} = \sum \frac12 mv^2 \]

Remember that kinetic energy depends on speed squared, not velocity.

This creates an important difference between momentum and kinetic energy.

So where does the missing kinetic energy go?

It hasn’t disappeared.

During an inelastic collision, some of the objects’ initial kinetic energy is transferred into other forms, including:

• Deformation of the objects.
• Internal energy and thermal energy.
• Vibrations within the objects.
• Sound produced during impact.

The total energy of a sufficiently complete isolated system is still conserved.

However, the total kinetic energy of the colliding objects does not have to remain constant.

There is also an important distinction between temporary energy storage and permanent energy loss from translational motion.

A compressed spring can temporarily store kinetic energy and return it afterwards. If the spring behaves ideally, the collision can still be elastic.

If some energy is irreversibly dissipated during the interaction, the final kinetic energy will be lower.

We will use this distinction in a more challenging question later.

04Momentum must be conserved in components

Now return to the original two-puck collision.

The pucks move away at different angles, so their momenta are not all in the same direction.

You cannot simply add their momentum magnitudes and expect to get the correct answer.

Momentum is a vector. For a two-dimensional collision, resolve it into horizontal and vertical components.

Choose:

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

If the net external impulse is negligible in both directions, conserve momentum separately:

\[ \sum p_{x,i}=\sum p_{x,f} \]

\[ \sum p_{y,i}=\sum p_{y,f} \]

For an object travelling at speed \(v\) at an angle \(\theta\) above the positive \(x\)-axis:

\[ v_x=v\cos\theta \]

\[ v_y=v\sin\theta \]

If the object travels below the positive \(x\)-axis, its vertical component is negative.

Kinetic energy is calculated differently. You use the full speed of each object:

\[ v^2=v_x^2+v_y^2 \]

The signs of the components matter for momentum, but kinetic energy is always non-negative.

05A reliable decision process for collision questions

When you encounter a collision problem in an HSC Physics exam, keep momentum and kinetic energy as two separate checks.

One subtle point is worth remembering: an external force does not automatically prevent momentum conservation.

What matters is the impulse produced by that force over the time interval.

For example, suppose a pair of trolleys experiences \(2.0\ \mathrm N\) of friction during a collision lasting \(0.010\ \mathrm s\).

The external impulse is:

\[ J=F\Delta t \]

\[ J=2.0(0.010)=0.020\ \mathrm{N\,s} \]

If the system initially has \(3.0\ \mathrm{kg\,m\,s^{-1}}\) of momentum, this impulse changes its momentum by only about \(0.67\%\).

Ignoring friction during the collision is therefore a reasonable approximation.

However, if the same friction force acts for \(0.50\ \mathrm s\) after the collision, the impulse becomes:

\[ J=2.0(0.50)=1.0\ \mathrm{N\,s} \]

That is no longer negligible compared with the original momentum.

The time interval matters just as much as the force itself.

For a broader explanation of choosing between momentum and kinetic-energy equations, see Momentum or Kinetic Energy? A Guide to Collision Problems.

For the specific case where colliding objects stick together, see Inelastic Collisions: Momentum and Kinetic Energy for HSC Physics.

07Questions and solutions

Question 1

A \(0.40\ \mathrm{kg}\) trolley moves to the right at \(3.0\ \mathrm{m\,s^{-1}}\) and collides with a stationary \(0.60\ \mathrm{kg}\) trolley. The trolleys stick together.

Assume the net external horizontal impulse during the collision is negligible.

Calculate:

(a) The common velocity immediately after the collision.

(b) The decrease in the total kinetic energy.

Solution 1

The joined trolleys move right at \(1.2\ \mathrm{m\,s^{-1}}\), and their total kinetic energy decreases by \(1.08\ \mathrm J\).

Part (a): Final velocity

Step 1: Apply conservation of momentum.

Take right as positive.

The initial momentum is:

\[ p_i=(0.40)(3.0)+(0.60)(0) \]

\[ p_i=1.20\ \mathrm{kg\,m\,s^{-1}} \]

Since the trolleys stick together, their combined mass is:

\[ m_{\text{total}}=0.40+0.60=1.00\ \mathrm{kg} \]

Step 2: Calculate the common final velocity.

\[ m_1u_1+m_2u_2 = (m_1+m_2)v \]

Substituting:

\[ 1.20=(1.00)v \]

\[ \boxed{v=+1.2\ \mathrm{m\,s^{-1}}} \]

The positive sign indicates motion to the right.

Part (b): Decrease in kinetic energy

Step 3: Calculate the initial kinetic energy.

\[ \begin{aligned} E_{k,i} &=\frac12(0.40)(3.0)^2\\ &=1.80\ \mathrm J \end{aligned} \]

The second trolley is stationary, so its initial kinetic energy is zero.

Step 4: Calculate the final kinetic energy.

\[ \begin{aligned} E_{k,f} &=\frac12(1.00)(1.2)^2\\ &=0.72\ \mathrm J \end{aligned} \]

Step 5: Calculate the decrease.

\[ \begin{aligned} E_{\text{decrease}} &=1.80-0.72\\ &=\boxed{1.08\ \mathrm J} \end{aligned} \]

The trolleys retain their total momentum, but \(1.08\ \mathrm J\) of their initial translational kinetic energy has been transferred into other forms.

The mistake would be to conserve kinetic energy simply because momentum is conserved.

Question 2

Two pucks collide on a horizontal air table.

Puck A has a mass of \(0.50\ \mathrm{kg}\), and puck B has a mass of \(0.25\ \mathrm{kg}\).

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

Immediately before the collision:

• Puck A has velocity \((+4.0,0)\ \mathrm{m\,s^{-1}}\).
• Puck B has velocity \((0,-2.0)\ \mathrm{m\,s^{-1}}\).

Immediately after the collision:

• Puck A has velocity \((+2.0,+1.0)\ \mathrm{m\,s^{-1}}\).
• Puck B has velocity \((+3.0,-4.0)\ \mathrm{m\,s^{-1}}\).

Assume the air table and guide produce negligible external impulse in the north-south direction. Any east-west external impulse must be determined from the data.

Calculate:

(a) The total initial momentum components.

(b) The total final momentum components.

(c) The net external impulse on the system.

(d) The decrease or increase in total translational kinetic energy.

Solution 2

The system receives an external impulse of \(0.25\ \mathrm{N\,s}\) west. Its north-south momentum is unchanged, while its total translational kinetic energy decreases by \(0.125\ \mathrm J\).

Part (a): Initial momentum

Step 1: Calculate the initial momentum of each puck.

For puck A:

\[ \vec p_A = 0.50(4.0,0) \]

\[ \vec p_A=(2.0,0) \]

For puck B:

\[ \vec p_B = 0.25(0,-2.0) \]

\[ \vec p_B=(0,-0.50) \]

Therefore, the total initial momentum is:

\[ \boxed{\vec p_i=(2.0,-0.50)} \]

The units are \(\mathrm{kg\,m\,s^{-1}}\).

Part (b): Final momentum

Step 2: Calculate the final momentum of each puck.

For puck A:

\[ \vec p_{A,f} = 0.50(2.0,1.0) \]

\[ \vec p_{A,f}=(1.0,0.50) \]

For puck B:

\[ \vec p_{B,f} = 0.25(3.0,-4.0) \]

\[ \vec p_{B,f}=(0.75,-1.0) \]

Add the components:

\[ \begin{aligned} p_{x,f}&=1.0+0.75=1.75\\ p_{y,f}&=0.50-1.0=-0.50 \end{aligned} \]

Therefore:

\[ \boxed{\vec p_f=(1.75,-0.50)} \]

Part (c): External impulse

Step 3: Apply the impulse-momentum theorem.

\[ \vec J_{\text{external}} = \vec p_f-\vec p_i \]

Substitute:

\[ \begin{aligned} J_x&=1.75-2.0=-0.25\\ J_y&=-0.50-(-0.50)=0 \end{aligned} \]

Therefore:

\[ \boxed{\vec J_{\text{external}} =(-0.25,0)\ \mathrm{N\,s}} \]

The external impulse acts west.

Notice that north-south momentum is conserved, but east-west momentum is not.

This is why you must consider momentum components separately when external impulses act in particular directions.

Part (d): Kinetic energy

Step 4: Calculate the initial kinetic energy.

For puck A:

\[ E_{k,A,i} = \frac12(0.50)(4.0)^2 =4.0\ \mathrm J \]

For puck B:

\[ E_{k,B,i} = \frac12(0.25)(2.0)^2 =0.50\ \mathrm J \]

Therefore:

\[ E_{k,i}=4.50\ \mathrm J \]

Step 5: Calculate the final kinetic energy.

For puck A:

\[ \begin{aligned} E_{k,A,f} &=\frac12(0.50)(2.0^2+1.0^2)\\ &=1.25\ \mathrm J \end{aligned} \]

For puck B:

\[ \begin{aligned} E_{k,B,f} &=\frac12(0.25)(3.0^2+4.0^2)\\ &=3.125\ \mathrm J \end{aligned} \]

Therefore:

\[ E_{k,f}=1.25+3.125 \]

\[ E_{k,f}=4.375\ \mathrm J \]

Step 6: Compare the energies.

\[ \begin{aligned} E_{\text{decrease}} &=4.50-4.375\\ &=\boxed{0.125\ \mathrm J} \end{aligned} \]

The system loses \(0.125\ \mathrm J\) of translational kinetic energy.

You cannot automatically attribute all of this decrease to deformation because the system also experiences an external impulse. More information about the guide and collision would be needed to establish the complete energy transfers.

Question 3

A \(0.50\ \mathrm{kg}\) cart A moves right at \(4.0\ \mathrm{m\,s^{-1}}\). It collides with a \(0.80\ \mathrm{kg}\) cart B moving left at an unknown speed.

The carts stick together and are observed to be stationary immediately after impact.

Assume the net external horizontal impulse is negligible.

Calculate:

(a) The initial velocity of cart B.

(b) The total initial kinetic energy.

(c) The decrease in total kinetic energy.

Explain why the final zero momentum does not mean the initial kinetic energy was also zero.

Solution 3

Cart B initially moves left at \(2.5\ \mathrm{m\,s^{-1}}\). The initial total kinetic energy is \(6.5\ \mathrm J\), all of which is removed from the carts’ translational motion during the perfectly inelastic collision.

Part (a): Initial velocity of cart B

Step 1: Use conservation of momentum.

The carts are stationary after impact, so:

\[ p_f=0 \]

Since momentum is conserved:

\[ p_i=0 \]

Take right as positive.

\[ m_Au_A+m_Bu_B=0 \]

Substitute:

\[ (0.50)(4.0)+(0.80)u_B=0 \]

\[ 2.0+0.80u_B=0 \]

Rearrange:

\[ u_B=\frac{-2.0}{0.80} \]

\[ \boxed{u_B=-2.5\ \mathrm{m\,s^{-1}}} \]

The negative sign confirms that cart B initially moves left.

Part (b): Initial kinetic energy

Step 2: Calculate the kinetic energy of each cart.

For cart A:

\[ \begin{aligned} E_{k,A} &=\frac12(0.50)(4.0)^2\\ &=4.0\ \mathrm J \end{aligned} \]

For cart B:

\[ \begin{aligned} E_{k,B} &=\frac12(0.80)(2.5)^2\\ &=2.5\ \mathrm J \end{aligned} \]

Therefore:

\[ \boxed{E_{k,i}=6.5\ \mathrm J} \]

Part (c): Decrease in kinetic energy

Step 3: Calculate the final kinetic energy.

Both carts are stationary after impact:

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

Therefore:

\[ \begin{aligned} E_{\text{decrease}} &=6.5-0\\ &=\boxed{6.5\ \mathrm J} \end{aligned} \]

The carts’ total initial momentum was zero because their momenta were equal in magnitude and opposite in direction.

However, kinetic energy is a scalar. The two positive kinetic-energy values add rather than cancel.

This is the important misconception: zero total momentum does not mean zero kinetic energy.

The carts can have substantial kinetic energy even when their combined momentum is zero.

When they stick together, their relative motion disappears, and their translational kinetic energy is transferred into other forms.

Question 4

Two deformable trolleys collide on a low-friction track.

During contact, trolley A exerts a force on trolley B, and trolley B exerts an equal and opposite force on trolley A at every instant.

A student argues:

The internal forces are equal and opposite, so they cancel. Therefore, the total kinetic energy of the two-trolley system must remain constant.

Explain why this reasoning is incorrect.

Your response should:

(a) Explain what happens to the total momentum of the two-trolley system.

(b) Distinguish the effect of the forces on momentum from their effect on kinetic energy.

(c) Identify where some of the initial kinetic energy can be transferred during the collision.

Solution 4

The student’s conclusion is incorrect. Equal-and-opposite internal forces cancel in the total impulse calculation, but they do not necessarily cancel in the total work calculation.

Part (a): Total momentum

Choose both trolleys as the system.

Newton’s third law gives:

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

Because these forces act over the same collision interval, their impulses are equal and opposite.

Therefore:

\[ \Delta\vec p_A+\Delta\vec p_B=0 \]

The total momentum of the two-trolley system remains unchanged, provided the net external impulse is negligible.

Part (b): Total kinetic energy

The mistake is assuming that equal-and-opposite forces must do equal-and-opposite work.

Work depends on both force and displacement:

\[ W=Fs\cos\theta \]

The contact forces act on different trolleys, and their points of application can move different distances as the bumpers compress.

For example, the force on the faster trolley opposes its motion and removes kinetic energy. The force on the slower trolley accelerates it, adding kinetic energy.

The energy removed from the faster trolley does not necessarily equal the energy gained by the slower one.

The difference is transferred into other energy stores.

Part (c): Other energy forms

Some of the original translational kinetic energy can become:

• Internal energy associated with deformation.
• Thermal energy.
• Vibrational energy.
• Sound.

The total energy of a sufficiently complete isolated system remains conserved.

The key distinction is that Newton’s third law explains why the internal impulses cancel. It does not require the total translational kinetic energy to remain constant.

Question 5

Two identical \(0.50\ \mathrm{kg}\) carts are positioned on a horizontal low-friction track.

Cart A moves right at \(4.0\ \mathrm{m\,s^{-1}}\), while cart B is stationary.

The carts have spring bumpers that are initially uncompressed. During impact, the bumpers compress, causing the carts to move together temporarily. They then separate as the spring expands.

Assume:

• There is negligible external horizontal impulse.
• The carts do not rotate.
• The spring initially contains no stored energy.
• At maximum compression, both carts have the same velocity.
• In the ideal case, no energy is dissipated.

Answer the following.

(a) Calculate the common velocity of the carts at maximum compression.

(b) Calculate their total translational kinetic energy at maximum compression. How much energy is temporarily stored in the spring?

(c) If the spring is ideal and returns all its stored energy, determine the final velocity of each cart. Is the collision elastic or inelastic?

(d) Now suppose \(0.50\ \mathrm J\) of the stored spring energy is irreversibly dissipated before the carts separate. Determine the final velocity of each cart and classify this second collision.

Solution 5

The ideal collision is elastic, even though the carts’ translational kinetic energy decreases temporarily at maximum compression. The missing kinetic energy is stored in the compressed spring and subsequently returned.

If \(0.50\ \mathrm J\) is dissipated, the collision becomes inelastic, and the final velocities are approximately \(0.268\ \mathrm{m\,s^{-1}}\) for cart A and \(3.73\ \mathrm{m\,s^{-1}}\) for cart B, both to the right.

This question tests an important distinction: you must compare kinetic energy immediately before and after the collision, not at an intermediate stage.

Part (a): Velocity at maximum compression

Step 1: Calculate the initial total momentum.

Take right as positive.

\[ \begin{aligned} p_i &=(0.50)(4.0)+(0.50)(0)\\ &=2.0\ \mathrm{kg\,m\,s^{-1}} \end{aligned} \]

At maximum compression, both carts have the same velocity.

The combined mass is:

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

Step 2: Apply conservation of momentum.

\[ p_i=p_f \]

\[ 2.0=(1.00)v \]

Therefore:

\[ \boxed{v=+2.0\ \mathrm{m\,s^{-1}}} \]

Both carts move right at \(2.0\ \mathrm{m\,s^{-1}}\).

Part (b): Kinetic energy at maximum compression

Step 3: Calculate the initial kinetic energy.

Only cart A is moving initially:

\[ \begin{aligned} E_{k,i} &=\frac12(0.50)(4.0)^2\\ &=4.0\ \mathrm J \end{aligned} \]

Step 4: Calculate the kinetic energy at maximum compression.

Both carts move together at \(2.0\ \mathrm{m\,s^{-1}}\):

\[ \begin{aligned} E_{k,\text{compression}} &=\frac12(1.00)(2.0)^2\\ &=2.0\ \mathrm J \end{aligned} \]

The decrease in translational kinetic energy is:

\[ \begin{aligned} \Delta E_k &=4.0-2.0\\ &=2.0\ \mathrm J \end{aligned} \]

Since the spring is assumed to be ideal and no energy has been dissipated, this \(2.0\ \mathrm J\) is stored as elastic potential energy in the compressed spring.

Part (c): Ideal spring and final velocities

Step 5: Apply conservation of kinetic energy between the initial and final states.

Because the spring returns all its stored energy:

\[ E_{k,f}=E_{k,i}=4.0\ \mathrm J \]

Let the final velocities of carts A and B be \(v_A\) and \(v_B\).

Momentum conservation gives:

\[ 0.50v_A+0.50v_B=2.0 \]

Therefore:

\[ v_A+v_B=4.0 \]

Kinetic energy conservation gives:

\[ \frac12(0.50)v_A^2 + \frac12(0.50)v_B^2 =4.0 \]

Simplifying:

\[ v_A^2+v_B^2=16 \]

Step 6: Solve the two equations.

We know:

\[ (v_A+v_B)^2=16 \]

Expanding:

\[ v_A^2+2v_Av_B+v_B^2=16 \]

Since \(v_A^2+v_B^2=16\):

\[ 2v_Av_B=0 \]

Therefore:

\[ v_Av_B=0 \]

One cart must finish stationary.

Combining this result with \(v_A+v_B=4.0\), the final velocities are:

\[ \boxed{v_A=0,\qquad v_B=+4.0\ \mathrm{m\,s^{-1}}} \]

Cart A stops, while cart B moves right at \(4.0\ \mathrm{m\,s^{-1}}\).

The final kinetic energy is equal to the initial kinetic energy.

Therefore, the collision is elastic.

Part (d): Some energy is dissipated

Step 7: Calculate the final kinetic energy.

The spring initially stores \(2.0\ \mathrm J\).

If \(0.50\ \mathrm J\) is irreversibly dissipated, only \(1.50\ \mathrm J\) is returned to the carts’ translational motion.

Therefore:

\[ E_{k,f}=4.0-0.50 \]

\[ \boxed{E_{k,f}=3.50\ \mathrm J} \]

Step 8: Apply momentum conservation again.

The external horizontal impulse remains negligible, so:

\[ 0.50v_A+0.50v_B=2.0 \]

Therefore:

\[ v_A+v_B=4.0 \]

Step 9: Use the new kinetic energy.

\[ \frac12(0.50)v_A^2 + \frac12(0.50)v_B^2 =3.50 \]

Simplifying:

\[ v_A^2+v_B^2=14 \]

Step 10: Solve for the final velocities.

Using:

\[ (v_A+v_B)^2 = v_A^2+2v_Av_B+v_B^2 \]

Substitute the known values:

\[ 16=14+2v_Av_B \]

Therefore:

\[ v_Av_B=1.0 \]

The two velocities satisfy:

\[ v^2-4.0v+1.0=0 \]

Using the quadratic formula:

\[ v=\frac{4.0\pm\sqrt{16-4}}{2} \]

\[ v=2.0\pm\sqrt3 \]

This gives:

\[ v=0.268\ \mathrm{m\,s^{-1}} \]

or:

\[ v=3.73\ \mathrm{m\,s^{-1}} \]

Cart A was initially moving towards cart B from behind. For the carts to separate, A must finish slower than B.

Therefore:

\[ \boxed{v_A=+0.268\ \mathrm{m\,s^{-1}}} \]

\[ \boxed{v_B=+3.73\ \mathrm{m\,s^{-1}}} \]

The final kinetic energy is \(3.50\ \mathrm J\), which is \(0.50\ \mathrm J\) less than the initial kinetic energy.

This second collision is inelastic.

The important lesson

At maximum compression, the carts’ translational kinetic energy decreased by \(2.0\ \mathrm J\). That alone did not make the collision inelastic.

In the ideal case, all \(2.0\ \mathrm J\) returned to kinetic energy before the carts separated.

Only the second case permanently reduced the final translational kinetic energy.

When solving collision problems, therefore, keep two questions separate:

• Is the net external impulse on my chosen system negligible? This determines whether I can conserve momentum.
• Is the total kinetic energy before the collision equal to the total kinetic energy after separation? This determines whether the collision is elastic.

Getting those two decisions right is the foundation for handling more complicated collisions, including angled impacts, spring interactions, and collisions involving external forces.