Why One Object’s Momentum Is Not Conserved in a Collision

A 0.50 kg trolley A moves to the right at 1.2 m/s and collides with a stationary 1.00 kg trolley B. After the collision, A is still moving right, but at only 0.20 m/s. Its momentum has changed from \(+0.60\) to \(+0.10\ \text{kg m s}^{-1}\).

Why hasn’t momentum been conserved?

Because trolley B exerts a force on A during the collision, giving A an impulse. This changes A’s momentum. Momentum is not generally conserved for an individual object during a collision. It is the total momentum of an appropriately chosen system that can be conserved, provided the net external impulse on that system is negligible.

So where did A’s missing \(+0.50\ \text{kg m s}^{-1}\) go?

It was transferred to trolley B. If B initially has zero momentum, it must gain \(+0.50\ \text{kg m s}^{-1}\). The total momentum of both trolleys remains unchanged.

01The experiment: measure both objects, not just one

The simplest way to understand this is to investigate what happens when two trolleys collide.

The question is straightforward:

Does the moving trolley retain its original momentum, or does some of its momentum change as a result of interacting with the second trolley?

Setting up the investigation

Consider a low-friction dynamics track positioned horizontally.

Place trolley A on the left and stationary trolley B on the right. Attach a small, visible marker to each trolley. Position a measuring scale parallel to the track and use a fixed overhead camera to record the collision.

The camera must capture both trolleys before, during, and after contact.

Measure the mass of each trolley using a balance. Then record their positions at known time intervals.

For example, if the camera records 60 frames per second, the time between consecutive frames is

\[ \Delta t=\frac{1}{60}=0.0167\text{ s} \]

If a trolley’s marker moves 0.020 m between two consecutive frames, its approximate velocity is

\[ v=\frac{\Delta x}{\Delta t} =\frac{0.020}{0.0167} =1.20\text{ m s}^{-1} \]

In practice, measuring displacement over several frames helps reduce uncertainty.

A level track is important because it reduces unwanted horizontal forces. Friction and air resistance should also be minimised, particularly during the short collision interval.

What should happen?

Before collecting the data, consider two possible predictions.

Prediction 1: Momentum of trolley A is conserved.

If this were true, A would retain its original momentum after the collision. Since its mass remains constant, its velocity would also remain unchanged.

Prediction 2: Momentum is transferred between the trolleys.

If this is correct, A can lose momentum while B gains momentum. The total momentum of A and B should remain constant if external impulses are negligible.

The second prediction is the one supported by the collision data.

Analysing illustrative results

Suppose the investigation produces the following illustrative measurements:

Take right as the positive direction.

Calculate the momentum of each trolley before and after the collision using

\[ p=mv \]

The results are:

Notice something important.

Trolley A loses \(0.50\ \text{kg m s}^{-1}\) of momentum, while trolley B gains exactly \(0.50\ \text{kg m s}^{-1}\).

The changes are equal in magnitude and opposite in direction.

For the combined system:

\[ \begin{aligned} p_{\text{initial}}&=0.60+0\\ &=0.60\ \text{kg m s}^{-1} \end{aligned} \]

After the collision:

\[ \begin{aligned} p_{\text{final}}&=0.10+0.50\\ &=0.60\ \text{kg m s}^{-1} \end{aligned} \]

The total momentum is unchanged, even though neither trolley has retained its original momentum.

Key experimental result

The individual momentum of trolley A changes, but the combined momentum of A and B remains constant under the stated assumptions.

These numbers are illustrative, not claimed experimental measurements. Real results will have some uncertainty due to friction, track alignment, timing, and position measurements.

Why tracking both trolleys matters

Imagine an incomplete version of this experiment in which you only measure trolley A.

You observe that it slows down after the collision. You can calculate its change in momentum, but you cannot establish where that momentum has gone.

The improved design measures both trolleys simultaneously.

This allows you to compare their individual momentum changes and test whether the combined momentum is conserved. It also makes it possible to identify external effects that might explain any discrepancy.

For a real investigation, checking the track’s alignment and measuring the motion over a short collision interval are important. Momentum conservation is an exact result for an isolated system, but experimental setups only approximate isolation.

02Why Newton’s third law does not conserve each object’s momentum

The explanation comes from Newton’s third law.

During the collision, trolley A pushes trolley B, and trolley B pushes trolley A.

These forces are equal in magnitude and opposite in direction:

\[ \vec F_{B\to A}=-\vec F_{A\to B} \]

However, there is a detail that students often miss.

The two forces act on different objects.

The force acting on A does not cancel the force acting on A from some other direction. The equal-and-opposite force acts on B.

The relationship between force and momentum is given by the impulse-momentum theorem:

\[ \vec J=\Delta\vec p \]

where:

• \(\vec J\) is impulse, measured in newton seconds (N s).
• \(\Delta\vec p\) is the change in momentum, measured in \(\text{kg m s}^{-1}\).

Since impulse is force multiplied by the time over which it acts, the two trolleys receive equal and opposite impulses during their interaction:

\[ \vec J_{B\to A}=-\vec J_{A\to B} \]

If other external impulses are negligible, this means

\[ \Delta\vec p_A=-\Delta\vec p_B \]

So when A loses momentum, B gains the corresponding momentum.

Adding their changes gives

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

This is why the total momentum of the two-trolley system remains constant.

The important distinction is that Newton’s third law explains why the internal impulses cancel when we consider both objects together. It does not say that either object’s momentum must remain unchanged.

03Choosing the correct system

Before applying conservation of momentum, decide which objects belong to your system.

A system is simply the collection of objects you have chosen to analyse. Changing that collection can change whether momentum is conserved.

Consider the same collision between trolleys A and B.

The third row is usually the most useful choice for HSC collision calculations.

For a more detailed explanation of system boundaries and external forces, see Closed Systems in Collisions: When Momentum Is Conserved.

What if an external force acts?

Momentum conservation is not automatically valid just because two objects are colliding.

The more general impulse-momentum relationship is

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

If the net external impulse is zero, the system’s total momentum is unchanged.

If the external impulse is non-zero, the system’s total momentum changes.

For a constant net external force:

\[ \vec J_{\text{external}} =\vec F_{\text{external}}\Delta t \]

This explains why friction can sometimes be ignored during a short collision but cannot necessarily be ignored over a much longer interval.

The relevant question is whether the external impulse is significant during the time being analysed.

Momentum components must be considered separately

Momentum is a vector quantity, so it has both magnitude and direction.

For a collision in two dimensions, choose perpendicular axes, usually \(x\) and \(y\).

If external impulses are negligible in both directions, conserve the total momentum components separately:

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

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

Here, \(p_x\) and \(p_y\) represent the horizontal and vertical components of momentum, respectively. The subscripts \(i\) and \(f\) represent initial and final values.

If there is an external impulse in one direction, you must account for it in that component. You cannot automatically assume both components are conserved.

For example, a smooth vertical wall can exert a horizontal impulse on a puck while exerting negligible impulse parallel to its surface.

Momentum and kinetic energy are separate checks

Another common misconception is that conserving momentum means conserving kinetic energy.

These are different physical quantities.

For a system of colliding objects, total kinetic energy is

\[ K=\sum\frac12 mv^2 \]

where \(K\) is kinetic energy in joules, \(m\) is mass in kilograms, and \(v\) is speed in metres per second.

After calculating the final velocities using momentum conservation, compare the total initial and final kinetic energies.

A decrease in kinetic energy does not mean energy has disappeared. It has been converted into other forms, such as thermal energy, sound, and deformation.

Total energy is still conserved.

04A contrasting case: a ball colliding with a wall

A fixed wall provides a useful comparison.

Suppose a ball travels towards a wall and rebounds in the opposite direction.

The ball’s momentum clearly changes because its velocity changes direction.

If the ball is your chosen system, the wall is outside the system. The wall exerts an external force on the ball, giving it an impulse.

But what happens to the momentum?

The wall receives an equal and opposite impulse. Since it is attached to a building and the Earth, the momentum is transferred to the much larger Earth system.

The wall’s movement is too small to notice, but that does not mean it receives no momentum.

If we instead choose the ball, wall, and Earth as the system, the momentum can be conserved across the interaction, assuming external impulses on that larger system are negligible.

This is the same principle as the two-trolley collision. The result depends on which objects are included in the system.

06Questions and solutions

Question 1

A \(0.40\text{ kg}\) trolley A travels right at \(2.5\text{ m s}^{-1}\) and collides with a stationary \(0.60\text{ kg}\) trolley B.

Immediately after the collision, A continues moving right at \(1.0\text{ m s}^{-1}\).

Assuming external horizontal impulse is negligible, calculate:

a) The final velocity of trolley B.

b) The change in momentum of trolley A.

Take right as positive.

Solution 1

Trolley B moves right at \(1.0\text{ m s}^{-1}\), and trolley A experiences a momentum change of \(-0.60\text{ kg m s}^{-1}\).

The key is to conserve the total momentum of both trolleys, rather than attempting to conserve A’s momentum separately.

Step 1: Calculate the initial total momentum.

\[ \begin{aligned} p_i&=m_Au_A+m_Bu_B\\ &=(0.40)(2.5)+(0.60)(0)\\ &=1.0\text{ kg m s}^{-1} \end{aligned} \]

Step 2: Apply conservation of momentum.

Let B’s final velocity be \(v_B\).

\[ \begin{aligned} p_i&=p_f\\ 1.0&=(0.40)(1.0)+(0.60)v_B \end{aligned} \]

Rearranging:

\[ \begin{aligned} 1.0-0.40&=0.60v_B\\ v_B&=\frac{0.60}{0.60}\\ &=1.0\text{ m s}^{-1} \end{aligned} \]

The positive result means B moves right.

Step 3: Calculate the change in A’s momentum.

\[ \Delta p_A=p_{A,f}-p_{A,i} \]

\[ \begin{aligned} \Delta p_A&=(0.40)(1.0)-(0.40)(2.5)\\ &=0.40-1.0\\ &=-0.60\text{ kg m s}^{-1} \end{aligned} \]

A loses \(0.60\text{ kg m s}^{-1}\) of momentum, while B gains \(0.60\text{ kg m s}^{-1}\).

Their momentum changes cancel, leaving the combined momentum unchanged.

Question 2

A \(0.25\text{ kg}\) ball travels right at \(8.0\text{ m s}^{-1}\) and collides with a \(0.50\text{ kg}\) trolley travelling left at \(1.0\text{ m s}^{-1}\).

After the collision, the ball rebounds left at \(2.0\text{ m s}^{-1}\).

Assume external impulse is negligible.

a) Calculate the final velocity of the trolley.

b) Calculate the momentum change of each object.

c) Determine whether kinetic energy is conserved.

Take right as positive.

Solution 2

The trolley moves right at \(4.0\text{ m s}^{-1}\). The ball receives an impulse of \(-2.5\text{ N s}\), while the trolley receives an impulse of \(+2.5\text{ N s}\). Kinetic energy is not conserved.

The important detail is that the ball rebounds. Its final velocity must therefore be negative.

Step 1: Calculate the initial total momentum.

\[ \begin{aligned} p_i&=m_Au_A+m_Bu_B\\ &=(0.25)(8.0)+(0.50)(-1.0)\\ &=2.0-0.50\\ &=1.50\text{ kg m s}^{-1} \end{aligned} \]

Step 2: Apply conservation of momentum.

The ball’s final velocity is \(-2.0\text{ m s}^{-1}\).

Let the trolley’s final velocity be \(v_B\).

\[ \begin{aligned} 1.50&=(0.25)(-2.0)+(0.50)v_B\\ 1.50&=-0.50+0.50v_B\\ 2.0&=0.50v_B\\ v_B&=4.0\text{ m s}^{-1} \end{aligned} \]

The trolley moves right.

Step 3: Calculate the momentum change of the ball.

\[ \begin{aligned} \Delta p_A&=p_{A,f}-p_{A,i}\\ &=(0.25)(-2.0)-(0.25)(8.0)\\ &=-0.50-2.0\\ &=-2.50\text{ kg m s}^{-1} \end{aligned} \]

Step 4: Calculate the momentum change of the trolley.

\[ \begin{aligned} \Delta p_B&=p_{B,f}-p_{B,i}\\ &=(0.50)(4.0)-(0.50)(-1.0)\\ &=2.0+0.50\\ &=+2.50\text{ kg m s}^{-1} \end{aligned} \]

The changes are equal in magnitude and opposite in direction, as expected when external impulse is negligible.

Step 5: Compare initial and final kinetic energy.

Initial kinetic energy:

\[ \begin{aligned} K_i&=\frac12(0.25)(8.0)^2\\ &\quad+\frac12(0.50)(1.0)^2\\ &=8.0+0.25\\ &=8.25\text{ J} \end{aligned} \]

Final kinetic energy:

\[ \begin{aligned} K_f&=\frac12(0.25)(2.0)^2\\ &\quad+\frac12(0.50)(4.0)^2\\ &=0.50+4.0\\ &=4.50\text{ J} \end{aligned} \]

The decrease is

\[ \Delta K=4.50-8.25=-3.75\text{ J} \]

The collision is inelastic. Some kinetic energy has been converted into other forms of energy.

The tempting mistake here is to assume that because the ball loses momentum, the entire system must lose momentum too. It doesn’t. The trolley receives the corresponding momentum change.

Question 3

Two pucks collide on a smooth horizontal air table.

Puck A has a mass of \(0.40\text{ kg}\) and initially travels right at \(3.0\text{ m s}^{-1}\).

Puck B has a mass of \(0.20\text{ kg}\) and initially travels upwards at \(2.0\text{ m s}^{-1}\).

After the collision, puck A has velocity components

\[ v_{Ax}=+1.0\text{ m s}^{-1} \]

\[ v_{Ay}=+1.0\text{ m s}^{-1} \]

Assume negligible external impulse in the plane.

Calculate:

a) The final horizontal and vertical velocity components of puck B.

b) The speed and direction of puck B.

c) Whether kinetic energy is conserved.

Solution 3

Puck B finishes moving right at \(4.0\text{ m s}^{-1}\), with no vertical velocity component. The collision is inelastic because total kinetic energy decreases.

This problem requires two separate momentum equations because the pucks move in two dimensions.

Step 1: Calculate the initial momentum components.

In the horizontal direction:

\[ \begin{aligned} p_{x,i}&=(0.40)(3.0)+(0.20)(0)\\ &=1.20\text{ kg m s}^{-1} \end{aligned} \]

In the vertical direction:

\[ \begin{aligned} p_{y,i}&=(0.40)(0)+(0.20)(2.0)\\ &=0.40\text{ kg m s}^{-1} \end{aligned} \]

Step 2: Calculate the final momentum components of A.

\[ p_{Ax,f}=(0.40)(1.0)=0.40 \]

\[ p_{Ay,f}=(0.40)(1.0)=0.40 \]

Both values are in \(\text{kg m s}^{-1}\).

Step 3: Calculate B’s final momentum components.

Conserve horizontal momentum:

\[ \begin{aligned} 1.20&=0.40+0.20v_{Bx}\\ 0.80&=0.20v_{Bx}\\ v_{Bx}&=4.0\text{ m s}^{-1} \end{aligned} \]

Conserve vertical momentum:

\[ \begin{aligned} 0.40&=0.40+0.20v_{By}\\ 0&=0.20v_{By}\\ v_{By}&=0 \end{aligned} \]

Step 4: Find B’s speed and direction.

Since its vertical velocity is zero:

\[ v_B=\sqrt{4.0^2+0^2} =4.0\text{ m s}^{-1} \]

It moves directly to the right, along the positive \(x\)-axis.

Step 5: Check kinetic energy.

Initial kinetic energy:

\[ \begin{aligned} K_i&=\frac12(0.40)(3.0)^2\\ &\quad+\frac12(0.20)(2.0)^2\\ &=1.80+0.40\\ &=2.20\text{ J} \end{aligned} \]

Final kinetic energy:

\[ \begin{aligned} K_f&=\frac12(0.40)(1.0^2+1.0^2)\\ &\quad+\frac12(0.20)(4.0)^2\\ &=0.40+1.60\\ &=2.00\text{ J} \end{aligned} \]

Kinetic energy decreases by \(0.20\text{ J}\).

Notice that B has transferred all of its initial vertical momentum to A. Meanwhile, A has transferred some horizontal momentum to B.

This is a useful reminder that momentum can be exchanged in different directions independently.

Question 4

Two trolleys collide on a horizontal track.

Their masses and measured velocities are:

The collision lasts \(0.030\text{ s}\).

During this interval, a constant horizontal resistive force of \(0.50\text{ N}\) acts to the left on the combined two-trolley system.

a) Calculate the total momentum before and after the collision.

b) Calculate the external impulse on the system.

c) Determine whether the measured momentum change is consistent with the stated resistive force.

d) Calculate the momentum change of each trolley. Explain why these changes are not exactly equal and opposite.

Take right as positive.

Solution 4

The data are consistent with the stated resistive force. The combined momentum decreases by \(0.015\text{ kg m s}^{-1}\), exactly matching the external impulse of \(0.015\text{ N s}\) to the left.

This question tests whether you can distinguish momentum transferred between objects from momentum changed by an external force.

Step 1: Calculate the initial total momentum.

\[ \begin{aligned} p_i&=(0.50)(2.0)+(0.75)(-0.40)\\ &=1.0-0.30\\ &=0.70\text{ kg m s}^{-1} \end{aligned} \]

Step 2: Calculate the final total momentum.

\[ \begin{aligned} p_f&=(0.50)(0.50)+(0.75)(0.58)\\ &=0.25+0.435\\ &=0.685\text{ kg m s}^{-1} \end{aligned} \]

Therefore:

\[ \begin{aligned} \Delta p_{\text{system}}&=p_f-p_i\\ &=0.685-0.70\\ &=-0.015\text{ kg m s}^{-1} \end{aligned} \]

The total momentum has decreased slightly.

Step 3: Calculate the external impulse.

Take right as positive, so the resistive force is negative.

\[ \begin{aligned} J_{\text{external}}&=F_{\text{external}}\Delta t\\ &=(-0.50)(0.030)\\ &=-0.015\text{ N s} \end{aligned} \]

Since \(1\text{ N s}=1\text{ kg m s}^{-1}\), this matches the measured momentum change.

Step 4: Calculate each trolley’s momentum change.

For trolley A:

\[ \begin{aligned} \Delta p_A&=(0.50)(0.50)-(0.50)(2.0)\\ &=0.25-1.0\\ &=-0.75\text{ kg m s}^{-1} \end{aligned} \]

For trolley B:

\[ \begin{aligned} \Delta p_B&=(0.75)(0.58)-(0.75)(-0.40)\\ &=0.435+0.30\\ &=+0.735\text{ kg m s}^{-1} \end{aligned} \]

Add the changes:

\[ \begin{aligned} \Delta p_A+\Delta p_B &=-0.75+0.735\\ &=-0.015\text{ kg m s}^{-1} \end{aligned} \]

This is exactly the external momentum change calculated earlier.

The tempting assumption is that the two trolleys must always have equal and opposite momentum changes. That would be true for their combined interaction if there were no other significant impulses. Here, the resistive force also acts on the system, so the total momentum changes.

The collision forces still form an equal-and-opposite Newton’s third law pair. The additional external impulse explains the difference.

Question 5

Two gliders collide on a smooth horizontal air table.

Glider A has a mass of \(0.40\text{ kg}\) and initially moves along the positive \(x\)-axis at \(3.0\text{ m s}^{-1}\).

Glider B has a mass of \(0.60\text{ kg}\) and is initially stationary.

During their collision, the external impulse in the plane of the table is negligible.

Immediately after the collision, glider A has velocity components

\[ v_{Ax}=+1.5\text{ m s}^{-1} \]

\[ v_{Ay}=+1.0\text{ m s}^{-1} \]

Glider B then travels towards a short, smooth vertical wall segment whose normal is parallel to the \(x\)-axis. Glider A misses the wall.

The collision between B and the wall is perfectly elastic in the direction perpendicular to the wall. Therefore, B’s horizontal velocity component reverses, while its vertical component remains unchanged.

Answer the following:

a) Calculate the horizontal and vertical velocity components of B immediately after the first collision.

b) Determine the total momentum of A and B after B rebounds from the wall.

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

d) Calculate the total kinetic energy before the first collision, after the first collision, and after the wall rebound.

e) Explain why momentum conservation can be applied to the first collision but not to the entire sequence if the chosen system contains only A and B.

Assume the wall is fixed to Earth, there are no other significant horizontal forces, and the wall exerts no impulse parallel to its surface.

Solution 5

Glider B leaves the first collision with velocity components \(+1.0\text{ m s}^{-1}\) horizontally and approximately \(-0.67\text{ m s}^{-1}\) vertically. After rebounding from the wall, the combined momentum of A and B is zero.

There is no violation of momentum conservation. The wall provides an external impulse to the A+B system, changing its total momentum during the second interaction.

This question is difficult because the same two gliders are involved throughout, but the external interactions change. You must analyse the two stages separately.

Step 1: Calculate the initial momentum components.

Initially, only A is moving.

\[ \begin{aligned} p_{x,i}&=(0.40)(3.0)\\ &=1.20\text{ kg m s}^{-1} \end{aligned} \]

There is no initial vertical momentum:

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

Step 2: Calculate A’s final momentum components after the first collision.

\[ \begin{aligned} p_{Ax,f}&=(0.40)(1.5)\\ &=0.60\text{ kg m s}^{-1} \end{aligned} \]

\[ \begin{aligned} p_{Ay,f}&=(0.40)(1.0)\\ &=0.40\text{ kg m s}^{-1} \end{aligned} \]

Step 3: Calculate B’s final momentum components.

During the first collision, external impulse is negligible, so total momentum is conserved in both directions.

Horizontal direction:

\[ \begin{aligned} 1.20&=0.60+p_{Bx,f}\\ p_{Bx,f}&=+0.60\text{ kg m s}^{-1} \end{aligned} \]

Vertical direction:

\[ \begin{aligned} 0&=0.40+p_{By,f}\\ p_{By,f}&=-0.40\text{ kg m s}^{-1} \end{aligned} \]

Step 4: Calculate B’s velocity components.

\[ v_{Bx}=\frac{0.60}{0.60} =+1.0\text{ m s}^{-1} \]

\[ v_{By}=\frac{-0.40}{0.60} =-0.67\text{ m s}^{-1} \]

Therefore, B travels right and down towards the wall.

Step 5: Analyse the wall collision.

The wall is smooth, so it exerts no impulse parallel to its surface.

Its normal is along the \(x\)-axis. Because the collision is perfectly elastic in that direction, B’s horizontal velocity component reverses.

Before the wall collision:

\[ \vec v_{B,\text{before}} =(+1.0,-0.67)\text{ m s}^{-1} \]

After the wall collision:

\[ \vec v_{B,\text{after}} =(-1.0,-0.67)\text{ m s}^{-1} \]

Glider A is unaffected by this second collision.

Step 6: Calculate the total momentum after the wall collision.

A’s momentum remains

\[ \vec p_A=(+0.60,+0.40) \text{ kg m s}^{-1} \]

B’s new momentum is

\[ \vec p_B=(-0.60,-0.40) \text{ kg m s}^{-1} \]

Adding the components:

\[ p_{x,\text{total}}=0.60-0.60=0 \]

\[ p_{y,\text{total}}=0.40-0.40=0 \]

Therefore,

\[ \boxed{\vec p_{\text{total, final}}=\vec 0} \]

Both gliders are still moving, but their momenta cancel.

This is another useful reminder that zero total momentum does not mean zero motion.

Step 7: Calculate the impulse from the wall.

The wall changes B’s horizontal momentum from \(+0.60\) to \(-0.60\text{ kg m s}^{-1}\).

Therefore:

\[ \begin{aligned} J_{\text{wall},x} &=\Delta p_{Bx}\\ &=-0.60-(+0.60)\\ &=-1.20\text{ N s} \end{aligned} \]

The vertical momentum is unchanged, so the vertical impulse is zero.

Thus, the wall gives B an impulse of \(1.20\text{ N s}\) to the left.

The wall and Earth receive the corresponding impulse to the right.

This explains why the total momentum of A and B changes between the initial state and the state after the wall rebound:

\[ \Delta\vec p_{A+B}=(-1.20,0)\text{ kg m s}^{-1} \]

The tempting mistake is to conserve the momentum of A+B across the entire sequence simply because those are the two gliders present at the beginning.

But during the second interaction, B collides with the wall, which is outside the chosen system. The external impulse is no longer negligible.

Step 8: Check kinetic energy before and after the first collision.

Initially, only A has kinetic energy:

\[ \begin{aligned} K_i&=\frac12(0.40)(3.0)^2\\ &=1.80\text{ J} \end{aligned} \]

After the first collision, A has kinetic energy

\[ \begin{aligned} K_{A,f} &=\frac12(0.40)(1.5^2+1.0^2)\\ &=0.65\text{ J} \end{aligned} \]

B has kinetic energy

\[ \begin{aligned} K_{B,f} &=\frac12(0.60) \left(1.0^2+0.67^2\right)\\ &\approx0.43\text{ J} \end{aligned} \]

The total is approximately

\[ K_f=0.65+0.43=1.08\text{ J} \]

The first collision decreases kinetic energy by approximately

\[ 1.80-1.08=0.72\text{ J} \]

It is therefore inelastic.

Step 9: Check kinetic energy after the wall rebound.

The wall reverses B’s horizontal velocity component but does not change its magnitude.

Therefore, B’s speed remains unchanged. Its kinetic energy is also unchanged.

The total kinetic energy after the wall rebound remains approximately

\[ K_{\text{after wall}}=1.08\text{ J} \]

The two stages demonstrate an important distinction:

• During the first collision, A and B form an approximately isolated system. Their total momentum is conserved, but their kinetic energy decreases.
• During the wall collision, A and B are not an isolated system. The wall supplies an external impulse, changing their total momentum, even though the idealised wall collision conserves kinetic energy.

If the wall and Earth were included in the system, momentum conservation could be applied to the larger system, assuming negligible external impulse.

07The next step: checking whether a collision is elastic

When solving a collision problem, keep two decisions separate.

First, select the system and determine whether its external impulse is negligible. If it is, conserve total momentum, resolving the components separately when necessary.

Second, compare initial and final kinetic energy to determine whether the collision is elastic or inelastic.

Never assume that the momentum of an individual object stays constant just because the total momentum of the system does.

For further practice with the second decision, see How to Check Whether a Collision Is Elastic from Measured Speeds.