How to Solve Collisions When Both Objects Are Moving

Two pucks meet on a level air table. Puck A has a mass of 0.20 kg and moves east at 3.0 m/s. Puck B has a mass of 0.30 kg and moves south at 2.0 m/s. They collide and stick together.

Before calculating anything, predict the direction of the joined pucks. Will they travel east, south, or diagonally?

They travel southeast, at 45° south of east. The reason is that their initial momenta combine as vectors, not as ordinary speeds.

Their total momentum is 0.60 kg m/s east and 0.60 kg m/s south. Dividing each component by their combined mass of 0.50 kg gives a final velocity of 1.20 m/s east and 1.20 m/s south.

This is the essential method for solving a collision when both objects are initially moving:

1. Choose both objects as your system.
2. Choose coordinate axes and assign positive and negative directions.
3. Calculate the initial momentum components of both objects.
4. Conserve momentum separately in each direction, provided the external impulse is negligible.
5. Solve for the unknown final velocity components.
6. Check kinetic energy separately to determine what happened during the collision.

The biggest mistake is treating the initial speeds as positive numbers and simply adding or subtracting them. You need to account for direction before doing the calculation.

01Why you must consider both objects as one system

During a collision, each object exerts a force on the other.

Object A pushes object B, while object B pushes object A. These forces are equal in magnitude and opposite in direction.

As a result, each object’s momentum changes. However, when you consider both objects together, these collision forces are internal forces. Their impulses cancel when calculating the change in total momentum.

If the net external impulse on the two-object system is negligible, its total momentum is conserved.

Mathematically:

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

Here:

• \(\vec p_{\text{system}}\) is the total momentum of both objects.
• \(\vec J_{\text{external}}\) is the net external impulse acting on the system.

When the external impulse is approximately zero:

\[ \boxed{\vec p_i=\vec p_f} \]

Notice that this equation applies to the combined system, not necessarily to either object individually.

For example, if you choose object A alone as your system, the collision force exerted by B is an external force. A’s momentum will generally change.

The correct system for an ordinary two-object collision is both objects together.

This does not mean external forces can always be ignored. If friction, a pusher, or another external force produces a significant impulse during the collision, you must account for it.

02Step 1: Choose coordinate axes and assign signs

For a two-dimensional collision, choose horizontal and vertical axes.

A convenient convention is:

The important thing is consistency. You can choose different positive directions, but you must use the same convention before and after the collision.

Consider two objects approaching a collision point:

• Object A moves east.
• Object B moves south.

Their initial velocity components might be:

\[ \vec u_A=(3.0,0)\ \text{m/s} \]

\[ \vec u_B=(0,-2.0)\ \text{m/s} \]

The negative sign on B’s vertical velocity tells us that it is travelling south.

A speed is always positive. Velocity components can be positive, negative, or zero.

If an object moves at an angle, resolve its velocity into components before applying momentum conservation.

For an angle \(\theta\) measured anticlockwise from the positive \(x\)-axis:

\[ u_x=u\cos\theta \]

\[ u_y=u\sin\theta \]

Here, \(u\) is the object’s speed, while \(u_x\) and \(u_y\) are its horizontal and vertical velocity components.

For example, an object travelling at 5.0 m/s, 30° north of east, has:

\[ u_x=5.0\cos30^\circ =4.33\ \text{m/s} \]

\[ u_y=5.0\sin30^\circ =2.50\ \text{m/s} \]

Both components are positive because the object travels east and north.

03Step 2: Conserve momentum separately in each direction

Momentum is defined as:

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

where:

• \(p\) is momentum, measured in kg m/s.
• \(m\) is mass, measured in kg.
• \(v\) is velocity, measured in m/s.

For two objects, A and B, the initial total momentum in the horizontal direction is:

\[ p_{ix}=m_Au_{Ax}+m_Bu_{Bx} \]

The final horizontal momentum is:

\[ p_{fx}=m_Av_{Ax}+m_Bv_{Bx} \]

Therefore, when external horizontal impulse is negligible:

\[ \boxed{ m_Au_{Ax}+m_Bu_{Bx} = m_Av_{Ax}+m_Bv_{Bx} } \]

You do exactly the same thing vertically:

\[ \boxed{ m_Au_{Ay}+m_Bu_{By} = m_Av_{Ay}+m_Bv_{By} } \]

These are the two equations you need for a two-dimensional collision.

Notice that both objects contribute to the initial momentum, even when they travel in completely different directions.

What if the objects stick together?

If the objects stick together after colliding, they share one final velocity.

Their combined mass is:

\[ m_{\text{total}}=m_A+m_B \]

The horizontal final velocity is:

\[ \boxed{ v_{fx}= \frac{m_Au_{Ax}+m_Bu_{Bx}} {m_A+m_B} } \]

The vertical final velocity is:

\[ \boxed{ v_{fy}= \frac{m_Au_{Ay}+m_Bu_{By}} {m_A+m_B} } \]

Once you have both components, calculate the final speed:

\[ v_f=\sqrt{v_{fx}^2+v_{fy}^2} \]

And the direction:

\[ \theta=\tan^{-1} \left|\frac{v_{fy}}{v_{fx}}\right| \]

Use the signs of the two components to identify the correct quadrant.

What if the objects separate after colliding?

You still use the same two momentum equations.

Suppose you know A’s final velocity but need to find B’s. Rearrange the horizontal equation:

\[ v_{Bx}= \frac{m_Au_{Ax}+m_Bu_{Bx}-m_Av_{Ax}} {m_B} \]

Similarly, vertically:

\[ v_{By}= \frac{m_Au_{Ay}+m_Bu_{By}-m_Av_{Ay}} {m_B} \]

You now have B’s final velocity components. From there, calculate its speed and direction.

The important distinction is that the final velocities do not need to be equal. They are equal only if the objects stick together.

04How an experiment tests momentum conservation

The momentum equations describe an idealised model. We can test that model using a collision experiment.

Imagine two pucks moving across a horizontal air table. The air table produces a thin cushion of air, reducing friction between the pucks and the surface.

Place a calibrated grid beneath the apparatus and position a camera directly above it.

Puck A starts to the west of a marked collision point and moves east. Puck B starts to the north and moves south. Both travel towards the same point.

Attach Velcro pads to the pucks so that they stick together when they collide.

What should the experiment show?

If the horizontal external impulse is negligible, the total horizontal momentum before and after the collision should be equal.

The same should be true vertically.

A simple first design might use a single photogate to measure the speed of one puck. However, that measurement is insufficient for an oblique collision. It does not establish both velocity components or account for the momentum of the other puck.

A better design uses overhead video tracking.

Track the centre of each puck over several frames before the collision. Then track the centre of the joined pair over several frames afterwards.

For each object, calculate:

\[ v_x=\frac{\Delta x}{\Delta t} \]

\[ v_y=\frac{\Delta y}{\Delta t} \]

This gives the horizontal and vertical velocity components needed for the momentum calculation.

Illustrative experimental data

Suppose the measured values are:

First, calculate the initial horizontal momentum:

\[ p_{ix}=(0.20)(3.00)+(0.30)(0) \]

\[ p_{ix}=+0.600\ \text{kg m/s} \]

The initial vertical momentum is:

\[ p_{iy}=(0.20)(0)+(0.30)(-2.00) \]

\[ p_{iy}=-0.600\ \text{kg m/s} \]

Now calculate the measured final momentum.

Horizontally:

\[ p_{fx}=(0.50)(1.18) =+0.590\ \text{kg m/s} \]

Vertically:

\[ p_{fy}=(0.50)(-1.18) =-0.590\ \text{kg m/s} \]

The measured values differ from the initial values by approximately 1.7% in each component.

This is consistent with momentum conservation within a 2% comparison tolerance.

The small discrepancy could result from friction, imperfect camera calibration, timing uncertainty, or errors in tracking the puck centres.

The observation supports the momentum-conservation model, although a real experiment cannot demonstrate perfect equality with unlimited precision.

Why the energy result is different

We can also compare kinetic energy before and after the collision.

Initially:

\[ K_i= \frac12(0.20)(3.00)^2+ \frac12(0.30)(2.00)^2 \]

\[ K_i=0.90+0.60 \]

\[ K_i=1.50\ \text{J} \]

After the collision, using the measured velocity components:

\[ K_f= \frac12(0.50)(1.18^2+1.18^2) \]

\[ K_f=0.696\ \text{J} \]

The decrease in kinetic energy is approximately:

\[ \Delta K=1.50-0.696 \]

\[ \boxed{\Delta K=0.804\ \text{J}} \]

This energy has not disappeared. Some has been transferred into deformation, sound, thermal energy, and vibrations.

The experiment can therefore support momentum conservation while showing that kinetic energy has decreased.

05Step 3: Check kinetic energy separately

This is where students often make an important mistake.

They remember that momentum is conserved and assume kinetic energy must also be conserved.

That is not true for every collision.

For two objects, the initial total kinetic energy is:

\[ K_i= \frac12m_Au_A^2+ \frac12m_Bu_B^2 \]

The final total kinetic energy is:

\[ K_f= \frac12m_Av_A^2+ \frac12m_Bv_B^2 \]

For a two-dimensional velocity, remember:

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

You can then classify the collision.

Total energy is always conserved. It is the kinetic energy of the moving objects that may decrease.

If the final kinetic energy is greater than the initial kinetic energy, investigate whether internal energy was released, external work was done, or the data and assumptions are incorrect.

Only use conservation of kinetic energy as an additional equation when the collision is known to be elastic.

For a separate worked explanation of this energy test, see How to Check Whether a Collision Is Elastic from Measured Speeds.

06A reliable exam method

When you encounter a collision where both objects are initially moving, follow this sequence.

1. Select the system: Usually both colliding objects together.
2. Check external impulse: Confirm that it is negligible or account for any stated external impulse.
3. Choose axes: Usually east as positive \(x\) and north as positive \(y\).
4. Resolve velocities: Find the initial and known final velocity components.
5. Calculate initial momentum: Include both objects in both component equations.
6. Conserve momentum: Write separate horizontal and vertical equations.
7. Find the unknowns: Calculate the final velocity components, then the speed and direction.
8. Check kinetic energy: Determine whether the collision is elastic or inelastic.

The following worked examples show how to apply this method in different situations.

07Worked examples

08Questions and solutions

Question 1

Two carts travel east along a straight, level track.

Cart A has a mass of 0.50 kg and travels at 2.4 m/s. Cart B has a mass of 0.30 kg and travels at 0.80 m/s.

Cart A catches up with B, and they collide and stick together.

Assuming negligible external impulse, calculate:

a) Their common final velocity.

b) The decrease in their total kinetic energy.

Solution 1

a) The carts move east together at 1.8 m/s.

Because both carts initially travel east, both momentum contributions are positive.

Choose east as the positive direction.

Initial momentum:

\[ p_i=(0.50)(2.4)+(0.30)(0.80) \]

\[ p_i=1.20+0.24 \]

\[ p_i=1.44\ \text{kg m/s} \]

The combined mass is:

\[ m_{\text{total}}=0.50+0.30 \]

\[ m_{\text{total}}=0.80\ \text{kg} \]

Using conservation of momentum:

\[ 1.44=0.80v_f \]

\[ \boxed{v_f=1.8\ \text{m/s east}} \]

b) The decrease in kinetic energy is 0.240 J.

Initial kinetic energy:

\[ K_i= \frac12(0.50)(2.4)^2+ \frac12(0.30)(0.80)^2 \]

\[ K_i=1.44+0.096 \]

\[ K_i=1.536\ \text{J} \]

Final kinetic energy:

\[ K_f=\frac12(0.80)(1.8)^2 \]

\[ K_f=1.296\ \text{J} \]

Therefore:

\[ \Delta K=1.536-1.296 \]

\[ \boxed{\Delta K=0.240\ \text{J}} \]

The carts stick together, so the collision is perfectly inelastic. Their total momentum is conserved, but some kinetic energy is converted into other forms.

Question 2

Two pucks collide on a horizontal, low-friction air table.

Puck A has a mass of 0.25 kg and travels east at 4.0 m/s.

Puck B has a mass of 0.75 kg and travels north at 2.0 m/s.

They collide and stick together.

Calculate:

a) The horizontal and vertical components of their common final velocity.

b) Their final speed and direction.

c) The decrease in total kinetic energy.

Assume negligible external horizontal impulse.

Solution 2

a) The final velocity components are 1.0 m/s east and 1.5 m/s north.

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

Initial horizontal momentum:

\[ p_{ix}=(0.25)(4.0)+(0.75)(0) \]

\[ p_{ix}=1.00\ \text{kg m/s} \]

Initial vertical momentum:

\[ p_{iy}=(0.25)(0)+(0.75)(2.0) \]

\[ p_{iy}=1.50\ \text{kg m/s} \]

The combined mass is:

\[ m_{\text{total}}=0.25+0.75=1.00\ \text{kg} \]

Since the pucks stick together, they share the same final velocity.

Horizontal component:

\[ 1.00=(1.00)v_{fx} \]

\[ \boxed{v_{fx}=1.0\ \text{m/s}} \]

Vertical component:

\[ 1.50=(1.00)v_{fy} \]

\[ \boxed{v_{fy}=1.5\ \text{m/s}} \]

b) The final speed is 1.80 m/s, at 56.3° north of east.

Calculate the magnitude:

\[ v_f=\sqrt{1.0^2+1.5^2} \]

\[ v_f=\sqrt{3.25} \]

\[ \boxed{v_f=1.80\ \text{m/s}} \]

Calculate the direction:

\[ \theta=\tan^{-1}\left(\frac{1.5}{1.0}\right) \]

\[ \boxed{\theta=56.3^\circ} \]

Both components are positive, so the final direction is north of east.

c) The kinetic energy decreases by 1.875 J.

Initial kinetic energy:

\[ K_i= \frac12(0.25)(4.0)^2+ \frac12(0.75)(2.0)^2 \]

\[ K_i=2.0+1.5 \]

\[ K_i=3.50\ \text{J} \]

Final kinetic energy:

\[ K_f=\frac12(1.00)(1.0^2+1.5^2) \]

\[ K_f=\frac12(3.25) \]

\[ K_f=1.625\ \text{J} \]

Therefore:

\[ \Delta K=3.50-1.625 \]

\[ \boxed{\Delta K=1.875\ \text{J}} \]

The pucks’ momentum components are conserved, but their kinetic energy decreases because they stick together.

Question 3

Two carts collide on a horizontal track and stick together.

Cart A has a mass of 0.40 kg and initially travels east at 2.0 m/s.

Cart B has a mass of 0.60 kg and initially travels west at 1.0 m/s.

During the collision, an external horizontal impulse of 0.10 N s acts eastwards on the two-cart system.

a) Calculate the final velocity of the joined carts.

b) Explain why using \(p_i=p_f\) without accounting for the external impulse gives an incorrect answer.

c) Explain why the collision forces between the carts do not cancel the external impulse.

Solution 3

a) The joined carts move east at 0.30 m/s.

Choose east as positive.

First, calculate the initial momentum:

\[ p_i=(0.40)(2.0)+(0.60)(-1.0) \]

\[ p_i=0.80-0.60 \]

\[ p_i=+0.20\ \text{kg m/s} \]

The external impulse is:

\[ J_{\text{external}}=+0.10\ \text{N s} \]

Using the impulse-momentum relationship:

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

Therefore:

\[ p_f=0.20+0.10 \]

\[ p_f=0.30\ \text{kg m/s} \]

The combined mass is:

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

Thus:

\[ 0.30=(1.00)v_f \]

\[ \boxed{v_f=0.30\ \text{m/s east}} \]

b) Momentum is not conserved for the two-cart system alone because an external impulse acts on it.

If you incorrectly assume that \(p_i=p_f\), you obtain:

\[ v_f=\frac{0.20}{1.00} =0.20\ \text{m/s east} \]

This is incorrect because it ignores the additional 0.10 N s of eastward impulse.

Momentum conservation for a system requires negligible net external impulse, not simply the fact that a collision has occurred.

c) The collision forces are internal, whereas the pusher’s force is external.

Cart A pushes cart B, and cart B pushes cart A with an equal and opposite force.

These internal forces cancel when considering the total momentum of both carts.

The pusher acts from outside the two-cart system. Its impulse therefore changes the system’s total momentum.

If the pusher and the rest of its supporting structure were included in a larger system, the momentum accounting would also include their corresponding changes.

Question 4

A student investigates a two-dimensional collision using an overhead camera and an air table.

Two pucks approach a collision point and stick together.

The camera records their centre positions at 0.10 s intervals. All positions are measured relative to the same coordinate origin.

The following illustrative data are obtained:

After the collision, the combined pucks have a mass of 0.50 kg.

Their measured positions are:

Answer the following:

a) Calculate the initial velocity components of both pucks.

b) Calculate the initial total momentum components.

c) Calculate the final total momentum components.

d) Determine whether the measurements support momentum conservation to within 2% in each direction.

e) Calculate the decrease in total kinetic energy.

f) Explain one limitation of the experimental method and why measuring only the speed of the joined pucks would be insufficient to establish momentum conservation.

Solution 4

a) The initial velocity of A is (3.0, 0) m/s, and the initial velocity of B is (0, −2.0) m/s.

Velocity is displacement divided by time.

For A:

\[ v_{Ax}=\frac{0.50-0.20}{0.10} \]

\[ v_{Ax}=3.0\ \text{m/s} \]

Its vertical position does not change, so:

\[ v_{Ay}=0 \]

Therefore:

\[ \boxed{\vec u_A=(3.0,0)\ \text{m/s}} \]

For B:

\[ v_{Bx}=\frac{0.50-0.50}{0.10}=0 \]

\[ v_{By}=\frac{0.00-0.20}{0.10} \]

\[ v_{By}=-2.0\ \text{m/s} \]

Therefore:

\[ \boxed{\vec u_B=(0,-2.0)\ \text{m/s}} \]

The negative vertical component indicates southward motion.

b) The initial total momentum is (0.60, −0.60) kg m/s.

Horizontal momentum:

\[ p_{ix}=(0.20)(3.0)+(0.30)(0) \]

\[ \boxed{p_{ix}=+0.60\ \text{kg m/s}} \]

Vertical momentum:

\[ p_{iy}=(0.20)(0)+(0.30)(-2.0) \]

\[ \boxed{p_{iy}=-0.60\ \text{kg m/s}} \]

c) The final total momentum is (0.59, −0.59) kg m/s.

First, calculate the final velocity components:

\[ v_{fx}=\frac{0.618-0.50}{0.10} \]

\[ v_{fx}=1.18\ \text{m/s} \]

\[ v_{fy}=\frac{-0.118-0.00}{0.10} \]

\[ v_{fy}=-1.18\ \text{m/s} \]

The combined mass is 0.50 kg.

Horizontal momentum:

\[ p_{fx}=(0.50)(1.18) \]

\[ \boxed{p_{fx}=+0.59\ \text{kg m/s}} \]

Vertical momentum:

\[ p_{fy}=(0.50)(-1.18) \]

\[ \boxed{p_{fy}=-0.59\ \text{kg m/s}} \]

d) Yes, the measurements support momentum conservation within 2% in both directions.

Calculate the percentage difference horizontally:

\[ \%\text{ difference} = \frac{|0.60-0.59|}{0.60}\times100 \]

\[ \boxed{\%\text{ difference}=1.67\%} \]

The vertical components have the same percentage difference:

\[ \boxed{\%\text{ difference}=1.67\%} \]

Both results fall within the stated 2% tolerance.

This supports the model, although the small discrepancy is consistent with experimental uncertainty.

e) The total kinetic energy decreases by approximately 0.804 J.

Initial kinetic energy:

\[ K_i= \frac12(0.20)(3.0)^2+ \frac12(0.30)(2.0)^2 \]

\[ K_i=0.90+0.60 \]

\[ K_i=1.50\ \text{J} \]

Final kinetic energy:

\[ K_f= \frac12(0.50)(1.18^2+(-1.18)^2) \]

\[ K_f=0.6962\ \text{J} \]

Therefore:

\[ \Delta K=1.50-0.6962 \]

\[ \boxed{\Delta K=0.804\ \text{J}} \]

The collision is perfectly inelastic because the pucks stick together. The decrease in kinetic energy is consistent with energy being transferred into other forms.

f) One limitation is uncertainty in measuring positions and time intervals.

The camera may have limited resolution, the calibration scale may be inaccurate, or the air table may have some residual friction.

Measuring only the speed of the joined pucks would also be insufficient because speed does not specify direction.

Momentum is a vector. To compare the initial and final momentum correctly, the experiment must determine both velocity components and account for both objects before the collision.

Question 5

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

Before colliding, puck A moves east at 4.0 m/s. Puck B moves at 3.0 m/s at an angle \(\theta\) north of east.

They collide on a horizontal, low-friction surface.

After the collision, puck A is observed to be stationary, while puck B moves away.

Assume:

• The net external impulse is negligible.
• No internal energy store, such as a compressed spring, releases energy during the collision.
• Kinetic energy may be converted into other forms.
• Use the idealised point-mass conservation model; detailed contact geometry is not being considered.

Answer the following:

a) Express the final velocity components of B in terms of \(\theta\).

b) Derive an expression for the change in total kinetic energy.

c) Find the angle \(\theta\) that would make the collision elastic.

d) Determine whether the stated outcome is consistent with the assumptions when \(\theta=60^\circ\) and when \(\theta=120^\circ\).

e) Explain why the claim that A stopping automatically means the collision is elastic is incorrect.

Solution 5

The collision is elastic only when \(\theta=90^\circ\). The proposed outcome is impossible at \(60^\circ\) under the stated assumptions, but it is consistent with an inelastic collision at \(120^\circ\).

The key is that B receives the combined initial momentum of both pucks, and its final kinetic energy depends on the angle between their initial velocities.

Step 1: Choose coordinate axes and resolve the initial velocities.

Let east be positive \(x\) and north be positive \(y\).

Puck A initially has:

\[ \vec u_A=(4.0,0) \]

Puck B initially has:

\[ \vec u_B=(3.0\cos\theta,3.0\sin\theta) \]

Both masses are 0.50 kg.

Step 2: Apply conservation of momentum.

The initial horizontal momentum is:

\[ p_{ix}=0.50(4.0)+0.50(3.0\cos\theta) \]

The initial vertical momentum is:

\[ p_{iy}=0.50(0)+0.50(3.0\sin\theta) \]

After the collision, A is stationary.

Therefore, B must carry the entire final momentum.

For the horizontal component:

\[ 0.50v_{Bx} = 0.50(4.0+3.0\cos\theta) \]

Dividing by 0.50:

\[ \boxed{v_{Bx}=4.0+3.0\cos\theta} \]

For the vertical component:

\[ 0.50v_{By}=0.50(3.0\sin\theta) \]

Therefore:

\[ \boxed{v_{By}=3.0\sin\theta} \]

The final velocity of B is:

\[ \boxed{ \vec v_B= (4.0+3.0\cos\theta,\ 3.0\sin\theta) } \]

This is the vector sum of the two initial velocities.

Step 3: Calculate the initial kinetic energy.

The initial kinetic energy is:

\[ K_i= \frac12(0.50)(4.0)^2+ \frac12(0.50)(3.0)^2 \]

\[ K_i=4.00+2.25 \]

\[ \boxed{K_i=6.25\ \text{J}} \]

Step 4: Derive the final kinetic energy.

Since A is stationary, only B contributes to the final kinetic energy.

\[ K_f=\frac12(0.50)v_B^2 \]

The final speed squared is:

\[ v_B^2=v_{Bx}^2+v_{By}^2 \]

Substitute the velocity components:

\[ v_B^2= (4.0+3.0\cos\theta)^2+ (3.0\sin\theta)^2 \]

Expand:

\[ v_B^2= 16+24\cos\theta+ 9\cos^2\theta+9\sin^2\theta \]

Using:

\[ \cos^2\theta+\sin^2\theta=1 \]

We obtain:

\[ v_B^2=25+24\cos\theta \]

Therefore:

\[ K_f=\frac12(0.50)(25+24\cos\theta) \]

\[ \boxed{K_f=6.25+6\cos\theta\ \text{J}} \]

Step 5: Find the change in kinetic energy.

Define the change as final minus initial:

\[ \Delta K=K_f-K_i \]

Substitute:

\[ \Delta K=(6.25+6\cos\theta)-6.25 \]

Therefore:

\[ \boxed{\Delta K=6\cos\theta\ \text{J}} \]

This is the important result.

The angle between the initial velocity vectors determines whether the final kinetic energy increases, decreases, or stays the same.

Step 6: Determine the angle for an elastic collision.

For an elastic collision:

\[ K_f=K_i \]

Therefore:

\[ \Delta K=0 \]

Using our expression:

\[ 6\cos\theta=0 \]

\[ \cos\theta=0 \]

Hence:

\[ \boxed{\theta=90^\circ} \]

The initial velocity vectors must be perpendicular for this particular collision to be elastic.

Step 7: Test the two proposed angles.

For \(\theta=60^\circ\):

\[ \Delta K=6\cos60^\circ \]

\[ \Delta K=6(0.5) \]

\[ \boxed{\Delta K=+3.0\ \text{J}} \]

The final kinetic energy would be:

\[ K_f=6.25+3.0 \]

\[ K_f=9.25\ \text{J} \]

This is an increase of 3.0 J.

The collision cannot produce this result under the stated assumptions because no external work or internal energy release is available.

It would require at least 3.0 J of additional energy from an internal source, such as a released spring.

For \(\theta=120^\circ\):

\[ \Delta K=6\cos120^\circ \]

\[ \Delta K=6(-0.5) \]

\[ \boxed{\Delta K=-3.0\ \text{J}} \]

The final kinetic energy is:

\[ K_f=6.25-3.0 \]

\[ \boxed{K_f=3.25\ \text{J}} \]

This decrease is consistent with an inelastic collision. Some kinetic energy has been converted into other forms.

Step 8: Evaluate the student’s claim.

The claim that A stopping automatically makes the collision elastic is incorrect.

When A stops, its initial momentum is transferred to B. However, the final kinetic energy depends on the magnitude of B’s resulting velocity.

The vector addition of the two initial velocities creates an angle-dependent term. When the initial velocities point in similar directions, their momentum contributions reinforce each other. When they point in opposing directions, they partly cancel.

That is why the same final condition, A stopping, can correspond to an elastic collision, an inelastic collision, or an outcome that requires an additional energy source.

The essential lesson is that momentum and kinetic energy answer different questions. Momentum determines the possible final motion; the energy comparison tells you whether that motion is consistent with the type of collision and the available energy.

Once you can calculate the final velocity vector reliably, the next useful decision is whether the collision was elastic or inelastic. Keep that as a separate calculation rather than assuming that conserving momentum automatically conserves kinetic energy.