How to Solve a Two-Dimensional Collision from Deflection Angles

Two pucks collide on a smooth horizontal surface. Before impact, puck A travels east while puck B is stationary. After impact, A moves 30° south of east, and B moves 45° north of east.

You know their masses, the initial speed, and both deflection angles. But how do you calculate their speeds after the collision?

The key is to conserve momentum in two perpendicular directions. Choose both pucks as one system, resolve their momentum into east-west and north-south components, and write a separate conservation equation for each direction. Solve the equations together, then check kinetic energy separately.

Before continuing, make a prediction: must the puck deflecting through 45° travel faster than the puck deflecting through 30°?

No. The angle alone doesn’t determine the speed. Each component depends on both mass and velocity, so a heavier puck can carry more momentum even when travelling more slowly.

01Choose the system and coordinate axes

Before using any equations, decide which objects belong to your system.

For a two-object collision, choose both colliding objects as one system. The forces they exert on each other during impact are internal forces. They do not change the total momentum of the combined system.

We also assume that the net external impulse during the short collision is negligible. For pucks on a level, low-friction surface, this is often a reasonable approximation.

There are two common choices for coordinate axes:

• The x-axis points east, and the y-axis points north.
• The x-axis follows the initial total momentum, and the y-axis is perpendicular to it.

The first choice is convenient when the initial momentum points east. The second can simplify problems where both objects are initially moving in different directions.

For our first example, use east as positive x and north as positive y.

This gives the following sign convention:

The signs matter. A velocity pointing south has a negative y-component, even though its speed is positive.

02Resolve momentum into components

Momentum is a vector quantity, so its direction matters.

The magnitude of momentum is:

\[ p=mv \]

where:

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

When an object travels at an angle \(\theta\) above the positive x-axis, its momentum has two components:

\[ p_x=mv\cos\theta \]

\[ p_y=mv\sin\theta \]

The angle is measured from the positive x-axis.

For an object travelling below the x-axis, its y-component is negative. You can handle this by using a negative angle.

For example, an object travelling 30° south of east has:

\[ p_x=mv\cos30^\circ \]

\[ p_y=-mv\sin30^\circ \]

Think of the momentum arrow casting two shadows: one on the east-west axis and one on the north-south axis. These are the two components of the same momentum vector, not two separate momenta.

This is useful because momentum conservation can now be applied independently along each axis.

03Apply conservation of momentum twice

The total momentum before a collision equals the total momentum afterwards, provided the net external impulse is negligible.

In vector form:

\[ \vec p_{\text{before}}=\vec p_{\text{after}} \]

In two dimensions, this gives two equations.

Horizontal momentum:

\[ \sum p_{x,\text{before}} = \sum p_{x,\text{after}} \]

Vertical momentum:

\[ \sum p_{y,\text{before}} = \sum p_{y,\text{after}} \]

These equations are independent. Both must be satisfied.

For a collision involving two objects, the equations become:

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

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

Here:

• \(m_A\) and \(m_B\) are the object masses.
• \(u\) represents initial velocity.
• \(v\) represents final velocity.
• The subscripts x and y identify the components.

Notice that these equations conserve momentum, not velocity. The masses must remain attached to their corresponding velocity components.

What if one object is initially stationary?

A stationary object has zero initial momentum in both directions.

If object A initially travels east and object B is stationary:

\[ p_{x,\text{before}}=m_Au_A \]

\[ p_{y,\text{before}}=0 \]

The two final momentum components must therefore satisfy:

\[ m_Av_A\cos\alpha+ m_Bv_B\cos\beta = m_Au_A \]

\[ -m_Av_A\sin\alpha+ m_Bv_B\sin\beta = 0 \]

Here, \(\alpha\) is the angle object A deflects below east, and \(\beta\) is the angle object B deflects above east.

The vertical equation can be rearranged:

\[ m_Av_A\sin\alpha = m_Bv_B\sin\beta \]

This is particularly useful. Since the initial vertical momentum is zero, the two final vertical momentum components must cancel.

It also shows why comparing angles alone is misleading. The masses and speeds determine how much momentum each object carries vertically.

04Check kinetic energy separately

There is an important distinction between conserving momentum and conserving kinetic energy.

Momentum is conserved for an isolated system, regardless of whether the collision is elastic or inelastic.

Kinetic energy, however, is only conserved in an elastic collision.

Calculate the total kinetic energy before and after impact using:

\[ K=\frac12mv^2 \]

Then compare:

• \(K_{\text{before}}=K_{\text{after}}\): elastic collision.
• \(K_{\text{before}}>K_{\text{after}}\): inelastic collision.
• \(K_{\text{before}}<K_{\text{after}}\): superelastic collision, where stored internal energy has been released.

The total energy of the system is still conserved. In an inelastic collision, some initial kinetic energy is transformed into other forms, such as sound, heat, and deformation.

Do not add kinetic energy conservation as another equation unless the question establishes that the collision is elastic.

For a more detailed method of comparing measured speeds and kinetic energies, see How to Check Whether a Collision Is Elastic from Measured Speeds.

05How an experiment produces the deflection angles

The mathematics makes more sense when you understand where the measurements come from.

Consider an experiment designed to investigate whether momentum is conserved during a two-dimensional collision.

Setting up the experiment

Place two low-friction pucks on a level horizontal air table. Puck A is launched towards stationary puck B.

Position an overhead camera directly above the table. Place a calibrated distance scale in the plane of motion so that distances can be measured from the recorded footage.

The camera records the motion before and after impact.

The experiment is trying to answer one question:

Does the total momentum vector remain unchanged when the objects change direction?

What should the experiment show?

If external horizontal impulses are negligible, the total horizontal and vertical momentum should each remain constant.

The camera provides two important measurements:

1. The direction of each puck’s motion, which gives its deflection angle.
2. The velocity of each puck, calculated from its displacement over a measured time interval.

The velocity can be determined from the gradient of its position-time data.

For example, suppose a puck’s centre moves 16 mm over four frame intervals in footage recorded at 240 frames per second.

The time interval is:

\[ \Delta t=\frac{4}{240}=0.0167\text{ s} \]

Convert the displacement into metres:

\[ \Delta x=0.016\text{ m} \]

The estimated speed is:

\[ v=\frac{\Delta x}{\Delta t} =\frac{0.016}{0.0167} \]

\[ v\approx0.96\text{ m/s} \]

This is a simple estimate. Tracking several frames and calculating the gradient of the position-time graph generally gives a more reliable velocity.

Importantly, the measurements before and after impact should be taken from separate sections of the motion. The velocity can change during the collision itself.

Why the experimental design matters

An initial design using an ordinary tabletop and a handheld stopwatch would introduce two problems.

First, friction acts on the pucks throughout their motion. Second, human reaction time makes accurate timing difficult.

A revised design uses a level, low-friction surface and overhead video tracking. This reduces unwanted external impulses and improves the precision of the velocity and angle measurements.

The prediction is that the measured total momentum before impact should agree with the measured total momentum afterwards, within experimental uncertainty.

If the horizontal components agree but the vertical components do not, investigate the angle measurements, camera alignment, and any external forces before rejecting the model.

There is also an important distinction between testing the model and applying it. If you use momentum conservation to calculate unknown speeds, you have assumed the model to obtain those speeds. To independently test conservation, measure the velocities from the video first, then compare the momentum totals.

06Worked examples

07When is there enough information to solve the problem?

Two-dimensional momentum conservation gives you two independent equations.

This means you can generally solve for two unknown quantities, provided the other information is known.

One common mistake is to assume that an unknown speed can always be calculated from the deflection angles.

It cannot.

Angles give you directions. To determine momentum magnitudes, you also need masses, initial momentum, measured speeds, or another valid independent condition.

If both objects are initially moving, the initial momentum must be calculated as a vector sum before solving the final components. For more examples of this situation, see How to Solve Collisions When Both Objects Are Moving.

08Questions and solutions

The following questions progress from a standard HSC-style calculation to more demanding problems involving experimental measurements, incomplete information, and the relationship between momentum and kinetic energy.

Question 1

A 0.20 kg puck A travels east at 4.0 m/s and collides with a stationary 0.30 kg puck B.

After the collision, puck A travels at 2.0 m/s, at an angle of 30° north of east.

Assuming negligible external horizontal impulse:

a) Calculate the final speed of puck B.

b) Determine its direction of travel.

c) Calculate the initial and final kinetic energies and classify the collision.

Solution 1

Puck B leaves at approximately 1.65 m/s, travelling 23.8° south of east. The collision is inelastic.

Step 1: Calculate the initial momentum.

Only puck A has initial momentum:

\[ p_{x,i}=0.20(4.0)=0.800 \]

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

Step 2: Calculate the final momentum components of puck A.

The magnitude of its momentum is:

\[ p_A=0.20(2.0)=0.400 \]

Resolve this into components:

\[ p_{Ax}=0.400\cos30^\circ =0.3464 \]

\[ p_{Ay}=0.400\sin30^\circ =0.200 \]

Both components are positive because puck A travels north of east.

Step 3: Calculate the momentum components of puck B.

Horizontal momentum:

\[ p_{Bx}=0.800-0.3464 \]

\[ p_{Bx}=0.4536\text{ kg m/s} \]

Vertical momentum:

\[ p_{By}=0-0.200 \]

\[ p_{By}=-0.200\text{ kg m/s} \]

The negative sign tells us that puck B travels south.

Step 4: Calculate the magnitude of puck B’s momentum.

Use Pythagoras:

\[ p_B=\sqrt{0.4536^2+0.200^2} \]

\[ p_B=0.4957\text{ kg m/s} \]

Since \(p=mv\):

\[ v_B=\frac{0.4957}{0.30} \]

\[ v_B=1.65\text{ m/s} \]

Step 5: Calculate its direction.

\[ \tan\theta=\frac{0.200}{0.4536} \]

\[ \theta=23.8^\circ \]

The negative vertical component means the direction is 23.8° south of east.

Step 6: Check kinetic energy.

Initially:

\[ K_i=\frac12(0.20)(4.0)^2 \]

\[ K_i=1.60\text{ J} \]

Finally:

\[ K_f= \frac12(0.20)(2.0)^2 + \frac12(0.30)(1.652)^2 \]

\[ K_f=0.810\text{ J} \]

The kinetic energy has decreased by approximately 0.790 J.

The collision is therefore inelastic. The fact that momentum is conserved does not mean kinetic energy must also be conserved.

Question 2

A 0.15 kg puck A moves east and collides with a stationary 0.25 kg puck B.

Video measurements after the collision give:

• Puck A: 2.0 m/s at 40° north of east.
• Puck B: 1.54 m/s at 30° south of east.

The measured speeds and angles are rounded.

a) Reconstruct the initial speed of puck A.

b) Determine whether the measured final vertical momentum is consistent with the initial direction of travel.

c) Calculate the change in kinetic energy and classify the collision.

Solution 2

Puck A’s reconstructed initial speed is approximately 3.75 m/s east. The small vertical momentum discrepancy is consistent with rounding, and the collision is inelastic.

Step 1: Check the vertical momentum.

The initial vertical momentum is zero because puck A initially travels east and puck B is stationary.

The final vertical momentum is:

\[ p_{y,f} = 0.15(2.0)\sin40^\circ - 0.25(1.54)\sin30^\circ \]

\[ p_{y,f}=0.000336\text{ kg m/s} \]

This is extremely close to zero.

The small residual is less than 0.1% of the horizontal momentum. Since the measurements are rounded, this is consistent with an initially horizontal total momentum.

If the residual were substantially larger, you would need to investigate measurement uncertainty or an external impulse.

Step 2: Calculate the final horizontal momentum.

\[ p_{x,f} = 0.15(2.0)\cos40^\circ + 0.25(1.54)\cos30^\circ \]

\[ p_{x,f}=0.5632\text{ kg m/s} \]

Step 3: Reconstruct the initial speed.

Since puck B was initially stationary:

\[ m_Au_A=p_{x,f} \]

\[ 0.15u_A=0.5632 \]

\[ u_A=3.75\text{ m/s} \]

The reconstructed initial velocity is therefore approximately 3.75 m/s east.

Step 4: Calculate initial kinetic energy.

\[ K_i=\frac12(0.15)(3.75)^2 \]

\[ K_i=1.06\text{ J} \]

Step 5: Calculate final kinetic energy.

\[ K_f= \frac12(0.15)(2.0)^2 + \frac12(0.25)(1.54)^2 \]

\[ K_f=0.596\text{ J} \]

The kinetic energy decreases by approximately:

\[ \Delta K=1.06-0.596 \]

\[ \Delta K=0.461\text{ J} \]

The collision is inelastic.

This question demonstrates why checking both momentum components matters. A small residual can result from rounding, but a significant discrepancy should not simply be ignored.

Question 3

Immediately before a collision, the total momentum of a two-object system is:

\[ \vec p_i= (1.20\hat i+0.90\hat j) \text{ kg m/s} \]

Here, \(\hat i\) represents the eastward direction and \(\hat j\) represents the northward direction.

After the collision:

• Object A has a mass of 0.50 kg and travels at 2.0 m/s, 30° south of east.
• Object B has a mass of 0.30 kg.

a) Calculate the final velocity of object B, including its speed and direction.

b) Explain whether the collision can be classified as elastic or inelastic from the information supplied.

Solution 3

Object B travels at approximately 4.80 m/s, 76.6° north of east. However, the information is insufficient to determine whether the collision is elastic.

Step 1: Calculate object A’s final momentum.

The magnitude is:

\[ p_A=0.50(2.0)=1.00\text{ kg m/s} \]

Its horizontal component is:

\[ p_{Ax}=1.00\cos30^\circ =0.866 \]

Its vertical component is:

\[ p_{Ay}=-1.00\sin30^\circ =-0.500 \]

Step 2: Find object B’s momentum components.

The total horizontal momentum must remain 1.20 kg m/s:

\[ p_{Bx}=1.20-0.866 \]

\[ p_{Bx}=0.334\text{ kg m/s} \]

The total vertical momentum must remain 0.90 kg m/s:

\[ p_{By}=0.90-(-0.500) \]

\[ p_{By}=1.40\text{ kg m/s} \]

Notice that subtracting a negative component increases the remaining vertical momentum.

Step 3: Calculate the magnitude of object B’s momentum.

\[ p_B=\sqrt{0.334^2+1.40^2} \]

\[ p_B=1.439\text{ kg m/s} \]

Therefore:

\[ v_B=\frac{1.439}{0.30} \]

\[ v_B=4.80\text{ m/s} \]

Step 4: Calculate its direction.

\[ \tan\theta=\frac{1.40}{0.334} \]

\[ \theta=76.6^\circ \]

Both components are positive, so object B travels 76.6° north of east.

Step 5: Assess whether the collision is elastic.

We cannot determine this from the information provided.

We know the initial total momentum, but we do not know the initial kinetic energy.

Different combinations of initial velocities can produce exactly the same total momentum while having different total kinetic energies.

Although the final kinetic energy can be calculated from the final velocities, there is no reliable initial kinetic energy for comparison.

This is why momentum conservation and kinetic energy conservation must be treated as separate questions.

Question 4

An experiment uses two pucks on a low-friction horizontal table.

Before impact:

• Puck A has a mass of 0.20 kg and travels east at 3.0 m/s.
• Puck B has a mass of 0.30 kg and is stationary.

An overhead camera records the following final velocities:

• Puck A: 2.0 m/s at 20° north of east.
• Puck B: 1.3 m/s at 30° south of east.

The collision lasts 0.015 s.

An independent test establishes that the horizontal drag force acting on the system is no greater than 0.05 N.

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

b) Calculate the change in system momentum.

c) Determine the average external force required to produce the measured change.

d) Decide whether the discrepancy can reasonably be explained by the measured drag force.

Solution 4

The measured final momentum differs from the initial momentum by approximately 0.128 kg m/s. Explaining this discrepancy would require an average external force of approximately 8.5 N, far greater than the measured drag force of 0.05 N.

Step 1: Calculate initial momentum.

Only puck A is moving:

\[ p_{x,i}=0.20(3.0)=0.600 \]

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

Therefore:

\[ \vec p_i=(0.600,0) \text{ kg m/s} \]

Step 2: Calculate the final momentum components of puck A.

\[ p_{Ax}=0.20(2.0)\cos20^\circ \]

\[ p_{Ax}=0.3759 \]

\[ p_{Ay}=0.20(2.0)\sin20^\circ \]

\[ p_{Ay}=0.1368 \]

Step 3: Calculate the final momentum components of puck B.

\[ p_{Bx}=0.30(1.3)\cos30^\circ \]

\[ p_{Bx}=0.3377 \]

Since B travels south:

\[ p_{By}=-0.30(1.3)\sin30^\circ \]

\[ p_{By}=-0.195 \]

Step 4: Calculate total final momentum.

Horizontal:

\[ p_{x,f}=0.3759+0.3377 \]

\[ p_{x,f}=0.7136 \]

Vertical:

\[ p_{y,f}=0.1368-0.195 \]

\[ p_{y,f}=-0.0582 \]

Therefore:

\[ \vec p_f=(0.7136,-0.0582) \text{ kg m/s} \]

Step 5: Calculate the change in momentum.

\[ \Delta\vec p=\vec p_f-\vec p_i \]

\[ \Delta p_x=0.7136-0.600 =0.1136 \]

\[ \Delta p_y=-0.0582-0 =-0.0582 \]

The magnitude of the change is:

\[ |\Delta\vec p| = \sqrt{0.1136^2+0.0582^2} \]

\[ |\Delta\vec p|=0.1276\text{ kg m/s} \]

Step 6: Calculate the average external force.

Newton’s second law in momentum form gives:

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

Using the collision duration of 0.015 s:

\[ F_x=\frac{0.1136}{0.015} =7.57\text{ N} \]

\[ F_y=\frac{-0.0582}{0.015} =-3.88\text{ N} \]

The magnitude is:

\[ F_{\text{avg}} = \sqrt{7.57^2+3.88^2} \]

\[ F_{\text{avg}}=8.51\text{ N} \]

Step 7: Compare this with the measured drag force.

The maximum impulse from the stated drag force is:

\[ J=F\Delta t \]

\[ J=0.05(0.015) \]

\[ J=0.00075\text{ N s} \]

The measured momentum discrepancy is approximately 0.128 N s.

This is roughly 170 times larger than the maximum drag impulse.

Therefore, the discrepancy cannot reasonably be explained by the stated drag force.

The measurements or assumptions need to be investigated. Possible problems include inaccurate angle measurements, camera calibration errors, incorrect velocity estimates, or an unaccounted external contact.

Importantly, the calculated 8.51 N is the average external force required to explain the discrepancy. It is not the force between the two pucks. Their contact forces are internal to the two-puck system.

Question 5

Two pucks collide on a frictionless horizontal surface.

Puck A has a mass of 0.20 kg and initially travels east. Puck B has a mass of 0.40 kg and is initially stationary.

After the collision, their momentum vectors are perpendicular. Their magnitudes are:

\[ p_A=0.16\text{ kg m/s} \]

\[ p_B=0.12\text{ kg m/s} \]

The vectors point in opposite vertical directions, so their resultant points east.

A student argues:

“The outgoing momentum vectors form a right angle, so the collision must be elastic.”

a) Calculate the initial speed of puck A.

b) Calculate the final speed of each puck.

c) Determine the initial and final kinetic energies. Is the student’s conclusion correct?

d) Keeping puck A’s mass and both outgoing momentum vectors unchanged, determine the mass puck B would need for the collision to be elastic.

Assume negligible external impulse and no release of stored internal energy in the elastic case.

Solution 5

The initial speed of A is 1.0 m/s, and the collision is inelastic for the stated masses. For the same outgoing momentum vectors to satisfy kinetic energy conservation, puck B would need a mass of 0.20 kg.

Step 1: Calculate the initial momentum using the perpendicular outgoing vectors.

The momentum of the system is conserved.

The two outgoing momentum vectors are perpendicular, so Pythagoras gives the magnitude of their resultant:

\[ p_i=\sqrt{p_A^2+p_B^2} \]

Substitute:

\[ p_i=\sqrt{0.16^2+0.12^2} \]

\[ p_i=\sqrt{0.0400} \]

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

Since puck B was initially stationary, all initial momentum belonged to puck A.

Step 2: Calculate the initial speed of puck A.

\[ p_i=m_Au_A \]

\[ 0.20=0.20u_A \]

\[ u_A=1.0\text{ m/s} \]

Step 3: Calculate both final speeds.

For puck A:

\[ v_A=\frac{p_A}{m_A} \]

\[ v_A=\frac{0.16}{0.20} \]

\[ v_A=0.80\text{ m/s} \]

For puck B:

\[ v_B=\frac{p_B}{m_B} \]

\[ v_B=\frac{0.12}{0.40} \]

\[ v_B=0.30\text{ m/s} \]

Step 4: Calculate the initial kinetic energy.

\[ K_i=\frac12m_Au_A^2 \]

\[ K_i=\frac12(0.20)(1.0)^2 \]

\[ K_i=0.100\text{ J} \]

Step 5: Calculate the final kinetic energy.

Puck A:

\[ K_A=\frac12(0.20)(0.80)^2 \]

\[ K_A=0.064\text{ J} \]

Puck B:

\[ K_B=\frac12(0.40)(0.30)^2 \]

\[ K_B=0.018\text{ J} \]

Total final kinetic energy:

\[ K_f=0.064+0.018 \]

\[ K_f=0.082\text{ J} \]

The kinetic energy lost is:

\[ \Delta K=0.100-0.082 \]

\[ \Delta K=0.018\text{ J} \]

Therefore, the collision is inelastic.

The student’s conclusion is incorrect. Perpendicular outgoing momentum vectors do not automatically prove that a collision is elastic.

Step 6: Determine the mass required for an elastic collision.

This is the most important part of the question.

Keep puck A’s mass and both outgoing momentum vectors unchanged. Only change puck B’s mass.

For an elastic collision:

\[ K_i=K_f \]

Using the perpendicular momentum vectors, the initial kinetic energy can be written as:

\[ K_i= \frac{p_A^2+p_B^2}{2m_A} \]

The final kinetic energy is:

\[ K_f= \frac{p_A^2}{2m_A} + \frac{p_B^2}{2m_B} \]

Equating these expressions:

\[ \frac{p_A^2+p_B^2}{2m_A} = \frac{p_A^2}{2m_A} + \frac{p_B^2}{2m_B} \]

Cancel the common term:

\[ \frac{p_B^2}{2m_A} = \frac{p_B^2}{2m_B} \]

Since \(p_B\) is not zero:

\[ \frac{1}{m_A}=\frac{1}{m_B} \]

Therefore:

\[ m_B=m_A \]

Since puck A has a mass of 0.20 kg:

\[ \boxed{m_B=0.20\text{ kg}} \]

For this mass, puck B’s final speed would be:

\[ v_B=\frac{0.12}{0.20} =0.60\text{ m/s} \]

Its final kinetic energy would be:

\[ K_B=\frac12(0.20)(0.60)^2 =0.036\text{ J} \]

The total final kinetic energy would then be:

\[ K_f=0.064+0.036 \]

\[ K_f=0.100\text{ J} \]

This matches the initial kinetic energy.

The surprising result is that the same perpendicular momentum vectors can describe either an elastic or an inelastic collision, depending on the masses.

The familiar right-angle result for elastic collisions involving an initially stationary object is therefore conditional. Equal masses are required in this particular perpendicular-momentum arrangement.

09The key decision when solving a two-dimensional collision

When you encounter a collision involving deflection angles, don’t start by searching for a formula containing those angles.

Start with the system and the initial momentum.

Then:

1. Choose positive x and y directions.
2. Calculate the initial momentum components.
3. Resolve each final momentum using its measured angle.
4. Apply horizontal and vertical momentum conservation separately.
5. Solve the simultaneous equations for the unknown quantities.
6. Check kinetic energy independently if the question asks whether the collision is elastic.

If your momentum components do not balance, investigate the signs, angle references, measurements, and external impulses before assuming the collision violates conservation.

Once the final velocities are known, the next useful step is deciding whether the kinetic energy has been conserved or transferred into other forms. That is a separate calculation, and one that often determines how you interpret the collision data.