How to Check Whether a Collision Is Elastic from Measured Speeds

One cart leaves a collision moving more slowly, while the other leaves moving faster. Has kinetic energy been lost, transferred, or conserved? You cannot tell by looking at either cart individually. You need to compare the total kinetic energy of both objects before and after the collision.

To check whether a collision is elastic, calculate the total kinetic energy before and after using the measured masses and speeds:

\[
K_i=\sum \frac12 m s_i^2
\]

\[
K_f=\sum \frac12 m s_f^2
\]

If \(K_i\) and \(K_f\) are equal within experimental uncertainty, the kinetic energy test is consistent with an elastic collision. You should also check momentum conservation in each relevant direction to confirm that your chosen system is behaving as an isolated system.

There is an important distinction: speeds are enough to calculate kinetic energy, but directions are needed to calculate momentum. If your measurements only provide speeds, you may be able to test kinetic energy without independently verifying momentum.

Let’s work through how to make that decision using actual collision measurements.

01What exactly are you checking?

Imagine two carts colliding on a horizontal, low-friction track.

Cart A approaches cart B. During the collision, the carts exert forces on each other. After contact, they move away with new velocities.

The important question is not whether either cart maintains its original speed. It is whether the total kinetic energy of the system remains unchanged.

Momentum and kinetic energy answer different questions

Momentum is a vector quantity. It depends on both mass and velocity, including direction.

For a one-dimensional collision:

\[
p=mv
\]

where:

• \(p\) is momentum in kg m s\(^{-1}\),
• \(m\) is mass in kg,
• \(v\) is signed velocity in m s\(^{-1}\).

Kinetic energy is a scalar quantity. It depends on mass and speed, but not direction.

\[
K=\frac12 ms^2
\]

where \(s\) is speed in m s\(^{-1}\), and \(K\) is measured in joules (J).

This gives us two separate checks.

For an isolated collision, momentum should be conserved whether the collision is elastic or inelastic.

Kinetic energy distinguishes between them:

• Elastic collision: total kinetic energy is conserved.
• Inelastic collision: total kinetic energy decreases.
• If kinetic energy increases, energy has been transferred into the motion from another source, such as stored internal energy.

Remember that total energy is always conserved. An inelastic collision does not destroy energy. Some kinetic energy is converted into other forms, such as sound, heating, vibration, and permanent deformation.

Why the system matters

Before calculating anything, decide which objects belong to your system.

For two colliding carts, the natural choice is usually both carts together.

During the collision, the force exerted by cart A on cart B is an internal force. The force exerted by B on A is equal and opposite. These internal forces do not change the total momentum of the two-cart system.

However, friction from the track is an external force.

If its impulse during the short collision is negligible, momentum is approximately conserved:

\[
\vec p_i=\vec p_f
\]

If the collision occurs in two dimensions, this means conserving momentum separately in each direction:

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

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

This is why choosing the correct system comes before substituting numbers.

02How do scientists measure the speeds?

Suppose you want to test whether two pucks collide elastically on a low-friction air table.

You need to measure:

• The mass of each puck.
• Each puck’s velocity before the collision.
• Each puck’s velocity after the collision.

The apparatus must capture both the speed and the direction of motion.

Measuring velocity with an overhead camera

A useful experimental arrangement is a high-frame-rate camera positioned above the air table.

The procedure is:

1. Measure each puck’s mass using a balance.
2. Place a calibrated grid or scale beneath the transparent surface.
3. Record the pucks approaching and moving away from the collision.
4. Track the centre of each puck in successive video frames.
5. Calculate velocity from the change in position over time.

You should use several frames before and after the collision rather than relying on a single position measurement.

The collision itself is a brief period of rapid acceleration. Exclude those frames when calculating the approximately constant velocities immediately before and after contact.

For motion in two dimensions, track the horizontal and vertical position separately.

A graph of position against time makes the measurement clearer: the gradient before contact gives the initial velocity component, while the gradient after contact gives the final velocity component.

What if you use photogates instead?

Photogates are useful for measuring speed in a one-dimensional collision.

A small flag of known length is attached to the cart. As the flag passes through a light beam, the photogate measures how long the beam is blocked.

The speed is calculated using:

\[
s=\frac{L}{\Delta t}
\]

where:

• \(L\) is the length of the flag in metres,
• \(\Delta t\) is the time for which the beam is blocked in seconds.

For example, suppose a 0.050 m flag blocks the beam for 0.025 s.

\[
s=\frac{0.050}{0.025}
\]

\[
\boxed{s=2.0\text{ m s}^{-1}}
\]

The photogate gives the speed, but there is a problem.

What happens if the cart rebounds and passes back through the same gate?

It can produce the same blocking time. The photogate alone does not tell you whether the cart is moving forwards or backwards.

A revised arrangement uses two photogates in a known order. The order in which the cart interrupts the beams reveals its direction.

For two-dimensional collisions, an overhead camera is generally more useful because it captures both velocity components.

The measurement method matters because an incorrect direction can produce an incorrect momentum calculation, even when the measured speed is accurate.

03Turning the measurements into a decision

Once you have the measurements, follow this sequence.

Step 1: Choose your system.

Include both colliding objects, and consider whether any external forces or additional objects provide a significant impulse.

Step 2: Set your direction convention.

For a one-dimensional collision, choose right as positive, for example.

An object travelling left at 3.0 m s\(^{-1}\) then has:

\[
v=-3.0\text{ m s}^{-1}
\]

Its speed remains positive:

\[
s=3.0\text{ m s}^{-1}
\]

Step 3: Check momentum conservation.

Calculate the total initial and final momentum.

For two objects:

\[
m_1u_1+m_2u_2=m_1v_1+m_2v_2
\]

For a two-dimensional collision, calculate the x- and y-components separately.

Step 4: Calculate total kinetic energy.

Add the kinetic energy of every object in the system.

\[
K_i=\frac12m_1u_1^2+\frac12m_2u_2^2
\]

\[
K_f=\frac12m_1v_1^2+\frac12m_2v_2^2
\]

Use the measured speeds, or square the velocity magnitudes.

Step 5: Compare the results.

If the total kinetic energy before and after is equal within the experimental uncertainty, the measurements are consistent with an elastic collision.

If momentum matches but kinetic energy decreases, the collision is inelastic.

If momentum does not match, investigate the system boundary, external impulse, measurement errors, and direction assignments before drawing a conclusion.

06How close do the kinetic energy measurements need to be?

In a theoretical HSC problem, you will usually be given values that allow you to determine whether kinetic energy is conserved.

In an actual experiment, measurements are never perfectly exact.

For example, a camera might record:

\[
K_i=0.450\pm0.010\text{ J}
\]

\[
K_f=0.454\pm0.010\text{ J}
\]

The measured difference is:

\[
\Delta K=0.454-0.450
\]

\[
\Delta K=0.004\text{ J}
\]

The difference is smaller than the uncertainty associated with the comparison.

If the quoted uncertainties are independent, the uncertainty in the difference is approximately:

\[
\sigma_{\Delta K}
=
\sqrt{(0.010)^2+(0.010)^2}
\]

\[
\sigma_{\Delta K}\approx0.014\text{ J}
\]

The 0.004 J difference is therefore consistent with experimental uncertainty.

You would not conclude that the collision is definitely inelastic just because the final measurement is slightly larger.

The correct conclusion is that the results are consistent with kinetic energy conservation.

There is no universal percentage difference that automatically determines whether an experimental collision is elastic. You must consider the precision and uncertainty of the actual measurements.

What if momentum does not match?

Before deciding that the collision is inelastic, check whether the momentum discrepancy has another explanation.

Common possibilities include:

• Friction or another external force acting during the collision.
• A missing object that receives momentum.
• An incorrect direction assigned to a velocity.
• Inaccurate mass measurements.
• Incorrect timing or position measurements.

If a significant external impulse acts on your chosen system, momentum will not necessarily be conserved.

The relationship is:

\[
\Delta\vec p=\vec J_{\mathrm{ext}}
\]

where \(\vec J_{\mathrm{ext}}\) is the net external impulse.

This is why a momentum mismatch should trigger an investigation of the system and apparatus, rather than an immediate conclusion about elasticity.

07Common mistakes when checking elastic collisions

One particularly important distinction is between speed and velocity.

An object travelling left at 4.0 m s\(^{-1}\) has:

\[
v=-4.0\text{ m s}^{-1}
\]

but its kinetic energy is:

\[
K=\frac12m(-4.0)^2
\]

\[
K=8m
\]

The negative velocity matters for momentum, but not for kinetic energy.

For a fuller explanation of why both conservation laws apply to an ideal elastic collision, see Elastic Collisions: Momentum and Kinetic Energy for HSC Physics.

08Questions and solutions

The following questions move from straightforward calculations to situations where the direction of motion, system boundaries, and vector relationships change the conclusion.

Question 1

Two carts collide on a horizontal, low-friction track.

Cart A has a mass of 0.50 kg and initially travels right at 2.0 m s\(^{-1}\). Cart B has a mass of 0.25 kg and is initially stationary.

After the collision:

• Cart A travels right at 0.80 m s\(^{-1}\).
• Cart B travels right at 2.4 m s\(^{-1}\).

Take right as positive.

Determine whether the collision is elastic. Show calculations for both momentum and kinetic energy.

Solution 1

The collision is inelastic. Momentum is conserved, but total kinetic energy decreases by 0.12 J.

Take right as positive.

Initial momentum:

\[
\begin{aligned}
p_i&=(0.50)(2.0)+(0.25)(0)\\
&=1.0\text{ kg m s}^{-1}
\end{aligned}
\]

Final momentum:

\[
\begin{aligned}
p_f&=(0.50)(0.80)+(0.25)(2.4)\\
&=0.40+0.60\\
&=1.0\text{ kg m s}^{-1}
\end{aligned}
\]

Momentum is conserved.

Now calculate the initial kinetic energy:

\[
K_i=\frac12(0.50)(2.0)^2
\]

\[
K_i=1.0\text{ J}
\]

The final kinetic energy is:

\[
\begin{aligned}
K_f&=\frac12(0.50)(0.80)^2\\
&\quad+\frac12(0.25)(2.4)^2\\
&=0.16+0.72\\
&=0.88\text{ J}
\end{aligned}
\]

Therefore:

\[
\Delta K=0.88-1.00
\]

\[
\boxed{\Delta K=-0.12\text{ J}}
\]

The collision is inelastic because kinetic energy has decreased, even though momentum is conserved.

The important point is that the final speed of cart B is greater than the initial speed of cart A. That does not mean the system has gained kinetic energy. The masses of the carts are different, so their contributions to total kinetic energy are different.

Question 2

Two carts collide head-on on a low-friction track.

Cart A has a mass of 0.20 kg and initially travels right at 5.0 m s\(^{-1}\).

Cart B has a mass of 0.30 kg and initially travels left at 1.0 m s\(^{-1}\).

After the collision:

• Cart A rebounds left at 1.0 m s\(^{-1}\).
• Cart B travels right at 3.0 m s\(^{-1}\).

Take right as positive.

Calculate the total momentum and kinetic energy before and after the collision.

Determine whether the collision is elastic and calculate the percentage decrease in kinetic energy.

Solution 2

The collision is inelastic. Momentum is conserved, but kinetic energy decreases by 1.20 J, approximately 45.3% of its initial value.

Take right as positive.

The signed initial velocities are:

\[
u_A=+5.0\text{ m s}^{-1}
\]

\[
u_B=-1.0\text{ m s}^{-1}
\]

The final velocities are:

\[
v_A=-1.0\text{ m s}^{-1}
\]

\[
v_B=+3.0\text{ m s}^{-1}
\]

Momentum check

Initial momentum:

\[
\begin{aligned}
p_i&=(0.20)(5.0)+(0.30)(-1.0)\\
&=1.00-0.30\\
&=0.70\text{ kg m s}^{-1}
\end{aligned}
\]

Final momentum:

\[
\begin{aligned}
p_f&=(0.20)(-1.0)+(0.30)(3.0)\\
&=-0.20+0.90\\
&=0.70\text{ kg m s}^{-1}
\end{aligned}
\]

Momentum is conserved.

Kinetic energy check

Initial kinetic energy:

\[
\begin{aligned}
K_i&=\frac12(0.20)(5.0)^2\\
&\quad+\frac12(0.30)(1.0)^2\\
&=2.50+0.15\\
&=2.65\text{ J}
\end{aligned}
\]

Final kinetic energy:

\[
\begin{aligned}
K_f&=\frac12(0.20)(1.0)^2\\
&\quad+\frac12(0.30)(3.0)^2\\
&=0.10+1.35\\
&=1.45\text{ J}
\end{aligned}
\]

The kinetic energy change is:

\[
\Delta K=1.45-2.65
\]

\[
\boxed{\Delta K=-1.20\text{ J}}
\]

Percentage decrease

\[
\begin{aligned}
\%\text{ decrease}
&=\frac{K_i-K_f}{K_i}\times100\\
&=\frac{2.65-1.45}{2.65}\times100\\
&\approx45.3\%
\end{aligned}
\]

Approximately 45.3% of the initial kinetic energy has been converted into other forms of energy.

Notice that cart A actually rebounds, while cart B leaves moving faster than it was initially. Neither observation tells you whether the collision is elastic.

Only the total kinetic energy calculation establishes that distinction.

Question 3

Two identical carts, each with a mass of 0.30 kg, collide on a straight, low-friction track.

Before the collision:

• Cart A travels right at 3.0 m s\(^{-1}\).
• Cart B travels left at 1.0 m s\(^{-1}\).

After the collision, two photogates record the following speeds:

• Cart A: 1.0 m s\(^{-1}\).
• Cart B: 3.0 m s\(^{-1}\).

However, the apparatus does not record the directions of the final velocities.

Determine whether the readings are consistent with an elastic collision, assuming the two-cart system is isolated.

Explain why the photogate measurements alone are insufficient to independently verify momentum conservation.

Solution 3

The readings are consistent with an elastic collision if cart A rebounds left and cart B moves right. However, the photogates alone cannot directly establish those directions.

Take right as positive.

The initial momentum is:

\[
\begin{aligned}
p_i&=(0.30)(3.0)+(0.30)(-1.0)\\
&=0.90-0.30\\
&=0.60\text{ kg m s}^{-1}
\end{aligned}
\]

The initial kinetic energy is:

\[
\begin{aligned}
K_i&=\frac12(0.30)(3.0)^2\\
&\quad+\frac12(0.30)(1.0)^2\\
&=1.35+0.15\\
&=1.50\text{ J}
\end{aligned}
\]

Now consider the final speed readings.

Because the photogates only measure speed, there are four possible combinations of signed velocities:

Only one combination conserves the initial momentum:

\[
v_A=-1.0\text{ m s}^{-1}
\]

\[
v_B=+3.0\text{ m s}^{-1}
\]

Using this direction assignment:

\[
\begin{aligned}
p_f&=(0.30)(-1.0)+(0.30)(3.0)\\
&=-0.30+0.90\\
&=0.60\text{ kg m s}^{-1}
\end{aligned}
\]

Momentum is conserved.

The final kinetic energy is:

\[
\begin{aligned}
K_f&=\frac12(0.30)(1.0)^2\\
&\quad+\frac12(0.30)(3.0)^2\\
&=0.15+1.35\\
&=1.50\text{ J}
\end{aligned}
\]

Therefore:

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

The collision is consistent with being elastic.

However, notice the limitation of the experiment. We inferred the final directions by assuming momentum conservation. We have not independently measured those directions.

An overhead video or a pair of directional photogates would provide an independent check.

This distinction is important in experimental physics: using a conservation law to infer a missing measurement is not the same as independently testing that conservation law.

Question 4

A 0.20 kg puck travels towards a rigid wall at 4.0 m s\(^{-1}\). It rebounds directly backwards at 4.0 m s\(^{-1}\).

A student calculates the momentum of the puck before and after the collision and concludes that momentum is not conserved. They therefore claim that the collision cannot be elastic.

Assume the wall is securely attached to the Earth and that energy transferred into sound, heating, and permanent deformation is negligible at the measurement precision.

Evaluate the student’s conclusion.

Solution 4

The student’s momentum calculation is correct for the puck alone, but their conclusion is incorrect. The collision can be approximately elastic because the puck’s kinetic energy is unchanged, while momentum is transferred to the wall and Earth.

Take motion towards the wall as positive.

The initial momentum of the puck is:

\[
p_i=(0.20)(4.0)
\]

\[
p_i=+0.80\text{ kg m s}^{-1}
\]

After the collision, the puck moves in the opposite direction:

\[
p_f=(0.20)(-4.0)
\]

\[
p_f=-0.80\text{ kg m s}^{-1}
\]

Therefore:

\[
\Delta p=-0.80-0.80
\]

\[
\boxed{\Delta p=-1.60\text{ kg m s}^{-1}}
\]

The puck’s momentum changes because the wall exerts an external impulse on the puck-only system.

However, consider the kinetic energy.

Before the collision:

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

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

After the collision:

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

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

The puck’s kinetic energy is unchanged.

To understand momentum conservation, expand the system to include the puck, wall, and Earth.

The wall and Earth receive an equal and opposite impulse. Their enormous combined mass means their resulting recoil speed is extremely small.

Under the stated assumptions, the collision is approximately elastic.

The key lesson is that momentum conservation depends on the system boundary. A puck by itself is not an isolated system during a collision with a wall.

Question 5

Two identical pucks, each with a mass of 0.20 kg, collide on a horizontal, low-friction air table.

Before the collision:

• Puck A has velocity components \(v_{Ax}=+4.0\text{ m s}^{-1}\) and \(v_{Ay}=0\).
• Puck B has velocity components \(v_{Bx}=+1.0\text{ m s}^{-1}\) and \(v_{By}=+1.0\text{ m s}^{-1}\).

After the collision, an overhead tracking system measures both puck speeds as 3.0 m s\(^{-1}\).

The camera also records that their final velocity vectors are perpendicular.

A student claims that the collision is elastic because the total kinetic energy is unchanged.

Determine whether the student’s conclusion is correct.

If the measured final speeds are accurate, calculate the angle between the final velocity vectors that would be required for momentum to be conserved.

Solution 5

The student’s conclusion is incorrect. The measured speeds conserve kinetic energy, but the measured 90° angle does not satisfy momentum conservation.

The collision is therefore inconsistent with the isolated elastic collision model.

Step 1: Check the initial kinetic energy.

Puck A initially has speed:

\[
s_A=4.0\text{ m s}^{-1}
\]

Puck B initially has speed:

\[
s_B=\sqrt{1.0^2+1.0^2}
\]

\[
s_B=\sqrt2\text{ m s}^{-1}
\]

The total initial kinetic energy is:

\[
\begin{aligned}
K_i&=\frac12(0.20)(4.0)^2\\
&\quad+\frac12(0.20)(1.0^2+1.0^2)\\
&=1.60+0.20\\
&=1.80\text{ J}
\end{aligned}
\]

Step 2: Check the final kinetic energy.

Both pucks have a final speed of 3.0 m s\(^{-1}\).

\[
\begin{aligned}
K_f&=\frac12(0.20)(3.0)^2\\
&\quad+\frac12(0.20)(3.0)^2\\
&=0.90+0.90\\
&=1.80\text{ J}
\end{aligned}
\]

Therefore:

\[
K_i=K_f
\]

Kinetic energy is conserved.

So far, the student’s reasoning looks correct.

But kinetic energy is only one part of the test.

Step 3: Calculate the initial momentum vector.

In the x-direction:

\[
\begin{aligned}
p_{x,i}&=0.20(4.0)+0.20(1.0)\\
&=1.00\text{ kg m s}^{-1}
\end{aligned}
\]

In the y-direction:

\[
\begin{aligned}
p_{y,i}&=0.20(0)+0.20(1.0)\\
&=0.20\text{ kg m s}^{-1}
\end{aligned}
\]

The initial momentum vector is:

\[
\vec p_i=(1.00,\ 0.20)
\]

Its magnitude squared is:

\[
|\vec p_i|^2=1.00^2+0.20^2
\]

\[
|\vec p_i|^2=1.04
\]

Step 4: Test the measured final directions.

Each puck has a final momentum magnitude of:

\[
p=mv
\]

\[
p=(0.20)(3.0)
\]

\[
p=0.60\text{ kg m s}^{-1}
\]

The two final momentum vectors are perpendicular.

Using the Pythagorean theorem, the magnitude squared of their resultant is:

\[
\begin{aligned}
|\vec p_f|^2&=0.60^2+0.60^2\\
&=0.72
\end{aligned}
\]

But the initial momentum magnitude squared was 1.04.

Therefore:

\[
|\vec p_f|\ne|\vec p_i|
\]

The final momentum cannot equal the initial momentum.

The measured 90° angle is inconsistent with momentum conservation.

Step 5: Calculate the angle required by momentum conservation.

Let \(\theta\) be the angle between the two final momentum vectors.

Because the pucks have equal masses, the angle between their velocity vectors is also \(\theta\).

Using the cosine rule for the resultant momentum:

\[
|\vec p_i|^2=p_1^2+p_2^2+2p_1p_2\cos\theta
\]

Substitute the known values:

\[
1.04=0.60^2+0.60^2+
2(0.60)(0.60)\cos\theta
\]

Simplifying:

\[
1.04=0.72+0.72\cos\theta
\]

Rearrange:

\[
0.32=0.72\cos\theta
\]

\[
\cos\theta=\frac{0.32}{0.72}
\]

\[
\cos\theta=\frac49
\]

Therefore:

\[
\boxed{\theta\approx63.6^\circ}
\]

For the measured speeds to be consistent with momentum conservation, the final velocity vectors must be separated by approximately 63.6°, not 90°.

This is the important conceptual distinction.

The kinetic energy measurements alone suggest an elastic collision. However, the measured directions make the complete data set impossible for an isolated collision.

The experiment would need to be investigated. Possible explanations include an inaccurate angle measurement, an omitted object, or an external impulse that was not accounted for.

A collision is not established as elastic simply because the total kinetic energy calculation works. The complete set of measurements must also be physically consistent with the momentum of the chosen system.

Once you can reliably classify collisions from measured data, the next useful step is to investigate what happens when momentum changes during a collision. The impulse-momentum relationship, \(\Delta p=F_{\mathrm{net}}\Delta t\), allows you to connect those momentum changes to the forces and interaction times responsible for them.