How to Find an Unknown Mass from Collision Data

A 0.80 kg trolley travels east at 2.40 m/s and collides with a second trolley at rest. The two trolleys lock together and move east at 1.50 m/s. What is the mass of the second trolley?

Before calculating, predict whether the second trolley is heavier or lighter than the first. It must be lighter. The combined mass is 1.6 times the original mass, so the second trolley has a mass of 0.48 kg.

The method is conservation of momentum. You calculate the total momentum before and after the collision, then rearrange the equation to find the unknown mass. For collisions in two dimensions, you do this separately for the horizontal and vertical components.

There is one important distinction: momentum is conserved in an isolated collision, but kinetic energy is not necessarily conserved. Keep those two ideas separate.

01How collision data reveal an unknown mass

Imagine investigating a collision between two trolleys on a horizontal dynamics track.

You know the mass of trolley A, but not trolley B. You can measure their velocities immediately before and after the collision. The challenge is to use those measurements to work backwards and determine the missing mass.

Setting up the experiment

Place the two trolleys on a level, low-friction track. Trolley A has a known mass, while trolley B has an unknown mass. A camera records the collision from the side, with a metre ruler positioned parallel to the track.

The arrangement is:

• Trolley A approaches trolley B.
• Trolley B initially rests on the track.
• The trolleys collide.
• Their positions are recorded immediately before and after impact.

The camera allows you to measure displacement over a known time interval. You can then calculate velocity using:

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

For example, if a trolley travels 0.240 m in 0.100 s, its average velocity is:

\[
v=\frac{0.240}{0.100}
=2.40\text{ m/s}
\]

Measure over short intervals immediately before and after the collision. This gives a better estimate of the velocities at impact than timing the trolley over its entire journey.

A position-time graph makes the same idea clear: the gradient of each straight-line section represents velocity.

What should the experiment show?

If the two-trolley system experiences negligible external impulse during the collision, its total momentum should be the same immediately before and after impact.

The measured data should therefore satisfy:

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

This is the relationship we use to find the unknown mass.

02Choosing the correct system

The most important decision in a collision problem happens before you substitute any numbers.

You must decide which objects belong to your system.

For two colliding trolleys, the best choice is usually the system containing both trolleys.

Why?

During the collision, trolley A pushes trolley B, and trolley B pushes trolley A. These forces are internal to the combined system. Their effects cancel when calculating the total momentum change.

External forces, such as friction from the track, can change the system’s momentum. However, during a short collision on a low-friction track, their impulse is often negligible.

If you choose only trolley A as your system, the force exerted by trolley B becomes an external force. You cannot simply assume that trolley A’s momentum is conserved.

The general rule is:

Include all interacting collision objects in the system, then consider whether the net external impulse is negligible.

If an external impulse is significant, use the impulse-momentum relationship instead:

\[
\Delta p=J_{\text{external}}
\]

Here, \(J_{\text{external}}\) is the net external impulse, measured in newton-seconds.

For an isolated system, or one with negligible external impulse:

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

03Finding an unknown mass in one dimension

For a one-dimensional collision, choose a positive direction before writing your equation.

For example, you might define east as positive. West is then negative.

Momentum is calculated using:

\[
p=mv
\]

where:

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

Velocity is a vector, so its direction matters. A trolley travelling west at 2.0 m/s has a velocity of \(-2.0\text{ m/s}\) if east is positive.

For two objects, conservation of momentum becomes:

\[
m_Au_A+m_Bu_B
=
m_Av_A+m_Bv_B
\]

Here:

• \(m_A\) and \(m_B\) are the two masses.
• \(u_A\) and \(u_B\) are their initial velocities.
• \(v_A\) and \(v_B\) are their final velocities.

Rearranging for the unknown mass \(m_B\):

\[
m_B(u_B-v_B)=m_A(v_A-u_A)
\]

Therefore:

\[
\boxed{
m_B=
\frac{m_A(u_A-v_A)}
{v_B-u_B}
}
\]

This formula is useful when both objects’ initial and final velocities are known.

All velocities must be signed. Do not substitute speed magnitudes if an object changes direction.

A useful shortcut: the objects stick together

When two objects stick together, they share the same final velocity.

If trolley B initially rests, then:

\[
u_B=0
\]

and:

\[
v_A=v_B=v
\]

The momentum equation simplifies to:

\[
m_Au_A=(m_A+m_B)v
\]

Rearranging:

\[
\boxed{
m_B=m_A\left(\frac{u_A}{v}-1\right)
}
\]

This shortcut is useful for perfectly inelastic collisions, where the objects remain joined after impact.

Do not use it when the objects separate after the collision.

04The important distinction: momentum versus kinetic energy

A common mistake is to assume that because momentum is conserved, kinetic energy must also be conserved.

These are different quantities.

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

Kinetic energy is conserved only in an elastic collision.

The kinetic energy of an object is:

\[
E_k=\frac12mv^2
\]

For a system containing two objects:

\[
E_{k,i}
=
\frac12m_Au_A^2
+
\frac12m_Bu_B^2
\]

\[
E_{k,f}
=
\frac12m_Av_A^2
+
\frac12m_Bv_B^2
\]

An elastic collision satisfies:

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

An inelastic collision has a decrease in total kinetic energy. Some kinetic energy is transferred into other forms, such as sound, thermal energy, and deformation.

In a perfectly inelastic collision, the objects stick together.

For the earlier trolley example:

Initial kinetic energy:

\[
E_{k,i}
=\frac12(0.80)(2.40)^2
\]

\[
E_{k,i}=2.304\text{ J}
\]

Final kinetic energy:

\[
E_{k,f}
=\frac12(1.28)(1.50)^2
\]

\[
E_{k,f}=1.440\text{ J}
\]

The decrease is:

\[
\Delta E_k=2.304-1.440
\]

\[
\Delta E_k=0.864\text{ J}
\]

The collision is therefore inelastic.

Notice what happened: momentum remained constant, but kinetic energy decreased.

Use momentum to find the unknown mass. Then check kinetic energy separately.

Unless the question explicitly establishes that the collision is elastic, do not assume kinetic energy is conserved.

05What happens when a trolley rebounds?

A collision becomes more interesting when one object changes direction.

This is where sign errors become common.

Consider a 0.50 kg trolley travelling east at 4.0 m/s. It collides with a stationary trolley of unknown mass.

After the collision:

• Trolley A rebounds west at 1.0 m/s.
• Trolley B moves east at 2.0 m/s.

Take east as positive.

The initial momentum is:

\[
p_i=0.50(4.0)=2.0
\]

The final momentum is:

\[
p_f=0.50(-1.0)+m_B(2.0)
\]

Equating the two:

\[
2.0=-0.50+2m_B
\]

\[
2.50=2m_B
\]

\[
\boxed{m_B=1.25\text{ kg}}
\]

The unknown trolley has a mass of 1.25 kg.

What if you accidentally substitute \(+1.0\text{ m/s}\) for trolley A’s final velocity?

You would get:

\[
2.0=0.50+2m_B
\]

\[
m_B=0.75\text{ kg}
\]

That is incorrect because trolley A is travelling in the opposite direction after the collision.

The negative sign is not optional. It represents a real change in momentum.

The collision is also not necessarily elastic just because a trolley rebounds. In this example:

\[
E_{k,i}=4.00\text{ J}
\]

\[
E_{k,f}=2.75\text{ J}
\]

The collision is inelastic despite the rebound.

06Finding an unknown mass in two dimensions

So far, the objects have moved along one straight line.

But what happens when two objects collide at an angle?

Imagine two pucks colliding on a smooth horizontal table. One puck approaches from the left, and the second is initially stationary. After impact, the pucks travel in different directions.

You cannot simply add their speeds.

Momentum is a vector, so you must resolve it into perpendicular components.

Choose:

• Positive \(x\): east, or right.
• Positive \(y\): north, or upwards.

Conservation of momentum must hold independently in each direction.

For the horizontal direction:

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

For the vertical direction:

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

If a velocity is given as a speed and angle, resolve it using:

\[
v_x=v\cos\theta
\]

\[
v_y=v\sin\theta
\]

The signs depend on the direction of travel.

07A reliable method for collision questions

Use this sequence whenever you need to calculate an unknown mass.

A few traps are worth remembering:

• If the objects stick together, they share one final velocity.
• If an object rebounds, its velocity changes sign.
• If the collision is two-dimensional, solve the components separately.
• If the unknown object’s velocity does not change, momentum data alone may not determine its mass.
• If the momentum equations give inconsistent results, investigate the assumptions or measurements rather than forcing an answer.
• Never assume kinetic energy is conserved just because momentum is conserved.

08Questions and solutions

The following questions move from a standard one-dimensional collision towards more demanding problems involving direction, components, experimental consistency, and competing models.

Question 1

A 0.60 kg trolley travels east at 3.0 m/s and collides with a stationary trolley of unknown mass.

The two trolleys stick together and move east at 1.8 m/s.

Calculate the mass of the second trolley. State whether kinetic energy is conserved.

Solution 1

The second trolley has a mass of 0.40 kg, and kinetic energy is not conserved.

Because the trolleys stick together, they share a final velocity. We can therefore use the simplified momentum equation.

Step 1: Identify the known quantities.

\[
m_A=0.60\text{ kg}
\]

\[
u_A=3.0\text{ m/s}
\]

\[
u_B=0,\qquad v=1.8\text{ m/s}
\]

Step 2: Apply conservation of momentum.

\[
m_Au_A=(m_A+m_B)v
\]

Substitute:

\[
0.60(3.0)=(0.60+m_B)(1.8)
\]

\[
1.80=1.08+1.8m_B
\]

\[
0.72=1.8m_B
\]

\[
\boxed{m_B=0.40\text{ kg}}
\]

Step 3: Compare kinetic energy.

Initially:

\[
E_{k,i}=\frac12(0.60)(3.0)^2
\]

\[
E_{k,i}=2.70\text{ J}
\]

Finally, the combined mass is 1.00 kg:

\[
E_{k,f}=\frac12(1.00)(1.8)^2
\]

\[
E_{k,f}=1.62\text{ J}
\]

The kinetic energy decreases by 1.08 J.

The collision is perfectly inelastic because the trolleys stick together. Momentum is conserved, but kinetic energy is transferred into other forms.

Question 2

Two trolleys collide along a straight horizontal track.

Trolley A has a mass of 0.40 kg and initially travels east at 5.0 m/s. Trolley B has an unknown mass and initially travels west at 1.0 m/s.

After the collision:

• Trolley A travels west at 2.0 m/s.
• Trolley B travels east at 3.0 m/s.

Calculate the mass of trolley B. Then calculate the total kinetic energy before and after the collision, and classify the collision.

Solution 2

Trolley B has a mass of 0.70 kg, and the collision is inelastic.

The important feature is that both trolleys initially move, and their directions change. We must include the negative velocities in the momentum equation.

Step 1: Choose east as positive.

The initial velocities are:

\[
u_A=+5.0\text{ m/s}
\]

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

The final velocities are:

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

\[
v_B=+3.0\text{ m/s}
\]

Step 2: Apply conservation of momentum.

\[
m_Au_A+m_Bu_B
=
m_Av_A+m_Bv_B
\]

Substitute:

\[
0.40(5.0)+m_B(-1.0)
\]

\[
=0.40(-2.0)+m_B(3.0)
\]

Simplify:

\[
2.0-m_B=-0.80+3m_B
\]

\[
2.80=4m_B
\]

\[
\boxed{m_B=0.70\text{ kg}}
\]

Step 3: Calculate initial kinetic energy.

Remember that kinetic energy uses speed squared, so the direction signs disappear.

\[
E_{k,i}
=
\frac12(0.40)(5.0)^2
+
\frac12(0.70)(1.0)^2
\]

\[
E_{k,i}=5.00+0.35
\]

\[
E_{k,i}=5.35\text{ J}
\]

Step 4: Calculate final kinetic energy.

\[
E_{k,f}
=
\frac12(0.40)(2.0)^2
+
\frac12(0.70)(3.0)^2
\]

\[
E_{k,f}=0.80+3.15
\]

\[
E_{k,f}=3.95\text{ J}
\]

The kinetic energy decreases by:

\[
\Delta E_k=5.35-3.95
\]

\[
\boxed{\Delta E_k=1.40\text{ J}}
\]

The collision is inelastic.

Notice the difference between the two calculations: momentum uses signed velocities, while kinetic energy uses squared speeds.

Question 3

A 0.30 kg puck A travels east at 4.0 m/s and collides with stationary puck B on a smooth horizontal surface.

After the collision:

• Puck A travels at 2.0 m/s, at 60° north of east.
• Puck B travels at 1.5 m/s, at 30° south of east.

Calculate the mass of puck B.

Verify that the horizontal and vertical momentum components are both conserved. Determine whether the collision is elastic.

Solution 3

Puck B has a mass of approximately 0.693 kg, and the collision is inelastic.

The most efficient approach is to use the vertical momentum equation first. Initially, there is no vertical momentum, so the two final vertical momentum components must cancel.

Step 1: Resolve the final velocities into components.

For puck A:

\[
v_{Ax}=2.0\cos60^\circ=1.0
\]

\[
v_{Ay}=2.0\sin60^\circ=1.732
\]

For puck B:

\[
v_{Bx}=1.5\cos30^\circ=1.299
\]

\[
v_{By}=-1.5\sin30^\circ=-0.750
\]

All values are in m/s.

Step 2: Apply conservation of vertical momentum.

Initially, both pucks have zero vertical velocity.

\[
0=m_Av_{Ay}+m_Bv_{By}
\]

Substitute:

\[
0=0.30(1.732)+m_B(-0.750)
\]

\[
0=0.5196-0.750m_B
\]

\[
m_B=\frac{0.5196}{0.750}
\]

\[
\boxed{m_B=0.6928\text{ kg}}
\]

Step 3: Verify horizontal momentum.

Initial horizontal momentum:

\[
p_{x,i}=0.30(4.0)
\]

\[
p_{x,i}=1.20\text{ kg m/s}
\]

Final horizontal momentum:

\[
p_{x,f}=0.30(1.0)+0.6928(1.299)
\]

\[
p_{x,f}=0.30+0.90
\]

\[
p_{x,f}=1.20\text{ kg m/s}
\]

Both components are conserved, so the calculated mass is consistent with the collision data.

Step 4: Check kinetic energy.

Initially:

\[
E_{k,i}=\frac12(0.30)(4.0)^2
\]

\[
E_{k,i}=2.40\text{ J}
\]

Finally:

\[
E_{k,f}
=
\frac12(0.30)(2.0)^2
+
\frac12(0.6928)(1.5)^2
\]

\[
E_{k,f}=0.60+0.7794
\]

\[
E_{k,f}=1.3794\text{ J}
\]

Kinetic energy has decreased, so the collision is inelastic.

The important point is that satisfying both momentum components does not require the collision to conserve kinetic energy.

Question 4

A 0.40 kg puck A travels east at 4.0 m/s and collides with stationary puck B on a smooth horizontal surface.

After the collision:

• Puck A continues east at 1.0 m/s.
• Puck B travels at 2.0 m/s, at 60° north of east.

A student uses horizontal momentum conservation and calculates the unknown mass as 1.2 kg.

Is this a valid solution if there is negligible external impulse in both horizontal directions?

Calculate the vertical momentum imbalance that would need to be supplied by an external impulse for the measured data to be correct.

Solution 4

No. There is no positive mass that satisfies both momentum components under the stated isolated-system assumption. The horizontal equation gives 1.2 kg, but the vertical equation does not balance.

This is an important experimental reasoning skill: satisfying one component of momentum conservation does not guarantee that the entire collision is physically consistent.

Step 1: Calculate the mass using horizontal momentum.

Initial horizontal momentum:

\[
p_{x,i}=0.40(4.0)=1.60
\]

Final horizontal momentum:

\[
p_{x,f}=0.40(1.0)+m_B(2.0\cos60^\circ)
\]

Equating:

\[
1.60=0.40+m_B(1.0)
\]

Therefore:

\[
\boxed{m_B=1.20\text{ kg}}
\]

So far, the calculation appears correct.

Step 2: Examine vertical momentum.

Initially, both pucks have zero vertical velocity.

Therefore:

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

After the collision, puck A still travels horizontally, so its vertical momentum is zero.

Puck B has vertical velocity:

\[
v_{By}=2.0\sin60^\circ
\]

\[
v_{By}=1.732\text{ m/s}
\]

Its vertical momentum is:

\[
p_{y,f}=1.20(1.732)
\]

\[
p_{y,f}=2.078\text{ kg m/s}
\]

This is not equal to the initial vertical momentum of zero.

Step 3: Determine the required external impulse.

Using the impulse-momentum relationship:

\[
J_{\text{external},y}
=
\Delta p_y
\]

\[
J_{\text{external},y}
=2.078-0
\]

\[
\boxed{J_{\text{external},y}=2.08\text{ N s north}}
\]

The measured data would require an external impulse of approximately 2.08 N s north.

However, the question states that external impulse is negligible in both directions.

Therefore, the data are inconsistent with the assumptions.

In a real investigation, you would check the camera measurements, the velocity directions, and whether an external force acted during the collision.

Do not simply report 1.2 kg as a valid answer. A physically consistent solution must satisfy every relevant momentum component.

Question 5

A 0.50 kg trolley A travels east at 4.0 m/s and collides with stationary trolley B, whose mass is unknown.

After the collision, trolley A rebounds west at 1.0 m/s. Trolley B moves east, but its final speed has not yet been measured.

Assume the two-trolley system experiences negligible external impulse.

A student argues that because trolley A rebounds, the collision must be elastic.

Part A: Use conservation of momentum to determine the relationship between the mass and final speed of trolley B. Explain why this information alone cannot determine its mass.

Part B: Suppose the collision actually is elastic. Use conservation of kinetic energy, together with momentum conservation, to calculate the predicted mass and final speed of trolley B.

Part C: A later video measurement shows that trolley B actually travels east at 2.5 m/s. Calculate its mass using momentum and determine whether the actual collision was elastic.

Explain why the student’s original argument was incorrect.

Solution 5

Momentum alone determines the product of trolley B’s mass and final speed, not its individual mass. An additional measurement or a justified energy condition is needed to determine both.

This question tests an important distinction between having enough information to conserve momentum and having enough information to calculate every unknown.

Part A: Apply momentum conservation.

Choose east as positive.

The initial momentum is:

\[
p_i=0.50(4.0)=2.0
\]

The final momentum is:

\[
p_f=0.50(-1.0)+m_Bv_B
\]

Equating:

\[
2.0=-0.50+m_Bv_B
\]

Therefore:

\[
\boxed{m_Bv_B=2.50\text{ kg m/s}}
\]

This equation gives the product of the unknown mass and speed.

It does not give either quantity individually.

For example, a larger mass travelling more slowly could produce the same momentum as a smaller mass travelling faster.

There are infinitely many possible combinations satisfying this equation.

Part B: Calculate the result if the collision is elastic.

An elastic collision conserves both momentum and kinetic energy.

Initial kinetic energy:

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

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

After the collision, trolley A has kinetic energy:

\[
E_{k,A,f}
=\frac12(0.50)(1.0)^2
\]

\[
E_{k,A,f}=0.25\text{ J}
\]

If the collision is elastic, trolley B must have the remaining kinetic energy:

\[
E_{k,B,f}=4.0-0.25
\]

\[
E_{k,B,f}=3.75\text{ J}
\]

Therefore:

\[
\frac12m_Bv_B^2=3.75
\]

Rearrange:

\[
m_Bv_B^2=7.50
\]

From Part A:

\[
m_Bv_B=2.50
\]

Divide the kinetic energy equation by the momentum equation:

\[
v_B=\frac{7.50}{2.50}
\]

\[
\boxed{v_B=3.0\text{ m/s}}
\]

Now calculate the mass:

\[
m_B=\frac{2.50}{3.0}
\]

\[
\boxed{m_B=0.833\text{ kg}}
\]

These are the predictions of the elastic-collision model.

Part C: Use the actual measured speed.

The video measurement gives:

\[
v_B=2.5\text{ m/s}
\]

Return to the momentum equation:

\[
m_Bv_B=2.50
\]

\[
m_B(2.5)=2.50
\]

\[
\boxed{m_B=1.00\text{ kg}}
\]

The actual unknown mass is 1.00 kg.

Now calculate the final kinetic energy:

\[
E_{k,f}
=
\frac12(0.50)(1.0)^2
+
\frac12(1.00)(2.5)^2
\]

\[
E_{k,f}=0.25+3.125
\]

\[
E_{k,f}=3.375\text{ J}
\]

The decrease in kinetic energy is:

\[
\Delta E_k=4.0-3.375
\]

\[
\boxed{\Delta E_k=0.625\text{ J}}
\]

The actual collision is inelastic.

The student’s mistake was assuming that a rebound proves a collision is elastic. A rebound only tells us that an object has changed direction. It does not tell us whether total kinetic energy has remained constant.

The elastic model predicted a mass of 0.833 kg and a speed of 3.0 m/s. The actual measured speed gives a mass of 1.00 kg instead.

Both momentum and the measured velocities are needed to identify the actual result. Kinetic energy then tells us which collision model applies.

09What to learn next

Once you can calculate an unknown mass from collision data, the next useful step is understanding what happens during the collision itself.

Momentum tells you how the motion changes, but it does not directly tell you how large the collision force was.

For that, you need the impulse-momentum relationship:

\[
J=\Delta p
\]

Impulse connects the change in momentum to the force acting over a time interval. It allows you to move from calculating an unknown mass to analysing collision forces, contact times, and force-time graphs.