How to Separate Collision Momentum from Frictional Stopping

Two pucks collide, stick together, and slide across a rough mat before stopping. You know their initial velocities and the distance they skid. Should you conserve momentum from the start of the collision all the way to the final stopping point?

No. Friction removes momentum from the moving pucks after the impact.

The correct method is to separate the motion into two stages:

1. Collision stage: Choose both pucks as your system. Conserve horizontal momentum components if the external impulse during the short collision is negligible. Find the velocity immediately after impact.
2. Frictional stopping stage: Start with that post-collision velocity. Use work-energy or kinematics to calculate how friction brings the pucks to rest.

Check kinetic energy separately across the collision. If the pucks stick, some of their original kinetic energy is converted into other forms during impact. You cannot assume all of it remains available for the stopping stage.

01Why the collision and stopping stages need different equations

Imagine recording the movement with a high-speed camera.

In one short interval, the pucks hit each other and stick. During that collision, each puck pushes on the other. Those forces are internal to the two-puck system.

Over the much longer stopping interval, the rough mat pushes backwards on the joined pucks through friction. This is an external force on the puck system.

That changes which conservation law you can use.

The important boundary is the instant the collision ends.

What actually stays conserved?

Predict what would happen if you doubled the roughness of the stopping mat but left the collision unchanged.

Would the joined pucks leave the collision at half their original speed?

Not in the ideal model. Their immediate post-collision velocity depends on the initial momentum and the masses. A rougher mat makes them stop over a shorter distance, but it does not change the earlier collision result.

This assumes the collision occurs on a sufficiently smooth surface and the mat does not exert a significant impulse during contact.

Momentum is conserved for an isolated system. During the stopping stage, the pucks are not an isolated horizontal system because the mat exerts an external force. The Earth receives the corresponding opposite impulse.

Total energy is still conserved, but the pucks’ kinetic energy is not.

02An experiment that separates the two stages

Suppose you want to investigate a sticking collision followed by frictional stopping.

A useful arrangement has a smooth approach section, a collision point, and a rough stopping section. All surfaces are horizontal.

The first design: measure the launch and final stopping point

Start with puck A moving towards stationary puck B. Give the pucks hook-and-loop collars so they remain attached after contact.

Measure A’s speed before impact, then measure where the joined pucks eventually stop.

What would this tell you?

You would know the starting motion and final position. But you could not reliably tell how much kinetic energy was converted during the collision and how much was removed by friction unless you also had enough information about the frictional force and the intermediate motion.

A short stopping distance might mean the collision removed a lot of kinetic energy. It might also mean the mat exerted strong friction.

The first design mixes two effects into one final observation.

The revised design: measure immediately after the collision

Add a second timing gate just after the collision, while the pucks are still on the smooth section.

Now you can measure three distinct states:

• The separate pucks’ velocities immediately before impact.
• The joined pucks’ velocity immediately after impact.
• The final resting position after travelling across the rough mat.

Measure the kinetic friction coefficient of the mat separately, for example by observing how a puck with a known initial speed slows on the same surface.

Gate placement is important. If the second gate is already partway across the rough mat, the speed it records is not the immediate post-collision speed.

What would the measurements show?

Consider these illustrative data, chosen to demonstrate the method rather than reported experimental results.

The reduced speed after impact is consistent with momentum being shared between the moving and stationary masses.

The subsequent stopping distance is consistent with the remaining kinetic energy being removed by friction.

Neither measurement on its own proves the entire model. You would also check whether the surface was level, whether the pucks remained attached, and whether friction acted before the second measurement.

The revised design is better because it tests the collision prediction and the stopping prediction separately.

03The equations belong to different time intervals

Stage A: select the collision system and conserve momentum

Choose both pucks as your system, not just the incoming puck.

Let:

• \(m_A\) and \(m_B\) be their masses in kilograms.
• \(\vec u_A\) and \(\vec u_B\) be their velocities immediately before collision.
• \(\vec v\) be their common velocity immediately after sticking.
• \(M=m_A+m_B\) be their combined mass.

Velocity and momentum are vectors. Choose east as positive \(x\), and north as positive \(y\).

Provided the external impulse during impact is negligible, momentum is conserved independently in each horizontal direction:

\[
\begin{aligned}
m_Au_{Ax}+m_Bu_{Bx}&=Mv_x\\
m_Au_{Ay}+m_Bu_{By}&=Mv_y
\end{aligned}
\]

Here, the subscripts \(x\) and \(y\) identify horizontal components. Each momentum term has units of \(\mathrm{kg\,m\,s^{-1}}\).

For movement along one straight line, you only need one equation:

\[
m_Au_A+m_Bu_B=Mv
\]

Use signed velocities. A puck moving west has a negative velocity if east is positive.

This equation describes the change across the impact, not the entire journey to rest.

If you need more practice with this particular step, see how to find the speed of two objects that stick together.

Stage B: calculate the work done by friction

After the collision, suppose the joined pucks slide across a stationary, horizontal mat.

The magnitude of kinetic friction is:

\[
F_f=\mu_kMg
\]

where:

• \(F_f\) is the friction force in newtons (\(\mathrm{N}\)).
• \(\mu_k\) is the dimensionless coefficient of kinetic friction.
• \(M\) is the combined mass in kilograms.
• \(g\) is gravitational acceleration, taken as \(9.8\ \mathrm{m\,s^{-2}}\).

Friction acts opposite the sliding velocity.

Let \(s\) be the speed immediately after impact, measured in \(\mathrm{m\,s^{-1}}\), and \(d\) the distance travelled on the rough mat in metres.

The pucks initially have kinetic energy \(\frac12 Ms^2\). Friction does negative work until their kinetic energy becomes zero.

\[
\frac12 Ms^2=\mu_kMgd
\]

The combined mass cancels:

\[
d=\frac{s^2}{2\mu_kg}
\]

This is the stopping-distance equation for constant kinetic friction on a stationary, level surface.

It also follows from \(v_f^2=v_i^2+2ad\), using final speed \(v_f=0\) and acceleration \(a=-\mu_kg\) along the direction of motion.

The cancellation of mass is useful: under these assumptions, two sliding objects with the same entry speed and friction coefficient have the same stopping distance, regardless of their masses.

It does not mean mass was irrelevant during the collision. Mass still determines the momentum shared between the objects and therefore their entry speed.

Stage C: check collision energy separately

The kinetic energy just before impact is:

\[
K_{\mathrm{pre}}
=\frac12m_Au_A^2+\frac12m_Bu_B^2
\]

Here \(u_A\) and \(u_B\) are the magnitudes of the initial velocities, so each is a non-negative speed.

The kinetic energy immediately after the pucks stick is:

\[
K_{\mathrm{post}}=\frac12Ms^2
\]

For a passive sticking collision, the decrease

\[
\Delta K_{\mathrm{collision}}
=K_{\mathrm{pre}}-K_{\mathrm{post}}
\]

represents kinetic energy converted into other forms during impact, assuming negligible external work during that short interval.

The mat then removes the remaining \(K_{\mathrm{post}}\) through friction.

You should not set \(K_{\mathrm{pre}}=\mu_kMgd\) unless the collision itself preserves kinetic energy and there are no other losses before the mat.

04Worked examples: solve the impact before the skid

05How to recognise when the usual shortcut fails

The method depends on what happens during contact and what forces act afterwards.

Collision on a smooth surface, then sliding on a rough surface

This is the standard two-stage model.

The external impulse during the collision is negligible, so total momentum is approximately conserved across impact.

Friction later changes momentum and removes kinetic energy.

Collision while friction is already acting

Suppose the pucks collide on a rough floor and remain in contact for a noticeable time.

Friction acts during the collision itself. You cannot automatically conserve the pucks’ horizontal momentum from the start to the end of contact.

Instead, use the impulse-momentum relationship:

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

Here \(\vec J_{\mathrm{ext}}\) is the external impulse, measured in newton seconds (\(\mathrm{N\,s}\)), equivalent to \(\mathrm{kg\,m\,s^{-1}}\).

If that impulse is negligible compared with the system’s momentum, the simpler conservation model remains a reasonable approximation. If it is significant, include it.

The objects rebound instead of sticking

The same stage boundary still applies, but there are now two post-collision velocities rather than one common velocity.

Each moving object may undergo a different frictional stopping stage.

You must check the total kinetic energy of the collision system rather than assuming that a slower-looking object means the collision was inelastic.

The surface moves

A stationary mat normally removes kinetic energy from a sliding puck.

A moving surface is different. Friction can accelerate an object in the laboratory frame if the surface moves faster than the object.

That requires a fresh force and energy analysis, not automatic use of the stationary-mat stopping equation.

The final practice question explores this case.

06Questions and solutions

Assume all surfaces are horizontal, air resistance is negligible, and the objects slide without rolling. Use \(g=9.8\ \mathrm{m\,s^{-2}}\) unless a question states otherwise. All numerical data are illustrative.

Question 1

A \(0.20\ \mathrm{kg}\) puck moves east at \(3.0\ \mathrm{m\,s^{-1}}\) and collides with a stationary \(0.40\ \mathrm{kg}\) puck. They stick together on a smooth surface and immediately enter a rough mat with \(\mu_k=0.30\).

(a) Find their velocity immediately after impact.

(b) Calculate the distance travelled across the mat before stopping.

(c) Calculate the kinetic energy converted during the collision and the kinetic energy removed during the stopping stage.

Solution 1

The joined pucks move east at \(1.0\ \mathrm{m\,s^{-1}}\), stop after approximately \(0.17\ \mathrm{m}\), and convert \(0.60\ \mathrm{J}\) of kinetic energy during impact. The mat then removes the remaining \(0.30\ \mathrm{J}\).

For part (a), conserve momentum during the collision:

\[
\begin{aligned}
(0.20)(3.0)+(0.40)(0)&=0.60v\\
0.60&=0.60v\\
v&=+1.0\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The positive sign means east.

For part (b), use work-energy for the stopping stage:

\[
\begin{aligned}
d&=\frac{v^2}{2\mu_kg}\\
&=\frac{1.0^2}{2(0.30)(9.8)}\\
&=0.170\ \mathrm{m}
\end{aligned}
\]

For part (c), calculate kinetic energy on each side of the collision:

\[
\begin{aligned}
K_{\mathrm{pre}}
&=\frac12(0.20)(3.0)^2\\
&=0.90\ \mathrm{J}\\
K_{\mathrm{post}}
&=\frac12(0.60)(1.0)^2\\
&=0.30\ \mathrm{J}
\end{aligned}
\]

The collision converts \(0.90-0.30=0.60\ \mathrm{J}\) into other forms. Friction subsequently removes the \(0.30\ \mathrm{J}\) remaining.

The tempting error is to use the original \(0.90\ \mathrm{J}\) to find the stopping distance. That would include energy already converted during impact.

Question 2

Puck A has mass \(0.40\ \mathrm{kg}\) and moves east at \(3.0\ \mathrm{m\,s^{-1}}\). Puck B has mass \(0.60\ \mathrm{kg}\) and approaches along the same east-west line with an unknown velocity.

They collide and stick on a smooth surface. Immediately afterwards, the joined pucks enter a stationary rough mat, slide \(0.200\ \mathrm{m}\) west, and stop.

The mat has \(\mu_k=0.25\).

(a) Determine B’s velocity immediately before impact, including its direction.

(b) Find the kinetic energy converted during the collision.

(c) Explain why the stopping distance by itself is not enough to determine the direction of the joined pucks’ velocity.

Solution 2

Puck B was travelling west at approximately \(3.65\ \mathrm{m\,s^{-1}}\). The collision converted about \(5.31\ \mathrm{J}\) of kinetic energy, while friction subsequently removed \(0.49\ \mathrm{J}\).

For part (a), first work backwards through the stopping stage.

The joined pucks’ speed just after impact follows from:

\[
\begin{aligned}
s^2&=2\mu_kgd\\
s&=\sqrt{2(0.25)(9.8)(0.200)}\\
&=0.990\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The skid is westwards, so the signed velocity is:

\[
v=-0.990\ \mathrm{m\,s^{-1}}
\]

Conserve momentum during the earlier collision:

\[
\begin{aligned}
m_Au_A+m_Bu_B&=Mv\\
(0.40)(3.0)+0.60u_B&=-0.990\\
1.20+0.60u_B&=-0.990\\
u_B&=\frac{-2.190}{0.60}\\
&=-3.65\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Therefore B was moving west at \(3.65\ \mathrm{m\,s^{-1}}\).

For part (b), the pre-collision kinetic energy is:

\[
\begin{aligned}
K_{\mathrm{pre}}
&=\frac12(0.40)(3.0)^2\\
&\quad+\frac12(0.60)(3.65)^2\\
&\approx5.80\ \mathrm{J}
\end{aligned}
\]

Immediately after impact:

\[
\begin{aligned}
K_{\mathrm{post}}
&=\frac12(1.00)(0.990)^2\\
&=0.490\ \mathrm{J}
\end{aligned}
\]

Hence:

\[
\begin{aligned}
\Delta K_{\mathrm{collision}}
&\approx5.80-0.490\\
&=5.31\ \mathrm{J}
\end{aligned}
\]

The later skid dissipates the remaining \(0.490\ \mathrm{J}\).

For part (c), the stopping-distance equation uses speed squared. It gives the magnitude of the post-collision velocity, not its direction.

Without knowing the direction of travel, there would be two possible signed post-collision velocities: approximately \(+0.990\) and \(-0.990\ \mathrm{m\,s^{-1}}\). The momentum calculation would then produce different values for B’s initial velocity.

The observed westward skid selects the negative velocity.

The trap is to treat the final speed as positive simply because speed is non-negative. Momentum calculations require velocity, including its sign.

Question 3

A \(0.30\ \mathrm{kg}\) puck A, moving east with unknown initial speed, collides with a stationary \(0.50\ \mathrm{kg}\) puck B. They stick on a smooth section just before a rough mat, with negligible travel between impact and mat entry.

A timing gate placed \(0.20\ \mathrm{m}\) into the mat records the joined pucks moving east at \(1.50\ \mathrm{m\,s^{-1}}\).

They continue east and stop another \(0.30\ \mathrm{m}\) beyond the gate.

Assume the coefficient of kinetic friction is constant but unknown. The external impulse during the short collision is negligible.

(a) Determine the friction coefficient of the mat.

(b) Find the joined pucks’ speed immediately after collision, and A’s speed immediately before collision.

(c) Determine the kinetic energy converted during the collision.

(d) Explain how using the timing-gate speed directly in the momentum equation would affect the estimate of A’s original speed.

Solution 3

The mat has \(\mu_k\approx0.383\). The joined pucks left the collision at approximately \(1.94\ \mathrm{m\,s^{-1}}\) east, and A approached at approximately \(5.16\ \mathrm{m\,s^{-1}}\) east. The collision converted \(2.50\ \mathrm{J}\) of kinetic energy.

The key is that the gate measures the speed after the joined pucks have already travelled \(0.20\ \mathrm{m}\) under friction.

For part (a), their combined mass is:

\[
M=0.30+0.50=0.80\ \mathrm{kg}
\]

At the gate, their kinetic energy is:

\[
\begin{aligned}
K_{\mathrm{gate}}
&=\frac12(0.80)(1.50)^2\\
&=0.90\ \mathrm{J}
\end{aligned}
\]

Friction removes this energy across the remaining \(0.30\ \mathrm{m}\).

\[
\begin{aligned}
F_f(0.30)&=0.90\\
F_f&=3.0\ \mathrm{N}
\end{aligned}
\]

Therefore:

\[
\begin{aligned}
\mu_k&=\frac{F_f}{Mg}\\
&=\frac{3.0}{(0.80)(9.8)}\\
&\approx0.383
\end{aligned}
\]

For part (b), work backwards through the first \(0.20\ \mathrm{m}\) of rough mat.

The friction work over this distance has magnitude:

\[
W_f=(3.0)(0.20)=0.60\ \mathrm{J}
\]

So the kinetic energy immediately after collision was:

\[
\begin{aligned}
K_{\mathrm{post}}
&=K_{\mathrm{gate}}+0.60\\
&=1.50\ \mathrm{J}
\end{aligned}
\]

Use this to recover the speed:

\[
\begin{aligned}
\frac12Ms^2&=1.50\\
s^2&=\frac{2(1.50)}{0.80}\\
s&=1.94\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

This is the speed to use in the momentum equation, not \(1.50\ \mathrm{m\,s^{-1}}\).

Only A was moving before the collision:

\[
\begin{aligned}
m_Au_A&=Ms\\
u_A&=\frac{(0.80)(1.9365)}{0.30}\\
&\approx5.16\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

Thus A approached at approximately \(5.2\ \mathrm{m\,s^{-1}}\) east to two significant figures.

For part (c), the kinetic energy immediately before the collision was:

\[
\begin{aligned}
K_{\mathrm{pre}}
&=\frac12(0.30)(5.164)^2\\
&\approx4.00\ \mathrm{J}
\end{aligned}
\]

The collision converted:

\[
\begin{aligned}
\Delta K_{\mathrm{collision}}
&=4.00-1.50\\
&=2.50\ \mathrm{J}
\end{aligned}
\]

The mat later removed \(1.50\ \mathrm{J}\), including \(0.60\ \mathrm{J}\) before the gate and \(0.90\ \mathrm{J}\) afterwards.

For part (d), using the gate speed as the collision-exit speed would give:

\[
u_A=\frac{(0.80)(1.50)}{0.30}
=4.0\ \mathrm{m\,s^{-1}}
\]

That underestimates the actual incoming speed by approximately \(23\%\).

This is an experimental timing error disguised as a momentum calculation. The momentum equation is appropriate, but the measured velocity belongs to the wrong stage.

Moving the gate immediately after the collision onto the smooth section, or using high-speed video to estimate the immediate exit velocity, would address the problem.

Question 4

Two pucks collide while sliding east on a rough, horizontal floor.

• Puck A has mass \(0.40\ \mathrm{kg}\) and initial velocity \(+3.0\ \mathrm{m\,s^{-1}}\).
• Puck B has mass \(0.60\ \mathrm{kg}\) and initial velocity \(+1.0\ \mathrm{m\,s^{-1}}\).
• Both have the same kinetic friction coefficient, \(\mu_k=0.25\), with the floor.
• They remain in contact for \(0.20\ \mathrm{s}\) before becoming a joined pair.
• Both pucks continue sliding east throughout contact.
• Their mutual contact forces are horizontal, so their normal forces remain equal to their individual weights.

After the collision, the joined pair continues across the same rough floor until it stops.

(a) Calculate the common velocity at the end of contact, accounting for the external friction impulse.

(b) Determine the subsequent stopping distance.

(c) A student calculates the difference between total kinetic energy immediately before and after contact and calls all of it ‘energy lost in the collision’. Explain the problem with this claim.

(d) Calculate the actual work done by floor friction during the \(0.20\ \mathrm{s}\) contact and determine how much kinetic energy was converted through the inelastic impact itself.

Solution 4

The joined pucks leave contact at \(1.31\ \mathrm{m\,s^{-1}}\) east and subsequently slide about \(0.350\ \mathrm{m}\). During contact, floor friction removes approximately \(0.762\ \mathrm{J}\), while the inelastic impact converts a further \(0.480\ \mathrm{J}\) into other forms.

This is not an isolated collision. The floor exerts a measurable external impulse during contact.

For part (a), choose both pucks as the system and take east as positive.

Their total initial momentum is:

\[
\begin{aligned}
p_i&=(0.40)(3.0)+(0.60)(1.0)\\
&=1.80\ \mathrm{kg\,m\,s^{-1}}
\end{aligned}
\]

Both slide east, so friction acts west on both.

The total friction force is:

\[
\begin{aligned}
F_f&=\mu_kMg\\
&=(0.25)(1.00)(9.8)\\
&=2.45\ \mathrm{N}
\end{aligned}
\]

During contact, the external impulse is:

\[
\begin{aligned}
J_{\mathrm{ext}}&=-F_f\Delta t\\
&=-(2.45)(0.20)\\
&=-0.490\ \mathrm{N\,s}
\end{aligned}
\]

Apply impulse-momentum:

\[
\begin{aligned}
p_f&=p_i+J_{\mathrm{ext}}\\
&=1.80-0.490\\
&=1.31\ \mathrm{kg\,m\,s^{-1}}
\end{aligned}
\]

Since the joined mass is \(1.00\ \mathrm{kg}\):

\[
v=+1.31\ \mathrm{m\,s^{-1}}
\]

Ignoring friction during contact would incorrectly give \(1.80\ \mathrm{m\,s^{-1}}\).

For part (b), calculate the later stopping distance:

\[
\begin{aligned}
d&=\frac{1.31^2}{2(0.25)(9.8)}\\
&=0.350\ \mathrm{m}
\end{aligned}
\]

This distance begins at the end of the contact interval, not when the pucks first touch.

For part (c), compare the kinetic energies.

Initially:

\[
\begin{aligned}
K_i&=\frac12(0.40)(3.0)^2\\
&\quad+\frac12(0.60)(1.0)^2\\
&=2.10\ \mathrm{J}
\end{aligned}
\]

At the end of contact:

\[
\begin{aligned}
K_f&=\frac12(1.00)(1.31)^2\\
&=0.858\ \mathrm{J}
\end{aligned}
\]

The total decrease is approximately \(1.242\ \mathrm{J}\).

But this is not entirely energy converted by the inelastic collision. Some kinetic energy is removed by the rough floor while the pucks are still in contact.

For part (d), use the motion of the system’s centre of mass to find the floor’s work.

The centre of mass is the mass-weighted average position of the two pucks. Since both have the same friction coefficient and slide east throughout, the total external friction force is constant.

The initial centre-of-mass velocity is:

\[
v_{\mathrm{cm},i}=\frac{p_i}{M}
=1.80\ \mathrm{m\,s^{-1}}
\]

Its acceleration is:

\[
a_{\mathrm{cm}}=-\mu_kg
=-2.45\ \mathrm{m\,s^{-2}}
\]

Over \(0.20\ \mathrm{s}\), its displacement is:

\[
\begin{aligned}
\Delta x_{\mathrm{cm}}
&=v_{\mathrm{cm},i}t+\frac12at^2\\
&=(1.80)(0.20)\\
&\quad-\frac12(2.45)(0.20)^2\\
&=0.311\ \mathrm{m}
\end{aligned}
\]

Because each puck experiences friction proportional to its mass, their combined friction work equals the total friction force multiplied by the centre-of-mass displacement:

\[
\begin{aligned}
W_f&=-\mu_kMg\Delta x_{\mathrm{cm}}\\
&=-(2.45)(0.311)\\
&=-0.762\ \mathrm{J}
\end{aligned}
\]

Therefore, the kinetic energy converted by the inelastic interaction itself is:

\[
\begin{aligned}
E_{\mathrm{impact}}
&=K_i-K_f-|W_f|\\
&=2.10-0.858-0.762\\
&=0.480\ \mathrm{J}
\end{aligned}
\]

The complete energy account is:

All the initial kinetic energy is accounted for.

The deeper point is that the word ‘collision’ describes a period of contact, not a guarantee that no external force acts. When friction overlaps the impact, momentum and energy must both account for that overlap.

Question 5

Extension challenge

A horizontal conveyor belt moves east at a constant speed of \(2.0\ \mathrm{m\,s^{-1}}\). A motor keeps the belt at that speed.

Two pucks moving along the belt collide and stick.

Immediately before impact:

• Puck A has mass \(0.40\ \mathrm{kg}\) and moves east at \(1.0\ \mathrm{m\,s^{-1}}\).
• Puck B has mass \(0.60\ \mathrm{kg}\) and moves west at \(2.0\ \mathrm{m\,s^{-1}}\).

All velocities are measured relative to the stationary laboratory.

Assume the collision is sufficiently short that the belt’s friction impulse during impact is negligible. After impact, the joined pucks experience kinetic friction with coefficient \(\mu_k=0.20\) until they move with the belt.

For this question, take \(g=10.0\ \mathrm{m\,s^{-2}}\). Assume the belt is level, the motor maintains constant speed, and motor losses are negligible.

(a) Find the joined pucks’ velocity immediately after collision and the kinetic energy converted during impact.

(b) Determine whether the joined pucks ever become momentarily stationary relative to the laboratory. If so, find when and where this occurs relative to the collision point.

(c) Find when the joined pucks stop sliding relative to the belt, their velocity at that moment, and their net displacement from the collision point.

(d) Calculate the change in the joined pucks’ kinetic energy between the end of the collision and the end of relative sliding. Explain why it does not violate energy conservation, and determine the work supplied by the belt motor and the energy dissipated by sliding friction during that interval.

Solution 5

The joined pucks initially move west at \(0.80\ \mathrm{m\,s^{-1}}\). They briefly become stationary \(0.16\ \mathrm{m}\) west of the collision point, then reverse and accelerate east until reaching the belt’s speed. At the end of sliding, they are \(0.84\ \mathrm{m}\) east of the collision point and have gained \(1.68\ \mathrm{J}\) of laboratory-frame kinetic energy.

The hinge is that friction opposes relative sliding, not necessarily motion relative to the laboratory.

For part (a), conserve momentum across the short collision. Take east as positive.

\[
\begin{aligned}
m_Au_A+m_Bu_B&=Mv\\
(0.40)(1.0)+(0.60)(-2.0)&=1.00v\\
0.40-1.20&=v\\
v&=-0.80\ \mathrm{m\,s^{-1}}
\end{aligned}
\]

The joined pucks initially travel west.

Before impact:

\[
\begin{aligned}
K_{\mathrm{pre}}
&=\frac12(0.40)(1.0)^2\\
&\quad+\frac12(0.60)(2.0)^2\\
&=0.20+1.20\\
&=1.40\ \mathrm{J}
\end{aligned}
\]

Immediately afterwards:

\[
\begin{aligned}
K_{\mathrm{post}}
&=\frac12(1.00)(0.80)^2\\
&=0.32\ \mathrm{J}
\end{aligned}
\]

The impact converts:

\[
1.40-0.32=1.08\ \mathrm{J}
\]

into other forms.

Now the collision has ended. The moving belt determines what happens next.

For part (b), consider the relative motion.

The belt moves east at \(+2.0\ \mathrm{m\,s^{-1}}\), while the joined pucks initially move west at \(-0.80\ \mathrm{m\,s^{-1}}\).

The belt surface is therefore moving east relative to the pucks.

Kinetic friction on the joined pucks acts east.

Its magnitude is:

\[
\begin{aligned}
F_f&=\mu_kMg\\
&=(0.20)(1.00)(10.0)\\
&=2.0\ \mathrm{N}
\end{aligned}
\]

So their acceleration during relative sliding is:

\[
a=\frac{F_f}{M}
=+2.0\ \mathrm{m\,s^{-2}}
\]

The pucks initially move west but accelerate east. Their westward speed decreases to zero.

Using \(v=u+at\):

\[
\begin{aligned}
0&=-0.80+2.0t\\
t&=0.40\ \mathrm{s}
\end{aligned}
\]

Their displacement during that interval is:

\[
\begin{aligned}
\Delta x
&=ut+\frac12at^2\\
&=(-0.80)(0.40)\\
&\quad+\frac12(2.0)(0.40)^2\\
&=-0.16\ \mathrm{m}
\end{aligned}
\]

They are momentarily stationary \(0.16\ \mathrm{m}\) west of the impact point.

But they are not stationary relative to the moving belt. At that instant, the belt still moves east past them at \(2.0\ \mathrm{m\,s^{-1}}\).

Friction therefore continues to accelerate them east.

For part (c), relative sliding stops when the joined pucks match the belt’s velocity:

\[
v=+2.0\ \mathrm{m\,s^{-1}}
\]

Their acceleration remains \(+2.0\ \mathrm{m\,s^{-2}}\) until that moment.

The elapsed time from the collision is:

\[
\begin{aligned}
2.0&=-0.80+2.0t\\
t&=1.40\ \mathrm{s}
\end{aligned}
\]

Their total displacement from the collision point is:

\[
\begin{aligned}
\Delta x
&=(-0.80)(1.40)\\
&\quad+\frac12(2.0)(1.40)^2\\
&=-1.12+1.96\\
&=+0.84\ \mathrm{m}
\end{aligned}
\]

Thus, after moving west, stopping momentarily, and reversing, they end up \(0.84\ \mathrm{m}\) east of the collision point.

They then continue east with the belt at \(2.0\ \mathrm{m\,s^{-1}}\), without further relative sliding.

For part (d), calculate the joined pucks’ final kinetic energy in the laboratory frame:

\[
\begin{aligned}
K_{\mathrm{final}}
&=\frac12(1.00)(2.0)^2\\
&=2.00\ \mathrm{J}
\end{aligned}
\]

Their kinetic energy immediately after collision was \(0.32\ \mathrm{J}\).

Therefore:

\[
\begin{aligned}
\Delta K&=2.00-0.32\\
&=+1.68\ \mathrm{J}
\end{aligned}
\]

This seems surprising. How can friction increase kinetic energy?

The belt is powered by a motor. Friction transfers energy from the moving belt to the joined pucks.

The friction force on the pucks points east throughout relative sliding. Their net laboratory displacement is \(0.84\ \mathrm{m}\) east.

So the net work done by friction on the pucks is:

\[
\begin{aligned}
W_{\mathrm{on\ pucks}}
&=F_f\Delta x\\
&=(2.0)(0.84)\\
&=+1.68\ \mathrm{J}
\end{aligned}
\]

That exactly matches their kinetic energy increase.

However, the motor must supply more than \(1.68\ \mathrm{J}\), because relative sliding also produces heat.

Over the \(1.40\ \mathrm{s}\) interval, the belt moves:

\[
d_{\mathrm{belt}}=(2.0)(1.40)
=2.80\ \mathrm{m}
\]

The pucks exert a \(2.0\ \mathrm{N}\) westward friction force on the belt, so the motor must do additional work to maintain the belt’s constant speed:

\[
\begin{aligned}
W_{\mathrm{motor}}
&=F_fd_{\mathrm{belt}}\\
&=(2.0)(2.80)\\
&=5.60\ \mathrm{J}
\end{aligned}
\]

Of this energy, \(1.68\ \mathrm{J}\) increases the kinetic energy of the pucks.

The remainder is dissipated as heat through sliding friction:

\[
\begin{aligned}
E_{\mathrm{thermal}}
&=5.60-1.68\\
&=3.92\ \mathrm{J}
\end{aligned}
\]

This also agrees with friction force multiplied by the total relative sliding distance. The belt moves \(2.80\ \mathrm{m}\) east while the pucks have a net eastward displacement of \(0.84\ \mathrm{m}\), giving:

\[
\begin{aligned}
d_{\mathrm{relative}}
&=2.80-0.84\\
&=1.96\ \mathrm{m}
\end{aligned}
\]

The dissipated energy is:

\[
E_{\mathrm{thermal}}
=(2.0)(1.96)=3.92\ \mathrm{J}
\]

The complete interpretation is now consistent:

• During the collision, the pucks lose \(1.08\ \mathrm{J}\) of kinetic energy through the inelastic impact.
• After the collision, friction transfers energy from the motor-driven belt to the pucks, increasing their laboratory-frame kinetic energy by \(1.68\ \mathrm{J}\).
• The motor supplies \(5.60\ \mathrm{J}\) during this later interval, of which \(3.92\ \mathrm{J}\) is dissipated through sliding.

The tempting approach is to assume friction must always slow an object in the laboratory frame.

That is true for ordinary kinetic friction on a stationary floor, but not for a powered moving belt. Friction opposes relative sliding at the contacting surfaces.

The other trap is the moment when the pucks become stationary. They have stopped relative to the laboratory for an instant, but they have not stopped relative to the belt.

The relevant stopping condition depends on which surface the object is moving against.

06The next decision: what does your measurement actually represent?

In a collision followed by frictional stopping, identify three moments before writing an equation: immediately before impact, immediately after impact, and the end of the skid.

The first two belong to the collision calculation. The second and third belong to the stopping calculation.

If you are given a stopping distance and a friction coefficient, work backwards to find the speed at the start of the rough section. If you are given incoming velocities and masses, work forwards through momentum to that speed. The two methods should agree when the model’s assumptions are valid.

If they do not, investigate the stage boundary before changing your algebra. A speed sensor might be too far from impact, friction might act during contact, or the objects might not behave as a simple sticking system.

The next useful distinction is between collisions that stick and collisions that rebound. For rebounds, the energy comparison is more informative because kinetic energy may or may not be conserved during impact. See how to check whether a collision is elastic from measured speeds.

In either case, the first question remains the same: at exactly which stage was each velocity measured?