Mechanics 203-SN1-RE • CEGEP, Montreal

Revision sheet: momentum and collisions (203-SN1-RE)

Conservation of momentum is the most robust law of 203-SN1-RE (formerly 203-NYA) mechanics: it holds for every collision, whatever the deformation, whatever the heat produced, as long as external forces stay negligible during the impact. That is also what makes it dangerous: it works so well that you forget to check whether kinetic energy is conserved too.

This sheet separates the two laws, says in which phase each one applies, and gathers the sign and direction errors that cost the most marks on collisions in one and two dimensions. It is written for CEGEP students in the Science program (Sciences de la nature).

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The thread of the chapter

Momentum is conserved in ALL collisions, kinetic energy almost never. The whole chapter is about knowing which law applies to which phase, and never applying both to the same one.

This chapter is part of 203-SN1-RE, Mechanics

The essentials

Two laws, two domains of validity

• $\vec{p}=m\vec{v}$ is a VECTOR: it changes sign with the direction of motion, and in two dimensions it is projected axis by axis.
• Conservation: $\vec{P}_{\text{before}}=\vec{P}_{\text{after}}$ for EVERY collision, elastic or not, as long as external forces are negligible during the impact.
• Kinetic energy, on the other hand, is conserved ONLY in a collision explicitly stated to be elastic. In all others, it decreases.
• Perfectly inelastic collision: the bodies leave together, $v'=\dfrac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}$ . It is the collision that dissipates the most energy.
• Elastic collision in one dimension: instead of the quadratic system, use the reversal of the relative velocity, $v_{1}-v_{2}=-(v_{1}'-v_{2}')$ , which is linear.

The coupling of the first trap, velocities to scale:

P=24{,}000

kg·m/s before and after. Kinetic energy, however, drops from

36

kJ to

24

kJ.

“External forces are negligible” is justified by the BREVITY of the collision: gravity and friction also act during the impact, but their impulse over a few milliseconds is negligible next to that of the collision.

Impulse: the same change by two routes

• $\vec{J}=\vec{F}\Delta t=\Delta\vec{p}=m\vec{v}_{f}-m\vec{v}_{i}$ , in newton-seconds, which equal kilogram-metres per second.
• Variable force: $J$ is the AREA under the curve $F(t)$ , and $F_{avg}=\dfrac{J}{\Delta t}$ .
• The same impulse can come from a large brief force or a small long one: this is the whole principle of the airbag, of knees bent on landing and of the baseball glove.
• REBOUND: if the velocity reverses, $\Delta p=m(-v)-m(+v)=-2mv$ , magnitude $2mv$ , never zero.
• Impulse is a vector: in two dimensions it is calculated component by component, and its direction is not that of $\vec{v}_{f}$ .

The two triangles have the same area, so the same impulse and the same change in velocity. Only the peak force changes, and it is the peak force that injures.

The word “average” in “average force exerted” always signals that you must go through impulse: it is the only way to get a force from a collision whose details are unknown.

Collision in two dimensions: one conservation per axis

• Choose two perpendicular axes and write conservation SEPARATELY on each: $P_{x}$ before $=P_{x}$ after, and likewise for $P_{y}$ .
• Magnitude and direction then come from the components: $P=\sqrt{P_{x}^{2}+P_{y}^{2}}$ and $\tan\alpha=\dfrac{P_{y}}{P_{x}}$ .
• Two perpendicular momenta add head to tail, like two forces, never by adding their magnitudes.

24{,}000

toward the east and

18{,}000

toward the north give

30{,}000

kg·m/s at

36.9°

, not

42{,}000

: it is the hypotenuse, not the sum.

A wreck that leaves close to the $x$ axis betrays a large $P_{x}$ : this is how the speed of a vehicle is recovered after an accident.

What to know about the centre of mass

• $x_{cm}=\dfrac{\sum m_{i}x_{i}}{\sum m_{i}}$ , and likewise for $y_{cm}$ .
• $\vec{P}_{\text{system}}=M_{\text{total}}\vec{v}_{cm}$ : conserving $P$ is exactly saying that $\vec{v}_{cm}$ does not change.
• No internal force, however violent, can move the centre of mass: not an explosion, not a collision, not an astronaut relaxing.
• Hence a result that comes up regularly: two people walking toward each other on a boat at rest meet at the centre of mass, never at the middle of the boat.
• For a system initially at rest, any motion of one part to the right forces a motion of the other to the left.

System at rest: the two carts leave with opposite momenta, the lighter one three times as fast, and the centre of mass, dashed, does not move.

When a problem says “the system is initially at rest” and then describes an internal event, the answer almost always starts with $P_{\text{total}}=0$ before and after.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Conserving kinetic energy in a collision where the bodies stick together

the whole question, 6 marks

What not to write

“The $8.0$ t railcar at $3.0$ m/s couples with a $4.0$ t railcar at rest: $\tfrac{1}{2}(8000)(3.0)^{2}=\tfrac{1}{2}(12000)v'^{2}$ , so $v'=2.45$ m/s.”

What to write

“Momentum is conserved: $(8000)(3.0)=(12000)v'$ , so $v'=2.00$ m/s. Kinetic energy, on the other hand, drops from $36$ kJ to $24$ kJ.”

Why: A coupling deforms, heats and makes noise: the kinetic energy lost has gone somewhere. Only momentum ignores what happens during the impact.

2. Treating momentum as a positive number in one dimension

the whole question, 5 marks

What not to write

“The two balls approach each other at $4$ m/s and $6$ m/s: $P=m(4)+m(6)=10m$ .”

What to write

“They travel in opposite directions, so with the axis pointing right: $P=m(+4)+m(-6)=-2m$ .”

Why: $\vec{p}$ is a vector, not an energy. Forgetting the sign turns a head-on collision into a chase and gives a final velocity three times too large.

3. Writing a zero change in momentum for a rebound

the whole question, 4 marks

What not to write

“The $0.15$ kg ball arrives at $20$ m/s and leaves at $20$ m/s: the speed has not changed, so $\Delta p=0$ and the force is zero.”

What to write

“The velocity has REVERSED: $\Delta p=0.15(-20)-0.15(+20)=-6.0$ kg·m/s, that is $6.0$ N·s in magnitude, not zero.”

Why: It is the same mistake as in kinematics with displacement: the MAGNITUDE of the velocity has not changed, but the vector has turned around. A wall that exerted no force would send nothing back.

4. Applying energy to the collision of a ballistic pendulum

the whole question, 8 marks

What not to write

“The bullet embeds in the hanging block, which rises $12$ cm: $\tfrac{1}{2}mv^{2}=(m+M)gh$ , so $v=\sqrt{\dfrac{2(m+M)gh}{m}}$ .”

What to write

“DURING the collision, momentum: $mv=(m+M)V$ . AFTER the collision, energy: $\tfrac{1}{2}(m+M)V^{2}=(m+M)gh$ . Two phases, two laws.”

Why: The collision here dissipates more than $99\%$ of the initial kinetic energy. Applying conservation of energy to the collision assumes nothing was lost, and underestimates the speed of the bullet by a factor greater than $10$ .

5. Conserving momentum during a phase where external forces act

the whole question, 6 marks

What not to write

“The cart hits the block then goes up the slope: I conserve momentum from the start to the top.”

What to write

“Conservation holds only DURING the impact, where the collision dominates everything. On the slope, gravity is an external force: switch back to energy or to Newton.”

Why: Conservation of $\vec{P}$ is not a universal law but a consequence of $\Sigma\vec{F}_{\text{ext}}=0$ . Over a long time, gravity supplies a perfectly measurable impulse, and $\vec{P}$ changes.

6. Adding magnitudes instead of projecting in a collision in two dimensions

the whole question, 8 marks

What not to write

“One car arrives at $15$ m/s eastward, the other at $20$ m/s northward: the total momentum corresponds to $35$ in the direction of the wreck.”

What to write

“We project: $P_{x}=m_{1}(15)$ and $P_{y}=m_{2}(20)$ , we conserve EACH component separately, then we go back to a magnitude and an angle.”

Why: Two perpendicular vectors do not add up as numbers. It is the same mistake as the river crossing of the kinematics chapter, transposed to collisions.

7. Believing that two balls that separate always make a ninety-degree angle

4 marks, and a wrong projection for the rest of the problem

What not to write

“After the collision, the two bodies necessarily leave at $90°$ from each other, it is the billiards rule.”

What to write

“This result requires TWO conditions: equal masses AND an elastic collision, with one of the two balls initially at rest. Outside these conditions, the angle is different from $90°$ .”

Elastic collision, target at rest, velocities to scale. Equal masses: right angle. Projectile twice as heavy: only

60°

, the two velocities close up.

Why: The angle comes from a dot product whose sign follows that of $m_{1}-m_{2}$ : zero if the masses are equal, positive if the projectile is heavier, hence a narrower angle.

8. Solving an elastic collision with the full quadratic system

no marks lost, but ten minutes and two roots to sort out

What not to write

“I write $m_{1}v_{1}=m_{1}v_{1}'+m_{2}v_{2}'$ and $\tfrac{1}{2}m_{1}v_{1}^{2}=\tfrac{1}{2}m_{1}v_{1}'^{2}+\tfrac{1}{2}m_{2}v_{2}'^{2}$ , then I substitute and solve the quadratic.”

What to write

“We replace the second equation by its linear consequence: $v_{1}-v_{2}=-(v_{1}'-v_{2}')$ . The system becomes two first-degree equations.”

Why: The quadratic solution always returns the trivial solution $v_{1}'=v_{1}$ , the one where the collision never took place. The linear form eliminates it at the outset and spares you choosing between two roots under exam pressure.

9. Believing that an internal explosion moves the centre of mass

4 marks on the reasoning questions

What not to write

“The rocket explodes in flight: the pieces fly off in all directions, so the centre of mass goes with the biggest piece.”

What to write

“No internal force changes $\vec{v}_{cm}$ : the centre of mass follows exactly the parabolic path the intact rocket would have followed.”

Why: Internal forces come in action-reaction pairs and cancel two by two in the sum. Only gravity, which is external, still acts, and it alone bends the path of the centre of mass.

Which method to choose

Which law to apply depending on the collision

The word in the problem that describes what happens to the bodies at the moment of contact

If “stay hooked together”, “embeds in”, “leave together”, “lock together” → perfectly inelastic collision: momentum alone, $v'=\dfrac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}$

Example: railcars coupling, bullet in a block
If “elastic collision”, “no energy lost”, “ideal billiard balls” → momentum AND reversal of the relative velocity

Example: $v_{1}-v_{2}=-(v_{1}'-v_{2}')$ , linear system
If the problem gives both final velocities and asks for the energy lost → calculate $K$ before and after, the difference is the answer

Example: from $36$ kJ to $24$ kJ, that is $33\%$ dissipated
If the problem describes an explosion, a recoil, a separation → momentum, with $P_{\text{before}}=0$ if the system started at rest

Example: astronaut throwing a tool, rifle recoiling
If the problem asks for an average force or an impact duration → impulse-momentum theorem, $F_{avg}\Delta t=\Delta p$

Example: $\Delta p=6.0$ kg·m/s in $0.010$ s gives $600$ N

No real collision is perfectly elastic: if the problem does not say so, do not assume it. The silence of the problem means you should use momentum only.

How many phases in the problem

The story of the problem, reread looking for the moments when the physics changes

If an object falls, then hits, then rebounds → three phases: energy or kinematics, impulse, then energy again

Example: a ball dropped from a height that comes back up less high
If a projectile embeds, then the assembly starts moving → two phases: momentum during the collision, energy or Newton afterward

Example: ballistic pendulum, cart that climbs a ramp after the impact
If an object slides, hits, then slides again with friction → three phases, and friction NEVER enters the momentum balance of the collision

Example: puck launched on a rough surface
If everything happens during the impact alone → a single phase, momentum, nothing else

Example: collision at an intersection, common velocity just after

Write the phases before calculating, numbered, with the law used next to each. It is the only habit that prevents you from carrying a conservation law from one end of the problem to the other.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Solving a two-phase problem

When to use it: A collision is followed by a motion, or preceded by a fall: as soon as the problem tells a story in several stages.

1 Split the story into numbered phases and write, next to each, the law that applies and why.
2 Set an axis and give it a direction, then express each velocity with its sign.
3 Collision phase: write conservation of momentum, justifying that external forces are negligible over such a brief time.
4 Next phase: write the energy balance or Newton's equation, starting from the velocity obtained in the previous step.
5 Work back up the chain to the requested unknown, then check that the result is plausible and give the percentage of energy lost if it is asked.

Concluding sentence

“During the collision, which is very brief, momentum is conserved: $mv=(m+M)V$ . After the collision, only gravity does work, so $\tfrac{1}{2}(m+M)V^{2}=(m+M)gh$ . Combining, $v=\dfrac{m+M}{m}\sqrt{2gh}=257$ m/s.”

The trap: Justifying the conservation. A paper that writes $mv=(m+M)V$ without the sentence “external forces are negligible during the impact” loses the rigour mark, and above all has no reason to refuse to use the same law during the rise.

Marking: Typically 1 mark for splitting into phases, 3 for the justified conservation, 3 for the energy balance, 1 for the conclusion and the order of magnitude.

Handling a collision in two dimensions

When to use it: Two bodies arrive along different directions, or leave along different directions.

1 Set two axes and draw a labelled diagram with the angles measured from the $x$ axis.
2 Write the components of each momentum before the collision: $p_{x}=mv\cos\theta$ and $p_{y}=mv\sin\theta$ .
3 Write conservation SEPARATELY on each axis, which gives two independent equations.
4 Solve the system, then return to the magnitude by Pythagoras and to the direction by the arctangent, checking the quadrant.

Concluding sentence

“Conservation along $x$ gives $m_{1}v_{1}=m_{1}v_{1}'\cos\alpha+m_{2}v_{2}'\cos\beta$ and along $y$ gives $0=m_{1}v_{1}'\sin\alpha-m_{2}v_{2}'\sin\beta$ ; solving the system gives the two velocities after the collision.”

The trap: The equation along $y$ when the initial momentum there is zero. It is not useless: it says that the two vertical components cancel exactly, which is often the most rewarding equation of the system.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Kinetic energy after never exceeds that before

Calculate $K$ on both sides, even when it is not asked. An energy that increases without an external source signals a sign error or a poorly chosen law.

$24$ kJ after against $36$ kJ before: plausible. $44$ kJ after: impossible without a compressed spring or an explosive.
The common velocity is bracketed by the two initial velocities

In a perfectly inelastic collision, the final velocity is always between the two initial velocities, closer to that of the heavier body.

A result of $v'=3.5$ m/s for a collision between $3.0$ m/s and $0$ is impossible: it must be between $0$ and $3.0$ .
Substituting back into the momentum balance

Plug the velocities found back into $P_{\text{before}}=P_{\text{after}}$ , with the signs. It is the most rewarding check of the chapter and takes ten seconds.

A final sum of $+2m$ against an initial sum of $-2m$ says a sign was lost along the way.
The order of magnitude of a projectile speed

A rifle bullet goes from $200$ to $900$ m/s, a baseball $40$ m/s, a car in town $14$ m/s. Compare before handing in.

A result of $v=25$ m/s for a $12$ g bullet that lifts a $2$ kg block by $12$ cm: ten times too low, energy has been applied to the collision.
The direction of each velocity reread on the diagram

Go back over the diagram and check that each sign matches an arrow. A negative velocity must point opposite to the axis, not elsewhere.

A negative final velocity for a body pushed by a heavier one in the same direction: the sign of the axis changed along the way.

The typical problem, taken apart

The ballistic pendulum, two phases and two laws

A bullet of mass $m=12.0$ g is fired horizontally into a block of wood of mass $M=2.00$ kg hanging from vertical strings. The bullet embeds immediately.

The assembly then rises $h=12.0$ cm above its initial position. Find the velocity of the bullet before the impact, then the percentage of kinetic energy dissipated in the collision. Take $g=9.80$ m/s $^{2}$ .

Two instants separated by the collision: the bullet disappears into the block, then the assembly rises by

h

. The first belongs to momentum, the second to energy.

Step 1

Phase 1, the collision: very brief, external forces negligible, bodies joined afterward. Law: conservation of momentum. Phase 2, the rise: only gravity does work, inextensible strings. Law: conservation of mechanical energy.

Why

Writing this split in black and white is what prevents the most costly error of the chapter. It is also worth a rigour mark by itself in most marking schemes.

Step 2

Phase 2 first, because it contains only one unknown: $\tfrac{1}{2}(m+M)V^{2}=(m+M)gh$ , so $V=\sqrt{2gh}=\sqrt{2(9.80)(0.120)}=1.53$ m/s.

Why

We work the problem backward. The mass cancels, which is the sign that the height reached depends only on the velocity just after the collision, never on how the masses are distributed.

Step 3

Phase 1 next: $mv=(m+M)V$ , so $v=\dfrac{(0.0120+2.00)(1.53)}{0.0120}=257$ m/s.

Why

The ratio $\dfrac{m+M}{m}$ is $168$ here: it is this amplification factor that makes the whole point of the device, since it turns a height measurable to the centimetre into a speed of several hundred metres per second.

Step 4

Kinetic energy before: $\tfrac{1}{2}(0.0120)(257)^{2}=396$ J. After the collision: $\tfrac{1}{2}(2.012)(1.53)^{2}=2.37$ J.

Why

The two numbers must be calculated separately, not estimated: it is the gap between them that justifies, after the fact, that applying energy to phase 1 was forbidden.

Step 5

Fraction dissipated: $1-\dfrac{2.37}{396}=0.994$ , that is $99.4\%$ . We directly recover $\dfrac{M}{m+M}=\dfrac{2.00}{2.012}$ .

Why

The formula $\dfrac{M}{m+M}$ holds for any perfectly inelastic collision on a stationary target: knowing it lets you check the full calculation in one line.

Step 6

Check: with the wrong method, $v=\sqrt{\dfrac{2(m+M)gh}{m}}=19.8$ m/s, thirteen times less than the correct value.

Why

Comparing the two results anchors the order of magnitude: a rifle bullet does not travel at $20$ m/s, and that remark alone is enough to reject the energy method applied to the collision.

The conclusion, written out

“Conservation of momentum during the impact, followed by conservation of energy during the rise, gives an initial bullet speed of $257$ m/s, and the collision dissipates $99.4\%$ of its kinetic energy.”

The classic mistake on this problem: Writing a single energy balance from the bullet to the highest point. You get $19.8$ m/s, an absurd value for a bullet, and you no longer have any way to calculate the energy dissipated, since you have just assumed there was none.

Learn by heart

• $\vec{p}=m\vec{v}$ is a VECTOR: sign in one dimension, components in two dimensions.
• $\vec{P}$ is conserved in ALL collisions; $K$ only in elastic collisions.
• $\vec{J}=\vec{F}\Delta t=\Delta\vec{p}$ , and $J$ is the area under the curve $F(t)$ .
• Rebound with reversal: $|\Delta p|=2mv$ , never zero.
• Perfectly inelastic collision: $v'=\dfrac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}$ , and the fraction of energy lost is $\dfrac{M}{m+M}$ on a stationary target.
• Elastic collision in one dimension: $v_{1}-v_{2}=-(v_{1}'-v_{2}')$ , linear, to be preferred to the quadratic system.
• The $90°$ angle after the collision requires EQUAL masses and an ELASTIC collision on a stationary target.
• No internal force moves the centre of mass: $\vec{v}_{cm}$ does not change in an explosion.

Frequently asked questions

Is momentum conserved in all collisions?

Yes, as long as external forces are negligible during the impact, which is the case whenever the collision is brief. This holds equally for an elastic collision and for one where the bodies stick together. Kinetic energy, however, is conserved only if the collision is explicitly stated to be elastic.

What is the difference between an elastic and an inelastic collision?

In both cases momentum is conserved. In an elastic collision, the total kinetic energy is the same before and after. In an inelastic collision, part of it has gone into deformation, heat and sound. The extreme case is the perfectly inelastic collision, where the bodies leave together and the loss is greatest.

How do you solve a ballistic pendulum problem?

In two steps. During the collision, momentum is conserved: the mass of the bullet times its speed equals the total mass times the common speed. After the collision, mechanical energy is conserved: the common speed is converted into height. Never apply conservation of energy to the collision itself, because it dissipates almost everything.

Why is the change in momentum of a rebound not zero?

Because momentum is a vector. A ball that arrives at twenty metres per second and leaves at twenty metres per second in the opposite direction has seen its vector turn around. The change is therefore twice the mass times the speed, not zero. It is this change that gives the force exerted by the wall.

When do two balls leave at ninety degrees?

Only if the two masses are equal, the collision is elastic and one of the balls was at rest before the impact. If the projectile is heavier than the target, the angle is narrower; if it is lighter, the angle is wider. Outside ideal billiards, this result must never be assumed.

Can an explosion move the centre of mass of a system?

No. The forces involved are internal and cancel in pairs by Newton's third law. The centre of mass therefore keeps following exactly the path it was on before, even if the pieces fly off in all directions. Only an external force, such as gravity, can change its motion.

Practise it

Corrected exercises: Momentum and collisions, 203-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys1-momentum-collisions. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: momentum and collisions (203-SN1-RE)

The essentials

Two laws, two domains of validity

Impulse: the same change by two routes

Collision in two dimensions: one conservation per axis

What to know about the centre of mass

The mistakes that cost marks

1. Conserving kinetic energy in a collision where the bodies stick together

2. Treating momentum as a positive number in one dimension

3. Writing a zero change in momentum for a rebound

4. Applying energy to the collision of a ballistic pendulum

5. Conserving momentum during a phase where external forces act

6. Adding magnitudes instead of projecting in a collision in two dimensions

7. Believing that two balls that separate always make a ninety-degree angle

8. Solving an elastic collision with the full quadratic system

9. Believing that an internal explosion moves the centre of mass

Which method to choose

Which law to apply depending on the collision

How many phases in the problem

How the answer is expected to be written

Solving a two-phase problem

Handling a collision in two dimensions

Check before you hand in

The typical problem, taken apart

The ballistic pendulum, two phases and two laws

Learn by heart

Frequently asked questions

Is momentum conserved in all collisions?

What is the difference between an elastic and an inelastic collision?

How do you solve a ballistic pendulum problem?

Why is the change in momentum of a rebound not zero?

When do two balls leave at ninety degrees?

Can an explosion move the centre of mass of a system?

Corrected exercises: Momentum and collisions, 203-SN1-RE

See also

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