PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Revision sheet: momentum, impulse and collisions (PHYS 101)

This sheet does not reprove the two conservation laws: they are already in your notes. It deals with what actually decides the mark, namely how to tell in one reading which of the two laws you are allowed to write, why a collision written with magnitudes makes a recoil disappear, and what an examiner expects to see on the page before the first number.

Every formula used here is on the PHYS 101 formula sheet that you annotate and bring to the exam, the one dimensional elastic collision included. What the sheet does not tell you is which line you are entitled to write, nor how to catch a misread line in ten seconds: the limit cases below are that check.

It is written for PHYS 101, the calculus free mechanics course at McGill University: no derivative and no integral appears anywhere, and every impulse is read as an area of triangles and rectangles. Once the sheet is read, the corrected exercise set of the same chapter puts each reflex to the test.

Updated on October 25, 2026

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

The moment two bodies touch, the momentum of the system is conserved and the kinetic energy almost never is. One equation is always available, the other is a hypothesis you have to be allowed to write. And the conservation line is written with SIGNED components, never with magnitudes, otherwise a recoil quietly disappears.

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

Before this chapter, you need

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Linear momentum
Impulse
Impulse momentum theorem
Force time graph area
Average force of impact
Conservation of momentum
Isolated system
Perfectly inelastic collision
Elastic collision limit cases
Recoil and explosion
Ballistic pendulum
Centre of mass

The essentials

On your formula sheet: this chapter's lines, and when each one holds

• $F=ma=\frac{\Delta p}{\Delta t}$ : $F$ is the NET force on ONE body, read from its free-body diagram, and the equation is written once per axis. The form $\frac{\Delta p}{\Delta t}$ gives the average force during a collision.
• $p=mv$ : A vector: two objects moving in opposite directions carry momenta of opposite signs.
• $p_{1,i}+p_{2,i}=p_{1,f}+p_{2,f}$ : An isolated system, or an axis along which the external forces cancel. It holds in EVERY collision, elastic or not.
• $I=\Delta p=F\Delta t$ : $F$ is the average net force over $\Delta t$ ; on a force-time graph the impulse is the area under the curve.
• $x_{COM}=\frac{m_{1}x_{1}+m_{2}x_{2}+\dots}{m_{1}+m_{2}+\dots}$ : Positions from one origin, with their signs. With no external force the centre of mass keeps its velocity.
• $v_{1f}=\frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1i}+\frac{2m_{2}}{m_{1}+m_{2}}v_{2i}$ : A 1D ELASTIC collision only, kinetic energy conserved, every velocity signed by your axis. Check: equal masses swap their velocities.
• $v_{2f}=\frac{2m_{1}}{m_{1}+m_{2}}v_{1i}+\frac{m_{2}-m_{1}}{m_{1}+m_{2}}v_{2i}$ : The same collision seen from body 2. Check: a target at rest that is much heavier than the projectile barely moves.

The sheet prints each equation, never the condition under which it holds: copy the condition next to its line on your own sheet before the exam. This sheet writes $K$ for kinetic energy where the formula sheet writes $KE$ .

Two laws, and only one of them is free

• Momentum of the system, ALWAYS conserved when no outside force acts during the contact: $m_{1}v_{1}+m_{2}v_{2} = m_{1}v_{1}'+m_{2}v_{2}'$ .
• Kinetic energy, conserved ONLY if the problem says elastic: $\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2} = \frac{1}{2}m_{1}v_{1}'^{2}+\frac{1}{2}m_{2}v_{2}'^{2}$ .
• Between the two extremes the energy after is neither conserved nor minimal, and no formula gives it: it is computed at the end, from velocities found by momentum.
• The lower limit is the perfectly inelastic case, where the two bodies move as one and there is no relative motion left to destroy.

Same collision,

0.600

kg at

4.00

m/s into

0.200

kg at rest: the dashed line is the

4.80

J before, and the three bars are the only kinetic energies the collision can end with.

Read the wording, not the picture. The words stick, couple, embed, wedge and remain together give ONE final velocity. The word elastic, and only it, buys you the second equation.

Momentum is a vector, so the sign comes first

• Choose an axis and write it on the paper before any number: to the right positive, for example.
• Every velocity then carries a sign, and $p = mv$ carries it too. Two bodies moving towards each other give momenta that SUBTRACT.
• In two dimensions there is one conservation equation per axis: $\sum p_{x}$ before $= \sum p_{x}$ after, and the same for $y$ . Never one equation on the magnitudes.
• Units: $1$ kg m/s $= 1$ N s. That equality is what lets an impulse be set equal to a change of momentum.

Two identical pucks: the momenta after, head to tail, close on the momentum before; along

x

they add to

0.600

kg m/s, along

y

they cancel.

The signed two line table, $v_{1}=+2.50$ and $v_{2}=-1.00$ , written before the equation, is the cheapest insurance of the chapter.

Impulse: three readings of one quantity

• By the force and the time: $J = F_{\text{av}}\,\Delta t$ .
• By the graph: $J$ is the AREA under the force time curve, cut into triangles and rectangles. In PHYS 101 it is never an integral.
• By the effect: $J = \Delta p = m v_{f} - m v_{i}$ , which is the impulse momentum theorem.
• Consequence used by every safety device: $\Delta p$ is fixed by the mass and the speeds, so multiplying the contact time by five divides the average force by five.

To scale: bouncing back at

15.0

m/s changes the momentum by

5.25

kg m/s, stopping dead by only

3.00

kg m/s, so the bounce is the harder contact.

A bounce is more violent than a stop: reversing $20.0$ m/s into $15.0$ m/s changes the momentum by $35.0$ units of speed, while stopping dead changes it by $20.0$ .

Explosion and recoil: a total of zero

• A system at rest that pushes itself apart keeps $\sum p = 0$ : $m_{1}v_{1}' + m_{2}v_{2}' = 0$ , so the two pieces leave in opposite directions.
• The speeds are in the inverse ratio of the masses: $\frac{v_{2}'}{v_{1}'} = -\frac{m_{1}}{m_{2}}$ . The light piece is the fast one.
• Kinetic energy is NOT conserved: it was zero before and is positive after; the spring, the powder or the legs supplied it.

Two carts pushed apart by a spring, velocities to scale: equal and opposite momenta, so the

1.0

kg cart leaves twice as fast as the

2.0

kg one.

A $4.50$ kg rifle firing a $9.00$ g bullet at $750$ m/s recoils at $1.50$ m/s, and the bullet carries $500$ times the kinetic energy of the rifle.

The elastic line of the formula sheet, in letters

• With the target at rest, $v_{2i} = 0$ , the sheet reduces to $v_{1f} = \frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1i}$ and $v_{2f} = \frac{2m_{1}}{m_{1}+m_{2}}v_{1i}$ . Each fraction is a pure number, so both answers are speeds.
• Equal masses: $v_{1f} = 0$ and $v_{2f} = v_{1i}$ , the two bodies swap velocities.
• Heavy projectile, $m_{1} \gg m_{2}$ : $v_{1f} \to v_{1i}$ and $v_{2f} \to 2v_{1i}$ . Light projectile, $m_{1} \ll m_{2}$ : $v_{1f} \to -v_{1i}$ and $v_{2f} \to 0$ , a bounce off a wall.
• Share of the kinetic energy handed to the target: $\frac{4m_{1}m_{2}}{(m_{1}+m_{2})^{2}}$ . All of it for equal masses, little of it at either extreme, and the same value when $m_{1}$ and $m_{2}$ swap.

Share of the kinetic energy handed to a target at rest:

1

for equal masses,

0.75

for a projectile three times heavier,

0.36

for one nine times heavier, as for one nine times lighter.

Run the three limit cases on any elastic formula you copy, read or derive. A line that fails one of them has been misread, and the check costs ten seconds.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The four kinds of encounter, and what each one lets you write

Wording of the problem	Momentum	Kinetic energy	What you may write
they stick, couple, embed	conserved	minimum possible, not conserved	one equation, one unknown
Example: $1.20$ kg at $2.50$ m/s into $0.800$ kg at rest: $v = 1.50$ m/s and $40.0$ per cent of the energy lost.
the word elastic appears	conserved	conserved	two equations, two unknowns
Example: $0.600$ kg at $4.00$ m/s into $0.200$ kg at rest: $2.00$ m/s and $6.00$ m/s, with $4.80$ J on both sides.
real bumpers, nothing said	conserved	between the two, unknown	momentum only, energy is an audit no energy equation
Example: Same gliders, but A leaves at $2.50$ m/s: momentum gives $4.50$ m/s for B, and only then $K_{f} = 3.90$ J. What to do: Write momentum, use the one extra velocity the problem hands you, and compute the energy at the very end. Never assume a fraction of energy is kept.
explosion, recoil, spring released	conserved, total zero	increases	the two momenta are opposite
Example: A $4.50$ kg rifle and a $9.00$ g bullet at $750$ m/s: recoil $1.50$ m/s, and the bullet takes $500$ times the energy.

The last column is what you are entitled to write down. Nothing in the first three columns is a calculation: they tell you which page of the toolbox is open.

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. Writing the conservation line with magnitudes instead of signed components

the whole question, 4 or 5 marks, and every later part with it

What not to write

A cart of $1.50$ kg at $4.00$ m/s meets one of $2.50$ kg at $2.00$ m/s head on and they stick, so $v = \frac{6.00+5.00}{4.00} = 2.75$ m/s.

What to write

Take the first direction positive: $v = \frac{(1.50)(+4.00)+(2.50)(-2.00)}{4.00} = \frac{1.00}{4.00} = 0.250$ m/s, so the pair barely drifts, and it drifts the way the LIGHT cart was going.

The same collision written two ways: the signed sum gives a slow drift at

0.250

m/s, the sum of magnitudes gives

2.75

m/s, eleven times too fast.

Why: Momentum is a vector. A head on collision is the only case where the two contributions subtract, and it is exactly the case examiners set. Eleven times too fast is the size of the error here, and no later line can recover from it.

2. Refusing to write momentum conservation because energy is lost

the question is left blank, so 4 or 5 marks

What not to write

The two carts stick, so energy is lost, so nothing is conserved and the final velocity cannot be found.

What to write

Momentum is conserved in EVERY collision of an isolated system, whatever happens to the energy: $(1.20)(2.50) = (2.00)v$ gives $v = 1.50$ m/s. The energy is then audited: $3.75$ J become $2.25$ J.

Why: The internal forces during the contact are an action reaction pair: their impulses are equal and opposite, so they cancel in the total and the sum cannot move. Energy has no such protection, because the same forces do work on deforming material.

3. Saying that an airbag reduces the impulse received by the driver

2 marks, and the sentence shows the theorem was not understood

What not to write

The airbag cushions the shock, so it reduces the impulse the driver receives.

What to write

The impulse is fixed: a driver of $70.0$ kg going from $15.0$ m/s to rest must receive $1050$ N s whatever the padding. The airbag stretches the contact from $0.080$ s to $0.40$ s, so the average force falls from $13125$ N to $2625$ N.

Why: $J = \Delta p$ depends only on the mass and the two velocities, neither of which the bag touches. The bag acts on the OTHER reading, $J = F_{\text{av}}\Delta t$ , where the same area is spread over five times the time. Bending the knees on landing and a helmet liner are the same sentence.

4. Reading the peak of a force time graph as the average force

2 marks on the impulse, and every velocity that follows from it

What not to write

The graph tops out at $400$ N over $14.0$ ms, so $J = (400)(0.0140) = 5.60$ N s.

What to write

The impulse is the AREA: a rising triangle $0.80$ N s, a plateau rectangle $2.40$ N s, a falling triangle $0.80$ N s, total $J = 4.00$ N s. The average force is then $\frac{4.00}{0.0140} = 286$ N, well under the peak.

The dashed rectangle of height

286

N has exactly the area of the trapezium,

4.00

N s; the rectangle drawn at the peak of

400

N would hold

5.60

N s.

Why: The two triangles are half empty, so a rectangle drawn at the peak overstates the area by $40$ per cent. The average force is the height of the rectangle of the SAME area, which is why it is the number to compare with a tolerance and the peak is the number that breaks the racket string.

5. Carrying energy conservation across the impact of a ballistic pendulum

the whole problem, 6 to 8 marks

What not to write

The bullet arrives with $450$ J, so $\frac{1}{2}mv^{2} = (m+M)gh$ and $h = 30.6$ m.

What to write

Momentum ACROSS the impact: $(0.0100)(300) = (1.50)V$ gives $V = 2.00$ m/s. Energy AFTER it: $h = \frac{V^{2}}{2g} = \frac{4.00}{19.6} = 0.204$ m.

Why: The embedding destroys $99.3$ per cent of the kinetic energy, $450$ J down to $3.00$ J, so an energy line written across it carries a number that no longer exists. The order is fixed: momentum through the collision, energy through the swing, never one law for the whole story.

6. Sliding from equal momentum to equal kinetic energy

2 marks, and often a wrong conclusion in the discussion question

What not to write

A ball of $0.150$ kg at $20.0$ m/s and a bowling ball of $6.00$ kg at $0.500$ m/s carry the same momentum, so they carry the same energy.

What to write

Both carry $3.00$ kg m/s, but $30.0$ J against $0.750$ J, a factor of $40$ . Momentum grows like $v$ , kinetic energy like $v^{2}$ .

Why: The link is $K = \frac{p^{2}}{2m}$ : at equal momentum the energies go like $1/m$ . This is also why a rifle bruises a shoulder while its bullet goes through a plank, the two carrying the same $6.75$ kg m/s.

7. Giving two bodies the same kinetic energy because they received the same push for the same time

the whole conceptual question, the kind that opens a midterm

What not to write

Two sleds of $20.0$ kg and $80.0$ kg are each pulled from rest by $40.0$ N for $3.00$ s, so they end with the same kinetic energy.

What to write

Same force and same time give the same IMPULSE, $120$ N s, hence the same momentum. Since $K = \frac{p^{2}}{2m}$ , the light sled has $360$ J and the heavy one $90.0$ J, four times less.

Same impulse,

120

N s each: the momentum bars are equal, the kinetic energy bars are not,

360

J against

90.0

J, the ratio of the masses turned upside down. Each group has its own scale.

Why: Equal TIME gives equal momentum; equal DISTANCE would give equal energy, through the work $W = Fd$ . The light sled covers $9.00$ m during the pull against $2.25$ m for the heavy one, so the same force does four times more work on it.

8. Dropping the minus sign of a recoil, so the rifle follows the bullet

1 to 2 marks, and the whole point of the question

What not to write

A rifle of $4.50$ kg fires a $9.00$ g bullet at $750$ m/s, so the rifle moves at $1.50$ m/s in the same direction.

What to write

The total was zero and stays zero: $0 = (0.00900)(+750)+(4.50)v$ , so $v = -1.50$ m/s. The minus sign IS the recoil, and it is the answer to the question.

Why: In an explosion the total momentum is zero, which is only possible if the two pieces have OPPOSITE signs. Writing the answer without its sign hides the single phenomenon the question is about, and the same slip sends two skaters off in the same direction after pushing each other apart.

9. Adding the magnitudes of two momenta in a two dimensional collision

3 marks, and the check that would have caught the rest

What not to write

Puck A leaves at $2.60$ m/s and puck B at $1.50$ m/s, both of mass $0.200$ kg, so the total momentum after is $(0.200)(2.60+1.50) = 0.820$ kg m/s.

What to write

One equation per axis. Along $x$ : $(0.200)(2.60)(0.866)+(0.200)(1.50)(0.500) = 0.600$ kg m/s, equal to the momentum before. Along $y$ : $0.260-0.260 = 0$ .

Why: Momentum adds head to tail, not as numbers. The two outgoing pucks pull in different directions, so their magnitudes never add up to the incoming momentum, and seeing $0.820$ where $0.600$ was expected is the sign that the components were skipped.

10. Letting a heavy projectile carry the light target along at its own speed

2 marks, and the energy audit of the next part fails without explanation

What not to write

A $0.900$ kg glider at $1.00$ m/s hits a $0.100$ kg glider at rest elastically, so the light glider is carried off at $1.00$ m/s.

What to write

From the formula sheet, $v_{2f} = \frac{2(0.900)}{1.000}(1.00) = 1.80$ m/s and $v_{1f} = \frac{0.800}{1.000}(1.00) = 0.800$ m/s: the light glider leaves at almost TWICE the speed of the projectile.

Final speeds over the incoming speed, against the mass ratio: the target (blue) climbs towards

2

and passes

1

as soon as

m_{1} > m_{2}

; the projectile (orange) goes from a full rebound at

-1

to carrying on.

Why: Seen from the heavy glider, the light one arrives at $1.00$ m/s and bounces back at $1.00$ m/s, like a ball off a wall; adding back the motion of the heavy glider gives close to $2v$ . The wrong answer also fails the audit: momentum then leaves $0.889$ m/s to the heavy glider, and only $0.406$ J of the $0.450$ J remain, a loss an elastic collision cannot have.

Which method to choose

Which law am I allowed to write

Read the wording of the collision, not the drawing. The words decide which equations exist, and how many unknowns you may carry.

If the words stick, couple, embed or remain together appear → momentum only, with ONE unknown final velocity

Example: $(1.20)(2.50) = (2.00)v$ gives $v = 1.50$ m/s

the energy after is then computed, never assumed, and it is always the smallest the collision can leave
If the word elastic appears, or the problem says no energy is lost → momentum AND kinetic energy, two equations

Example: $0.600$ kg at $4.00$ m/s into $0.200$ kg at rest gives $2.00$ m/s and $6.00$ m/s

with one body at rest, the formula sheet's 1D elastic line reduces to $v_{1f} = \frac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1i}$ and $v_{2f} = \frac{2m_{1}}{m_{1}+m_{2}}v_{1i}$ , which saves five minutes; check it on equal masses, where the two bodies must swap velocities
If nothing is said, but one final velocity is given → momentum for the other velocity, then the energy as an AUDIT

Example: A leaves at $2.50$ m/s, so B leaves at $4.50$ m/s and $K_{f} = 3.90$ J against $4.80$ J
If the system starts at rest and pushes itself apart → total momentum zero, so $m_{1}v_{1} = -m_{2}v_{2}$

Example: a $60.0$ kg skater at $1.20$ m/s sends a $45.0$ kg skater the other way at $1.60$ m/s

here kinetic energy INCREASES, paid for by a spring, a muscle or a charge, so no energy equation is available either
If angles appear, or the motion is not along one line → one momentum equation per axis, solved starting with the axis that has a zero

Example: $0 = (0.200)v_{A}'\sin 30^{\circ} - (0.200)v_{B}'\sin 60^{\circ}$ gives $v_{A}' = 1.732\,v_{B}'$ in one line

A ballistic pendulum, a bullet in a block sliding on a rough floor and a dart on a cart hitting a spring all fall in the first branch, then continue with an energy problem AFTER the impact. Two stages, two different laws.

Where does the impulse come from

Look at what the problem hands you, and pick the reading of $J$ that uses it. All three give the same number.

If a force and a contact time are given → $J = F_{\text{av}}\Delta t$

Example: $2625$ N during $0.40$ s gives $1050$ N s
If a force time GRAPH is given → cut the area into triangles and rectangles

Example: $0.80 + 2.40 + 0.80 = 4.00$ N s

convert the milliseconds first: leaving them in ms multiplies the answer by one thousand
If velocities before and after are given → $J = \Delta p = m v_{f} - m v_{i}$ , with signs

Example: $(0.150)(-15.0) - (0.150)(+20.0) = -5.25$ N s
If a time is asked and the force is known → same theorem read backwards, $\Delta t = J / F_{\text{av}}$

Example: $1050$ N s at $13125$ N lasts $0.080$ s

The three readings are the same quantity, which is why an exam question can give you a graph and ask for a speed, or give you two speeds and ask for a force.

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.

Writing up a collision in one dimension

When to use it: Two bodies meet along a line and at least one final velocity is unknown.

1 Draw the axis and state the convention in words: to the right is positive.
2 Make a two line table of the SIGNED data, before and after, with units, and mark the unknowns.
3 Name the kind of collision by quoting the wording, then say which laws that allows: momentum always, kinetic energy only if elastic.
4 Write the conservation of momentum in letters first, then substitute the signed numbers.
5 Solve, then audit the energy if the question asks for it, and check that the result is not larger than the energy before.
6 Conclude in a sentence that gives the value, the unit and the direction.

Concluding sentence

Taking the direction of the first cart as positive, the pair moves off at $v = 0.250$ m/s, that is $0.250$ m/s in the direction the light cart was travelling.

Marking: 1 mark for the axis and the signed data, 2 for the conservation line, 1 for the algebra, 1 for a conclusion that carries a direction.

Writing up a two stage problem, collision then motion

When to use it: A bullet embeds in a block that then rises, a dart hits a cart that then slides on a rough floor, a pendulum is struck.

1 Split the problem explicitly on the page: stage 1, the contact; stage 2, what happens afterwards.
2 Stage 1: momentum only, and say in one line why energy cannot be used, namely that the bodies end up moving together.
3 Carry ONE number across the boundary, the common velocity just after the impact.
4 Stage 2: energy, with no momentum, since an outside force now acts, the string or the friction.
5 Check the boundary number against common sense before continuing, a block at $2.00$ m/s and not at $300$ m/s.

Concluding sentence

The block and bullet leave the impact at $V = 2.00$ m/s, and the swing that follows raises them by $h = 0.204$ m, that is an angle of $33.9^{\circ}$ .

Check before you hand in

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

Does the final velocity sit where it has to

After a perfectly inelastic collision, the common velocity always lies BETWEEN the two initial velocities, closer to the heavier body. Anything outside that range is a sign error.

$+4.00$ m/s and $-2.00$ m/s can only give something between $-2.00$ and $+4.00$ : a $2.75$ m/s answer is possible, a $6.00$ m/s answer is not.
Is the energy after inside its band

Except in an explosion, the kinetic energy after can never exceed the energy before, and never fall below the perfectly inelastic value. Compute the two limits in ten seconds and check that your answer is inside.

$4.80$ J before and $3.60$ J for the stuck case: any answer outside $[3.60 , 4.80]$ is wrong before it is read.
Put the numbers back into the momentum line

Recompute the total momentum AFTER, with signs, and compare with the total before. It is two multiplications and it catches a sign, a swapped mass and a slipped decimal at once.

$(0.600)(2.00)+(0.200)(6.00) = 2.40$ kg m/s, the same as $(0.600)(4.00)$ before.
Does the light one move faster

In every recoil and every explosion, the lighter piece leaves faster, in exactly the inverse ratio of the masses, and it carries far more of the energy.

$60.0$ kg at $1.20$ m/s against $45.0$ kg at $1.60$ m/s: the ratio $1.60/1.20$ is $60.0/45.0$ .
Do the units survive

An impulse is in N s, which is kg m/s. If a line mixes grams with kilograms or milliseconds with seconds, the answer is out by a factor of one thousand, which is visible without a calculator.

A tennis impulse of $4000$ N s instead of $4.00$ N s means the milliseconds were never converted.
Run the three limit cases

Before using an elastic formula, put in equal masses, then a huge projectile, then a tiny one. The equal case must swap the velocities, the huge projectile must carry on and launch the target at about twice its speed, the tiny one must bounce back.

$\frac{2m_{1}}{m_{1}+m_{2}}v$ gives $v$ for equal masses and tends to $2v$ for a huge $m_{1}$ : right. A misread $\frac{2m_{2}}{m_{1}+m_{2}}v$ gives $0$ there, a target that never moves, and fails.

The typical problem, taken apart

A head on collision where the heavy cart is turned around

On a straight track, a cart of mass $1.50$ kg moves to the right at $4.00$ m/s and meets a cart of mass $2.50$ kg moving to the left at $2.00$ m/s.

The two couple on contact and move off together.

Find their common velocity, the kinetic energy lost, and say what answer the usual mistake would have produced.

The two velocities carry the sign of the axis drawn at the bottom right:

+4.00

m/s for the light cart,

-2.00

m/s for the heavy one, so the total is a DIFFERENCE and not a sum.

Step 1

Axis: to the right positive. Data: $m_{1} = 1.50$ kg, $v_{1} = +4.00$ m/s, $m_{2} = 2.50$ kg, $v_{2} = -2.00$ m/s.

Why

The signed table exists so the minus sign is written ONCE, in a place where it cannot be forgotten. Every mark lost in this chapter is lost because this line was skipped.

Step 2

The carts couple, so there is one final velocity and momentum alone applies: $m_{1}v_{1}+m_{2}v_{2} = (m_{1}+m_{2})v$ .

Why

Naming the kind of collision before writing anything is what tells you how many equations you have. Here the word couple buys one equation and forbids the energy one.

Step 3

$(1.50)(+4.00)+(2.50)(-2.00) = 6.00-5.00 = 1.00$ kg m/s, so $v = \dfrac{1.00}{4.00} = +0.250$ m/s.

Why

The two momenta nearly cancel, which is the whole interest of the example: a head on meeting of comparable momenta leaves a system almost at rest.

Step 4

The sign is positive, so the pair drifts to the RIGHT at $0.250$ m/s: the heavy cart has been turned around.

Why

Reading the sign out loud is part of the answer, not a decoration. A conclusion without a direction loses the last mark even when the number is right.

Step 5

$K_{i} = \frac{1}{2}(1.50)(4.00)^{2}+\frac{1}{2}(2.50)(2.00)^{2} = 12.0+5.00 = 17.0$ J and $K_{f} = \frac{1}{2}(4.00)(0.250)^{2} = 0.125$ J.

Why

The audit comes last because it uses the velocity just found. Losing $99.3$ per cent of the energy is not an error: in a head on coupling, almost all of it goes into the deformation.

Step 6

The magnitude version would have given $\dfrac{6.00+5.00}{4.00} = 2.75$ m/s, eleven times too fast.

Why

Quoting the wrong answer next to the right one is how the reflex sticks. It also fails the first check of the sheet: $2.75$ m/s is fine as a range, but it would mean the collision barely slowed the light cart, which no head on impact does.

The conclusion, written out

Taking the right as positive, the coupled carts move off at $v = +0.250$ m/s, that is $0.250$ m/s to the right, and the collision has destroyed $16.9$ J of the $17.0$ J available.

Learn by heart

• $p = mv$ carries the SIGN of the velocity. Two bodies meeting head on give momenta that subtract.
• Momentum of an isolated system is conserved in every collision. Kinetic energy is conserved only if the word elastic appears.
• $J = F_{\text{av}}\Delta t =$ area under the force time graph $= \Delta p$ . The three are one quantity, and $1$ N s $= 1$ kg m/s.
• Stretch the contact time by five and the average force is divided by five, while the impulse does not move: airbag, helmet, bent knees.
• Perfectly inelastic means ONE final velocity, $v = \dfrac{m_{1}v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}$ , and the smallest kinetic energy the collision can leave.
• Elastic with a target at rest: $v_{1}' = \dfrac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1}$ and $v_{2}' = \dfrac{2m_{1}}{m_{1}+m_{2}}v_{1}$ . Equal masses swap velocities, a heavy projectile sends a light target off at almost $2v$ , a light projectile bounces back at almost $-v$ .
• Same push for the same time means the same momentum, never the same kinetic energy: $K = \dfrac{p^{2}}{2m}$ gives the lighter body more.
• Explosion from rest: $m_{1}v_{1} = -m_{2}v_{2}$ , and since $K = \dfrac{p^{2}}{2m}$ the light piece takes almost all the energy.
• Ballistic pendulum: momentum ACROSS the impact, then $h = \dfrac{V^{2}}{2g}$ for the swing. Never one energy line for both.
• In two dimensions, one conservation equation per axis, and start with the axis whose total is zero.
• Centre of mass: $v_{\text{cm}} = \dfrac{\sum m_{i}v_{i}}{\sum m_{i}}$ is the total momentum over the total mass, so a collision never changes it.

Frequently asked questions

Is momentum conserved when two objects stick together?

Yes, always, as long as no outside force acts during the contact. The two bodies push each other with equal and opposite forces, so the two impulses cancel and the total cannot change. What is lost is kinetic energy, which goes into deformation, heat and sound. The two statements are true at the same time, and confusing them is the reason many students leave the question blank when the only available equation was right there.

What is the difference between impulse and force?

A force is what acts at an instant, an impulse is the force multiplied by the time it acts, and it is the impulse that changes the momentum. On a graph of force against time, the impulse is the area under the curve. The same impulse can therefore be delivered by a huge force during a very short time or by a gentle force during a long one, which is exactly the trade an airbag makes.

Why does an airbag reduce the force but not the impulse?

Because the impulse is decided by the driver alone. A given mass going from a given speed to rest must receive a fixed change of momentum, whatever is in the way. The bag cannot change that number, so it changes the other factor: it stretches the contact from about eight hundredths of a second to four tenths, five times longer, and the average force comes out five times smaller.

How do I know if a collision is elastic?

The wording tells you, or the numbers do. If the problem uses the word elastic, or says that no energy is lost, you may write the energy equation. If you are given all four velocities instead, add up the kinetic energies before and after: equal totals mean elastic, a smaller total after means some was lost. Never assume elastic because the surface is frictionless, which is a different question entirely.

How do I solve a ballistic pendulum problem?

In two separate stages, with a different law in each. The bullet embedding in the block is a collision, so use conservation of momentum and find the speed of the pair just after impact. The swing that follows loses no energy, so use conservation of energy to turn that speed into a height. Only one number crosses from the first stage to the second, and writing a single energy line for the whole problem gives a height a hundred and fifty times too large.

Why does the centre of mass keep moving at the same speed during a collision?

Because its velocity is the total momentum divided by the total mass, and a collision changes neither of those two things. Whatever the two bodies do, bounce, stick or reverse, the point that represents the system rolls along a straight line at constant speed. On video analysis software this is the cleanest visual proof that no outside push was involved in the experiment.

Practise it

Corrected exercises: Momentum, impulse and collisions, PHYS 101

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

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Work, energy and power Next sheet Torque, rotation and equilibrium

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/phys101-momentum-collisions. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Looking for a PHYS 101 tutor in Montreal?

Get in touch for a first session, in person in Montreal or online. Ten years of tutoring McGill and Concordia students, and a physics course taught without a single derivative when the course asks for none.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: momentum, impulse and collisions (PHYS 101)

The essentials

On your formula sheet: this chapter's lines, and when each one holds

Two laws, and only one of them is free

Momentum is a vector, so the sign comes first

Impulse: three readings of one quantity

Explosion and recoil: a total of zero

The elastic line of the formula sheet, in letters

The rules in table form

The four kinds of encounter, and what each one lets you write

The mistakes that cost marks

1. Writing the conservation line with magnitudes instead of signed components

2. Refusing to write momentum conservation because energy is lost

3. Saying that an airbag reduces the impulse received by the driver

4. Reading the peak of a force time graph as the average force

5. Carrying energy conservation across the impact of a ballistic pendulum

6. Sliding from equal momentum to equal kinetic energy

7. Giving two bodies the same kinetic energy because they received the same push for the same time

8. Dropping the minus sign of a recoil, so the rifle follows the bullet

9. Adding the magnitudes of two momenta in a two dimensional collision

10. Letting a heavy projectile carry the light target along at its own speed

Which method to choose

Which law am I allowed to write

Where does the impulse come from

How the answer is expected to be written

Writing up a collision in one dimension

Writing up a two stage problem, collision then motion

Check before you hand in

The typical problem, taken apart

A head on collision where the heavy cart is turned around

Learn by heart

Frequently asked questions

Is momentum conserved when two objects stick together?

What is the difference between impulse and force?

Why does an airbag reduce the force but not the impulse?

How do I know if a collision is elastic?

How do I solve a ballistic pendulum problem?

Why does the centre of mass keep moving at the same speed during a collision?

Corrected exercises: Momentum, impulse and collisions, PHYS 101

See also

Looking for a PHYS 101 tutor in Montreal?