Chapter 4 · Laws of Motion

4.2 Newton's Third Law and Conservation of Momentum

Forces come in equal and opposite pairs acting on two different bodies. From that one fact follows the conservation of linear momentum — the principle behind recoil, explosions, rockets and every collision.

Stand on a skateboard and push against a wall: you roll away from it. Fire a rifle and it kicks back into your shoulder. A swimmer pushes water backward and moves forward; a rocket throws hot gas downward and rises, with nothing outside it to push against. In every case a body gets a push by pushing something else. Forces are never one-way: whenever one body pushes or pulls another, the second pushes or pulls back on the first.

Newton's third law states this exactly, and it has a remarkable consequence. Because the two forces of a pair are equal and opposite and last for the same time, the momentum one body gains is exactly the momentum the other loses. However violent the interaction — a collision, an explosion, a push-off between two skaters — the total momentum of the bodies does not change, provided nothing outside them interferes. This conservation of linear momentum lets us find speeds after a collision or explosion without knowing anything about the complicated forces during it.

Newton's third law

📐 Law 4.4 — Newton's third law of motion

To every action there is always an equal and opposite reaction. If body $A$ exerts a force $\mathbf{F}_{BA}$ on body $B$, then $B$ exerts a force $\mathbf{F}_{AB}$ on $A$ with

$$\mathbf{F}_{AB}=-\mathbf{F}_{BA}.$$

(Here $\mathbf{F}_{AB}$ means the force on $A$ by $B$.) The two forces of a pair are equal in magnitude, opposite in direction, act along the same line, and act on different bodies.

The words "action" and "reaction" can mislead, so be precise about what the law says and does not say.

✏️ Example 1 — A book on a table
A $1.5\ \text{kg}$ book rests on a table. (a) Name every force on the book and give its size. (b) Name the reaction to each. (c) Are the weight of the book and the push of the table on it an action–reaction pair?
(a) Its weight $W=mg=1.5\times9.8=14.7\ \text{N}$ down (the Earth's pull), and the normal force $N$ of the table, up. The book is in equilibrium, so $N=14.7\ \text{N}$.
(b) Reaction to $W$: the book pulls the Earth up with $14.7\ \text{N}$ (acting on the Earth, at its centre). Reaction to $N$: the book presses down on the table with $14.7\ \text{N}$ (acting on the table).
(c) No. They act on the same body (the book) and are of different kinds (gravity and contact). They are equal here only because the book is not accelerating — the first law, not the third.
Check Put the book in a lift accelerating upward and $N$ becomes larger than $W$ (Section 4.4); each third-law pair stays exactly equal.
✏️ Example 2 — The horse-and-cart puzzle
A horse pulls a cart. "The cart pulls back on the horse with an equal force, so they cancel and nothing can move." Resolve the puzzle using these data: the horse ($500\ \text{kg}$) pulls the cart ($300\ \text{kg}$) with $600\ \text{N}$; friction from the road on the cart's wheels is $150\ \text{N}$. Find the acceleration, and the forward force the ground must exert on the horse's hooves.
Idea The pull on the cart and the pull on the horse act on different bodies. To find how the cart moves, use only the forces on the cart.
Cart $F_\text{net}=600-150=450\ \text{N}$, so $a=\dfrac{450}{300}=1.5\ \text{m/s}^2$.
Horse It has the same acceleration. On it act the cart's backward pull $600\ \text{N}$ and the ground's forward push $P$ on its hooves (the reaction to the horse pushing the ground backward): $P-600=500\times1.5$, so $P=1350\ \text{N}$.
Check Treat horse + cart as one system: the $600\ \text{N}$ pair is now internal and drops out; $1350-150=800\times1.5$ ✓. The motion is driven by the ground, not by the horse–cart pair.
✏️ Example 3 — Pushing two blocks
Blocks of mass $4.0\ \text{kg}$ and $2.0\ \text{kg}$ sit side by side on a smooth floor. A horizontal force of $12\ \text{N}$ pushes the $4.0\ \text{kg}$ block, which pushes the $2.0\ \text{kg}$ block. Find (a) the acceleration, (b) the contact force between the blocks. (c) What is the contact force if the $12\ \text{N}$ push is applied to the $2.0\ \text{kg}$ block from the other side?
(a) Both blocks together: $a=\dfrac{12}{4.0+2.0}=2.0\ \text{m/s}^2$ (the contact forces are internal).
(b) The only horizontal force on the $2.0\ \text{kg}$ block is the contact force $C$: $C=2.0\times2.0=4.0\ \text{N}$. By the third law the $2.0\ \text{kg}$ block pushes back on the $4.0\ \text{kg}$ block with $4.0\ \text{N}$.
(c) Now $C$ must accelerate the $4.0\ \text{kg}$ block: $C=4.0\times2.0=8.0\ \text{N}$.
Check In (b), for the $4.0\ \text{kg}$ block: $12-4.0=4.0\times2.0$ ✓. The contact force is larger when it has to push the heavier block.
Interactive 1 — Push-Off on Ice

Two skaters stand together at rest on frictionless ice and push each other apart for $0.50\ \text{s}$. During the push the red and blue arrows are the force on each skater — always equal and opposite (third law). Afterwards they glide apart at steady speeds (first law). Make one skater much heavier: the forces stay equal, but the accelerations, speeds and distances do not. The dashed line marks the balance point of the pair.

50 kg75 kg300 N
force on each—
left skater after push—
right skater after push—
total momentum—

Conservation of linear momentum

To use the third law we need to be clear about which bodies we are looking at.

📘 Definition 4.4 — System, internal and external forces

A system is any collection of bodies we choose to study together. Forces that bodies of the system exert on each other are internal forces; forces exerted on bodies of the system by anything outside it are external forces. A system on which the net external force is zero is called isolated.

Let two bodies $A$ and $B$ interact — collide, push off, or are blown apart — during a time $\Delta t$, with no external force. By the third law, at every instant the force on $A$ by $B$ is the negative of the force on $B$ by $A$, so their impulses over $\Delta t$ are also equal and opposite. By the impulse–momentum theorem (Result 4.3),

$$\Delta\mathbf{p}_A=\mathbf{F}_{AB}\,\Delta t,\qquad \Delta\mathbf{p}_B=\mathbf{F}_{BA}\,\Delta t=-\mathbf{F}_{AB}\,\Delta t=-\Delta\mathbf{p}_A .$$

Hence $\Delta\mathbf{p}_A+\Delta\mathbf{p}_B=\mathbf{0}$: the momentum lost by one is gained by the other, and the total does not change. (If the force varies during the interaction, the same argument works with the areas under the two force–time graphs, which are mirror images.)

📐 Result 4.5 — Law of conservation of linear momentum

If the net external force on a system is zero, the total linear momentum of the system remains constant:

$$\mathbf{p}_\text{total}=m_1\mathbf{v}_1+m_2\mathbf{v}_2+\dots=\text{constant}.$$

For two bodies, $m_1\mathbf{u}_1+m_2\mathbf{u}_2=m_1\mathbf{v}_1+m_2\mathbf{v}_2$ (before = after). The law holds component by component: if the external force has no component along some direction, the momentum along that direction is conserved even when other components change.

Two remarks make the law far more useful than it first looks. First, for a very short interaction — a collision, a gunshot, an explosion — even forces that are external (gravity, friction from the ground) give only a tiny impulse in the short time $\Delta t$, while the internal impulsive forces are huge. So momentum just before and just after is conserved to a very good approximation. Second, the law needs nothing about the internal forces: we can find the result of a collision without knowing how the bodies deform, bounce or stick.

🧭 Problem-solving strategy — Using conservation of momentum
  1. Choose the system so that the forces you do not know are internal (the bullet and the block; both skaters; all fragments of the shell).
  2. Check that the net external force is zero, or negligible during the short interaction, at least along one direction.
  3. Choose a positive direction (or $x$ and $y$ axes) and write every velocity with its sign or components.
  4. Write total momentum before = total momentum after, and solve. In two dimensions do this separately for $x$ and $y$.
✏️ Example 4 — Recoil of a rifle
A $4.0\ \text{kg}$ rifle fires a $20\ \text{g}$ bullet at $400\ \text{m/s}$. (a) Find the recoil speed of the rifle if it is free to move. (b) Find the recoil speed if the rifle is held firmly against the shoulder of a $76\ \text{kg}$ shooter standing on smooth ice, so that rifle and shooter move together. (c) Why does holding the rifle firmly reduce the kick?
System Rifle + bullet, initially at rest: total momentum $0$. The explosion forces are internal.
(a) $0=0.020\times400+4.0\,V$, so $V=-\dfrac{8.0}{4.0}=-2.0\ \text{m/s}$: $2.0\ \text{m/s}$ backward.
(b) Now the recoiling mass is $4.0+76=80\ \text{kg}$: $V=-\dfrac{8.0}{80}=-0.10\ \text{m/s}$.
(c) The same backward momentum $8.0\ \text{kg m/s}$ is shared by a twenty-times larger mass, so the recoil speed — and the jolt as it is stopped — is much smaller. A loosely held rifle reaches $2.0\ \text{m/s}$ before it slams into the shoulder.
Check Momenta $+8.0$ and $-8.0\ \text{kg m/s}$ add to zero ✓.
✏️ Example 5 — A bullet embeds in a block
A $10\ \text{g}$ bullet moving horizontally at $300\ \text{m/s}$ hits a $1.99\ \text{kg}$ wooden block at rest on a smooth table and stays inside it. (a) Find the velocity of the block just after the impact. (b) If instead the block had been sliding towards the bullet at $0.50\ \text{m/s}$, what would the final velocity be?
(a) $0.010\times300+1.99\times0=(0.010+1.99)\,v$, so $v=\dfrac{3.0}{2.0}=1.5\ \text{m/s}$ in the bullet's direction.
(b) Take the bullet's direction as $+$: $0.010\times300+1.99\times(-0.50)=2.0\,v$, so $v=\dfrac{3.0-0.995}{2.0}=1.0\ \text{m/s}$, still in the bullet's direction.
Note Bodies that stick together after colliding are said to undergo a perfectly inelastic collision; Chapter 5 shows that kinetic energy is lost in it, while momentum is still conserved.
Check In (a), momentum before $3.0\ \text{kg m/s}$; after, $2.0\times1.5=3.0\ \text{kg m/s}$ ✓.
✏️ Example 6 — A shell bursts into three pieces
A $4.0\ \text{kg}$ object at rest explodes into three pieces. A $1.0\ \text{kg}$ piece flies off due east at $12\ \text{m/s}$ and another $1.0\ \text{kg}$ piece due north at $12\ \text{m/s}$. Find the velocity of the third piece.
Set up $x$ east, $y$ north. Total momentum before: $\mathbf{0}$. The third piece has mass $4.0-2.0=2.0\ \text{kg}$ and velocity $(v_x,v_y)$.
x $1.0\times12+0+2.0\,v_x=0$, so $v_x=-6.0\ \text{m/s}$.
y $0+1.0\times12+2.0\,v_y=0$, so $v_y=-6.0\ \text{m/s}$.
Answer $v=\sqrt{6.0^2+6.0^2}=8.5\ \text{m/s}$, directed $45^\circ$ south of west — exactly opposite to the resultant momentum of the first two pieces.
Check First two pieces: momentum $12\sqrt2=17.0\ \text{kg m/s}$ north-east; third: $2.0\times8.49=17.0\ \text{kg m/s}$ south-west ✓.
Interactive 2 — Collision Cart Lab

Two carts on a smooth track collide. Choose the masses, the starting velocities (negative = moving left) and the bumper: soft clay makes them stick together; a spring makes them bounce apart. The lower-left graph shows the contact force on each cart during the collision — mirror images, by the third law, so their areas (impulses) are equal and opposite. The bars on the right compare the momenta before and after: the individual momenta change, the total does not.

2 kg3 m/s1 kg0 m/s
$v_A$ after—
$v_B$ after—
total momentum before / after—
impulse on A / on B—
⚠️ Common mistakes

1. Cancelling an action–reaction pair. The two forces of a pair act on different bodies, so they can never balance each other. Weight and normal force on a book balance, but they are not a third-law pair.

2. "The bigger body pushes harder." A truck and a mosquito in a collision exert equal and opposite forces on each other. The mosquito suffers more because its mass is tiny, so the same force gives it an enormous acceleration.

3. Losing the signs. Momentum is a vector. In head-on problems choose a positive direction and give velocities in the opposite direction a minus sign (Example 5(b)); in two dimensions conserve each component separately (Example 6).

4. Conserving momentum when an external force acts for a long time. A ball falling under gravity does not conserve its momentum. Conservation applies to an isolated system, or approximately during a very short interaction.

5. Conserving speed or kinetic energy instead. In a collision where bodies stick, momentum is conserved but kinetic energy is not (Chapter 5). Never add speeds without directions.

Exercises

Take $g=9.8\ \text{m/s}^2$. Surfaces are smooth (frictionless) unless stated otherwise.

I
Action and reaction.
  • 1 A girl stands still on a floor. Name the two forces on her and the reaction to each, saying which body each reaction acts on.
  • 2 Explain, using the third law, why a rifle recoils when it is fired.
  • 3 A swimmer pushes water backward with her hands and feet. Explain how this moves her forward.
  • 4 A $0.50\ \text{kg}$ ball falls freely towards the Earth. Find the force of the ball on the Earth and the resulting acceleration of the Earth ($M_E=5.97\times10^{24}\ \text{kg}$).
  • 5 Two spring balances are hooked together and pulled apart by two students. One balance reads $20\ \text{N}$. What does the other read? Why?
  • 6 "When a mosquito hits a moving lorry's windscreen, the lorry exerts a much larger force on the mosquito than the mosquito does on the lorry." Correct this statement.
1 Her weight (Earth on girl, down) — reaction: girl pulls the Earth up, acting on the Earth. The normal force (floor on girl, up) — reaction: girl presses down on the floor, acting on the floor.
2 The expanding gases push the bullet forward; the bullet (through the gases) pushes the rifle backward with an equal and opposite force, so the rifle moves back.
3 She exerts a backward force on the water; by the third law the water exerts an equal forward force on her, which accelerates her forward.
4 The Earth pulls the ball with $mg=4.9\ \text{N}$, so the ball pulls the Earth up with $4.9\ \text{N}$. $a_E=\dfrac{4.9}{5.97\times10^{24}}=8.2\times10^{-25}\ \text{m/s}^2$ — utterly unnoticeable.
5 $20\ \text{N}$: the hooks exert equal and opposite forces on each other (third law), and each light balance transmits the same pull along its length.
6 The forces are equal and opposite. The mosquito's tiny mass gives it a huge acceleration (and it is crushed); the lorry's huge mass gives it a negligible one.
II
Contact forces and pushes.
  • 1 Blocks of $3.0\ \text{kg}$ and $1.0\ \text{kg}$ are in contact on a smooth floor. A horizontal force of $8.0\ \text{N}$ is applied to the $3.0\ \text{kg}$ block. Find the acceleration and the contact force between them.
  • 2 Repeat part 1 with the force applied to the $1.0\ \text{kg}$ block from the other side.
  • 3 Three blocks of $1.0$, $2.0$ and $3.0\ \text{kg}$ are in contact in a row on a smooth floor. A $12\ \text{N}$ force pushes the $1.0\ \text{kg}$ block. Find the acceleration and both contact forces.
  • 4 A $60\ \text{kg}$ skater and a $40\ \text{kg}$ skater, at rest, push each other with a force of $120\ \text{N}$ for $0.50\ \text{s}$. Find each one's speed afterwards.
  • 5 An engine pulls three wagons, each of mass $1.0\times10^4\ \text{kg}$, with a force of $4.5\times10^4\ \text{N}$ on a level track (ignore friction). Find the acceleration and the force in the coupling between the first and second wagons and between the second and third.
  • 6 In part 5, what force does the second wagon exert on the first? Give magnitude and direction.
1 $a=\dfrac{8.0}{4.0}=2.0\ \text{m/s}^2$; contact force $=1.0\times2.0=2.0\ \text{N}$.
2 $a=2.0\ \text{m/s}^2$; contact force $=3.0\times2.0=6.0\ \text{N}$.
3 $a=\dfrac{12}{6.0}=2.0\ \text{m/s}^2$. Between $1.0$ and $2.0\ \text{kg}$: it pushes $5.0\ \text{kg}$, $C_1=10\ \text{N}$. Between $2.0$ and $3.0\ \text{kg}$: $C_2=3.0\times2.0=6.0\ \text{N}$.
4 Impulse on each $=120\times0.50=60\ \text{N s}$: $v_{60}=1.0\ \text{m/s}$, $v_{40}=1.5\ \text{m/s}$, in opposite directions.
5 $a=\dfrac{4.5\times10^4}{3.0\times10^4}=1.5\ \text{m/s}^2$. First–second coupling pulls two wagons: $2.0\times10^4\times1.5=3.0\times10^4\ \text{N}$; second–third pulls one: $1.5\times10^4\ \text{N}$.
6 The first wagon pulls the second forward with $3.0\times10^4\ \text{N}$, so the second pulls the first backward with $3.0\times10^4\ \text{N}$. (Check on the first wagon: $4.5\times10^4-3.0\times10^4=1.0\times10^4\times1.5$ ✓.)
III
Conservation of momentum in one dimension.
  • 1 A $5.0\ \text{kg}$ gun fires a $25\ \text{g}$ bullet at $300\ \text{m/s}$. Find the recoil velocity of the gun.
  • 2 A $20\ \text{g}$ bullet at $250\ \text{m/s}$ embeds itself in a $980\ \text{g}$ block at rest. Find their common velocity.
  • 3 A $2.0\ \text{kg}$ cart moving at $3.0\ \text{m/s}$ hits a $1.0\ \text{kg}$ cart at rest and they stick together. Find their velocity.
  • 4 A $1.5\ \text{kg}$ cart moving right at $4.0\ \text{m/s}$ meets a $2.5\ \text{kg}$ cart moving left at $2.0\ \text{m/s}$; they couple together. Find the velocity afterwards.
  • 5 A $60\ \text{kg}$ boy running at $4.0\ \text{m/s}$ jumps onto a $40\ \text{kg}$ trolley at rest. Find the speed of boy and trolley.
  • 6 A $50\ \text{kg}$ girl stands at rest in a $100\ \text{kg}$ boat on still water. She throws a $2.0\ \text{kg}$ ball horizontally at $15\ \text{m/s}$. Find the recoil speed of girl and boat (ignore water resistance).
  • 7 A bomb at rest explodes into two fragments whose masses are in the ratio $1:3$. The lighter fragment moves off at $24\ \text{m/s}$. Find the velocity of the heavier one.
  • 8 A $9.0\ \text{kg}$ shell moving horizontally at $20\ \text{m/s}$ bursts into two equal halves. One half comes to rest. Find the velocity of the other.
1 $V=-\dfrac{0.025\times300}{5.0}=-1.5\ \text{m/s}$: $1.5\ \text{m/s}$ opposite to the bullet.
2 $v=\dfrac{0.020\times250}{1.0}=5.0\ \text{m/s}$.
3 $v=\dfrac{2.0\times3.0}{3.0}=2.0\ \text{m/s}$ (the default of Interactive 2 with the clay bumper).
4 $1.5(4.0)+2.5(-2.0)=4.0\,v$: $v=\dfrac{6.0-5.0}{4.0}=0.25\ \text{m/s}$ to the right.
5 $v=\dfrac{60\times4.0}{100}=2.4\ \text{m/s}$.
6 $0=2.0\times15+150\,V$: $V=-0.20\ \text{m/s}$, i.e. $0.20\ \text{m/s}$ opposite to the throw.
7 $m(24)+3m\,v=0$: $v=-8.0\ \text{m/s}$, i.e. $8.0\ \text{m/s}$ opposite to the lighter fragment.
8 $9.0\times20=4.5\times0+4.5\,v$: $v=40\ \text{m/s}$ in the original direction.
IV
Two dimensions, rockets and harder problems.
  • 1 A $3.0\ \text{kg}$ body at rest explodes into three equal pieces. Two of them move off at $10\ \text{m/s}$, one along $+x$ and one along $+y$. Find the velocity of the third.
  • 2 A $60\ \text{kg}$ man stands at one end of a $120\ \text{kg}$ boat, $3.0\ \text{m}$ long, at rest on still water. He walks to the other end. How far does the boat move, and in which direction? (Ignore water resistance.)
  • 3 A rocket in deep space ejects $50\ \text{kg}$ of gas per second at $2000\ \text{m/s}$ relative to itself. Find the thrust. If the rocket's mass at that moment is $5000\ \text{kg}$, find its acceleration. What would its acceleration be at lift-off from the Earth, moving vertically?
  • 4 A $10\ \text{g}$ bullet passes through a $1.0\ \text{kg}$ block at rest, entering at $400\ \text{m/s}$ and leaving at $200\ \text{m/s}$. Find the velocity of the block.
  • 5 Two carts of $1.0\ \text{kg}$ and $3.0\ \text{kg}$ are held at rest with a compressed spring between them, then released. The lighter cart moves off at $6.0\ \text{m/s}$. Find the speed of the heavier one, and the impulse the spring gave each.
  • 6 (Modelling) An astronaut, $100\ \text{kg}$ with her suit, is floating at rest $20\ \text{m}$ from her space station with a broken jet pack. She throws a $2.0\ \text{kg}$ spanner directly away from the station at $10\ \text{m/s}$. How fast does she move, in which direction, and how long does she take to reach the station?
1 Each piece is $1.0\ \text{kg}$. $\mathbf{v}_3=-(10\,\hat{\mathbf{i}}+10\,\hat{\mathbf{j}})\ \text{m/s}$: speed $10\sqrt2=14\ \text{m/s}$, at $45^\circ$ below the $-x$ axis (i.e. $225^\circ$ from $+x$).
2 Total momentum stays zero: $60\,v_m=-120\,v_b$ at every instant, so the displacements satisfy $60\,d_m=120\,d_b$ in opposite directions, with $d_m+d_b=3.0\ \text{m}$. So $d_m=2.0\ \text{m}$ and the boat moves $1.0\ \text{m}$ in the direction opposite to his walk.
3 Thrust $=\dfrac{dm}{dt}\,v=50\times2000=1.0\times10^5\ \text{N}$; $a=\dfrac{1.0\times10^5}{5000}=20\ \text{m/s}^2$. At lift-off gravity also acts: $a=20-9.8=10.2\ \text{m/s}^2$ upward.
4 $0.010\times400=0.010\times200+1.0\,v$: $v=2.0\ \text{m/s}$.
5 $1.0\times6.0=3.0\,v$: $v=2.0\ \text{m/s}$ the other way. Impulse on each: $6.0\ \text{N s}$, in opposite directions.
6 $0=2.0\times10+100\,V$: $V=-0.20\ \text{m/s}$, i.e. $0.20\ \text{m/s}$ towards the station. Time $=\dfrac{20}{0.20}=100\ \text{s}$.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Using Newton's second and third laws, show that the total linear momentum of two bodies that collide is conserved when no external force acts on them.

Third law During the collision (time $\Delta t$) body $A$ exerts $\mathbf{F}_{BA}$ on $B$ and $B$ exerts $\mathbf{F}_{AB}=-\mathbf{F}_{BA}$ on $A$. [1]
Second law $\Delta\mathbf{p}_A=\mathbf{F}_{AB}\,\Delta t$ and $\Delta\mathbf{p}_B=\mathbf{F}_{BA}\,\Delta t$. [1]
Conclusion Adding, $\Delta\mathbf{p}_A+\Delta\mathbf{p}_B=\mathbf{0}$, so $m_A\mathbf{u}_A+m_B\mathbf{u}_B=m_A\mathbf{v}_A+m_B\mathbf{v}_B$. [1]
Exam tip State "no external force" explicitly, and note that the argument holds even if the forces vary, because the pair is equal at every instant.

E2. NEET A $5.0\ \text{kg}$ gun fires a $50\ \text{g}$ bullet with a speed of $200\ \text{m/s}$. The recoil speed of the gun is (a) $0.50\ \text{m/s}$ (b) $1.0\ \text{m/s}$ (c) $2.0\ \text{m/s}$ (d) $20\ \text{m/s}$

Idea Gun + bullet start at rest, so $MV=mv$.
Working $V=\dfrac{0.050\times200}{5.0}=2.0\ \text{m/s}$.
Answer (c).
Exam tip (d) comes from leaving the bullet's mass in grams. Convert to kilograms first.

E3. JEE Main NEET A body of mass $3m$ at rest explodes into three equal pieces. Two pieces fly off with equal speeds $v$ at right angles to each other. The speed of the third piece is (a) $\dfrac{v}{\sqrt2}$ (b) $v$ (c) $\sqrt2\,v$ (d) $2v$

Idea Total momentum is zero, so the third piece's momentum cancels the resultant of the other two.
Working The two perpendicular momenta $mv$ have resultant $\sqrt2\,mv$. The third piece (mass $m$) has momentum $\sqrt2\,mv$, so speed $\sqrt2\,v$.
Answer (c), directed opposite to the resultant of the first two ($135^\circ$ from each).
Exam tip (d) adds magnitudes as if the momenta were parallel — momentum adds as vectors.

E4. JEE Main (Numerical value) A $2.0\ \text{kg}$ block moving at $6.0\ \text{m/s}$ on a smooth track runs into a $4.0\ \text{kg}$ block moving at $3.0\ \text{m/s}$ in the same direction, and they stick together. Find their common speed (in m/s).

Momentum $2.0\times6.0+4.0\times3.0=6.0\,v$, so $v=\dfrac{24}{6.0}=4.0\ \text{m/s}$.
Answer $4$.
Exam tip The common speed must lie between the two initial speeds ($3.0$ and $6.0$) — a quick sanity check.

E5. JEE Advanced (One or more correct) Two skaters of masses $50\ \text{kg}$ and $75\ \text{kg}$ stand at rest on smooth ice and push each other apart. Afterwards the $50\ \text{kg}$ skater moves at $3.0\ \text{m/s}$. Which statements are correct? (A) The $75\ \text{kg}$ skater moves at $2.0\ \text{m/s}$ in the opposite direction. (B) During the push the force on the $50\ \text{kg}$ skater is larger than the force on the $75\ \text{kg}$ skater. (C) Each skater receives an impulse of magnitude $150\ \text{N s}$. (D) The total momentum of the two skaters is zero at every instant.

(A) $50\times3.0=75\,v$: $v=2.0\ \text{m/s}$, opposite ✓.
(B) Third law: the forces are equal and opposite at every instant ✗.
(C) $J=\Delta p=50\times3.0=150\ \text{N s}$ for each (opposite directions) ✓.
(D) No external horizontal force, and the total started at zero ✓ (so their balance point stays fixed, as Interactive 1 shows).
Answer (A), (C), (D).

E6. Board · Assertion–Reason Assertion (A): For a book resting on a table, the weight of the book and the normal force of the table on it form an action–reaction pair. Reason (R): The weight of the book and the normal force on it are equal in magnitude and opposite in direction. Choose: (a) A and R are true and R explains A (b) A and R are true but R does not explain A (c) A is true, R is false (d) A is false, R is true

A False: both forces act on the same body, and they are of different kinds. The reaction to the weight is the book's pull on the Earth.
R True for a book at rest — by the first law, since its acceleration is zero.
Answer (d).
Exam tip "Equal and opposite" is not enough for a third-law pair: the two forces must act on two different bodies.
🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. How do accident investigators work out how fast a car was going before a crash, from the marks it left afterwards?

Why it matters. Forensic engineers in police and insurance investigations around the world reconstruct collisions to decide responsibility and to design safer roads. The method combines two ideas from this chapter: momentum is conserved during the brief crash, and after it the locked wreckage skids to rest with a deceleration set by friction (Section 4.3 shows this deceleration is $\mu_kg$).

What to do.

  1. A $1500\ \text{kg}$ car runs into the back of a $1000\ \text{kg}$ car waiting at a red light. The cars lock together and skid $12\ \text{m}$ in a straight line. Take $\mu_k=0.70$ for tyres on dry asphalt. Find the speed of the wreck just after the crash from $v^2=2\mu_kg\,s$.
  2. Use conservation of momentum to find the speed of the first car just before the crash, in km/h. The speed limit was $50\ \text{km/h}$ — was it broken?
  3. The value of $\mu_k$ is uncertain: wet roads give about $0.45$. Recalculate, and state your answer as a range. Which input matters most?
  4. Search for how investigators measure $\mu_k$ at a crash site (a test skid or a drag sled), and summarise in three sentences.

Tools: Calculator or spreadsheet; optionally a toy-car collision on a ramp filmed with a phone.

Hand in: Your calculation with a clear free-body/momentum sketch, the speed range, and a one-paragraph report as an investigator would write it.

Going further. At a crossroads, a car travelling east collides with one travelling north and they skid together in a direction $30^\circ$ north of east. Show how conserving the $x$- and $y$-components separately gives both pre-crash speeds.

Instructor note: $v=\sqrt{2(0.70)(9.8)(12)}=12.8\ \text{m/s}$ after the crash; before, $v_1=\dfrac{2500}{1500}\times12.8=21.4\ \text{m/s}\approx77\ \text{km/h}$ — well over the limit. With $\mu_k=0.45$: $10.3\ \text{m/s}$ after and $17.1\ \text{m/s}\approx62\ \text{km/h}$ before, still over $50$. Since $v\propto\sqrt{\mu_k s}$, a $10\%$ error in $\mu_k$ gives only a $5\%$ error in speed; the mass ratio enters linearly.

🧪 Self-check quiz

0/5 correct

Q1. The action and reaction forces of Newton's third law

Q2. A $4.0\ \text{kg}$ rifle fires a $10\ \text{g}$ bullet at $400\ \text{m/s}$. The recoil speed of the rifle is

Q3. The total linear momentum of a system is conserved when

Q4. A $2.0\ \text{kg}$ cart moving at $3.0\ \text{m/s}$ hits a $1.0\ \text{kg}$ cart at rest and they stick. Their common speed is

Q5. A horse pulls a cart and the cart pulls back on the horse with an equal force. The cart nevertheless accelerates forward because

← 4.1 Inertia, Momentum and Newton's First and Second Laws4.3 Equilibrium of a Particle, Common Forces and Friction →