Momentum and Collisions

Beginner

Why is a slowly rolling shopping trolley full of bottles so hard to stop, while a fast ping-pong ball is easy to catch? Four steps explain it.

Step 1. Multiply an object's mass by its velocity and you get its momentum, measured in kilogram metres per second. Like velocity, momentum has a direction. A heavy, slow object and a light, fast one can carry the same amount: a 4 kg cart at 1 m/s has as much momentum as a 1 kg cart at 4 m/s.

Step 2. To change momentum you must push for some time, and the longer the push lasts, the gentler it can be. That is why you bend your knees when you jump down from a wall: the stop lasts longer, so the force on you is smaller.

Step 3. In a collision the two objects push on each other equally hard, in opposite directions, for exactly the same time. So the momentum one loses is precisely what the other gains, and the total stays the same, as long as nothing outside, such as friction or a wall, joins in. This is the conservation of momentum.

Step 4. The animation shows one crash twice: a light, fast cart hits a heavy cart at rest. With a springy bumper the light cart bounces back and the heavy one rolls away; with a sticky pad they roll on together, more slowly. The total momentum is the same either way. Only the energy of motion differs: the springy bounce keeps it, the sticky crash turns part of it into heat, sound and dents.

Check yourself: a firework shell at rest bursts into two equal halves. One half flies east. What does the other do? Answer: it flies west at the same speed, so the total momentum stays zero.

Intermediate

Momentum: mass times velocity

The momentum of a body is its mass multiplied by its velocity:
$p=mv$

It is measured in kilogram metres per second (kg·m/s). Momentum has a direction: pick a positive direction along the track, and anything moving the other way has negative momentum. Mass and speed can stand in for each other: the 1 kg cart in the animation, arriving at 4 m/s, carries $1\cdot 4=4\mathrm{kg}\cdot m/s$, as much as both carts together, 4 kg, would carry at 1 m/s.

Momentum is not kinetic energy. $K=\frac{1}{2}m{v}^2$ has no direction and grows with the square of the speed: the 1 kg cart at 4 m/s has 8 J, but the same 4 kg·m/s carried by 4 kg at 1 m/s is only 2 J.

Impulse and the force–time graph

A force changes momentum, and the change depends on how long it acts. For a steady force, Newton's second law $F=ma=m\Delta v/\Delta t$ becomes $F\Delta t=\Delta p$. The product $F\Delta t$ is the impulse $J$, in newton seconds (1 N·s = 1 kg·m/s); when the force varies, the impulse is the area under the force–time graph.

The same area can be tall and narrow or low and wide. A 70 kg person jumping down from 1 m lands at about 4.4 m/s, so the floor must remove about 310 kg·m/s. Landing stiff-legged stops the body in roughly 0.02 s, which takes an average net force near 15 kN, over 20 times body weight; bending the knees stretches the stop to 0.3 s and cuts the force to about 1 kN.

Why the total momentum is conserved

When two carts collide, each pushes on the other. By Newton's third law the two forces are equal and opposite at every instant, and they act for the same time, the duration of the contact. Their force–time graphs are mirror images, so the impulses are equal and opposite: whatever momentum cart 2 gains, cart 1 loses. The total is the same before and after:
$m_1{v}_1+m_2{v}_2=m_1{v}_1^{\prime }+m_2{v}_2^{\prime }$

Primes mark velocities after the collision, and every velocity keeps its sign. The law does not care whether the forces between the carts are springy, sticky or explosive. It needs only that no outside force acts along the track, or that outside forces are too weak or brief to matter: a rolling friction of 0.3 N acting during the 0.16 s contact changes the momentum by about 0.05 kg·m/s, about 1 % of the total.

Elastic and inelastic collisions

Momentum conservation is one equation, but a head-on collision has two unknown final velocities; the second condition concerns the kinetic energy. In an elastic collision it is conserved, as with steel balls or spring bumpers. In an inelastic collision part of it becomes heat, sound and deformation. In a perfectly inelastic collision the bodies stick and share one velocity.

For these two cases the final velocities have closed formulas; the elastic pair assumes that body 2 starts at rest.

• $v^{\prime }=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$
• $v_1^{\prime }=\frac{(m_1-m_2)v_1}{m_1+m_2},v_2^{\prime }=\frac{2m_1{v}_1}{m_1+m_2}$

The first needs only momentum conservation; the second uses both laws, and it implies that elastic bodies separate at the speed at which they approached. Equal masses swap velocities, as in Newton's cradle. A light body hitting a much heavier one bounces back at almost its full speed, like a ball off a wall; a heavy body hitting a light one hardly slows, and the light one flies off at nearly $2v_1$.

Worked example: the two tracks in the animation

Both tracks start alike: $m_1=1\mathrm{kg}$ arrives at $v_1=4m/s$ and $m_2=3\mathrm{kg}$ waits at rest. The total momentum is $1\cdot 4=4\mathrm{kg}\cdot m/s$ and the kinetic energy $\frac{1}{2}\cdot 1\cdot 4^2=8\mathrm{J}$.

On track B the sticky pad joins the carts, which then share one velocity:
$v^{\prime }=\frac{1\cdot 4+3\cdot 0}{1+3}=1m/s$
Their kinetic energy is now $\frac{1}{2}\cdot 4\cdot 1^2=2\mathrm{J}$: 6 J went into heat and squashing the pad. On track A the springy bumper makes the collision elastic:
$v_1^{\prime }=\frac{(1-3)\cdot 4}{1+3}=-2m/s$
$v_2^{\prime }=\frac{2\cdot 1\cdot 4}{1+3}=2m/s$
Check: $1\cdot (-2)+3\cdot 2=4\mathrm{kg}\cdot m/s$ and $\frac{1}{2}\cdot 1\cdot 2^2+\frac{1}{2}\cdot 3\cdot 2^2=8\mathrm{J}$. The carts approached at 4 m/s and separate at $2-(-2)=4m/s$.

Watch the arrows. The blue $p_1$ arrow starts 4 units long, one unit per kg·m/s; on track B it shrinks to 1, while on track A it shrinks through zero and returns pointing left, 2 units long. The gold $p_2$ arrow grows to 3 units on B and to 6 on A. Head to tail they always make 4 units, and both readouts show $p_1+p_2=4.0\mathrm{kg}\cdot m/s$ throughout, even during the contact.

The bounce takes twice the impulse: in the same 0.16 s, cart 2 gains 6 kg·m/s on A, an average force of $6/0.16=37.5\mathrm{N}$, but 3 kg·m/s on B, 18.75 N.

Worked example: a rear-end crash

A 1500 kg car at 20 m/s runs into the back of a 1000 kg car moving the same way at 10 m/s, and they lock together. During the 0.1 s crash road friction is negligible, so momentum is conserved:
$v^{\prime }=\frac{1500\cdot 20+1000\cdot 10}{1500+1000}=16m/s$

The kinetic energy falls from $\frac{1}{2}\cdot 1500\cdot 20^2+\frac{1}{2}\cdot 1000\cdot 10^2=350\mathrm{kJ}$ to $\frac{1}{2}\cdot 2500\cdot 16^2=320\mathrm{kJ}$: 30 kJ goes into crumpled metal.

Now let the small car come the other way, at $v_2=-10m/s$. The wreck moves at $(1500\cdot 20-1000\cdot 10)/2500=8m/s$ and keeps only $\frac{1}{2}\cdot 2500\cdot 8^2=80\mathrm{kJ}$: 270 kJ is lost, nine times as much, because the cars close at 30 m/s instead of 10 m/s. Both feel the same average force of 180 kN, yet the small car's velocity changes by 18 m/s in 0.1 s, a deceleration of about 18 g, and the big car's by only 12 m/s.

Common mistakes

• Forgetting the sign. On track A the small cart rebounds at $-2m/s$; counted as positive, the momenta would add up to $2+6=8\mathrm{kg}\cdot m/s$ instead of 4.
• Believing that a sticky collision destroys momentum. The joined carts on track B still carry 4 kg·m/s; what disappears is kinetic energy, 6 of the 8 J.
• Thinking that the heavier or faster body pushes harder. The forces are equal and opposite, 37.5 N on each cart on track A; the lighter cart just changes its velocity more, by 6 m/s against 2 m/s.
• Ignoring an outside force that keeps acting. A 0.5 kg ball falling for 1 s gains about 4.9 kg·m/s because gravity pulls it; only the ball and the Earth together keep a constant total. Keep both colliding bodies in the system.
• Using the elastic formulas for every collision. They assume no loss of kinetic energy and a target at rest; on track B they would predict $-2$ and $+2m/s$ instead of the shared 1 m/s.

Where you meet it

Airbags, helmets and crumple zones lengthen collisions so that the same change of momentum needs a smaller force. Guns recoil: a 4 kg rifle firing a 10 g bullet at 800 m/s kicks back at $0.01\cdot 800/4=2m/s$. Rockets throw exhaust backwards to move forwards, with no air to push against. Next, Work, Energy and Power follows the kinetic energy that collisions keep or lose, and Angular Momentum and Its Conservation applies the same idea to spinning bodies.

Advanced

The second law in momentum form

This section rewrites Newton's second law in terms of momentum. That is the form that generalises: it handles impacts whose force nobody can measure, systems of many bodies and, later in the lesson, rockets.

The general form of the second law is not $F=ma$ but $\vec{F}=d\vec{p}/dt$, with $\vec{p}=m\vec{v}$: the net force on a body equals the rate at which its momentum changes. For a constant mass the two forms agree, because the mass can be taken outside the derivative: $d(m\vec{v})/dt=md\vec{v}/dt=m\vec{a}$. The momentum form is also the one that survives beyond Newtonian mechanics: in special relativity $\gamma m\vec{v}$ replaces $m\vec{v}$, and in electromagnetism the fields themselves carry momentum.

Multiply both sides by $dt$ and add up over the whole contact, from the instant it begins to the instant it ends. The left side becomes the impulse $\vec{J}$ and the right side the total change of momentum. This is the impulse–momentum theorem:
$\vec{J}=\int \vec{F}dt=\Delta \vec{p}$

Its value lies in what it does not need. The force during an impact rises and falls within a fraction of a second and is rarely known in detail, yet only its integral decides the motion afterwards. A 46 g golf ball leaves the club at 70 m/s after about 0.5 ms of contact. Its momentum changes by $0.046\cdot 70\approx 3.2\mathrm{kg}\cdot m/s$, so the average force is $0.046\cdot 70/0.0005\approx 6400\mathrm{N}$, whatever the shape of the pulse. Because the theorem is a vector equation, it holds for each component separately: a horizontal impulse changes only $p_x$ and leaves $p_y$ alone.

Internal forces cancel in pairs

Why does the total momentum of two colliding carts stay the same, however hard they push on each other? Because every push comes with an equal and opposite push back. In the animation a 1 kg cart 1 arrives at 4 m/s and hits a 3 kg cart 2 at rest; on track A they meet through a springy bumper, on track B through a sticky pad. Figure 1 shows track A during the contact; the general proof follows it.

Legend

$t$	Time, the same axis for all three graphs.	$\mathrm{s}$
$\Delta t$	Duration of the contact.	$\mathrm{s}$
$F$	Force of the bumper on cart 2; cart 1 feels $-F$. Constant here; it is the slope of the $p_2$ ramp.	$\mathrm{N}$
$p$	Momentum along the track: the vertical axis of the middle graph.	$\mathrm{kg}\cdot m/s$
$p_1$	Momentum of cart 1: from 4 down to $-2$ during the contact.	$\mathrm{kg}\cdot m/s$
$p_2$	Momentum of cart 2: from 0 up to 6 during the contact.	$\mathrm{kg}\cdot m/s$
$P$	Total momentum $p_1+p_2$ (dashed): flat at 4 throughout.	$\mathrm{kg}\cdot m/s$
$K$	Total kinetic energy of the two carts.	$\mathrm{J}$
$A$	Track A, springy bumper: $K$ dips to 2 J mid-contact, then returns to 8 J.
$B$	Track B, sticky pad (dashed): $K$ falls to 2 J and stays there.

Figure 1 follows the contact, which lasts $\Delta t$; the time axis $t$ is stretched so that the contact fills the middle of the picture. Read it from top to bottom. The top strip is the force: the bumper pushes cart 2 forward with $F$ and cart 1 backward with $-F$. By the third law the two are equal and opposite at every instant. In this figure the force is also constant while the contact lasts.

The middle graph shows the momenta. Because $dp/dt$ equals the force, the slope of each momentum graph is the force on that cart, and a constant force draws a straight ramp. $p_2$ climbs with slope $F$ from 0 to 6, and $p_1$ falls with slope $-F$ from 4 to $-2\mathrm{kg}\cdot m/s$: cart 1 bounces back. Cart 2 gains 6 kg·m/s in $\Delta t=0.16\mathrm{s}$, so $F=6/0.16=37.5\mathrm{N}$. Now look at the dashed line $P=p_1+p_2$. The two ramps are mirror images, so whatever one cart gains the other loses at the same moment, and $P$ stays at 4 kg·m/s through the whole collision, not only before and after it. The expert layer of the animation plots the same curves.

The flat line does not depend on the force being constant. A real spring pushes harder the more it is squeezed, which bends the ramps into S-shaped curves; they are still mirror images, and their sum is still flat.

The same cancellation works for any number of bodies. The force on body $i$ is the sum of the forces ${\vec{F}}_{ij}$ exerted on it by each other body $j$ of the system (internal forces) and of whatever acts from outside (external forces). Add up $d{\vec{p}}_i/dt$ over all bodies. Every internal pair appears twice in the sum, once as ${\vec{F}}_{ij}$ and once as ${\vec{F}}_{ji}$, and the third law makes ${\vec{F}}_{ij}+{\vec{F}}_{ji}=0$. Only the external forces survive. With $\vec{P}$ the total momentum, $M$ the total mass, ${\vec{v}}_{\mathrm{cm}}$ the velocity of the centre of mass and ${\vec{F}}_{\mathrm{ext}}$ the sum of all external forces:
$\frac{d\vec{P}}{dt}={\vec{F}}_{\mathrm{ext}}$
$\vec{P}={\sum }_i{m}_i{\vec{v}}_i=M{\vec{v}}_{\mathrm{cm}}$
The second line is the definition of the centre of mass, ${\vec{r}}_{\mathrm{cm}}={\sum }_i{m}_i{\vec{r}}_i/M$, differentiated once.

So the total momentum changes only through external forces, and the centre of mass moves like a single particle of mass $M$ pushed by their sum. If ${\vec{F}}_{\mathrm{ext}}=0$, $\vec{P}$ is constant, however complicated the internal forces are. If only one component of ${\vec{F}}_{\mathrm{ext}}$ is zero, for instance the horizontal one when gravity acts vertically, that component of $\vec{P}$ is still conserved.

Kinetic energy can hide, momentum cannot

The bottom graph of Figure 1 shows the total kinetic energy $K$ of the two carts, and it behaves quite differently from $P$. On track A, $K$ dips during the contact, down to 2 J at the instant both carts move at 1 m/s, then climbs back to 8 J as the bumper pushes them apart. The missing 6 J was not lost: for a moment it was stored in the squeezed spring of the bumper. On track B, with the sticky pad, $K$ falls to 2 J and stays there (dashed): the 6 J has gone into deforming and warming the pad.

Track B also needs less force. The pair leaves together at 1 m/s, so cart 2 gains only $3\cdot 1=3\mathrm{kg}\cdot m/s$, and over the same 0.16 s the average force is $3/0.16=18.75\mathrm{N}$, half that of track A. The two graphs together carry the main message: energy can be parked inside the bodies, as spring energy or as heat, but momentum has nowhere to hide, and it balances at every instant.

Conservation from symmetry

Momentum conservation can also be derived without the third law, from two statements about symmetry. This matters because the third law is not exact (see Limits of validity below), while momentum conservation has never been seen to fail; the symmetry argument shows where the law really comes from.

Legend

$S$	Frame of the ground (top row).
$S^{\prime }$	Frame of a car moving with the right-hand ball, at $-v$ relative to the ground (bottom row).
$m$	Mass of each of the two identical balls; stuck together they have $2m$.	$\mathrm{kg}$
$v$	Speed of each ball in $S$; $+v$ is the shift of every velocity from $S$ to $S^{\prime }$. All arrows share one scale, so $2v$ is twice as long as $v$.	$m/s$

Figure 2 shows one collision seen from two frames of reference, before it on the left and after it on the right. The first assumption is mirror symmetry. In the top row, in the frame $S$ of the ground, two identical balls of mass $m$ approach each other at $v$ and $-v$ and stick. Nothing in this situation prefers left to right, so the stuck pair, of mass $2m$, cannot move either way: it stays at rest.

The second assumption is Galilean relativity: an observer moving at constant velocity finds the same laws of physics, with every velocity shifted by the same amount. Take as $S^{\prime }$ the frame of a car keeping pace with the right-hand ball, which moves at $-v$ relative to the ground. Seen from that car, every velocity is larger by $v$, as the arrow $+v$ between the rows says. The bottom row follows: the right ball is at rest, the left ball arrives at $2v$, and the pair, which stood still in $S$, leaves at $v$.

So a mass $m$ at $2v$ striking an equal mass at rest gives $2m$ at $v$, and the bottom row balances: $m\cdot 2v=2m\cdot v$. Speeds alone do not balance ($2v$ in, $v$ out), and neither does kinetic energy ($2m{v}^2$ in, $m{v}^2$ out); the product of mass and velocity does. Building bodies of other masses from equal units and repeating the same change of frame extends the result to any pair of masses, and the quantity that always comes out the same is $\sum mv$.

The argument also works backwards. Any two-body collision, watched from the frame that moves with the centre of mass, has total momentum zero: the bodies arrive with equal and opposite momenta, like the top row. For the animation's carts that frame moves at 1 m/s. In it cart 1 brings $+3$ and cart 2 brings $-3\mathrm{kg}\cdot m/s$, and on track B the stuck pair simply stops.

The general statement is Noether's theorem (1918): every continuous symmetry of the laws of physics comes with a conserved quantity. If the laws are the same everywhere, so that moving the whole apparatus to another place changes nothing (space is homogeneous), momentum is conserved. Invariance under shifts in time gives conservation of energy, and invariance under rotations conservation of angular momentum. Because it rests on a symmetry rather than on the third law, momentum conservation holds even where the third law fails.

Energy in the centre-of-mass frame

How much kinetic energy can a collision remove, and what are the final velocities? Both questions become simple in the frame that moves with the centre of mass. This section splits the kinetic energy into a part that no collision can touch and a part that it can, and finds the final velocities without solving a quadratic equation.

Legend

$v$	Velocity: positions on the axis are velocities.	$m/s$
$V$	Velocity of the centre of mass (dashed line), written $v_{\mathrm{cm}}$ in the equations.	$m/s$
$m_1$	Mass of cart 1 (one block wide).	$\mathrm{kg}$
$m_2$	Mass of cart 2 (three blocks wide).	$\mathrm{kg}$
$v_1$	Velocity of cart 1 before the collision.	$m/s$
$v_2$	Velocity of cart 2 before the collision.	$m/s$
$u_1$	Velocity of cart 1 relative to the centre of mass, $v_1-V$.	$m/s$
$u_2$	Velocity of cart 2 relative to the centre of mass, $v_2-V$.	$m/s$
$v_1^{\prime }$	Velocity of cart 1 after the collision.	$m/s$
$v_2^{\prime }$	Velocity of cart 2 after the collision.	$m/s$
$u_1^{\prime }$	Velocity of cart 1 relative to the centre of mass after the collision: $-e{u}_1$, here $-u_1$.	$m/s$
$u_2^{\prime }$	Velocity of cart 2 relative to the centre of mass after the collision: $-e{u}_2$, here $-u_2$.	$m/s$

Figure 3 places the velocities of the animation's carts on one axis $v$: cart 1, $m_1=1\mathrm{kg}$, at $v_1=4m/s$, and cart 2, $m_2=3\mathrm{kg}$, at $v_2=0$; the width of each block shows its mass. The dashed line marks the velocity of the centre of mass, called $V$ in the figure and $v_{\mathrm{cm}}$ in the equations:
$v_{\mathrm{cm}}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}=\frac{1\cdot 4+3\cdot 0}{4}=1m/s$

Measured from the dashed line, the carts move at $u_1=v_1-v_{\mathrm{cm}}=+3m/s$ and $u_2=v_2-v_{\mathrm{cm}}=-1m/s$: the two arrows of the top row. In this frame the total momentum is zero, $m_1{u}_1+m_2{u}_2=1\cdot 3+3\cdot (-1)=0$, so the arrows are inversely proportional to the masses, and the heavy cart sits three times closer to $V$ than the light one.

A collision cannot move the dashed line, because the internal forces leave the total momentum, and with it $v_{\mathrm{cm}}$, unchanged. It can only change $u_1$ and $u_2$, and afterwards they must still satisfy $m_1{u}_1^{\prime }+m_2{u}_2^{\prime }=0$. Both therefore change by the same factor. The carts separate instead of passing through each other, so the factor is negative or zero, and we write it as $-e$:
$u_i^{\prime }=-e{u}_i$
The number $e$ is the coefficient of restitution. Since $u_1-u_2=v_1-v_2=v_{\mathrm{rel}}$, the relative velocity, it can also be read as $e=-v_{\mathrm{rel}}^{\prime }/v_{\mathrm{rel}}$: the fraction of the relative motion that survives the collision, from 1 for an elastic collision to 0 for a sticky one.

The lower row of Figure 3 is the elastic case, $e=1$: $u_1^{\prime }$ and $u_2^{\prime }$ are the old arrows turned around, so the final velocities $v_1^{\prime }$ and $v_2^{\prime }$ are the mirror images of $v_1$ and $v_2$ in the dashed line. To return to the frame of the track, add $v_{\mathrm{cm}}$ back:
$v_i^{\prime }=v_{\mathrm{cm}}+u_i^{\prime }=v_{\mathrm{cm}}-e(v_i-v_{\mathrm{cm}})$
$v_i^{\prime }=(1+e)v_{\mathrm{cm}}-e{v}_i$
For $e=1$ this gives $v_1^{\prime }=2\cdot 1-4=-2m/s$ and $v_2^{\prime }=2\cdot 1-0=+2m/s$, the result on track A. A smaller $e$ shortens both arrows by the same factor; at $e=0$ they vanish, and both carts leave together at $V$, as on track B.

Now the kinetic energy. Substitute $v_i=v_{\mathrm{cm}}+u_i$ into $K={\sum }_i\frac{1}{2}{m}_i{v}_i^2$ and expand the square:
$K=\frac{1}{2}M{v}_{\mathrm{cm}}^2+v_{\mathrm{cm}}{\sum }_i{m}_i{u}_i+{\sum }_i\frac{1}{2}{m}_i{u}_i^2$
The middle term is zero, because ${\sum }_i{m}_i{u}_i=0$. What remains is
$K=\frac{1}{2}M{v}_{\mathrm{cm}}^2+K_{\mathrm{rel}}$
with $M=m_1+m_2$. The first term is the energy of the centre of mass, and no internal force can change it, because the momentum fixes $v_{\mathrm{cm}}$. The second, $K_{\mathrm{rel}}$, is the energy of motion relative to the centre of mass, and a collision acts only on it. For two bodies it depends only on the relative velocity. From $m_1{u}_1+m_2{u}_2=0$ and $u_1-u_2=v_{\mathrm{rel}}$ follow $u_1=m_2{v}_{\mathrm{rel}}/M$ and $u_2=-m_1{v}_{\mathrm{rel}}/M$; then $\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{m}_2(m_2+m_1)v_{\mathrm{rel}}^2/M^2$, which simplifies to
$K_{\mathrm{rel}}=\frac{1}{2}\mu v_{\mathrm{rel}}^2,\mu =\frac{m_1{m}_2}{m_1+m_2}$
where $\mu$ is called the reduced mass.

Since $K_{\mathrm{rel}}$ is proportional to $v_{\mathrm{rel}}^2$, a collision multiplies it by $e^2$. For the animation $M=4\mathrm{kg}$, $v_{\mathrm{cm}}=1m/s$, $\mu =0.75\mathrm{kg}$ and $v_{\mathrm{rel}}=4m/s$. The 8 J therefore splits into $\frac{1}{2}\cdot 4\cdot 1^2=2\mathrm{J}$ that no collision can touch and $\frac{1}{2}\cdot 0.75\cdot 4^2=6\mathrm{J}$ of relative motion. An elastic bumper gives all 6 J back. Sticking ($e=0$) removes all 6 J, the largest loss that momentum conservation allows, and the 2 J left is what track B keeps.

Worked problem: a half-elastic bumper

Fit the carts with a bumper of $e=0.5$: same masses, and cart 1 again arrives at 4 m/s on cart 2 at rest. Find (a) the final velocities, (b) the kinetic energy lost, (c) the average force if the contact still lasts 0.16 s, and (d) the value of $e$ that leaves cart 1 at rest.

(a) The centre of mass still moves at $v_{\mathrm{cm}}=1m/s$, and $1+e=1.5$:
$v_1^{\prime }=(1+e)v_{\mathrm{cm}}-e{v}_1=1.5\cdot 1-0.5\cdot 4=-0.5m/s$
$v_2^{\prime }=(1+e)v_{\mathrm{cm}}-e{v}_2=1.5\cdot 1-0=1.5m/s$
Cart 1 bounces back slowly. Momentum checks: $1\cdot (-0.5)+3\cdot 1.5=4\mathrm{kg}\cdot m/s$, as before.

(b) The 2 J of the centre of mass stays, and the 6 J of relative motion is multiplied by $e^2=0.25$: $K=2+0.25\cdot 6=3.5\mathrm{J}$, so $8-3.5=4.5\mathrm{J}$ is lost. Directly: $\frac{1}{2}\cdot 1\cdot 0.5^2+\frac{1}{2}\cdot 3\cdot 1.5^2=0.125+3.375=3.5\mathrm{J}$.

(c) Cart 2 gains $3\cdot 1.5=4.5\mathrm{kg}\cdot m/s$. With $v_2=0$ the formula gives $v_2^{\prime }=(1+e)v_{\mathrm{cm}}$, so this is $(1+e)$ times the 3 kg·m/s of the sticky case. Over 0.16 s it takes an average force of $4.5/0.16\approx 28\mathrm{N}$, between the 18.75 N of track B ($e=0$) and the 37.5 N of track A ($e=1$).

(d) Cart 1 stops when $v_1^{\prime }=0$, that is when $(1+e)v_{\mathrm{cm}}=e{v}_1$, or $1+e=4e$, so $e=1/3$. Cart 2 then carries all 4 kg·m/s, at $4/3\approx 1.33m/s$, and $K=2+6/9\approx 2.67\mathrm{J}$. In general, with the target at rest, the striker stops when $e=m_1/m_2$. A striker that stops dead is therefore the sign of an elastic collision only between equal masses; here it takes a rather lossy bumper.

Glancing collisions

In two or three dimensions momentum conservation is one equation per component. For two bodies in a plane, momentum gives two equations and energy, if the collision is elastic, a third, but the two final velocities have four components. The missing piece of information is the geometry of the impact, and a collision that is not head-on has to settle directions as well as speeds.

Legend

$m$	Mass of each of the two identical balls.	$\mathrm{kg}$
$b$	Offset: distance between the path of ball 1 and the parallel line through the centre of ball 2.	$\mathrm{m}$
${\vec{v}}_1$	Velocity of ball 1 before the collision; ball 2 is at rest.	$m/s$
${\vec{v}}_1^{\prime }$	Velocity of ball 1 after the collision, across the line of centres.	$m/s$
${\vec{v}}_2^{\prime }$	Velocity of ball 2 after the collision, along the line of centres.	$m/s$

Figure 4 shows an off-centre elastic collision of two identical balls of mass $m$, ball 2 at rest. Ball 1 comes in with velocity ${\vec{v}}_1$ along a path that misses the centre of ball 2 by the offset $b$. At the moment of contact the line joining the two centres is therefore tilted away from the path. For smooth balls without spin, the contact force acts along this line of centres, and nothing acts across it.

Split each velocity into a component along the line of centres and one across it. Across the line there is no force, so those components do not change. Along the line the collision is a head-on elastic collision of equal masses, which swaps the two velocities: ball 1 hands its whole component along the line to ball 2. So ball 2 leaves along the line of centres with ${\vec{v}}_2^{\prime }$, and ball 1 keeps only its component across the line, ${\vec{v}}_1^{\prime }$, which is perpendicular to ${\vec{v}}_2^{\prime }$. If the balls have radius $R$, the line of centres makes an angle $\phi$ with the incoming path, with $\sin \phi =b/(2R)$, and the speeds afterwards are $v_2^{\prime }=v_1\cos \phi$ and $v_1^{\prime }=v_1\sin \phi$. A head-on hit ($b=0$) passes everything to ball 2; a graze ($b$ close to $2R$) leaves ball 1 almost undisturbed.

The triangle on the right reaches the right angle without any geometry of the contact. Momentum conservation, with the equal masses cancelled, gives ${\vec{v}}_1={\vec{v}}_1^{\prime }+{\vec{v}}_2^{\prime }$: the three velocities close into a triangle. Energy conservation gives $v_1^2=v_1^{\prime 2}+v_2^{\prime 2}$. Squaring the first equation gives $v_1^2=v_1^{\prime 2}+v_2^{\prime 2}+2{\vec{v}}_1^{\prime }\cdot {\vec{v}}_2^{\prime }$, and comparing it with the second leaves ${\vec{v}}_1^{\prime }\cdot {\vec{v}}_2^{\prime }=0$. The two legs meet at a right angle: after a glancing elastic collision of equal masses, one of them at rest, the balls leave at right angles to each other (in a head-on hit ball 1 simply stops). A snooker ball struck without spin nearly does this, and so do the tracks of slow protons striking protons at rest.

Rockets and variable mass

A rocket changes its own mass as it burns fuel, so $\vec{F}=m\vec{a}$ cannot be applied to it directly. The safe route is momentum conservation for the rocket together with the exhaust it throws out.

Take a rocket of mass $m$ moving at $v$, far from other forces, that ejects exhaust backwards at speed $w$ relative to itself. In a short time $dt$ its mass changes by $dm$ (a negative amount), its velocity by $dv$, and a mass $-dm$ of exhaust leaves at velocity $v-w$. Momentum before and after must agree:
$mv=(m+dm)(v+dv)+(-dm)(v-w)$
Multiplying out, $mv$ appears on both sides, the two terms $vdm$ cancel, and the product $dmdv$ of two small changes is negligible. What remains is $mdv=-wdm$. Dividing by $m$ and integrating from the initial mass $m_0$ to the final mass $m_f$ gives
$\Delta v=w\ln \frac{m_0}{m_f}$

Only the exhaust speed and the mass ratio matter, and the mass ratio enters only through a logarithm. With $w=3km/s$ and nine tenths of the launch mass as propellant, $m_0/m_f=10$ and $\Delta v=3\ln 10\approx 6.9km/s$. That is short of the 7.8 km/s of a low orbit even before the losses to gravity and air drag. Dropping empty tanks along the way lowers the mass that is left to accelerate, which is why rockets are built in stages.

Limits of validity

Newtonian momentum $mv$ is an approximation, valid for speeds far below the speed of light, $c\approx 3.00\times 10^{8}m/s$. The quantity that is really conserved is $\gamma mv$, with $\gamma =1/\sqrt{1-v^2/c^2}$. At $0.6c$, $\gamma =1.25$, so the true momentum is already 25 % larger than $mv$, and in particle collisions kinetic energy can even become the mass of new particles. For the carts at 4 m/s the correction is of order $v^2/c^2\approx 2\times 10^{-16}$.

The third law is not exact either. Two charges moving at right angles to each other exert magnetic forces on each other that are not equal and opposite, so their combined mechanical momentum changes. The total balances only when the momentum stored in the electromagnetic field is counted as well. Light itself carries momentum, $p=E/c$ for light of energy $E$. Sunlight above the atmosphere delivers 1361 W/m²; on a sheet that absorbs it, this gives a pressure of $1361/(3.00\times 10^8)\approx 4.5\times 10^{-6}N/{\mathrm{m}}^2$, about 4.5 µN on each square metre, enough to steer a solar sail.

In quantum mechanics momentum is tied to wavelength, $p=h/\lambda$, and its conservation has held in every interaction ever observed. It has even predicted particles. In beta decay the emitted electron seemed to break the conservation of energy, because its energy varied from one decay to the next. In 1930 Pauli proposed that an unseen neutral particle carries off the difference. Later measurements of the recoiling nucleus showed that it carries off momentum too: the electron and the nucleus did not fly apart with equal and opposite momenta. The neutrino was detected in 1956.

Where this leads

Invariance under rotations leads to Angular Momentum and Its Conservation, and the rotational form of the second law to Rotational Dynamics and Torque. The energy side of collisions is in Work, Energy and Power. Gas pressure is the momentum that countless molecular impacts deliver to each square metre of the walls each second, as Kinetic Theory of Gases shows, and $\gamma mv$, with the conversion of energy into mass, belongs to Special Relativity. In Particles and Forces: the Standard Model, the balance of momentum and energy reveals particles that leave no track.