Collisions in One and Two Dimensions

Two cars meet at an intersection, crumple, and slide away together as one mangled object. You do not know the force between them. You do not know how long the crash lasted, or how the bumpers deformed, or where the heat went. And yet, if you know the two masses and the two velocities going in, you can tell someone exactly how fast the wreck slides and in which direction.

That is the strange gift of momentum. In the previous lesson you found that impulse equals change in momentum, and that when two objects push only on each other, their pushes are equal and opposite at every instant, so the total momentum of the pair never changes. A collision is the extreme case: enormous forces, tiny duration, complicated internal details, and none of it matters. This lesson is about turning that fact into answers.

What survives a collision, and what does not

During a collision, the two bodies exert forces on each other. By Newton's third law those forces are equal and opposite at every moment, so the impulses they deliver are equal and opposite too. Whatever momentum one body gains, the other loses. Add them and the total does not move:

$m_1\vec v_{1i} + m_2\vec v_{2i} = m_1\vec v_{1f} + m_2\vec v_{2f}$

Here $m$ is mass, $\vec v_i$ is a velocity just before the collision and $\vec v_f$ just after. The arrows matter: this is a vector equation, and in a plane it is really two equations, one for the $x$ components and one for the $y$ components.

Outside forces like gravity or friction do act during the crash, but a collision lasts milliseconds and those forces are small compared to the bang. Their impulse is negligible, so you treat total momentum as exactly conserved across the collision itself. Before and after, gravity is back in charge.

Kinetic energy gets no such protection. Bending metal, heating rubber, and making noise all cost kinetic energy, and nothing in the laws says you get it back. So kinetic energy is a quantity you check, not a quantity you assume.

Three kinds of collision

Sort a collision before you solve it, because the sorting decides your equations.

Perfectly inelastic. The bodies leave with one common velocity: they stick, embed, couple, or crumple into a single lump. Momentum alone fixes the answer. This is the most energy a collision can lose while still conserving momentum.

Elastic. Kinetic energy is conserved as well. Steel balls, billiard balls, atoms, subatomic particles.

Partially inelastic. Everything between. You need a third piece of information, almost always the coefficient of restitution $e$, defined as the separation speed divided by the approach speed:

$e = \frac{|v_{2f} - v_{1f}|}{|v_{1i} - v_{2i}|}$

$e = 1$ is elastic, $e = 0$ is perfectly inelastic, and a dropped tennis ball is about $0.75$.

Here is a useful way to see why the perfectly inelastic case loses the most. Split the kinetic energy into the energy of the whole system drifting along plus the energy of the two bodies moving relative to each other. The drifting part is locked to the total momentum, which cannot change, so it cannot be lost. Only the relative-motion part is available to be destroyed. Sticking together kills all of it.

Worked example 1: the cost of sticking

Take that same collision: $m_1 = 2.0\ \mathrm{kg}$ at $u = 6.0\ \mathrm{m/s}$ into $m_2 = 4.0\ \mathrm{kg}$ at rest. How much kinetic energy disappears?

Momentum first. $p = (2.0)(6.0) = 12\ \mathrm{kg\,m/s}$, and after the collision the whole $6.0\ \mathrm{kg}$ moves at $v$, so $6.0v = 12$ and $v = 2.0\ \mathrm{m/s}$.

Now the energies. Before:

$K_i = \tfrac12 (2.0)(6.0)^2 = \tfrac12 (2.0)(36) = 36\ \mathrm{J}$

After:

$K_f = \tfrac12 (6.0)(2.0)^2 = \tfrac12 (6.0)(4.0) = 12\ \mathrm{J}$

So $36 - 12 = 24\ \mathrm{J}$ went into crumpling and heat. Two thirds of the kinetic energy gone, and momentum never flinched. That gap between the two bookkeepings is the whole point of the topic.

The same machinery handles a bullet in a block. A $20\ \mathrm{g}$ bullet at $300\ \mathrm{m/s}$ into a $1.98\ \mathrm{kg}$ block gives $(0.020)(300) = 6.0\ \mathrm{kg\,m/s}$, shared by $2.00\ \mathrm{kg}$, so the block leaves at $3.0\ \mathrm{m/s}$. If the block then swings up on a string, switch bookkeeping: the collision is over, energy is now conserved, and $\tfrac12 v^2 = gh$ gives the rise. Momentum through the bang, energy after it. Never both at once.

Head-on elastic collisions without the quadratic

For an elastic collision in one dimension you have two statements: momentum,

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

and energy,

$\tfrac12 m_1u_1^2 + \tfrac12 m_2u_2^2 = \tfrac12 m_1v_1^2 + \tfrac12 m_2v_2^2$

where $u$ means before and $v$ means after. Solving a quadratic simultaneously with a linear equation is slow and invites sign errors. There is a much better route.

Move all the $m_1$ terms to one side of each equation. Momentum gives $m_1(u_1 - v_1) = m_2(v_2 - u_2)$. Energy, after cancelling the halves and factoring each difference of squares, gives $m_1(u_1 - v_1)(u_1 + v_1) = m_2(v_2 - u_2)(v_2 + u_2)$. Divide the second by the first (legal as long as the collision actually changed something) and everything with a mass in it cancels:

$u_1 + v_1 = v_2 + u_2 \quad\Longrightarrow\quad u_1 - u_2 = -(v_1 - v_2)$

In words: in an elastic collision the relative velocity reverses. They separate exactly as fast as they approached. That is the energy equation in linear form, and it is what $e = 1$ means.

Pair it with momentum for the common case $u_2 = 0$, and the algebra is two lines:

$v_1 = \frac{m_1 - m_2}{m_1 + m_2}u_1, \qquad v_2 = \frac{2m_1}{m_1 + m_2}u_1$

Read the limits as behaviour, not formulas. Equal masses: $v_1 = 0$ and $v_2 = u_1$, so the incoming body stops dead and hands over everything. Light hits heavy ($m_1 \ll m_2$): $v_1 \to -u_1$, a bounce off a wall, with the heavy target barely stirring. Heavy hits light ($m_1 \gg m_2$): $v_1 \to u_1$ and $v_2 \to 2u_1$, the plough goes through and the light body leaves at twice the speed of the thing that hit it.

Worked example 2: how much energy a neutron loses

A neutron of mass $1.0\ \mathrm{u}$ hits a stationary carbon nucleus of mass $12\ \mathrm{u}$ head-on and elastically. What fraction of its kinetic energy does it lose? This is the real question behind a nuclear reactor's moderator: you want to slow neutrons down fast.

Start with the formula. $v_1 = \dfrac{1 - 12}{1 + 12}u_1 = -\dfrac{11}{13}u_1$.

Kinetic energy goes as speed squared, so the fraction of energy the neutron keeps is

$\frac{K_f}{K_i} = \left(\frac{v_1}{u_1}\right)^2 = \left(\frac{11}{13}\right)^2 = \frac{121}{169} = 0.716$

So it loses $1 - 0.716 = 0.284$, about 28%. Note that the sign of $v_1$ vanished when you squared, which is why you may not answer this one by feel: the neutron bounces backwards, yet keeps most of its energy.

The general version is worth carrying. The fraction of kinetic energy transferred from $m_1$ to a stationary $m_2$ in a head-on elastic collision is

$\frac{\Delta K}{K_i} = \frac{4m_1m_2}{(m_1+m_2)^2}$

which equals $1$ only when $m_1 = m_2$. Check it here: $4(1)(12)/169 = 48/169 = 0.284$. Matched. And it explains why reactors moderate with hydrogen or deuterium rather than lead: you want the target to weigh about what the neutron weighs.

Bouncing, and the coefficient of restitution

A ball dropped from height $h$ arrives at the floor with speed $u = \sqrt{2gh}$, where $g = 9.8\ \mathrm{m/s^2}$. The floor does not move, so the approach speed is $u$ and the separation speed is $eu$. Rising with speed $eu$ it reaches

$h' = \frac{(eu)^2}{2g} = e^2h$

Each bounce multiplies the height by $e^2$. That also gives you a clean way to measure $e$: drop from $2.0\ \mathrm{m}$, see it return to $1.28\ \mathrm{m}$, and $e = \sqrt{1.28/2.0} = \sqrt{0.64} = 0.80$.

The total path length is a geometric series. The ball falls $h$, then goes up and down $e^2h$, then up and down $e^4h$, and so on:

$d = h + 2h(e^2 + e^4 + e^6 + \cdots) = h + 2h\frac{e^2}{1-e^2} = h\,\frac{1+e^2}{1-e^2}$

For $h = 2.0\ \mathrm{m}$ and $e = 0.75$: $e^2 = 0.5625$, so $d = 2.0 \times 1.5625 / 0.4375 = 7.14\ \mathrm{m}$. Infinitely many bounces, finite distance, and in fact finite time too.

Two dimensions: split into components, always

Off-centre collisions stay in a plane. Conserve $p_x$ and $p_y$ separately (two independent equations) and if the collision is elastic you get a third from kinetic energy. There are four unknown final components, so a 2D problem must hand you one more fact, usually one outgoing angle.

The standard setup: point the $x$ axis along the incoming velocity, so the initial $y$ momentum is zero. Then

$m_1u_1 = m_1v_1\cos\theta_1 + m_2v_2\cos\theta_2, \qquad 0 = m_1v_1\sin\theta_1 - m_2v_2\sin\theta_2$

with the two bodies going to opposite sides of the axis.

One result deserves memorising. Equal masses, one at rest, elastic: the two leave at exactly 90° to each other. Momentum says $\vec u = \vec v_1 + \vec v_2$. Square both sides as vectors: $u^2 = v_1^2 + v_2^2 + 2\vec v_1\cdot\vec v_2$. Energy says $u^2 = v_1^2 + v_2^2$. Subtract and $\vec v_1 \cdot \vec v_2 = 0$, so the velocities are perpendicular. Pool players live by this. If you see an angle that is not 90° between equal masses, energy was lost.

When the masses are unequal, the incoming body can only be deflected so far. For $m_1 > m_2$ the maximum deflection angle satisfies $\sin\theta_{\max} = m_2/m_1$. A $5.0\ \mathrm{kg}$ ball on a $2.0\ \mathrm{kg}$ ball can turn at most $\arcsin(0.40) = 23.6^\circ$: a heavy thing cannot be knocked sideways by a light one, no matter how you aim.

Worked example 3: finding a recoil speed

A $4.0\ \mathrm{kg}$ mass moving east at $5.0\ \mathrm{m/s}$ strikes a $6.0\ \mathrm{kg}$ mass at rest. Afterwards the $4.0\ \mathrm{kg}$ mass moves at $3.0\ \mathrm{m/s}$ at $40^\circ$ north of east. How fast does the $6.0\ \mathrm{kg}$ mass go?

Write down the momentum components before anything else. East is $x$, north is $y$.

Before: $p_x = (4.0)(5.0) = 20\ \mathrm{kg\,m/s}$, $p_y = 0$.

After, for the light mass: $p_x = (4.0)(3.0)\cos 40^\circ = 12 \times 0.766 = 9.19$, and $p_y = (4.0)(3.0)\sin 40^\circ = 12 \times 0.643 = 7.71$.

The heavy mass must carry the rest:

$p_{2x} = 20 - 9.19 = 10.81, \qquad p_{2y} = 0 - 7.71 = -7.71$

Its momentum magnitude is $\sqrt{10.81^2 + 7.71^2} = \sqrt{116.9 + 59.4} = \sqrt{176.3} = 13.28\ \mathrm{kg\,m/s}$, so

$v_2 = \frac{13.28}{6.0} = 2.21\ \mathrm{m/s}$

directed $\arctan(7.71/10.81) = 35.5^\circ$ south of east. Nobody told you this collision was elastic and you never needed to know. Two component equations and one supplied angle were enough.

The same recipe handles explosions, which are collisions run backwards. A stationary $9.0\ \mathrm{kg}$ shell bursting into pieces starts with zero total momentum, so the final momenta must add to zero: a $3.0\ \mathrm{kg}$ piece east at $8.0$ carries $24$, a $2.0\ \mathrm{kg}$ piece north at $6.0$ carries $12$, and the last $4.0\ \mathrm{kg}$ piece must carry $\sqrt{24^2+12^2} = 26.83$ the other way, giving $6.71\ \mathrm{m/s}$.

Worked example 4: the superball trick

A tiny superball rests on top of a basketball. Both are dropped from $h = 1.0\ \mathrm{m}$. The basketball bounces elastically off the floor and then hits the superball elastically. How high does the superball go, if its mass is negligible?

Both balls arrive travelling down at $v = \sqrt{2gh}$. The basketball bounces and is now travelling up at $v$ while the superball is still coming down at $v$. That is the picture to draw.

Now use the relative-velocity rule, which is easier than the mass formulas here. Approach speed between the two balls is $v + v = 2v$. Elastic, so the separation speed is also $2v$. The basketball is enormously heavier, so it keeps moving up at $v$ essentially unchanged. For the superball to be separating from it at $2v$, the superball must be moving up at $v + 2v = 3v$.

Triple the speed means nine times the height:

$h' = \frac{(3v)^2}{2g} = 9\frac{v^2}{2g} = 9h = 9.0\ \mathrm{m}$

This is what "shift to the frame of the heavy body" buys you. In that frame the light ball simply reverses, and coming back to the ground frame adds the heavy body's velocity twice over.

The same manoeuvre solves the spring problem. A $1.0\ \mathrm{kg}$ block at $6.0\ \mathrm{m/s}$ hits a $2.0\ \mathrm{kg}$ block carrying a spring. The spring stores the most energy at the instant of maximum compression, and that is the instant the two blocks move at the same speed: otherwise the gap is still closing or already opening. Common speed is $v = (1.0)(6.0)/3.0 = 2.0\ \mathrm{m/s}$. Energy before is $\tfrac12(1.0)(36) = 18\ \mathrm{J}$; kinetic energy at that instant is $\tfrac12(3.0)(2.0)^2 = 6.0\ \mathrm{J}$; the missing $12\ \mathrm{J}$ is in the spring.

Recognising a collision problem

Some signals are obvious: "collide", "strike", "explode", "bounce", "embeds". The useful skill is spotting the ones that do not say so.

Reach for momentum conservation when the problem gives you masses and velocities on both sides of a short, violent event and tells you nothing about the force or the duration. If you find yourself wishing you knew the force, that is the signal: the fact that you cannot know it is exactly why momentum is the tool.

Look for the second equation in the wording. "Stick together", "embed", "couple", "come to rest relative to each other" all mean one common final velocity. "Elastic", "billiard balls", "no energy lost" mean the relative velocity reverses. A number given for restitution, or a pair of drop and rebound heights, means partially inelastic.

Split the problem at the collision. Anything before it (a slide down a ramp, a free fall) and anything after it (a swing, a compression, a rise) is an energy or kinematics problem. The collision itself is a momentum problem. The two halves meet at a single velocity.

If the geometry is not a straight line, go to components immediately. Do not try to be clever with magnitudes and angles until the momentum bookkeeping is written as two scalar equations.

"Maximum compression", "maximum stored energy", "closest approach" all mean the same thing: the moment the two bodies move at a common velocity. Compute that velocity from momentum and put the missing kinetic energy where the problem says it went.

One idea runs under all of this: a system of colliding bodies has a velocity that never changes, no matter what the bodies do to each other. That velocity belongs to a point, and the next lesson gives it a name and a set of tools: center of mass and systems of particles.