HookWhy a car that crushes is safer than a car that doesn't
In 1952 the Mercedes engineer Béla Barényi patented an idea that contradicted every instinct about safety: build the ends of the car weak. His 'crumple zone' reached production in 1959, and the reasoning behind it is one line of this section's algebra. A 1,200 kg car hitting a wall at 13 m s\(^{-1}\) (about 30 mph) must lose momentum \(mv-mu=15{,}600\) kg m s\(^{-1}\) — that figure is not negotiable. What is negotiable is the time over which it is lost. A rigid car stops in perhaps \(0.05\) seconds, so the average force is \(15{,}600\div 0.05\approx 312\) kN. A crumpling front end stretches the stop to \(0.15\) seconds, and the force drops to about \(104\) kN — a third. Same momentum change, three times the time, one third the force on the passengers. Airbags, cycle helmets, and a cricketer drawing back their hands as they catch are all the same equation wearing different clothes.
That equation is impulse equals change in momentum, and it anchors everything here. When two bodies collide, the impulses they exert on each other are equal and opposite, so the total momentum of the system survives the collision untouched — conservation of momentum, in one dimension or two. What conservation alone cannot tell you is how bouncy the collision was; that needs a second, experimental law and its coefficient of restitution. And when the force varies during the impact, the product \(Ft\) matures into the integral \(\int F\,dt\). Four leaves, one bookkeeping principle.
ModelConservation of momentum, in one dimension and two
Momentum is \(m\mathbf{v}\) — a vector, and the vector nature is the whole game. During a collision each body receives an impulse from the other; by Newton's third law those impulses are equal and opposite, so their sum is zero and the total momentum of the system is identical before and after: \(m_1\mathbf{u}_1+m_2\mathbf{u}_2=m_1\mathbf{v}_1+m_2\mathbf{v}_2\). No external impulse, no change — which is why the principle holds during the instant of impact even when gravity acts, because a finite force over a vanishing instant contributes vanishing impulse.
In one dimension the vector discipline reduces to signs: fix a positive direction, write every velocity with its sign, and let the algebra track the directions. A negative answer is not an error — it is the mathematics telling you the body ended up moving the other way, and you should say so in words. In two dimensions, resolve: momentum is conserved separately in each of two perpendicular directions, giving two scalar equations from one vector law. Column-vector velocities make this near-automatic, and coalescence problems — bodies sticking together — are the friendliest case, because after impact there is only one velocity to find.
A particle of mass \(2\) kg moving with velocity \(\binom{4}{1}\) m s\(^{-1}\) collides with a particle of mass \(3\) kg moving with velocity \(\binom{-1}{2}\) m s\(^{-1}\). They coalesce. Find the velocity of the combined particle.
Total momentum before: \(2\binom{4}{1}+3\binom{-1}{2}=\binom{8}{2}+\binom{-3}{6}=\binom{5}{8}\) kg m s\(^{-1}\).
Conservation: the combined mass is \(5\) kg, so \(5\mathbf{v}=\binom{5}{8}\), giving \(\mathbf{v}=\binom{1}{1.6}\) m s\(^{-1}\).
Interpret: speed \(=\sqrt{1^2+1.6^2}\approx 1.89\) m s\(^{-1}\), at \(\arctan(1.6)\approx 58.0^{\circ}\) to the \(x\)-direction. Kinetic energy is not conserved here — coalescence always destroys some — but momentum, both components, is.
MechanismNewton's experimental law: how bouncy was it?
Conservation of momentum is one equation, but a collision between two bodies has two unknown final velocities. The missing information is empirical: Newton measured colliding spheres and found that the speed of separation is proportional to the speed of approach. The constant of proportionality is the coefficient of restitution \(e\), a property of the pair of materials, with \(0\le e\le 1\): \(e=1\) is a perfectly elastic bounce (kinetic energy survives), \(e=0\) means no separation at all — the bodies coalesce. For direct collisions the working equation is \(v_2-v_1=e(u_1-u_2)\), separation speed on the left, approach speed on the right, and the two most common exam errors are swapping the roles or mixing the sign convention between this equation and the momentum equation. Fix one positive direction and use it in both.
Against a fixed smooth surface the law simplifies: the wall does not move, so the rebound speed is \(e\) times the approach speed — measured perpendicular to the surface. For oblique impacts with a smooth wall, resolve: the component parallel to the wall is unchanged (smooth means no friction, so no impulse along the surface), while the perpendicular component reverses and shrinks by the factor \(e\). One consequence worth knowing: a ball dropped from height \(h\) rebounds to \(e^2h\), because height depends on the square of speed. In every case with \(e<1\), kinetic energy falls — and 'find the loss in kinetic energy' is a routine final part, computed as \(\frac{1}{2}\sum mu^2-\frac{1}{2}\sum mv^2\).
Sphere A (mass \(2\) kg, speed \(5\) m s\(^{-1}\)) catches up and collides directly with sphere B (mass \(1\) kg, speed \(1\) m s\(^{-1}\)) moving the same way; \(e=0.5\). Find both final velocities and the kinetic energy lost.
Momentum (positive = direction of motion): \(2(5)+1(1)=2v_A+v_B\), so \(2v_A+v_B=11\).
Restitution: \(v_B-v_A=e(u_A-u_B)=0.5(5-1)=2\).
Solve: substituting \(v_B=v_A+2\): \(3v_A+2=11\), so \(v_A=3\) m s\(^{-1}\) and \(v_B=5\) m s\(^{-1}\), both still moving forwards — and consistently, B now moves faster than A, so they separate.
Energy: before, \(\frac{1}{2}(2)(25)+\frac{1}{2}(1)(1)=25.5\) J; after, \(\frac{1}{2}(2)(9)+\frac{1}{2}(1)(25)=21.5\) J. Loss \(=4\) J — positive, as it must be whenever \(e<1\).
MechanismImpulse: the bridge between force and momentum
Impulse is what a force accomplishes over time. For a constant force, \(\mathbf{I}=\mathbf{F}t\); its effect is the change in momentum, \(\mathbf{F}t=m\mathbf{v}-m\mathbf{u}\). The units, newton seconds, are literally kg m s\(^{-1}\) — impulse and momentum are the same currency. This is the crumple-zone equation from the introduction: for a fixed momentum change, force and time trade off exactly inversely, which is the entire engineering of survivable collisions.
Because impulse is a vector, the equation resolves like any other. When a ball strikes a smooth surface obliquely, the impulse acts along the normal — so only the normal component of momentum changes, which is precisely why the parallel component sails through unchanged. In component questions, compute \(m\mathbf{v}-m\mathbf{u}\) as column vectors and report magnitude and direction; the classic slip is finding \(m\mathbf{v}-m\mathbf{u}\) with a sign error in one component, so write both momenta down explicitly before subtracting. In a collision between two bodies, the impulse on one is the exact negative of the impulse on the other — computing one and quoting it for both (with the sign flipped) is legitimate and quick, and 'find the magnitude of the impulse between the spheres' is answerable from either body: \(I=|m_1v_1-m_1u_1|\).
A ball of mass \(0.5\) kg has velocity \(\binom{6}{-2}\) m s\(^{-1}\) just before a strike and \(\binom{2}{4}\) m s\(^{-1}\) just after. Find the impulse and its magnitude.
Impulse: \(\mathbf{I}=m\mathbf{v}-m\mathbf{u}=0.5\binom{2}{4}-0.5\binom{6}{-2}=\binom{1}{2}-\binom{3}{-1}=\binom{-2}{3}\) N s.
Magnitude: \(|\mathbf{I}|=\sqrt{(-2)^2+3^2}=\sqrt{13}\approx 3.61\) N s, at \(\arctan\!\left(\frac{3}{2}\right)\approx 56.3^{\circ}\) above the negative \(x\)-direction. Note the impulse does not point along either velocity — it points along the change, which is the direction the striking force acted.
CaseVariable forces: when Ft becomes an integral
Real impacts never deliver constant force. A crash pulse, a bat on a ball, a rocket burn — the force rises, peaks and falls, and the impulse is the accumulation \(I=\int F\,dt\): the area under the force–time graph. The constant-force formula \(Ft\) is just the rectangle case of this integral, and the 'average force' quoted in safety engineering is the constant force that would enclose the same area — total impulse divided by contact time, which is exactly the calculation the crumple-zone numbers in the introduction ran.
The exam version is one dimension and clean calculus: you are given \(F(t)\), often a polynomial that is zero at the start and end of contact (the force switches on and off smoothly), and asked for the impulse over the contact and the resulting velocity change via \(I=mv-mu\). The physics content is knowing what to integrate and between which limits — the contact interval, usually where \(F(t)\ge 0\); after that it is E-grade integration. This is also the honest version of every headline number about car safety: legislators specify crash pulses as force–time (or acceleration–time) curves, and the area constraint is why no clever design can reduce the impulse — only spread it.
During a kick lasting \(0.2\) s, the force on a stationary \(0.16\) kg ball is modelled by \(F(t)=600t(0.2-t)\) N. Find the impulse and the ball's speed as it leaves the kick.
Impulse: \[I=\int_0^{0.2}600\,(0.2t-t^2)\,dt=600\left[0.1t^2-\frac{t^3}{3}\right]_0^{0.2}=600\left(0.004-\frac{0.008}{3}\right)=0.8\text{ N s}.\]
Velocity: \(I=mv-mu\) with \(u=0\) gives \(v=\frac{0.8}{0.16}=5\) m s\(^{-1}\).
Check the model: \(F(0)=F(0.2)=0\) — the force switches on and off smoothly — and the peak force \(F(0.1)=6\) N acts only instantaneously; the average force is \(0.8\div 0.2=4\) N, the constant force that would deliver the same impulse in the same time.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Every collision answer should open the same way: a before/after diagram with masses, velocities and a declared positive direction. Mark schemes award the momentum equation and the restitution equation separately, so write them as two labelled lines — momentum from conservation, then \(v_2-v_1=e(u_1-u_2)\) with approach and separation identified — and solve simultaneously. Keep the same sign convention in both equations; the single most common error in this topic is a restitution equation whose signs disagree with the momentum line beside it. Interpret negative answers in words ('A rebounds, moving away from the wall at 2 m s\(^{-1}\)') and, where the situation allows a physical check, run it: after a direct collision the rear body cannot still be moving faster than the front one.
Quote units on every numerical answer — N s for impulse, kg m s\(^{-1}\) for momentum, J for energy — and give non-exact answers to three significant figures unless told otherwise. In two-dimensional work, resolve early and keep components in columns; the impulse is \(m\mathbf{v}-m\mathbf{u}\) computed componentwise, with magnitude and direction stated at the end. For variable-force parts, define the contact interval before integrating and show the evaluated bracket — the method mark sits on visible limits. And on 'find the loss in kinetic energy' finishes, compute total KE before and after separately and subtract; a positive loss is a built-in sanity check, because a collision with \(e<1\) that appears to gain energy is telling you a velocity is wrong.