Comparing competing mechanics models

A mechanics model is a short list of assumptions plus the Newton-law equations that follow from them: which bodies are treated as point masses, which forces are kept in $\sum \vec F = m \vec a$, and which conservation laws are therefore available. Two models of the same event can both be correct Newtonian physics and still disagree, because they keep different terms — for example a rigid body versus a deforming one, or constant friction versus speed-dependent drag. Comparing models means stating each model's assumptions, computing its prediction for one measurable quantity, and testing that prediction against data from the event. Ideal models predict limiting values: for a given momentum change, the shortest possible collision time and the largest possible force. For skid distance the bound goes the other way: for a fixed initial speed, neglecting drag gives an upper bound on skid distance, while including drag makes the skid shorter. Real accident reconstruction brackets the truth between two models — an upper and a lower bound — instead of trusting one number. Whenever two models disagree, the disagreement itself indicates which physical effect (deformation, friction, rotation, external forces) the simpler model has discarded.

A car of mass $m = 1400\ \text{kg}$ travels at $v = 18\ \text{m/s}$ and comes to rest in a crash. Two models are compared: A — point mass with deformation, stopping time $\Delta t = 0.12\ \text{s}$; B — rigid body, stopping time $\Delta t = 0.030\ \text{s}$. Compute the average force in each model and the relative difference.

A car of mass $m = 1500\ \text{kg}$ begins a skidding stop from $v_0 = 30\ \text{m/s}$. Model 1: point mass with constant friction force, deceleration $a_0 = \mu g = 7.0\ \text{m/s}^2$. Model 2: quadratic air drag $F_{\text{drag}} = -k v^{2}$ is added, with $k/m = c = 8.0\times 10^{-4}\ \text{m}^{-1}$. Find the stopping distance in each model and the relative difference.

Reconstruction of a two-car collision. Car 1 ($m_1 = 1300\ \text{kg}$) travels east and strikes car 2 ($m_2 = 1700\ \text{kg}$), which stands at a light with its brakes applied. A witness states that car 1 was moving at $v_1 = 14\ \text{m/s}$ at impact. Model A: isolated system, perfectly inelastic collision; the wreck then slides with $\mu = 0.65$. Model A predicts $v' = \frac{m_1 v_1}{m_1 + m_2} = 6.07\ \text{m/s}, \qquad d = \frac{v'^{2}}{2\mu g} = 2.9\ \text{m}.$ The measured skid of the wreck is $11\ \text{m}$. Identify which assumption of model A the evidence contradicts, decide whether external friction from car 2's brakes can explain the discrepancy, and propose one measurement that would test your conclusion.

Safe self-study activity in a simulation. Real crash tests must not be attempted — work only with a simulator (for example PhET Collision Lab or any equivalent two-cart 1D simulator).

1. Set $m_1 = 1.0\ \text{kg}$, $m_2 = 2.0\ \text{kg}$, $v_2 = 0$ and run the collision with elasticity 1.00, 0.50 and 0.00.
2. For each run record $v_1$, $v_1'$, $v_2'$ and compute the prediction of the perfectly inelastic model $v' = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}.$
3. Tabulate the residual $v_2' - v'$ against elasticity and note that the point-mass momentum-conservation model is exact only at elasticity 0.
4. Switch to a braking model: compute stopping distances $d = v_0^{2}/(2\mu g)$ for $\mu = 0.7$ and $v_0 = 10,\ 20,\ 30\ \text{m/s}$, and compare with a numerical integration that adds drag $c v^{2}$ with $c = 8.0\times 10^{-4}\ \text{m}^{-1}$. Record at which speed the constant-friction model becomes a poor description and why.

Show that comparing models means comparing predictions, not formulas: the same change of momentum comes from a short, tall force pulse (rigid model) or a long, low one (crumpling model), and data choose between them.

A two-panel sketch. Left — force pulses $F(t)$ with equal areas $\Delta p$: a narrow tall rectangle for the rigid model and a wide low one for the crumpling model. Right — a bar chart of predicted and measured skid lengths for the two models, with the measured value drawn as a horizontal line so that it is visible which model the data support.

A $1200\ \text{kg}$ car hits a barrier at $v = 15\ \text{m/s}$ ($54\ \text{km/h}$). The crumple zone compresses by $d = 0.60\ \text{m}$; a rigid car would stop in $d = 0.05\ \text{m}$. Model A uses constant acceleration and Newton's second law; Model B uses the work–energy theorem; Model C uses impulse–momentum. All three must give the same average force if the assumptions match: $F = \frac{mv^2}{2d}$. For the crumple zone this gives $F \approx 2.3 \times 10^5\ \text{N}$, and for the rigid car $F \approx 2.7 \times 10^6\ \text{N}$.

The scenario forces you to choose which forces to superpose: during the crash, the barrier's normal force dominates, while air drag and rolling friction are negligible. You compare models by checking that each one applies $\sum \vec F = m\vec a$ to the same interval, and you see that the work–energy and impulse–momentum forms are not new physics but rearrangements of Newton's laws for a constant net force.

A football of mass $m = 0.43\ \text{kg}$ is kicked at $v_0 = 30\ \text{m/s}$ at $30^\circ$ to the horizontal. Model 1 ignores air drag and predicts a range $R = \frac{v_0^2 \sin 2\theta}{g} \approx 79\ \text{m}$ and flight time $t = \frac{2v_0\sin\theta}{g} \approx 3.1\ \text{s}$. Model 2 adds a drag force opposite the velocity, $F_D = \frac12 C_D \rho A v^2$, with $C_D \approx 0.25$, $\rho = 1.2\ \text{kg/m}^3$, $A \approx 0.038\ \text{m}^2$; at $v = 30\ \text{m/s}$ this gives $F_D \approx 5.1\ \text{N}$, comparable to the weight $mg \approx 4.2\ \text{N}$, and predicts a range of about $65$–$70\ \text{m}$.

You compare the two models by using Newton's second law with force superposition: in Model 1 only gravity acts, so $\vec a = \vec g$ is constant; in Model 2 gravity and drag act together, so the acceleration changes direction and magnitude as $v$ changes. The decision whether to keep or drop the drag term is a model choice, checked against measured range and time.

A phone of mass $m = 0.20\ \text{kg}$ falls from height $h = 1.2\ \text{m}$. Before impact its speed is $v = \sqrt{2gh} \approx 4.9\ \text{m/s}$. Model 1 treats the impact as a constant net force over contact time $\Delta t$ and uses impulse–momentum: $F_{\text{avg}} = \frac{mv}{\Delta t} + mg$. On a carpet $\Delta t = 0.050\ \text{s}$ gives $F_{\text{avg}} \approx 22\ \text{N}$; on tile $\Delta t = 0.005\ \text{s}$ gives $F_{\text{avg}} \approx 198\ \text{N}$. Model 2 (point-mass free fall) ignores the impact and cannot predict damage at all.

Here the competing models differ in the time interval they describe. During contact, the force superposition is $\vec N + m\vec g = m\vec a$; if you assume the net force is constant, Newton's second law integrated over $\Delta t$ becomes the impulse–momentum theorem $F_{\text{net}}\Delta t = \Delta p$. Comparing the carpet and tile cases shows that the same change in momentum produces a force inversely proportional to contact time.