System Dynamics and Collision Theory

The linear momentum of a body is $\vec{p} = m\vec{v}$. Newton's Second Law can be written in its most general form as $\vec{F}_{net} = d\vec{p}/dt$, which also covers situations where mass changes (rockets, variable-mass systems).

Conservation of momentum: if the net external force on a system is zero, the total momentum $\vec{p}_{tot} = \sum m_i \vec{v}_i$ is constant. This holds even during violent collisions where internal forces are enormous — the internal action–reaction pairs cancel by Newton's Third Law.

The impulse $\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}$ is the change in momentum produced by a force acting over a time interval. During a short collision the average force can be very large while $\Delta t$ is very small; their product (the impulse) equals the change in momentum.

The center of mass (CM) of a system of particles is the mass-weighted average position: $\vec{r}_{CM} = \sum m_i\vec{r}_i / M$. It is the unique point that behaves exactly like a point mass of total mass $M$ subject to the total external force — regardless of how complex the internal motions are.

The two cardinal equations of mechanics govern the motion of any system:
1st cardinal equation (translational): $\vec{F}_{ext} = M\vec{a}_{CM}$. The CM accelerates exactly as if all the mass were concentrated there and all external forces were applied there.
2nd cardinal equation (rotational): $\vec{\tau}_{ext} = d\vec{L}/dt$. The net external torque about a fixed point (or about the CM) equals the rate of change of angular momentum. Internal forces and torques contribute nothing to either equation.

In an elastic collision both total momentum and total kinetic energy are conserved simultaneously. This gives two equations for the two unknown final velocities (in 1D), yielding a unique solution.

The formulas for body 2 initially at rest are derived by solving the system of conservation equations. Two notable limiting cases:

• Equal masses ($m_1 = m_2$): $v_1' = 0$, $v_2' = v_1$ — the velocities exchange completely (billiard ball behaviour).
• Heavy projectile ($m_1 \gg m_2$): $v_1' \approx v_1$, $v_2' \approx 2v_1$ — the heavy body is barely affected, the light body rebounds at nearly twice the original speed.
• Light projectile ($m_1 \ll m_2$): $v_1' \approx -v_1$, $v_2' \approx 0$ — the light body bounces back elastically; the heavy body barely moves.

Head-on vs oblique collision. A collision is head-on (central) when the bodies' velocities are directed along the line joining the two centres: the motion stays on a single line — the 1D case above. It is oblique when the velocities make an angle with that line (a glancing or off-centre hit): the bodies scatter in a plane — the 2D case. A collision is central only if the centres approach exactly along the line of centres; in all other cases it is oblique.

2D (oblique) elastic collision — general case. In two dimensions the collision is no longer head-on: the bodies separate along different directions. In the most general case both masses have arbitrary initial velocities $\vec{v}_1$ and $\vec{v}_2$. Momentum conservation is vectorial and holds component by component; together with kinetic energy: $$m_1v_{1x} + m_2v_{2x} = m_1v_{1x}' + m_2v_{2x}' \quad\text{(p along x)}$$ $$m_1v_{1y} + m_2v_{2y} = m_1v_{1y}' + m_2v_{2y}' \quad\text{(p along y)}$$ $$\tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2 = \tfrac{1}{2}m_1v_1'^2 + \tfrac{1}{2}m_2v_2'^2 \quad\text{(KE)}$$
Warning — the system is underdetermined. There are four final unknowns (two velocity components each, i.e. two speeds and two angles) but only three conservation equations. One piece is missing: the collision geometry (the impact parameter, i.e. how head-on the hit is). Without it the 2D problem has no unique solution — unlike 1D, where 2 equations and 2 unknowns suffice.

How it is solved: line of centres. For smooth spheres (frictionless) each velocity is decomposed along two directions at contact: the normal $\hat{n}$ (the line joining the two centres) and the tangent $\hat{t}$. Then:

• the tangential components are unchanged (no force in that direction): $v_{1t}' = v_{1t}$, $v_{2t}' = v_{2t}$;
• the normal components transform by the 1D elastic formulas above (the impact force acts only along $\hat{n}$).

This is the 4th piece that closes the problem: the angle of the line of centres sets how grazing or head-on the collision is.

What the impact parameter $b$ is. It is the sideways offset between the path of the incoming ball's centre and the target's centre. It tells how off-centre the hit is:

• $b = 0$: the incoming ball aims straight at the target's centre → central (head-on, 1D) collision;
• $0 < b < 2r$: oblique (glancing) collision;
• $b \geq 2r$: the balls do not even touch (no collision).

What it is for in problems. The impact parameter is exactly the missing "fourth piece": it fixes the tilt of the line of centres through $\sin\varphi = b/(2r)$. Read it off the right triangle in the figure: the hypotenuse is the centre-to-centre distance at contact ($2r$), the vertical leg is $b$. Recipe to solve an oblique collision:

1. from the given $b$ (and the radii) get $\varphi = \arcsin\!\big(b/(2r)\big)$ — the direction of the line of centres $\hat{n}$;
2. resolve the velocities along $\hat{n}$ (normal) and $\hat{t}$ (tangent, perpendicular);
3. apply the 1D collision to the normal components only; leave the tangential ones unchanged;
4. recombine the vectors to get the final magnitudes and directions.

Special case — stationary target. If $m_2$ is initially at rest and the $x$ axis is taken along $\vec{v}_1$, the three equations simplify (with $\theta_1, \theta_2$ the deflection angles on opposite sides of the $x$ axis): $$m_1v_1 = m_1v_1'\cos\theta_1 + m_2v_2'\cos\theta_2$$ $$0 = m_1v_1'\sin\theta_1 - m_2v_2'\sin\theta_2$$ $$\tfrac{1}{2}m_1v_1^2 = \tfrac{1}{2}m_1v_1'^2 + \tfrac{1}{2}m_2v_2'^2$$ (choosing $x$ along $\vec{v}_1$ is merely convenient: it zeroes the initial $y$-momentum. The problem stays underdetermined until $\theta_1$ is fixed by the geometry.)

Notable property — equal masses, stationary target. If $m_1 = m_2$ and $m_2$ is initially at rest, the final velocities are always perpendicular: $\theta_1 + \theta_2 = 90°$, whatever the impact parameter. Vector proof: with $m_1=m_2=m$, momentum conservation gives $\vec{v}_1 = \vec{v}_1' + \vec{v}_2'$ and KE conservation gives $v_1^2 = v_1'^2 + v_2'^2$. Squaring the first ($v_1^2 = v_1'^2 + v_2'^2 + 2\,\vec{v}_1'\cdot\vec{v}_2'$) and comparing with the second yields $\vec{v}_1'\cdot\vec{v}_2' = 0$, so the two vectors are orthogonal (provided both bodies keep moving). Example: if $m_1$ deflects by $\theta_1 = 30°$, then $\theta_2 = 90° - 30° = 60°$. This is why in billiards, after an off-centre hit between equal balls, the two balls leave at nearly $90°$. Note: with $m_2$ moving, or unequal masses, this orthogonality no longer holds.

In an inelastic collision momentum is conserved but kinetic energy is not — some KE is converted into internal energy (deformation, heat, sound). The missing kinetic energy is not "lost" but transformed into other forms.

In a perfectly inelastic collision the two bodies stick together and move with a single common velocity $v'$ after impact. This maximises the kinetic energy lost, yet momentum is still conserved.

The coefficient of restitution $e$ characterises the "bounciness" of a collision:

• $e = 1$: perfectly elastic (no KE lost)
• $e = 0$: perfectly inelastic (maximum KE lost)
• $0 < e < 1$: partially inelastic (real collisions)

$e$ is defined as the ratio of the relative speed of separation to the relative speed of approach along the line of impact.

The moment of inertia $I = \int r^2\,dm$ is the rotational analogue of mass: it measures how the mass is distributed around the rotation axis. A larger $I$ means more torque is needed to produce a given angular acceleration, exactly as a larger $m$ requires more force for a given linear acceleration.

Unlike mass, $I$ depends on the choice of axis. The Parallel Axis Theorem (Steiner's theorem) relates the moment about any axis to the moment about the parallel axis through the CM: $I = I_{CM} + Md^2$, where $d$ is the perpendicular distance between the axes.

The rotational equations mirror the translational ones: $\tau = I\alpha$ (analogous to $F = ma$) and $KE_{rot} = \tfrac{1}{2}I\omega^2$ (analogous to $\tfrac{1}{2}mv^2$). For a body rolling without slipping, total KE = translational KE + rotational KE.

Standard results: solid cylinder $I = \tfrac{1}{2}MR^2$; hollow cylinder $I = MR^2$; solid sphere $I = \tfrac{2}{5}MR^2$; thin rod about centre $I = \tfrac{1}{12}ML^2$.