Impulse $\vec{J} = \int \vec{F}\,dt = \Delta\vec{p} = m\vec{v}_f - m\vec{v}_i$. For constant force: $J = F\Delta t$. Unit: N·s $=$ kg·m/s. Key insight: same impulse can be produced by large force for short time or small force for long time. Applications: bouncing ball (impulse = change in momentum including direction reversal), catching a cricket ball (spreading force over longer time reduces peak force and injury), crumple zones (increase collision time → reduce peak deceleration force on occupants).

When $\vec{F}_{ext} = 0$: $\vec{p}_{total} = $ constant. For collision of two bodies: $m_1\vec{v}_1 + m_2\vec{v}_2 = m_1\vec{v}_1' + m_2\vec{v}_2'$. For ball-floor collision: treat floor as having infinite mass. Floor momentum change is negligible. Impulse from floor on ball $= $ magnitude of change in ball's momentum. Coefficient of restitution $e = v_{rel,after}/v_{rel,before} = v_2/v_1$ for ball and floor. Here $e = (10\sqrt{2})/(20\sqrt{2}) = 0.5$. $e = \sqrt{h_2/h_1} = \sqrt{10/40} = 1/2$.

Elastic collision ($e=1$): both KE and momentum conserved. Head-on: $v_1' = (m_1-m_2)u_1/(m_1+m_2)$, $v_2' = 2m_1u_1/(m_1+m_2)$. Equal masses: velocities exchanged. Inelastic ($0 < e < 1$): momentum conserved, KE not. Energy lost $= \frac{1}{2}\mu(u_{rel})^2(1-e^2)$ where $\mu = m_1m_2/(m_1+m_2)$ = reduced mass. Perfectly inelastic ($e=0$): stick together. Max KE loss. Energy lost $= \frac{1}{2}\mu u_{rel}^2$.

Thrust force = $v_{rel} \times dm/dt$ (where $v_{rel}$ = exhaust speed, $dm/dt$ = mass ejection rate). Rocket equation (Tsiolkovsky): $\Delta v = v_{rel}\ln(m_0/m_f)$. $m_0/m_f$ = mass ratio (initial/final). Multiple stages: each stage has own mass ratio, total $\Delta v$ = sum. Specific impulse $I_{sp} = F/(g \times dm/dt) =$ thrust efficiency measure. Hydrogen-oxygen engines: $I_{sp} \approx 450$ s. No external medium needed (works in vacuum).

$\vec{F}_{ext} = M\vec{a}_{cm}$. CM moves as if all mass were concentrated there and all external forces applied there. If $\vec{F}_{ext} = 0$: CM moves at constant velocity (or remains at rest). Internal forces (between parts of system) do NOT affect CM motion. Example: exploding shell — fragments fly apart, but CM continues along original parabolic trajectory. Walking person — internal (muscle) forces do work, but floor reaction provides the external force that accelerates CM forward.

For rocket (losing mass): $F_{thrust} = v_{rel}|dm/dt|$. Net: $F_{thrust} - mg = ma$. For conveyor belt (gaining mass): force needed to keep belt at constant speed $= v \times dm/dt$. For chain falling on table: extra force on table due to falling links = $\lambda v^2$ where $\lambda$ = linear density, $v$ = speed. Water jet hitting surface: force $= \rho A v^2$ (for flat surface, assuming water stops) or $2\rho Av^2$ (if water bounces back elastically).

Coefficient of restitution $e$: $e = $ relative speed of separation / relative speed of approach. For ball bouncing off floor: $e = v_2/v_1$ (since floor is stationary). Height relationship: if ball dropped from $H$ and rebounds to $h$: $e = \sqrt{h/H}$. After $n$ bounces: height $h_n = e^{2n} H$. For perfectly elastic ball: $e=1$, never loses height. Real balls: $e < 1$ (energy absorbed in deformation). Steel ball on steel plate: $e \approx 0.9$. Tennis ball: $e \approx 0.75$. Clay ball: $e \approx 0$.

During collision: force of A on B $= -$ force of B on A (Newton 3rd law). Equal in magnitude, opposite in direction, acting on different objects. At all times during collision. Consequence: impulse on A from B $= -$ impulse on B from A. So $\Delta p_A = -\Delta p_B$, thus total $\Delta p = 0$. This is why momentum is conserved in all collisions (as long as no external forces act during the brief collision time). The internal forces between colliding objects are always equal and opposite — they can't create net momentum change for the system.