9.2 Newton’s Laws, Energy & Momentum

9.2 Newton’s Laws, Energy & Momentum

Quick Answer: Most “why does the aircraft accelerate?” questions reduce to net force, work–energy, or momentum. Memorise $\Sigma \vec{F} = m\vec{a}$, $W = Fs\cos\theta$, $K = \tfrac{1}{2}mv^{2}$, $U_g = mgh$, and $\vec{J} = \Delta \vec{p}$, then practise free-body diagrams for fluency.

Kinematics describes how motion changes; this section explains why. Engineering selection papers probe whether you can connect forces to energy and collisions without mixing units—skills you will use later for loads, landing gear impulses, and structural energy absorption.

Newton’s Laws of Motion

First law (inertia): A body remains at rest or in uniform straight-line motion unless a net external force acts. No net force ⇒ zero acceleration, not necessarily “no forces.”

Second law:

$$\sum \vec{F} = m\vec{a}$$

Or, equivalently, net force equals rate of change of momentum: $\sum \vec{F} = d\vec{p}/dt$ with $\vec{p} = m\vec{v}$.

Third law: If A exerts $\vec{F}_{AB}$ on B, then B exerts $-\vec{F}_{AB}$ on A. The pair acts on two different objects—never cancel each other on the same free-body diagram.

Weight vs Mass

An object of mass $50\,\text{kg}$ has weight $W = 50 \times 9.8 = 490\,\text{N}$ near Earth.

Free-Body Diagram Checklist

1. Isolate one body.
2. Draw weight $mg$ downward.
3. Draw normal force perpendicular to the surface.
4. Draw friction opposite the tendency to slide.
5. Draw tension along strings/cables.
6. Apply $\sum F_x = ma_x$ and $\sum F_y = ma_y$.

Worked Example — Elevator

A $80\,\text{kg}$ passenger stands in a lift accelerating upward at $2.0\,\text{m/s}^{2}$. Apparent weight (normal force $N$):

$$N - mg = ma \implies N = m(g+a) = 80(9.8+2.0) = 944\,\text{N}$$

If the lift accelerates downward at $2.0\,\text{m/s}^{2}$, $N = m(g-a) = 624\,\text{N}$. In free fall, $a = g$ and $N = 0$ (weightlessness).

Friction

Static friction $f_s \le \mu_s N$ (equality at impending slip). Kinetic friction $f_k = \mu_k N$ once sliding starts. Usually $\mu_s > \mu_k$.

On a horizontal surface with applied force $F$ just overcoming static friction:

$$F_{\min} = \mu_s mg$$

On an incline of angle $\theta$, component down the plane is $mg\sin\theta$; normal is $mg\cos\theta$. Object slides if $mg\sin\theta > \mu_s mg\cos\theta$, i.e. $\tan\theta > \mu_s$.

Work, Energy & Power

Work by a constant force:

$$W = Fs\cos\theta$$

where $\theta$ is the angle between force and displacement. If force is perpendicular to motion, $W = 0$ (ideal centripetal force does no work).

Kinetic energy:

$$K = \tfrac{1}{2}mv^{2}$$

Gravitational potential energy near Earth (constant $g$):

$$U = mgh$$

(choose a reference height; only $\Delta U$ matters).

Work–energy theorem: net work on a particle equals change in kinetic energy:

$$W_{\text{net}} = \Delta K = K_f - K_i$$

Power:

$$P = \frac{W}{t} = Fv$$

(for force parallel to velocity). Units: watt (W).

Conservation of Mechanical Energy

If only conservative forces (gravity, ideal springs) do work,

$$K_i + U_i = K_f + U_f$$

With friction, mechanical energy decreases:

$$K_i + U_i = K_f + U_f + W_{\text{friction}}$$

(where $W_{\text{friction}}$ is the positive energy dissipated as heat, or treat friction work as negative on the system).

Worked Example — Energy on a Ramp

A $2.0\,\text{kg}$ block slides from rest down a frictionless ramp of height $5.0\,\text{m}$. Speed at bottom:

$$mgh = \tfrac{1}{2}mv^{2} \implies v = \sqrt{2gh} = \sqrt{2\times 9.8\times 5} = \sqrt{98} \approx 9.9\,\text{m/s}$$

Mass cancels—an MCQ favourite.

Worked Example — With Friction

Same block, but friction does $-30\,\text{J}$ of work over the path.

$$mgh + W_{\text{fric}} = \tfrac{1}{2}mv^{2}$$ $$(2)(9.8)(5) - 30 = \tfrac{1}{2}(2)v^{2}$$ $$98 - 30 = v^{2} \implies v = \sqrt{68} \approx 8.2\,\text{m/s}$$

Momentum & Impulse

Linear momentum:

$$\vec{p} = m\vec{v}$$

Impulse:

$$\vec{J} = \int \vec{F}\,dt \approx \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p}$$

A soft landing gear extends $\Delta t$, reducing peak force for the same momentum change—an aviation link that does not require aircraft-specific data.

Conservation of Momentum

For a system with negligible net external impulse,

$$\sum \vec{p}_{\text{initial}} = \sum \vec{p}_{\text{final}}$$

Elastic collision (1-D, equal mass): velocities exchange. Completely inelastic: bodies stick; $m_1u_1 + m_2u_2 = (m_1+m_2)v$.

Kinetic energy is conserved in perfectly elastic collisions; momentum is conserved in both elastic and inelastic (isolated system), but kinetic energy is not conserved in inelastic cases.

Worked Example — Inelastic Catch

A $0.20\,\text{kg}$ dart at $30\,\text{m/s}$ embeds in a $1.8\,\text{kg}$ block at rest on a frictionless surface.

$$(0.20)(30) + 0 = (2.0)v \implies v = 3.0\,\text{m/s}$$

Initial KE $= \tfrac{1}{2}(0.20)(30)^{2} = 90\,\text{J}$; final KE $= \tfrac{1}{2}(2.0)(3)^{2} = 9\,\text{J}$. Most energy dissipated—still valid momentum conservation.

Connecting the Three Pillars

Self-Timed Practice Note

Build formula fluency with progressively timed self-practice, but do not infer an official PAF item count, subject battery, or clock from coaching summaries.

Next: circular motion, gravitation, and SHM extend these same force and energy ideas to rotating and oscillating systems.