M2 Mechanics: Work, Energy, Impulses and Collisions

M2 Mechanics: Work, Energy, Impulses and Collisions

From Forces to Conservation: A New Perspective on Motion

In the 17th century, Newton’s laws revolutionised mechanics: if you know all the forces, you can trace every instant of motion. But by the late 1600s, scientists noticed something remarkable — certain quantities remain constant throughout complex motions, even when forces change dramatically.

The Historical Quest: Beyond Forces

Imagine a pendulum swinging back and forth, or a roller coaster plunging down a track. Using $F=ma$ at every instant is possible but exhausting. Forces change direction, speeds vary, paths curve. Is there a shortcut?

Leibniz, the Bernoullis, and later Joule discovered that for many systems, you can ignore the messy details of how motion unfolds and focus on what is conserved:

• “Vis viva” (living force) $K = mv^2$ — what we now call kinetic energy
• Potential energy $U$ from position in a gravitational or elastic field
• Total mechanical energy $E = K + U$ stays constant if only conservative forces act

Instead of solving differential equations step by step, you write one equation: $E_{\text{initial}} = E_{\text{final}}$. This conservation principle became one of the most powerful tools in all of physics.

The Core Insight

• Energy methods answer: “What is possible?” and “What is the final state?” without tracking every intermediate moment.
• Force methods answer: “What happens at this instant?” and give the entire trajectory.

A Parallel Revolution: Momentum and Impulse

Around the same time, scientists studying collisions faced another puzzle: when two objects crash, forces are enormous and fleeting. How do you predict the outcome without knowing the exact force history?

Descartes (1644) and Huygens (1669) realised that the total quantity of motion — mass × velocity — is conserved in collisions (in the absence of external forces). Newton later formalised this as momentum conservation.

Instead of $F(t)$, consider $I = \int F\,dt$, the accumulated effect. This transforms intractable collision problems into simple before-and-after algebra.

Why Two Approaches?

• Energy: Focus on work, scalar quantities, and changes in speed/height. Best when you care about “how fast” or “how high.”
• Momentum/Impulse: Focus on vector quantities and interactions. Best for collisions, explosions, or when external forces act briefly.

Work and Energy

The Concept of Work: Quantifying Energy Transfer

In the 1800s, engineers building steam engines needed to measure how effectively a force could lift a weight or push a piston. They coined the term work to quantify the energy transferred by a force along a displacement.

Key realisation: Only the component of force along the direction of motion contributes. Pushing perpendicular to motion does no work (think of carrying a suitcase horizontally — gravity does no work because displacement is perpendicular to the weight).

When to choose the energy method

• Use energy when you care about initial vs final states (speed, height, distance) more than the path.
• If only conservative forces (e.g., gravity) do work, total mechanical energy is constant.
• With friction or drives, include their work using $W_{\text{nc}}$ and apply the work–energy principle.

Definition of Work

For a constant force of magnitude $F$ making angle $\theta$ with the displacement $\vec{s}$, the work done $W$ is:

$W = \vec{F} \cdot \vec{s} = Fs\cos\theta$

Only the component along the direction of motion does work.

Example: Pulling at an angle

A constant force of 20 N is applied at $30^\circ$ above the horizontal to move a crate 5 m along a smooth floor.

$W = 20 \times 5 \times \cos(30^\circ) = 100 \times \frac{\sqrt{3}}{2} = 86.6 \text{ J}$

Constant-Force Work: Component × Distance

Work measures energy transfer. When you push a box, only the force component parallel to the motion does useful work.

Sign convention:

• $W > 0$: force aids motion (engine pushing a car forward)
• $W < 0$: force opposes motion (friction, air resistance)
• $W = 0$: force perpendicular to motion

Example: Work against friction on a rough floor

A 12 kg box is pushed 8 m on a horizontal rough surface with coefficient of friction $\mu = 0.25$. The pushing force is horizontal and just large enough to keep constant speed.

Normal force: $N = mg = 12 \times 9.8 = 117.6 \text{ N}$

Friction: $f = \mu N = 0.25 \times 117.6 = 29.4 \text{ N}$

Work by push: $W_{\text{push}} = 29.4 \times 8 = 235.2 \text{ J}$

Work by friction: $W_f = -29.4 \times 8 = -235.2 \text{ J}$

Net work: $W_{\text{net}} = 0$ (constant speed, $\Delta K = 0$)

Kinetic and Potential Energy

Two forms of mechanical energy: energy of motion (kinetic) and energy of position (potential).

Kinetic energy — the energy of motion:

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

Why squared? From work: $W = F \cdot s = ma \cdot s$. Using $v^2 = u^2 + 2as$ with $u = 0$ gives $W = \frac{1}{2}mv^2$. The work done to accelerate from rest equals the kinetic energy gained.

Gravitational potential energy — stored energy of position:

$U = mgh$

When you lift a mass $m$ to height $h$, you do work $W = mgh$ against gravity. This energy is “stored” and can be recovered when the object falls.

Standard Forms

• Kinetic energy: $K = \dfrac{1}{2}mv^2$
• Gravitational potential (near Earth): $U = mgh$

Conservation of Mechanical Energy

When only conservative forces act (gravity, springs, electrostatic forces), the total mechanical energy $E = K + U$ remains constant throughout the motion.

A force is conservative if the work it does is independent of path, depending only on start and end positions. Gravity is conservative: lifting a book 1 m requires the same work whether you lift it straight up or carry it up a winding staircase.

Energy Conservation — no non-conservative work

If only gravity (and other conservative forces) do work on the system:

$K_i + U_i = K_f + U_f$

Example: Smooth track drop

A bead of mass $m$ starts from rest at height $H$ on a smooth track and descends to height $h$.

$mgH + 0 = mgh + \frac{1}{2}mv^2$

$v = \sqrt{2g(H - h)}$

Work–Energy Principle

When non-conservative forces (like friction) are present, they do work $W_{\text{nc}}$ that changes the total mechanical energy:

$\Delta(K + U) = W_{\text{nc}}$

• $W_{\text{nc}} > 0$: external agent adds energy
• $W_{\text{nc}} < 0$: energy dissipated (friction, air resistance)
• $W_{\text{nc}} = 0$: pure conservative system

Example: Block up a rough incline

A block of mass $m$ is pulled up an incline of angle $\alpha$ by a constant force $P$ parallel to the plane over distance $s$. Coefficient of friction is $\mu$; the block starts from rest.

• Work by $P$: $Ps$
• Work by gravity: $-mg s\sin\alpha$
• Work by friction: $-\mu mg\cos\alpha \cdot s$

$\frac{1}{2}mv^2 = Ps - mgs\sin\alpha - \mu mgs\cos\alpha$

$v^2 = \frac{2s}{m}(P - mg\sin\alpha - \mu mg\cos\alpha)$

Example: Braking distance

A car of mass $m$ moves on level ground at speed $v_0$. The brakes lock and the tyres slide with coefficient of kinetic friction $\mu$ until rest.

$\Delta K = W_f \implies 0 - \frac{1}{2}mv_0^2 = -\mu mgd$

$d = \frac{v_0^2}{2\mu g}$

Power

Power measures how fast energy is transferred.

Power

• Average power: $\bar{P} = \dfrac{W}{t}$
• Instantaneous power: $P = \vec{F} \cdot \vec{v} = Fv\cos\theta$

James Watt (1780s) needed to compare steam engines to horses. He defined power as work per unit time, creating the unit “horsepower” (1 hp ≈ 746 W).

A car climbing a hill at constant speed needs power $P = F_{\text{engine}} \times v$ to balance resistive forces (gravity component, friction, air resistance).

Practice Exercise: Multi-Stage Motion

A block of mass 5 kg starts from rest at point A, at the top of a smooth curved ramp of height $h_1 = 4$ m. It slides down to point B, then travels across a rough horizontal surface BC of length $d = 6$ m with $\mu = 0.3$. Finally, it moves up a smooth incline CD at $\alpha = 30^\circ$.

(a) Speed at B:

$mgh_1 = \frac{1}{2}mv_B^2 \implies v_B = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 4} = 8.85 \text{ m/s}$

(b) Work by friction from B to C:

$W_f = -\mu mgd = -0.3 \times 5 \times 9.8 \times 6 = -88.2 \text{ J}$

$\frac{1}{2}mv_C^2 = \frac{1}{2}mv_B^2 + W_f = 5 \times 9.8 \times 4 - 88.2 = 196 - 88.2 = 107.8$

$v_C = \sqrt{\frac{2 \times 107.8}{5}} = 6.57 \text{ m/s}$

(c) Maximum height on the incline:

$\frac{1}{2}mv_C^2 = mgh \implies h = \frac{v_C^2}{2g} = \frac{107.8}{5 \times 9.8} = 2.20 \text{ m}$

Impulses and Collisions

The Story of Collisions: From Billiard Balls to Conservation Laws

In the mid-1600s, natural philosophers were puzzled by a simple question: When two objects collide, what determines the outcome?

Christiaan Huygens (1669) noticed a striking pattern: even though forces during impact were unknown and impossibly brief, the total “quantity of motion” (mass × velocity) before and after collision remained the same — provided no external forces interfered.

The Impulse–Momentum Breakthrough

Newton’s insight (1687): Define momentum $\vec{p} = m\vec{v}$ and impulse $I = \int F\,dt$. Then Newton’s second law integrates to:

$I = \Delta p = mv_{\text{final}} - mv_{\text{initial}}$

This transforms an intractable force problem into a simple before-and-after comparison.

Impulse–Momentum Principle

The impulse $I$ is the “total push” delivered over the contact time:

$I = \int_{t_i}^{t_f} F\, dt$

Newton’s second law $F = ma = m\dfrac{dv}{dt}$ integrates to:

$\int F\, dt = m \int dv = m(v - u) = \Delta p$

So impulse equals the change in momentum: $I = mv - mu$.

Why this is powerful: Even if $F(t)$ is complicated, the integral $I$ depends only on before-and-after velocities.

Example: Why airbags help

A 70 kg passenger moves at 15 m/s. The car stops in a collision.

Stopping time 0.04 s: $F_{\text{avg}} = \dfrac{70 \times 15}{0.04} = 26{,}250 \text{ N}$

Stopping time 0.20 s: $F_{\text{avg}} = \dfrac{70 \times 15}{0.20} = 5{,}250 \text{ N}$

The airbag increases stopping time by 5×, reducing peak force by 5×!

Real-world engineering: Safety by design

Airbags and crumple zones increase collision time $\Delta t$, so $F_{\text{avg}} = I/\Delta t$ decreases for the same momentum change. Lower peak force → reduced injury.

Conservation of Linear Momentum

When two objects interact, Newton’s third law says forces are equal and opposite: $\vec{F}_{12} = -\vec{F}_{21}$.

Therefore impulses are also equal and opposite, so $\Delta p_1 = -\Delta p_2$, meaning $\Delta(p_1 + p_2) = 0$.

For an isolated system (no external forces), total momentum is constant:

$p_{\text{total, before}} = p_{\text{total, after}}$

Momentum Conservation — isolated system

Along a straight line with no external impulse:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Example: Sticky collision

Two trolleys of masses $m_1$ and $m_2$ move with speeds $u_1 > u_2$. They stick together after impact.

$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v \implies v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$

Newton’s Law of Restitution

Momentum conservation gives one equation, but for two unknowns ($v_1$ and $v_2$) we need a second relationship. This comes from the coefficient of restitution $e$:

$e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}$

Law of Restitution

For direct (head-on) impact along a common line:

$e = \frac{\text{speed of separation}}{\text{speed of approach}}, \quad 0 \leq e \leq 1$

• $e = 1$: perfectly elastic (no energy lost)
• $e = 0$: perfectly inelastic (objects stick together)
• $0 < e < 1$: real-world collisions

Example: Head-on impact with $e$

Given $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ and $e = \dfrac{v_2 - v_1}{u_1 - u_2}$, solve simultaneously for $v_1$ and $v_2$.

Example: Bounce from a wall

A ball strikes a smooth vertical wall head-on with speed $u$ and rebounds with coefficient of restitution $e$.

Speed after impact: $v = eu$ (in the opposite direction)

Impulse on the ball: $I = m(v - (-u)) = m(eu + u) = mu(1 + e)$

Loss of Kinetic Energy in Impact

Momentum is always conserved in isolated collisions, but kinetic energy is conserved only if $e = 1$.

Where does the energy go? Heat, sound, permanent deformation, internal vibrations.

Energy Loss Formula — head-on

For two masses $m_1, m_2$ with relative speed $u_1 - u_2$ before impact and coefficient $e$:

$\Delta K = \frac{1}{2}\,\mu\,(u_1 - u_2)^2\,(1 - e^2), \quad \mu = \frac{m_1 m_2}{m_1 + m_2}$

where $\mu$ is the reduced mass.

Physical insight:

• If $e = 1$: no energy lost (elastic)
• If $e = 0$: maximum energy lost for given approach speed
• Larger relative speed → more energy dissipated

Example: How much energy is lost?

Let $m_1 = 2$ kg, $m_2 = 3$ kg, $u_1 = 6$ m/s, $u_2 = 1$ m/s, $e = 0.6$.

$\mu = \frac{2 \times 3}{2 + 3} = 1.2 \text{ kg}$

$\Delta K = \frac{1}{2} \times 1.2 \times (6 - 1)^2 \times (1 - 0.6^2) = 0.6 \times 25 \times 0.64 = 9.6 \text{ J}$

Practice Exercise: Two-Stage Collision

Three smooth spheres A, B, C with masses $m_A = 2$ kg, $m_B = 3$ kg, $m_C = 5$ kg lie in a line. A moves at 6 m/s toward B (at rest). C is also at rest. $e_1 = 0.5$ (A–B), $e_2 = 0.8$ (B–C).

Stage 1: A collides with B

Momentum: $2(6) + 3(0) = 2v_A + 3v_B \implies 2v_A + 3v_B = 12$

Restitution: $e_1 = 0.5 = \dfrac{v_B - v_A}{6 - 0} \implies v_B - v_A = 3$

Solving: $v_A = 0.6$ m/s, $v_B = 3.6$ m/s (both forward)

(a) A does not rebound — it continues forward at 0.6 m/s.

(b) Impulse on A: $I = 2(0.6 - 6) = -10.8 \text{ N·s}$, magnitude = 10.8 N·s.

Stage 2: B collides with C

Momentum: $3(3.6) + 5(0) = 3v_B' + 5v_C \implies 3v_B' + 5v_C = 10.8$

Restitution: $e_2 = 0.8 = \dfrac{v_C - v_B'}{3.6 - 0} \implies v_C - v_B' = 2.88$

Solving: $v_B' = 0.45$ m/s, $v_C = 3.33$ m/s.

(d) Total kinetic energy lost:

Initial KE: $\frac{1}{2}(2)(6^2) = 36$ J

After Stage 1: $\frac{1}{2}(2)(0.6^2) + \frac{1}{2}(3)(3.6^2) = 0.36 + 19.44 = 19.8$ J

After Stage 2: $\frac{1}{2}(2)(0.6^2) + \frac{1}{2}(3)(0.45^2) + \frac{1}{2}(5)(3.33^2) = 0.36 + 0.304 + 27.72 = 28.38$ J

Total loss: $36 - 28.38 = 7.62$ J.