Physics by Lamhi: Not Your Boring Physics

CLASS 11 · CHAPTER 4 · MECHANICS

DETAILED NOTES

Laws of Motion

The complete chapter, section by section: from Aristotle's mistake to Newton's three laws, momentum, friction, and circular motion on level and banked roads, explained in plain language with original worked examples. For the interactive friction simulation, see the concept page.

4.1 From Aristotle to Galileo: The Law of Inertia

It seems obvious from everyday experience that keeping something moving takes a continuous push, a toy car needs its string constantly pulled, or it stops. This was essentially Aristotle’s view: an external force is required to keep a body in motion. It is also wrong, and finding the flaw took nearly two thousand years.

The flaw is friction. The toy car stops because the floor’s friction opposes it, and the child’s pull exists only to cancel that friction, not to sustain motion by itself. Galileo saw this by imagining away the friction entirely: rolling a ball down one incline and up another, he noted it always climbed back to nearly its starting height, and reasoned that in the ideal, frictionless limit it would climb to exactly that height, however gentle the second slope, meaning on a perfectly flat frictionless surface it would simply keep rolling forever.

This is the law of inertia: a state of rest and a state of uniform straight-line motion are physically equivalent, and in both, the net external force is zero. Inertia is a body’s built-in resistance to having that state changed.

4.2 Newton's First Law of Motion

Newton took Galileo’s insight and stated it as a law:

Every body continues to be in its state of rest or of uniform motion in a straight line, unless compelled by some external force to act otherwise.

Equivalently: if the net external force on a body is zero, its acceleration is zero, and acceleration can be nonzero only if a net force acts. This runs both ways in practice. Given the forces, the first law predicts the motion (a spaceship with engines off and nothing nearby must move at constant velocity). Given the motion, it lets you infer the forces: a book lying still on a table tells you the table’s normal force RR must exactly cancel its weight WW, not because of any separate law linking RR and WW, but purely because the book is observed to be unaccelerated.

4.3 Momentum, Newton's Second Law and Impulse

The first law only covers the special case of zero net force. The second law covers everything else, by relating force to momentum, p=mv\mathbf{p}=m\mathbf{v}:

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

F=dpdt=d(mv)dt=ma\mathbf{F} = \dfrac{d\mathbf{p}}{dt} = \dfrac{d(m\mathbf{v})}{dt} = m\mathbf{a}

choosing the constant of proportionality to be 1, which is what defines the SI unit of force: one newton is the force that gives a 1 kg mass an acceleration of 1 m/s². A few things worth holding onto about this law: it is a vector equation, so it applies separately to each of the x-, y-, and z-components; it is a local and instantaneous relation, meaning the force here and now determines the acceleration here and now, with no memory of the body’s past motion, a ball at the top of its throw is momentarily at rest but is not force-free, gravity and the acceleration gg act on it exactly as they did a moment earlier; and it applies to a system of particles too, with F\mathbf{F} as the total external force (internal forces are excluded entirely).

4.3.1 Impulse

Sometimes a large force acts for a very short time, a bat striking a ball, a foot striking the ground, and the force and the exact duration are hard to measure separately, but their product is easy to find because it equals the change in momentum:

Impulse
J=F Δt=Δp\textcolor{#e08a1e}{J} = F\,\Delta t = \Delta p

Newtonian mechanics treats an impulsive force as nothing special, just an ordinary force that happens to be large and brief.

Worked example

Force from a sudden stop

A 50 g hailstone falls vertically and strikes a roof at 20 m/s, coming to rest (no bounce) in 0.01 s. Find the average force on the roof.

The change in the hailstone’s momentum:

Δp=mΔv=0.05×(0−20)=−1.0 kg m/s\Delta p = m\Delta v = 0.05\times(0-20) = -1.0\text{ kg m/s}

so the average force the roof exerts on the hailstone is:

F=ΔpΔt=−1.00.01=−100 N (i.e. 100 N upward, decelerating it)F = \dfrac{\Delta p}{\Delta t} = \dfrac{-1.0}{0.01} = -100\text{ N (i.e. 100 N upward, decelerating it)}

By Newton’s third law (next section), the hailstone exerts 100 N downward on the roof, briefly, which is exactly why hail can dent a roof despite each stone weighing almost nothing.

4.4 Newton's Third Law of Motion

To every action, there is always an equal and opposite reaction.

Stated more carefully: the force on body A due to body B is equal and opposite to the force on body B due to A, FAB=−FBA\mathbf{F}_{AB} = -\mathbf{F}_{BA}. Three points that the words “action” and “reaction” tend to obscure:

  • The two forces are simultaneous, not cause-and-effect; either one may be called the action and the other the reaction.
  • They act on two different bodies, never on the same body, so they can never cancel each other out for either body individually. If you push a wall, the wall pushes back on you, not on itself.
  • Two forces on the same body that happen to be equal and opposite (like a book’s weight and the table’s normal force on it) are not a third-law pair, they are just a coincidence of that body being in equilibrium. A genuine third-law pair is always “force on A by B” and “force on B by A”, on two different objects.

4.5 Conservation of Momentum

Combine the second and third laws and something powerful falls out. For two colliding bodies A and B, Newton’s third law gives FAB=−FBA\mathbf{F}_{AB}=-\mathbf{F}_{BA}, and the second law turns each side into a rate of change of momentum over the same contact time Δt\Delta t:

pA′+pB′=pA+pB\mathbf{p}_A' + \mathbf{p}_B' = \mathbf{p}_A + \mathbf{p}_B

The total momentum after the interaction equals the total momentum before it, for any pair of mutual forces, elastic or inelastic. Extended to any number of particles, this is the law of conservation of momentum: in an isolated system, with no external force, mutual internal forces can shuffle momentum between individual particles, but the total never changes, because every internal push has an equal and opposite partner that cancels it out overall. Firing a bullet is the classic case: bullet and gun start with zero total momentum, and they must end with zero total momentum too, so the gun recoils backward with exactly the momentum the bullet carries forward.

4.6 Equilibrium of a Particle

A particle is in equilibrium when the net external force on it is zero, which by the first law means it is either at rest or moving with constant velocity. For two forces, equilibrium simply requires F1=−F2\mathbf{F}_1 = -\mathbf{F}_2. For three or more concurrent forces, it requires the vector sum to vanish:

F1+F2+F3+⋯=0\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 + \cdots = 0

which geometrically means the forces, drawn head to tail, close up into a polygon with no gap left over. In components, this is three independent scalar conditions, one per axis, all of which must hold simultaneously.

4.7 Common Forces and Friction

Only two forces are truly fundamental at the scale of ordinary mechanics: gravity, and the electrical forces between charged particles inside matter. Every other everyday force, normal reaction, tension, the spring force F=−kxF=-kx, buoyancy, air resistance, and friction, is really electrical in origin, arising from how atoms and molecules resist being pushed into or pulled apart from each other, just expressed at a much simpler, macroscopic level because tracking the underlying electrical interactions directly would be hopeless.

4.7.1 Static and kinetic friction

Push a heavy box and, up to a point, nothing happens, some opposing force is exactly cancelling your push. This is static friction fsf_s: a self-adjusting reaction that matches whatever force is trying to cause sliding, right up to a maximum:

fs≤μsNf_s \le \mu_s N

Push past that ceiling and the box starts to slide; friction then drops to a roughly constant kinetic value, fk=μkNf_k = \mu_k N, with μk\mu_k reliably smaller than μs\mu_s. Both coefficients depend only on the two surfaces in contact, not on the contact area. The concept page has an interactive version of exactly this switch-over, along with the common mistake of assuming static friction is always μsN\mu_s N rather than a self-adjusting maximum, worked through in full at Laws of Motion.

4.8 Circular Motion: Level and Banked Roads

A body moving in a circle of radius RR at speed vv needs a centripetal acceleration v2/Rv^2/R toward the centre, which by the second law needs a matching centripetal force:

fc=mv2Rf_c = \dfrac{mv^2}{R}

“Centripetal force” is not a new, separate kind of force, it is simply the name for whichever real force happens to be providing that inward pull: tension for a whirled stone, gravity for an orbiting planet, and for a car turning on a flat road, friction.

On a level road, the only horizontal force available is friction, so the car cannot turn faster than the speed at which the required centripetal force would exceed the maximum static friction μsN=μsmg\mu_s N = \mu_s mg:

vmax=μsRgv_{max} = \sqrt{\mu_s R g}

which, notice, does not depend on the car’s mass at all. A banked road tilts the normal force itself inward, so it can supply part or all of the centripetal force without relying on friction. Balancing forces along the vertical and horizontal directions and eliminating the normal force NN gives the maximum safe speed on a bank of angle θ\theta:

vmax=Rg(μs+tan⁡θ1−μstan⁡θ)v_{max} = \sqrt{Rg\left(\dfrac{\mu_s + \tan\theta}{1-\mu_s\tan\theta}\right)}

which is always greater than the flat-road value for the same μs\mu_s. Setting μs=0\mu_s = 0 in the same derivation gives the single speed at which the bank alone, with zero friction, supplies exactly the needed centripetal force, the optimum speed:

Optimum banked speed
v0=Rgtan⁡θ\textcolor{#e08a1e}{v_0} = \sqrt{Rg\tan\theta}

Driving at exactly v0v_0 causes the least wear on the tyres, since no sideways friction is needed at all.

Worked example

Optimum and maximum speed on a banked curve

A curve of radius 100 m is banked at 10°, with μs=0.2\mu_s = 0.2 between the tyres and the road. Find the optimum speed and the maximum speed before slipping. (g=9.8 m/s2g = 9.8\text{ m/s}^2)

The optimum speed needs no friction at all:

v0=Rgtan⁡θ=100×9.8×tan⁡10∘≈13.1 m/sv_0 = \sqrt{Rg\tan\theta} = \sqrt{100\times9.8\times\tan10^\circ} \approx \textcolor{#e08a1e}{13.1\text{ m/s}}

and the maximum speed uses the full banked-road formula:

vmax=100×9.8×0.2+tan⁡10∘1−0.2tan⁡10∘≈19.6 m/sv_{max} = \sqrt{100\times9.8\times\dfrac{0.2+\tan10^\circ}{1-0.2\tan10^\circ}} \approx \textcolor{#e08a1e}{19.6\text{ m/s}}

Between these two speeds, friction is doing some of the work, below v0v_0 it acts up the slope, and above it, down the slope, right up until vmaxv_{max}, past which the car slides outward.

— From the NCERT exercises

A few of the chapter’s own exercise questions, with original worked solutions. (g = 10 m/s² throughout, as the chapter’s exercises specify.)

NCERT Exercise 4.6

A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m/s to 3.5 m/s in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?

Solution

The acceleration is constant, so:

a=3.5−2.025=0.06 m/s2a = \dfrac{3.5-2.0}{25} = 0.06\text{ m/s}^2F=ma=3.0×0.06=0.18 N, in the direction of motionF = ma = 3.0\times0.06 = \textcolor{#e08a1e}{0.18\text{ N, in the direction of motion}}

since the force is speeding the body up without changing its direction, it must point the same way the body is already moving.

NCERT Exercise 4.13

A man of mass 70 kg stands on a weighing scale in a lift. What is the reading of the scale when the lift is (a) moving upward with a uniform speed of 10 m/s, (b) moving downward with a uniform acceleration of 5 m/s², (c) moving upward with a uniform acceleration of 5 m/s², and (d) falling freely under gravity?

Solution

The scale reads the normal force NN on the man, found from N−mg=maN - mg = ma with upward taken as positive (so a=0a=0 for uniform speed, and aa negative for a downward acceleration):

(a)    a=0  ⇒  N=mg=700 N(b)    a=−5  ⇒  N=m(g−5)=70×5=350 N(c)    a=+5  ⇒  N=m(g+5)=70×15=1050 N(d)    a=−g  ⇒  N=m(g−g)=0\begin{gathered} \text{(a)}\;\; a=0 \;\Rightarrow\; N = mg = 700\text{ N} \\[6px] \text{(b)}\;\; a=-5 \;\Rightarrow\; N = m(g-5) = 70\times5 = 350\text{ N} \\[6px] \text{(c)}\;\; a=+5 \;\Rightarrow\; N = m(g+5) = 70\times15 = 1050\text{ N} \\[6px] \text{(d)}\;\; a=-g \;\Rightarrow\; N = m(g-g) = \textcolor{#e08a1e}{0} \end{gathered}

Case (d) is why an astronaut, or anyone in true free fall, feels weightless: the scale (or floor) needs to supply no force at all when it is falling exactly as fast as they are.

NCERT Exercise 4.21

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev/min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

Solution

The angular speed first, then tension via §4.8’s fc=mv2/R=mω2Rf_c = mv^2/R = m\omega^2R:

ω=2π×4060≈4.19 rad/s\omega = \dfrac{2\pi\times40}{60} \approx 4.19\text{ rad/s}T=mω2R=0.25×(4.19)2×1.5≈6.6 NT = m\omega^2 R = 0.25\times(4.19)^2\times1.5 \approx \textcolor{#e08a1e}{6.6\text{ N}}

For the maximum speed, set the tension to its limit and solve for v:

vmax=Tmax Rm=200×1.50.25≈34.6 m/sv_{max} = \sqrt{\dfrac{T_{max}\,R}{m}} = \sqrt{\dfrac{200\times1.5}{0.25}} \approx \textcolor{#e08a1e}{34.6\text{ m/s}}
Want the interactive version instead? Try the friction simulation → or take a practice paper →