📌 Snapshot
- Newton's laws, friction and circular motion explained from first principles with worked CUET numericals.
- Understand inertia, momentum and impulse, and why a cricketer pulls his hands back to catch a fast ball.
- Learn why action and reaction never cancel, and use momentum conservation for recoil and collisions.
- Handle friction, banked roads and free body diagrams without guessing which formula applies.
📖 Detailed Notes
Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 4.
Why do you lurch forward when the bus brakes?
Nobody pushed you. The bus slowed down and you did not. That small, familiar jolt contains the whole of Newtonian mechanics: a body keeps doing what it is doing until something forces it to stop, how hard you must push depends on how much "stuff" is moving, and every push you give comes back at you. This chapter turns those three observations into three precise laws, and then uses them on friction, collisions and cars taking a bend.
From Aristotle to Galileo: why things stop
Aristotle said that a body needs a continuous push to keep moving. It looks reasonable, because a toy car you stop pushing does come to rest. But the car stops because the floor pushes back with friction, not because motion needs a cause. Galileo tested this with two inclined planes facing each other. A ball released from one plane climbs nearly to the same height on the other. Make the second plane gentler and the ball travels farther to regain that height. Make it horizontal, remove friction, and the ball would roll forever at constant velocity.
That is the law of inertia: rest and uniform motion in a straight line are equally natural states, and neither needs a force to maintain it.
Newton's first law and inertia
First law. Every body continues in its state of rest or of uniform motion in a straight line unless an external force compels it to change that state. Put another way, if the net external force on a body is zero, its acceleration is zero, and a non-zero acceleration needs a net external force.
Inertia is the resistance of a body to any change in its state of rest or of uniform motion, and mass is its measure. A heavier cart is harder to start and harder to stop.
| Situation | What happens | The explanation |
|---|---|---|
| Bus starts suddenly | Passengers fall backward | Your body tends to stay at rest while your feet are carried forward by the floor (inertia of rest) |
| Bus brakes suddenly | Passengers fall forward | Your body tends to keep moving while friction slows your feet (inertia of motion) |
| Book resting on a table | No acceleration | Weight down and normal reaction up are balanced, so the net force is zero |
Notice the last row. Weight and normal reaction are equal and opposite because the book is observed to be at rest, and the first law then demands zero net force. It is not a consequence of the third law.
Momentum and the second law
Momentum is $p = mv$, a vector in the direction of velocity. It captures how hard a moving body is to stop: a heavy lorry and a light bicycle at the same speed need different efforts, and so do a fast and a slow bicycle.
Second law. The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.
$$F = \dfrac{dp}{dt}$$
With mass constant, $dp/dt = m\,dv/dt$, so
$$F = ma$$
The constant of proportionality is chosen as 1 in SI units, which defines the newton: 1 N = 1 kg m s⁻², the force that gives a 1 kg mass an acceleration of 1 m s⁻².
| Feature of the second law | What it means |
|---|---|
| F = 0 implies a = 0 | Contains the first law as a special case |
| It is a vector equation | Three independent component equations, along x, y and z |
| F is the net external force | Internal forces between parts of a system cancel in pairs |
| For a system, a is the acceleration of the centre of mass | Works for a single particle, a rigid body or a collection of particles |
| It is a local law | The acceleration at this instant depends only on the force at this instant, not on the history |
Force is needed even to change only the direction of momentum. A stone whirled in a circle at constant speed has changing momentum, and the string supplies the force.
Worked example: a bullet in a block. A 20 g bullet moving at 200 m s⁻¹ penetrates 10 cm into a wooden block and stops. Find the average resistive force.
| Step | What we do | Result |
|---|---|---|
| 1 | Convert units: m = 0.020 kg, s = 0.10 m | |
| 2 | Retardation from $v^2 = u^2 + 2as$ with v = 0 | $a = u^2/2s = 40\,000/0.2 = 2 \times 10^5$ m s⁻² |
| 3 | Second law | $F = ma = 0.020 \times 2 \times 10^5 = 4000$ N |
| 4 | Check with energy: kinetic energy $1/2\,mu^2$ = 400 J, and F × s = 400 J | F = 400 / 0.10 = 4000 N |
The common wrong step: using the mass as 20 (grams) in a formula that needs kilograms. Newtons come only from kilograms, metres and seconds.
Impulse
When a large force acts for a very short time (a bat hitting a ball, a collision), neither F nor the time is easy to measure, but their product is. Impulse is the force multiplied by the time it acts, and it equals the change in momentum:
$$J = F\,\Delta t = \Delta p$$
Its unit is N s, which equals kg m s⁻¹. The same impulse can come from a big force over a short time or a small force over a long time, and that is exactly why a cricketer draws his hands back while catching.
Worked example: the cricketer's catch. A 0.16 kg ball moving at 30 m s⁻¹ is caught and brought to rest. Find the average force on the hands if the catch takes 0.02 s (hands rigid) and if it takes 0.1 s (hands drawn back).
| Step | What we do | Result |
|---|---|---|
| 1 | Change in momentum (ball stops) | $\Delta p = 0.16 \times 30 = 4.8$ N s |
| 2 | Rigid hands | F = 4.8 / 0.02 = 240 N |
| 3 | Hands drawn back | F = 4.8 / 0.1 = 48 N |
The impulse is the same, but a five times longer catch means a five times smaller force.
The common wrong step: forgetting that a ball which rebounds has its momentum reversed. If the ball in this example flew back at 30 m s⁻¹ the change would be $2mv = 9.6$ N s, twice as large, not 4.8 N s.
Beyond the textbook. For the exam: $F = ma$ for constant mass, and $F = dp/dt$ in general. In reality: the general form is the one that works when the mass changes. A rocket burns fuel and throws gas backward at high speed, so its mass falls as it moves, and F = ma alone would give the wrong answer. Momentum is conserved between the rocket and its exhaust, so the gas goes backward and the rocket forward with no need for air to push against. Why it matters: it is why rockets work in a vacuum, which a "pushing against the air" picture cannot explain.
Newton's third law
Third law. To every action there is an equal and opposite reaction. In terms of two bodies: the force on A due to B is equal and opposite to the force on B due to A.
$$F_{AB} = -F_{BA}$$
Three things about this law decide most exam questions.
| Fact | Consequence |
|---|---|
| The two forces act on different bodies | They never cancel on a single body |
| They occur simultaneously | "Action" and "reaction" are just labels; neither causes the other |
| They are the same kind of force | A gravitational pull is paired with a gravitational pull, a contact push with a contact push |
If action and reaction cancelled, nothing could ever accelerate. They do not cancel because one acts on the horse and the other on the cart. The horse moves forward because the ground pushes it forward (friction) harder than the cart pulls it back.
Pause and predict: a heavy truck hits a fly on the windscreen. Which feels the larger force, the truck or the fly?
Check your answer: equal. The truck pushes the fly with the same force the fly pushes the truck. The fly is destroyed because the same force on a tiny mass gives it an enormous acceleration, and on the truck a very small one.
A book on a table has two force pairs that are easy to confuse. The Earth pulls the book down (weight), and its pair is the book pulling the Earth up. The table pushes the book up (the normal reaction), and its pair is the book pushing the table down. The weight and the normal reaction on the book are not a third-law pair: they act on the same body.
Conservation of linear momentum
Take two bodies that interact only with each other. The force on A is the negative of the force on B at every instant, so $dp_a/dt = -dp_b/dt$, and therefore $d(p_a + p_b)/dt = 0$. The total momentum of an isolated system is conserved. This is the second law and the third law working together, and it holds for any interaction between the bodies, even one too complicated to describe.
Two applications are examination staples. In recoil, a gun and a bullet start at rest, so the total momentum is zero and afterwards the two momenta are equal and opposite. In collisions, momentum is conserved whether the collision is elastic (kinetic energy also conserved) or inelastic (some kinetic energy lost as heat and sound).
Worked example: gun and bullet. A 5 kg gun fires a 50 g bullet at 400 m s⁻¹. Find the recoil speed of the gun and compare the two kinetic energies.
| Step | What we do | Result |
|---|---|---|
| 1 | Initial momentum is zero, so $m v_b + M V = 0$ | $V = -(0.050 \times 400)/5 = -4$ m s⁻¹ |
| 2 | Interpret | The gun recoils at 4 m s⁻¹, opposite to the bullet |
| 3 | Kinetic energies | Bullet: $1/2 \times 0.05 \times 400^2 = 4000$ J. Gun: $1/2 \times 5 \times 4^2 = 40$ J |
The momenta are equal in magnitude (20 kg m s⁻¹), but the light bullet carries 100 times the kinetic energy. The common wrong step: thinking "momentum is conserved, so the gun does not move". The total is zero, which means the two parts have opposite momenta, not that each is zero.
Equilibrium of concurrent forces
A particle is in equilibrium when the net force on it is zero. With two forces they must be equal and opposite. With three concurrent forces, their vectors drawn head to tail close into a triangle, and with n forces they close into a polygon. In practice you resolve the forces into x and y components and set each sum to zero.
For example, a 6 kg mass hangs from a rope and a horizontal 50 N force is applied at the rope's midpoint. In equilibrium the vertical rope carries the weight (60 N, taking g = 10 m s⁻²), and the upper rope supplies both a vertical component of 60 N and a horizontal component of 50 N. So $\tan \theta = 50/60$ for the angle $\theta$ with the vertical, giving $\theta \approx 40^\circ$.
Common forces and free body diagrams
| Force | Direction | Key rule |
|---|---|---|
| Gravity (weight) | Toward the centre of the Earth | Acts at a distance; W = mg |
| Normal reaction N | Perpendicular to the contact surface | Adjusts to whatever the surface can support |
| Tension T | Along the string, pulling | The same throughout an inextensible, massless string |
| Spring force | Opposite to the displacement | $F = -kx$, where x is measured from the natural length and k is the force constant |
| Friction | Along the contact surface | Discussed below |
To solve any force problem, follow the same routine. Draw a clear picture and choose the body or system you will analyse. Draw only the forces acting on that body (never the forces it exerts on others), choose axes, apply F = ma along each axis, and use the third law when you move between bodies.
Worked example: two blocks joined by a string. Blocks of 2 kg (A) and 3 kg (B) rest on a smooth table, joined by a light string. A horizontal pull of 10 N is applied to B, directly away from A. Find the acceleration and the tension in the string.
| Step | What we do | Result |
|---|---|---|
| 1 | Treat both blocks as one system. The string is internal, so only the 10 N is external | a = 10 / (2 + 3) = 2 m s⁻² |
| 2 | Look at block A alone. The only horizontal force on it is the tension | $T = m_A a = 2 \times 2 = 4$ N |
| 3 | Check on block B: 10 − T = 3 × 2 | T = 4 N |
The common wrong step: saying the tension is 10 N. The string pulls A with 4 N and pulls B backward with 4 N, and only the remaining 6 N of the 10 N accelerates B.
Friction
Friction is the contact force parallel to the surfaces. It comes in three varieties, and the differences are the examinable part.
| Static friction | Kinetic (sliding) friction | Rolling friction | |
|---|---|---|---|
| When | Surfaces are at relative rest but a push tries to slide them | Surfaces are sliding past each other | A body rolls without slipping |
| Value | Self-adjusting, equal to the applied push up to a maximum: $f_s \le \mu_s N$ | $f_k = \mu_k N$, nearly independent of speed | Much smaller than sliding friction |
| Direction | Opposes the impending relative motion | Opposes the actual relative motion | Opposes rolling |
| Coefficient | $\mu_s$, the larger of the two | $\mu_k$, usually less than $\mu_s$ | Smaller still |
Both coefficients depend only on the nature of the two surfaces and not on the apparent area of contact. As you increase a push on a resting block, the static friction grows with it until the maximum $\mu_s N$, then the block starts to slide and the friction drops to the smaller $f_k$. That is why it takes more effort to start a heavy box moving than to keep it moving.
A useful result: put a block on an incline and raise the angle slowly. The block begins to slide when $mg \sin \theta$ equals $\mu_s mg \cos \theta$, so $\mu_s = \tan \theta_{\max}$, the angle of repose. The mass cancels out.
Worked example: pushing a crate. A 10 kg crate rests on a floor with $\mu_s = 0.5$ and $\mu_k = 0.4$ (g = 10 m s⁻²). Find the friction and the acceleration for a horizontal push of 40 N and of 60 N.
| Step | What we do | Result |
|---|---|---|
| 1 | Normal reaction on the level floor | N = mg = 100 N |
| 2 | Maximum static friction | $\mu_s N = 50$ N |
| 3 | Push of 40 N is less than 50 N | The crate stays at rest, and friction equals the push, 40 N |
| 4 | Push of 60 N exceeds 50 N | The crate slides, so friction is $f_k = \mu_k N = 40$ N |
| 5 | Net force and acceleration | a = (60 − 40) / 10 = 2 m s⁻² |
The common wrong step: using 50 N ($\mu_s N$) at the 40 N push, which would imply the crate accelerates backward, or using 50 N once the crate is sliding, which gives a = 1 m s⁻². Static friction equals $\mu_s N$ only at the verge of slipping.
Beyond the textbook. For the exam: friction opposes motion. In reality: it opposes the relative sliding of the two surfaces, and that sometimes means it pushes you forward. When you walk, your foot pushes backward on the ground, so the ground's static friction on you points forward. The same is true for the driving wheels of a car, and it is why you cannot walk on ice. Why it matters: it removes the apparent paradox that friction both stops cars and makes them go.
Beyond the textbook. For the exam: friction does not depend on the area of contact. In reality: the surfaces actually touch only at microscopic bumps, and the total real contact area is set by the load pressing them together, which is why friction follows N. Soft materials such as tyre rubber deviate from the simple law. Why it matters: the laws of friction are good approximations rather than exact laws, so always use the stated coefficients in problems.
Circular motion: centripetal force
A body moving in a circle of radius R at speed v needs an inward acceleration $v^2/R$, so it needs a net inward force
$$F_c = \dfrac{mv^2}{R}$$
Centripetal force is not a new kind of force. It is the name for whichever real force supplies the inward push: tension for a stone on a string, gravity for a satellite, friction or the normal reaction for a car on a bend. Never draw it as an extra force on a free body diagram.
Level road. On a flat road the only horizontal force on the car is the static friction from the road, so the car stays on the circle only if $mv^2/R \le \mu_s mg$:
$$v_{\max} = \sqrt{\mu_s R g}$$
The mass of the car cancels, so a heavy truck and a small car have the same limit.
Banked road. If the road is tilted inward by angle θ, the normal reaction itself has an inward horizontal component $N \sin \theta$. At one special speed, the optimum speed, the normal reaction alone provides all the centripetal force and friction is not needed. Resolving gives $N \cos \theta = mg$ and $N \sin \theta = mv^2/R$, so
$$\tan \theta = \dfrac{v^2}{Rg}, \quad \text{that is} \quad v_o = \sqrt{Rg \tan \theta}$$
At speeds above this, friction acts down the slope; below it, friction acts up the slope. The maximum safe speed uses friction at its limit:
$$v_{\max}^2 = \dfrac{Rg(\mu_s + \tan \theta)}{1 - \mu_s \tan \theta}$$
The optimum speed is not the maximum speed: it is the speed at which friction is not needed.
Worked example: a car on a bend. (a) A car goes round a level circular road of radius 40 m with $\mu_s = 0.4$. What is the greatest safe speed? (b) A road of radius 100 m is banked so that cars can take it at 20 m s⁻¹ without any friction. Find the banking angle (g = 10 m s⁻²).
| Step | What we do | Result |
|---|---|---|
| (a) 1 | Friction supplies the centripetal force | $v_{\max} = \sqrt{0.4 \times 40 \times 10} = \sqrt{160}$ |
| (a) 2 | Evaluate | 12.6 m s⁻¹, which is 12.65 × 3.6 ≈ 45.5 km h⁻¹ |
| (b) 1 | No friction, so $\tan \theta = v^2/(Rg)$ | $\tan \theta = 400/1000 = 0.4$ |
| (b) 2 | Angle | $\theta = \tan^{-1} 0.4 \approx 22^\circ$ |
The common wrong step: including the car's mass in the answer. It cancels in both parts, and the same speed limit applies to a bicycle and a bus.
Exam traps
| The trap | Why it is tempting | The truth |
|---|---|---|
| Action and reaction cancel, so nothing moves | "Equal and opposite" sounds like zero | They act on different bodies, so they cannot cancel on either one |
| Weight and the normal reaction are a third-law pair | They are equal and opposite on a resting book | They act on the same body; the first law explains the equality |
| Friction is always $\mu N$ | One formula is easy to remember | Static friction is anything from 0 up to $\mu_s N$; only kinetic friction and the limiting case equal $\mu N$ |
| Friction always opposes motion | Sliding blocks show it | It opposes relative motion of the surfaces, and can push a walker forward |
| $\mu_k$ is larger than $\mu_s$ | Moving feels harder | $\mu_k$ is usually smaller, which is why starting is harder than continuing |
| If v = 0 then F = 0 | A resting body feels balanced | At the top of a throw, v = 0 but a = g |
| Centripetal force is an additional force on the diagram | The name sounds like a force | It is the resultant of the real forces pointing to the centre |
| Optimum speed is the maximum speed on a banked road | Both formulas contain tan θ | The optimum speed needs no friction; the maximum uses limiting friction |
Formula sheet
| Quantity | Formula | SI unit | When to use |
|---|---|---|---|
| Momentum | $p = mv$ | kg m s⁻¹ | Collisions, impulse |
| Second law | $F = dp/dt = ma$ | N | Constant mass: $F = ma$ |
| Impulse | $J = F\,\Delta t = \Delta p$ | N s | Short, large forces |
| Third law | $F_{AB} = -F_{BA}$ | N | Any pair of interacting bodies |
| Conservation of momentum | $\Sigma p$ before = $\Sigma p$ after | kg m s⁻¹ | Recoil, collisions, isolated systems |
| Spring force | $F = -kx$ | N | Springs within their elastic limit |
| Static friction | $f_s \le \mu_s N$ | N | Body at rest |
| Kinetic friction | $f_k = \mu_k N$ | N | Body sliding |
| Angle of repose | $\tan \theta = \mu_s$ | Dimensionless | Block about to slide on a slope |
| Centripetal force | $F_c = mv^2/R$ | N | Any circular motion |
| Level road, maximum speed | $v = \sqrt{\mu_s R g}$ | m s⁻¹ | Flat bend |
| Banked road, no friction | $\tan \theta = v^2/(Rg)$ | Dimensionless | Optimum speed |
| Banked road, maximum speed | $v^2 = Rg(\mu_s + \tan\theta)/(1 - \mu_s\tan\theta)$ | m² s⁻² | Friction at its limit |
60-second revision
| Topic | Lock this in |
|---|---|
| First law | Zero net force means zero acceleration; inertia is the resistance to change of velocity |
| Second law | $F = dp/dt$; for constant mass $F = ma$; 1 N = 1 kg m s⁻² |
| Impulse | $F\,\Delta t = \Delta p$; a longer time of impact means a smaller force |
| Third law | Equal and opposite, simultaneous, on different bodies, same kind of force |
| Momentum | Conserved for an isolated system; gun and bullet have equal and opposite momenta |
| Friction | $f_s \le \mu_s N$ (self-adjusting); $f_k = \mu_k N$; $\mu_k < \mu_s$; rolling is far smaller |
| Angle of repose | $\tan \theta = \mu_s$, independent of mass |
| Centripetal force | $mv^2/R$, supplied by a real force, never an extra one |
| Level road | $v_{\max} = \sqrt{\mu_s R g}$, independent of mass |
| Banked road | Optimum $v = \sqrt{Rg \tan \theta}$ (no friction); maximum speed needs friction |
Where this chapter connects
- Previous: Motion in a Plane gave you centripetal acceleration $v^2/R$ and vector resolution, which this chapter turns into forces.
- Next: Work, Energy and Power uses F and displacement to define work, and revisits the bullet-in-block calculation as a work–energy statement.
- Later in Class XI: System of Particles and Rotational Motion extends conservation of momentum to the centre of mass, and Oscillations builds simple harmonic motion from the restoring force F = −kx.
🎯 Practice MCQs
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Q1. Match List I with List II. List I: (i) A passenger is thrown backward when a bus starts suddenly (ii) The force on the hands is reduced when a cricketer draws his hands back while catching (iii) A gun recoils when it fires (iv) A car slips on a level bend at high speed. List II: (p) Conservation of momentum (q) Friction is insufficient to supply the centripetal force (r) Inertia of rest (s) Longer time of impact for the same change in momentum
▸ Show answer & explanation
Answer: A
The passenger's body tends to stay at rest while the bus moves forward, which is inertia of rest. A longer catch time reduces the force for the same impulse. Gun recoil follows from conservation of momentum. A car slips when $v^2$ exceeds $\mu_s R g$, meaning friction cannot supply $mv^2/R$. The other options mismatch at least two of these.
Q2. Consider these statements about friction. (i) Static friction is self-adjusting up to a maximum value $\mu_s N$. (ii) The coefficient of kinetic friction is usually less than the coefficient of static friction. (iii) Rolling friction is usually greater than sliding friction. (iv) Friction always acts in the direction opposite to the motion of the body. Which are correct?
▸ Show answer & explanation
Answer: A
(i) and (ii) are correct. (iii) is false because rolling friction is much smaller than sliding friction, which is why wheels and ball bearings reduce friction. (iv) is false because friction opposes relative motion of the surfaces, and for a walking person it acts forward.
Q3. A bullet of mass 50 g moving at 200 m s⁻¹ is stopped by a wall in 0.01 s. The average force exerted by the wall on the bullet is
▸ Show answer & explanation
Answer: B
Impulse = change in momentum = 0.050 × 200 = 10 N s. Average force = 10 / 0.01 = 1000 N. Options A and C come from slipping a factor of ten in the time.
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Q4. A ball of mass 0.2 kg strikes a wall normally at 15 m s⁻¹ and rebounds along the same line at the same speed. The magnitude of the impulse given to the ball is
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Answer: C
Take the initial direction as positive. $\Delta p = (-0.2 \times 15) - (0.2 \times 15) = -6$ kg m s⁻¹, so the impulse has magnitude 6 N s. Option B forgets that the momentum reverses, and option A assumes that equal speeds mean no change.
Q5. A gun of mass 4 kg fires a bullet of mass 20 g with a muzzle speed of 300 m s⁻¹. The recoil speed of the gun is
▸ Show answer & explanation
Answer: B
The total momentum stays zero, so $4V = 0.020 \times 300 = 6$, giving $V = 1.5$ m s⁻¹ opposite to the bullet. Option C uses the bullet's momentum, 6 kg m s⁻¹, as if it were a speed, and D divides by 0.4 instead of 4.
Q6. A book rests on a horizontal table. Which of the following forces is the third-law reaction to the weight of the book (the Earth's gravitational pull on it)?
▸ Show answer & explanation
Answer: C
A third-law pair consists of two forces of the same kind acting on two different bodies. The pull of the Earth on the book is paired with the pull of the book on the Earth. The normal reaction (A) acts on the same body as the weight, so it is not a pair, and (B) is the pair of the normal reaction.
Q7. Blocks of 2 kg and 3 kg rest on a smooth horizontal table, joined by a light string. A horizontal force of 10 N pulls the 3 kg block directly away from the 2 kg block. The tension in the string is
▸ Show answer & explanation
Answer: A
The common acceleration is a = 10 / 5 = 2 m s⁻². The tension alone accelerates the 2 kg block, so T = 2 × 2 = 4 N. Option B is the net force on the 3 kg block (10 − 4), option C ignores the 2 kg block, and D divides the force equally.
Q8. A 6 kg mass hangs from a rope. A horizontal force of 50 N is applied at the midpoint of the rope, and the system is in equilibrium. Taking g = 10 m s⁻², the angle the upper half of the rope makes with the vertical is closest to
▸ Show answer & explanation
Answer: A
At the midpoint the upper rope's vertical component balances the 60 N weight and its horizontal component balances the 50 N force. So $\tan \theta = 50/60 = 0.83$, giving $\theta \approx 40^\circ$ ($\tan 40^\circ \approx 0.84$). Option B corresponds to $\tan \theta = 6/5$, with the ratio inverted.
Q9. A 5 kg block rests on a rough horizontal floor with $\mu_s = 0.4$ and $\mu_k = 0.3$. A horizontal force of 12 N is applied to it. Taking g = 10 m s⁻², the friction force on the block is
▸ Show answer & explanation
Answer: C
The maximum static friction is $\mu_s N = 0.4 \times 50 = 20$ N, which is more than 12 N, so the block does not move and static friction exactly balances the push: 12 N. Option A is the limiting value, and option B is the kinetic friction that would act only if the block were sliding.
Q10. A block just begins to slide down a rough plane when the plane is tilted to 30° with the horizontal. The coefficient of static friction between block and plane is
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Answer: B
At the point of slipping, $mg \sin \theta = \mu_s mg \cos \theta$, so $\mu_s = \tan 30^\circ = 1/\sqrt{3} \approx 0.58$, independent of the block's mass. Option A is sin 30°, option C is cos 30°, and option D is tan 60°.
Q11. A road of radius 90 m is banked at 45°. The speed at which a car can take the bend with no help from friction is (g = 10 m s⁻²)
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Answer: B
$v = \sqrt{Rg \tan \theta} = \sqrt{90 \times 10 \times 1} = \sqrt{900} = 30$ m s⁻¹. Option D is v² without the square root.
Q12. Assertion (A): On a level circular road, the maximum safe speed of a car does not depend on its mass. Reason (R): The friction that supplies the centripetal force has a maximum $\mu_s mg$, and the mass m also appears in the required $mv^2/R$.
▸ Show answer & explanation
Answer: A
The condition $mv^2/R \le \mu_s mg$ has m on both sides, so it cancels and leaves $v^2 \le \mu_s R g$. The reason states exactly this cancellation, so it explains the assertion.
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