Laws of Motion ISC Class 11: Notes, Friction and Circular Motion

What Laws of Motion is about

Weightage: Laws of Motion belongs to unit 3 of the ISC Class XI Physics syllabus. CISCE weights units 1-3 together (Physical World and Measurement; Kinematics; Laws of Motion) at 23 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Physics syllabus, examination year 2028, Class XI unit weightage

Key concepts

Force, inertia and Newton's first law

Newton's first law: every body stays at rest, or keeps moving uniformly in a straight line, until an external unbalanced force changes that state. It defines force (what changes the state of motion) and inertia (the tendency to resist the change). Mass is the measure of inertia. Kinds: inertia of rest (a passenger falls back when a bus starts), inertia of motion (falls forward when it brakes) and inertia of direction (pushed outward on a curve). A frame in which the first law holds is an inertial frame; a frame accelerating relative to it is non-inertial. Balanced forces give zero acceleration, not necessarily zero velocity.

Worked example

Explain why (a) dust leaves a carpet when it is beaten with a stick, and (b) a passenger in a turning car seems to be thrown outward.

Try it first, then show the solutionSolution

1. (a) The carpet is suddenly set in motion; the dust, by inertia of rest, stays where it was and separates from the carpet.
2. (b) By inertia of direction the passenger tends to keep moving in a straight line, while the car turns beneath; the car must push the passenger inward to make them turn.

Answer: Both are effects of inertia: of rest and of direction

Momentum, Newton's second law and impulse

Linear momentum $\vec{p}=m\vec{v}$ (kg m s⁻¹). Second law: the rate of change of a body's momentum is proportional to the net force on it and is in the direction of that force: $\vec{F}=\frac{d\vec{p}}{dt}$. For constant mass, $\vec{F}=m\frac{d\vec{v}}{dt}=m\vec{a}$: 1 N is the force that gives 1 kg an acceleration of 1 m s⁻². F = ma is the classical special case; near the speed of light the mass changes, $m=\frac{m_0}{\sqrt{1-v^2/c^2}}$, and only F = dp/dt holds. Impulse $\vec{J}=\int \vec{F}dt={\vec{F}}_{av}\Delta t=\Delta \vec{p}$ (N s), the area under the F-t graph. For a given change of momentum, a longer contact time means a smaller force: catching a ball with moving hands, crumple zones and airbags.

Worked example

A 0.15 kg cricket ball moving at 20 m s⁻¹ is hit straight back at 30 m s⁻¹. The bat is in contact for 0.01 s. Find the impulse and the average force.

Try it first, then show the solutionSolution

1. Taking the final direction as positive: Δp = 0.15(30) − 0.15(−20) = 7.5 N s
2. $F_{av}=\frac{\Delta p}{\Delta t}=\frac{7.5}{0.01}=750N$

Answer: 7.5 N s; 750 N (in the direction of the return)

Newton's third law and conservation of linear momentum

Third law: to every action there is an equal and opposite reaction. Forces occur in pairs, ${\vec{F}}_{AB}=-{\vec{F}}_{BA}$, acting on different bodies, so they never cancel each other. Conservation of momentum: if no external force acts on a system, its total momentum is constant. Proof: during an interaction ${\vec{F}}_{AB}\Delta t=-{\vec{F}}_{BA}\Delta t$, so $\Delta {\vec{p}}_A=-\Delta {\vec{p}}_B$ and $\Delta ({\vec{p}}_A+{\vec{p}}_B)=0$. Conversely, the third law can be derived from conservation of momentum. Applications: recoil of a gun ($MV=mv$), explosions (fragments' momenta add to zero if at rest), rockets (exhaust gas pushed back, rocket pushed forward), swimming and walking.

Worked example

A 4 kg rifle fires a 20 g bullet at 400 m s⁻¹. Find the recoil velocity of the rifle.

Try it first, then show the solutionSolution

1. Initial momentum = 0, so 0 = (0.02)(400) + 4V
2. V = −8/4 = −2 m s⁻¹

Answer: 2 m s⁻¹ opposite to the bullet

Equilibrium of concurrent forces; tension and normal reaction

Forces whose lines of action meet at one point are concurrent. A particle is in equilibrium when the resultant force is zero: $\sum F_x=0$ and $\sum F_y=0$. For three concurrent forces in equilibrium, each force is equal and opposite to the resultant of the other two, and they can be drawn as a closed triangle. Lami's theorem: $\frac{F_1}{\sin \alpha }=\frac{F_2}{\sin \beta }=\frac{F_3}{\sin \gamma }$, where each angle is between the other two forces. Tension is the pulling force in a string (the same throughout a light string over a smooth pulley). Normal reaction is the contact force perpendicular to a surface; it is not always equal to mg.

Worked example

A 100 N weight hangs from two strings that make 30° and 60° with a horizontal ceiling. Find the tension in each string.

Try it first, then show the solutionSolution

1. Horizontal: $T_1\cos 30^{\circ }=T_2\cos 60^{\circ }$, so $T_1=0.577T_2$.
2. Vertical: $T_1\sin 30^{\circ }+T_2\sin 60^{\circ }=100$, so $(0.289+0.866)T_2=100$.
3. $T_2=86.6N$; $T_1=50N$

Answer: 50 N in the 30° string, 86.6 N in the 60° string

Free body diagrams and connected bodies

A free body diagram (FBD) isolates one body and shows every force acting on it (not forces it exerts): weight mg, normal reaction N, tension T, friction f and any applied force. Choose axes (along the acceleration is easiest) and write $\sum F=ma$ for each body. For bodies tied by a light inextensible string over a light frictionless pulley, both move with the same magnitude of acceleration and the tension is the same on both sides. Atwood machine (masses $m_1>m_2$ over one pulley): $a=\frac{(m_1-m_2)g}{m_1+m_2}$, $T=\frac{2m_1{m}_{2}g}{m_1+m_2}$. Lift problems: apparent weight R = m(g + a) going up with acceleration a, m(g − a) going down, zero in free fall. (Systems with two pulleys are not required.)

Worked example

Masses of 5 kg and 3 kg hang from the ends of a light string over a smooth pulley. Find the acceleration and the tension. (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. 5 kg: 5g − T = 5a; 3 kg: T − 3g = 3a
2. Adding: 2g = 8a, so a = 2.45 m s⁻²
3. T = 3(9.8 + 2.45) = 36.75 N (check: $\frac{2\times 5\times 3\times 9.8}{8}=36.75$)

Answer: 2.45 m s⁻²; 36.75 N

Friction: static, kinetic and rolling; laws of friction

Friction opposes relative motion (or attempted motion) between surfaces in contact. Classical view: interlocking of surface irregularities. Modern view: temporary adhesion (cold welding) at the few points of real contact, which must be broken. Static friction adjusts to match the applied force up to the limiting friction $f_s^{max}={\mu }_{s}N$. Kinetic friction $f_k={\mu }_{k}N$ acts once sliding starts; ${\mu }_k<{\mu }_s$. Laws of limiting friction: it acts opposite to the impending motion; it is proportional to the normal reaction; it is independent of the area of contact (for the same N); it depends on the nature of the surfaces. Rolling friction is much smaller than sliding friction (wheels, ball bearings). Lubrication (oil, grease, air cushions) reduces friction by keeping surfaces apart. Angle of friction λ: angle between the resultant of limiting friction and N, and N: $\tan \lambda ={\mu }_s$. Angle of repose θ: the steepest incline on which a body rests without sliding: $\tan \theta ={\mu }_s$, so θ = λ.

Worked example

A block placed on a plank just begins to slide when the plank is tilted to 30°. Find the coefficient of static friction.

Try it first, then show the solutionSolution

1. At the angle of repose, $mg\sin \theta ={\mu }_{s}mg\cos \theta$
2. ${\mu }_s=\tan 30^{\circ }=0.577$

Answer: 0.58

Motion on a rough inclined plane; pulling and pushing a roller

On an incline of angle θ: weight components are $mg\sin \theta$ down the slope and $mg\cos \theta$ into it, so $N=mg\cos \theta$ and kinetic friction is ${\mu }_{k}mg\cos \theta$. Sliding down: $a=g(\sin \theta -{\mu }_k\cos \theta )$. Pushed up (friction now down the slope): force needed for constant speed $F=mg(\sin \theta +{\mu }_k\cos \theta )$; to hold it from sliding down, the least force is $mg(\sin \theta -{\mu }_s\cos \theta )$. Roller: pulling with force F at angle θ above the horizontal gives $N=mg-F\sin \theta$, while pushing at θ below the horizontal gives $N=mg+F\sin \theta$. Pulling reduces N and hence friction, so it is easier to pull a lawn roller than to push it.

Worked example

A 2 kg block slides down a 30° incline with ${\mu }_k=0.2$. Find its acceleration, and the force along the incline needed to push it up at constant speed. (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. a = 9.8(sin 30° − 0.2 cos 30°) = 9.8(0.5 − 0.173) = 3.20 m s⁻²
2. F = 2 × 9.8(0.5 + 0.173) = 13.2 N

Answer: 3.2 m s⁻² down; 13.2 N up the incline

Dynamics of circular motion: level and banked roads

Angular displacement θ, angular velocity $\omega =\frac{d\theta }{dt}$ and angular acceleration $\alpha =\frac{d\omega }{dt}$ describe rotation; v = ωr and the centripetal acceleration is $\frac{v^2}{r}={\omega }^{2}r$. The centripetal force $F_c=\frac{m{v}^2}{r}$ is not a new kind of force: it is the net inward force supplied by friction, tension, gravity or a normal reaction. Level road: friction supplies it, so $\frac{m{v}^2}{r}\le {\mu }_{s}mg$ and $v_{max}=\sqrt{{\mu }_{s}rg}$. Banked road (outer edge raised by angle θ): with no friction, the horizontal component of N supplies $F_c$: $N\sin \theta =\frac{m{v}^2}{r}$, $N\cos \theta =mg$, so $\tan \theta =\frac{v^2}{rg}$. With friction, $v_{max}=\sqrt{\frac{rg(\mu +\tan \theta )}{1-\mu \tan \theta }}$. Railway track: the outer rail is raised by h above the inner; for track width l and a small angle, $h\approx l\tan \theta =\frac{l{v}^2}{rg}$, so the flanges do not press on the outer rail. A cyclist leans inward at angle θ with $\tan \theta =\frac{v^2}{rg}$.

Worked example

(a) Find the maximum safe speed on a level curve of radius 50 m with ${\mu }_s=0.4$. (b) At what angle should a curve of radius 100 m be banked for 72 km h⁻¹ with no friction? (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. (a) $v_{max}=\sqrt{0.4\times 50\times 9.8}=14m{s}^{-1}$ (50.4 km h⁻¹)
2. (b) v = 20 m s⁻¹; $\tan \theta =\frac{400}{100\times 9.8}=0.408$
3. θ = 22.2°

Answer: (a) 14 m s⁻¹ (b) about 22°

Motion in a vertical circle

For a body of mass m on a string of length r whirled in a vertical circle, gravity changes the speed around the circle. At the top: $T_{top}+mg=\frac{m{v}_{top}^2}{r}$; the string stays taut only if $T_{top}\ge 0$, so the minimum speed at the top is $\sqrt{gr}$. At the bottom: $T_{bottom}-mg=\frac{m{v}_{bottom}^2}{r}$. Energy conservation between top and bottom: $v_{bottom}^2=v_{top}^2+4gr$, so the minimum speed at the bottom for a complete circle is $\sqrt{5gr}$, and with these speeds $T_{bottom}-T_{top}=6mg$. A bucket of water swung fast enough does not spill at the top because the water's weight is less than the centripetal force needed.

Worked example

A 0.2 kg stone on a 1 m string is whirled in a vertical circle. Find the least speed at the top and at the bottom for it to complete the circle, and the tension at the bottom in that case. (g = 9.8 m s⁻²)

Try it first, then show the solutionSolution

1. $v_{top}=\sqrt{gr}=\sqrt{9.8}=3.13m{s}^{-1}$
2. $v_{bottom}=\sqrt{5gr}=\sqrt{49}=7m{s}^{-1}$
3. $T_{bottom}=\frac{0.2\times 49}{1}+0.2\times 9.8=9.8+1.96=11.76N$ (= 6mg, since $T_{top}=0$)

Answer: 3.13 m s⁻¹; 7 m s⁻¹; 11.76 N

Equilibrium of a rigid body: the ladder problem

A rigid body is in equilibrium when (1) the vector sum of forces is zero (no translation) and (2) the sum of the moments (torques) about any point is zero (no rotation). For three coplanar forces keeping a rigid body in equilibrium, their lines of action must meet at one point (or be parallel). Ladder against a smooth wall on rough ground: forces are the weight W at the middle, the wall's normal reaction $R_w$ (horizontal, as the wall is smooth), the ground's normal reaction $R_g$ (vertical) and friction f at the foot (towards the wall). Then $R_g=W$, $f=R_w$, and moments about the foot give $R_{w}h=W\frac{d}{2}$, where h is the height of the top and d the distance of the foot from the wall. The ladder does not slip if $f\le \mu R_g$.

Worked example

A uniform ladder 5 m long and weighing 200 N leans against a smooth wall with its foot 3 m from the wall. Find the friction at the foot and the least coefficient of friction needed.

Try it first, then show the solutionSolution

1. Height of the top: $h=\sqrt{5^2-3^2}=4m$; $R_g=200N$
2. Moments about the foot: $R_w\times 4=200\times 1.5$, so $R_w=75N$ and f = 75 N
3. ${\mu }_{min}=\frac{75}{200}=0.375$

Answer: 75 N; μ ≥ 0.375

Physics in real life: Laws of Motion

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Circular Motion, Banking and Vertical Circle
• Friction and Motion on an Incline

Laws of Motion formula sheet

Second law

$\vec{F}=\frac{d\vec{p}}{dt}=m\vec{a}(m\text{constant})$

1 N = 1 kg m s⁻².

Impulse

$\vec{J}={\vec{F}}_{av}\Delta t=\Delta \vec{p}$

Area under the F-t graph; unit N s.

Conservation of momentum

$m_1{\vec{u}}_1+m_2{\vec{u}}_2=m_1{\vec{v}}_1+m_2{\vec{v}}_2$

No external force; recoil MV = −mv.

Lami's theorem

$\frac{F_1}{\sin \alpha }=\frac{F_2}{\sin \beta }=\frac{F_3}{\sin \gamma }$

Three concurrent forces in equilibrium; angle opposite each force.

Atwood machine

$a=\frac{(m_1-m_2)g}{m_1+m_2},T=\frac{2m_1{m}_{2}g}{m_1+m_2}$

Light string, smooth pulley.

Apparent weight in a lift

$R=m(g\pm a)$

+ accelerating up; − accelerating down; 0 in free fall.

Friction

$f_s\le {\mu }_{s}N,f_k={\mu }_{k}N,\tan \lambda =\tan {\theta }_{repose}={\mu }_s$

${\mu }_k<{\mu }_s$.

Rough incline

$a_{down}=g(\sin \theta -{\mu }_k\cos \theta ),F_{up}=mg(\sin \theta +{\mu }_k\cos \theta )$

Friction opposes the motion.

Level and banked road

$v_{max}=\sqrt{\mu rg};\tan \theta =\frac{v^2}{rg};v_{max}=\sqrt{\frac{rg(\mu +\tan \theta )}{1-\mu \tan \theta }}$

Rail: $h\approx \frac{l{v}^2}{rg}$.

Vertical circle

$v_{top}\ge \sqrt{gr},v_{bottom}\ge \sqrt{5gr},T_{bottom}-T_{top}=6mg$

Using $v_b^2=v_t^2+4gr$.

Laws of Motion: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Laws of Motion

The question patterns to expect, and the method that works for each.

Tricks to remember Laws of Motion

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Action and reaction on different bodies

  Say 'A on B, B on A': if both forces are on the same body, they are not a third-law pair.

  Use it when: Explaining why action and reaction do not cancel.

• Pull is easier than push

  Pull lifts (N = mg − F sin θ); push presses (N = mg + F sin θ).

  Use it when: Lawn-roller questions.

• Up the slope, add; down the slope, subtract

  Going up: sin θ + μ cos θ. Sliding down: sin θ − μ cos θ.

  Use it when: Incline numericals.

• 5-gr-6

  Vertical circle: top √(gr), bottom √(5gr), tension difference 6mg.

  Use it when: Vertical-circle questions.

• Banking: tan = v² over rg

  "Tan the curve, v-squared over r-g".

  Use it when: Angle of banking and cyclist lean.

Laws of Motion: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Laws of Motion

The slips examiners see most often, and what to write instead.

• ✗Mistake: Taking the normal reaction as mg every time

  On an incline, in a lift, or when a force is applied at an angle, N differs from mg, and friction μN changes with it.

  Wrong working

  Block on a 30° incline: f = μmg.

  Right working

  f = μmg cos 30°; pulling at θ: N = mg − F sin θ.

  ✓Do this: Find N from the FBD before writing f = μN.

  Usually costs: 1-2 marks.

• ✗Mistake: Treating centripetal force as an extra force on the FBD

  Students add 'mv²/r' as one more arrow alongside N, mg and friction, counting the same force twice.

  Wrong working

  On a banked road: N sin θ + mv²/r = 0.

  Right working

  The net inward force equals mv²/r: N sin θ = mv²/r.

  ✓Do this: Draw only real forces; then set their inward resultant equal to mv²/r.

  Usually costs: 2-3 marks.

• ✗Mistake: Cancelling action and reaction

  Students argue that since the horse pulls the cart and the cart pulls the horse equally, nothing can move.

  Wrong working

  The forces are equal and opposite, so the net force is zero and the cart cannot move.

  Right working

  The two forces act on different bodies; the cart moves because the horse's pull on it exceeds the friction on the cart.

  ✓Do this: Write 'on which body?' next to every force.

  Usually costs: 1-2 marks.

• ✗Mistake: Forgetting signs in momentum changes

  When a ball rebounds, students subtract the speeds instead of adding them.

  Wrong working

  0.15 kg ball, 20 m s⁻¹ in and 30 m s⁻¹ out: Δp = 0.15(30 − 20) = 1.5 N s.

  Right working

  Δp = 0.15(30 − (−20)) = 7.5 N s.

  ✓Do this: Choose a positive direction and give every velocity a sign.

  Usually costs: 1 mark.

• ✗Mistake: Using $\sqrt{gr}$ as the speed at the bottom of a vertical circle

  Students apply the top condition everywhere.

  Wrong working

  Least speed at the bottom = √(gr).

  Right working

  Least speed at the top = √(gr); at the bottom = √(5gr), using $v_b^2=v_t^2+4gr$.

  ✓Do this: Write energy conservation between top and bottom first.

  Usually costs: 1-2 marks.

Practice questions on Laws of Motion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The SI unit of impulse is:
   (a) N
   (b) N s
   (c) N s⁻¹
   (d) kg m s⁻²

   Hint

   Force × time.

   Answer

   (b) N s (same as kg m s⁻¹)

2. Q2 · 1 mark · MCQExam level

   A rocket moves forward because of:
   (a) conservation of energy
   (b) conservation of linear momentum
   (c) the air pushing on it
   (d) Newton's first law

   Hint

   Gases backward, rocket forward.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   State Newton's first law of motion. Why is it called the law of inertia?

   Hint

   Rest or uniform motion.

   Answer

   A body continues in its state of rest or uniform straight-line motion unless an external unbalanced force acts on it. It describes inertia, the tendency to resist a change in that state.

4. Q4 · 1 mark · Very short answerExam level

   Why are ball bearings used in machines?

   Hint

   Rolling versus sliding.

   Answer

   They replace sliding friction by much smaller rolling friction, reducing wear and energy loss.

5. Q5 · 2 marks · Short answerExam level

   State two laws of limiting friction.

   Hint

   Direction, proportionality, area.

   Answer

   Any two: it acts opposite to the impending motion; it is proportional to the normal reaction (f = μN); it is independent of the area of contact; it depends on the nature of the surfaces in contact.

6. Q6 · 2 marks · Short answerHarder exam level

   A 50 N weight hangs from two strings, each making 60° with the vertical. Find the tension in each string.

   Hint

   Vertical components share the weight.

   Answer

   2T cos 60° = 50, so T = 50 N.

7. Q7 · 2 marks · Short answerHarder exam level

   A cyclist goes round a curve of radius 20 m at 10 m s⁻¹. At what angle to the vertical must he lean? (g = 9.8 m s⁻²)

   Hint

   Same as banking.

   Answer

   tan θ = v²/rg = 100/196 = 0.51, θ = 27.0°.

8. Q8 · 2 marks · Short answerHarder exam level

   A 3 kg shell at rest explodes into two pieces of 1 kg and 2 kg. The 1 kg piece moves at 20 m s⁻¹. Find the velocity of the other piece and the energy released.

   Hint

   Total momentum stays zero.

   Answer

   2v = 1 × 20, v = 10 m s⁻¹ opposite. Energy = ½(1)(400) + ½(2)(100) = 300 J.

9. Q9 · 3 marks · Short answerHarder exam level

   A 500 kg lift is pulled up by a cable. Find the tension when it (a) accelerates upward at 1 m s⁻², (b) moves at constant speed, (c) accelerates downward at 1 m s⁻². (g = 9.8 m s⁻²)

   Hint

   T − mg = ma.

   Answer

   (a) 500 × 10.8 = 5400 N (b) 4900 N (c) 500 × 8.8 = 4400 N.

10. Q10 · 3 marks · Short answerHarder exam level

    A 0.5 kg stone tied to a 1 m string moves in a vertical circle with speed 5 m s⁻¹ at the top. Find the tension at the top, the speed at the bottom and the tension at the bottom. (g = 9.8 m s⁻²)

    Hint

    $v_b^2=v_t^2+4gr$.

    Answer

    $T_{top}=0.5\times 25-4.9=7.6N$; $v_b^2=25+39.2=64.2$, $v_b=8.01$ m s⁻¹; $T_{bottom}=0.5\times 64.2+4.9=37.0N$ (difference 29.4 N = 6mg).

11. Q11 · 2 marks · Short answerHarder exam level

    A 20 kg roller is moved by a force of 100 N at 30° to the horizontal. Find the normal reaction when it is (a) pulled and (b) pushed. (g = 9.8 m s⁻²)

    Hint

    Vertical component F sin 30° = 50 N.

    Answer

    (a) N = 196 − 50 = 146 N (b) N = 196 + 50 = 246 N; pulling gives less friction.

12. Q12 · 5 marks · Long answerHarder exam level

    (a) Starting from F = dp/dt, show that F = ma for constant mass. [1]
    (b) Using Newton's second and third laws, prove the law of conservation of linear momentum for two colliding bodies. [2]
    (c) A 20 g bullet at 400 m s⁻¹ embeds in a 1.98 kg wooden block at rest on a smooth table. Find their common velocity. [2]

    Hint

    Forces during contact are equal and opposite for the same time.

    Answer

    (a) F = d(mv)/dt = m dv/dt = ma. (b) During contact Δt, $F_{AB}=-F_{BA}$, so $F_{AB}\Delta t=-F_{BA}\Delta t$, i.e. $\Delta p_B=-\Delta p_A$, so $p_A+p_B$ is unchanged. (c) 0.02 × 400 = 2.00v, v = 4 m s⁻¹.

13. Q13 · 3 marks · Short answerHOTS / competency

    A curve of radius 100 m is banked at 15°. If ${\mu }_s=0.2$, find the maximum safe speed. (tan 15° = 0.268, g = 9.8 m s⁻²)

    Hint

    Friction acts down the slope at maximum speed.

    Answer

    $v^2=\frac{rg(\mu +\tan \theta )}{1-\mu \tan \theta }=\frac{980\times 0.468}{0.946}=485$, v = 22.0 m s⁻¹ (about 79 km h⁻¹).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.State Newton's second law in terms of momentum.Show answer

   The rate of change of momentum is proportional to the force and in its direction: F = dp/dt.

2. 2.Define impulse.Show answer

   Force × time of action = change in momentum (N s).

3. 3.Relation between angle of repose and μ?Show answer

   tan θ = μ_s.

4. 4.Is kinetic friction more or less than limiting friction?Show answer

   Less: μ_k < μ_s.

5. 5.Maximum speed on a level curve?Show answer

   $\sqrt{\mu rg}$.

6. 6.Angle of banking without friction?Show answer

   tan θ = v²/(rg).

7. 7.Minimum speed at the top of a vertical circle?Show answer

   √(gr).

8. 8.Is the conical pendulum in the ISC syllabus?Show answer

   No, it is excluded.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state Newton's three laws exactly?
• ☐Can you find impulse and average force with correct signs?
• ☐Can you solve a recoil or explosion problem?
• ☐Can you draw FBDs and solve an Atwood machine?
• ☐Can you show that the angle of repose equals the angle of friction?
• ☐Can you find acceleration down and force up a rough incline?
• ☐Can you derive the angle of banking and the maximum speed on a level road?
• ☐Can you find the tension at the top and bottom of a vertical circle?

Laws of Motion: frequently asked questions