Laws of Motion | ISC Class 11 Physics Notes

This note covers force and inertia, Newton’s laws of motion, momentum and impulse, conservation of momentum, force diagrams, connected bodies, equilibrium and ladders, friction and inclined planes, angular motion, centripetal force, banked roads and vertical circles.

What do force, inertia and Newton’s first law mean?

A force is an interaction that can change a body’s state of motion. Contact forces require physical contact; gravitational attraction can act across a separation.

Velocity describes the rate of change of position, including direction. Acceleration is the rate of change of velocity. A vector has magnitude and direction, whereas a scalar has magnitude alone. Force, velocity and acceleration are vectors; mass is a scalar.

Definition: Inertia is a body’s resistance to a change in its state of rest or uniform straight-line motion. Mass measures this inertia: a larger mass requires a larger force for the same acceleration.

How does the first law describe motion?

Newton’s first law states that a body remains at rest or moves with constant velocity unless a net external force changes that state. An external force comes from outside the chosen body or collection of bodies, called the system.

Let Σ mean vector summation, F⃗ an external force and a⃗ acceleration; the arrow marks a vector. Then ΣF⃗ = 0 implies a⃗ = 0. Zero acceleration allows rest or constant velocity.

On smooth double inclined planes, the final height is nearly the same as the initial height, a little less but never greater. Equal heights belong to the ideal frictionless case. When a bus starts, the feet move with its floor while the upper body initially resists the change.

How does Newton’s second law connect force and momentum?

Linear momentum is the product of mass and velocity. With p⃗ denoting momentum, m mass and v⃗ velocity, p⃗ = mv⃗. Its direction follows velocity. The SI unit of momentum is kilogram metre per second, written kg m s⁻¹.

SI means the International System of Units. The symbols kg, m and s used in units mean kilogram, metre and second respectively; the variable m in equations means mass. The SI unit of acceleration is metre per second squared, m s⁻².

Derivation: the constant-mass form of the second law

1. The rate of change of momentum is proportional to the net applied force and has its direction. Write F⃗ = k dp⃗/dt, where F⃗ now denotes the net external force, k is a proportionality constant and t is time.
2. The notation d/dt means an instantaneous time derivative, or rate of change. Choose the unit of force so that k = 1; thus F⃗ = dp⃗/dt.
3. For a fixed mass, dp⃗/dt = m dv⃗/dt. Since dv⃗/dt = a⃗, the momentum law reduces to the familiar acceleration law.

F⃗ = ma⃗ holds for constant mass in classical mechanics. Acceleration follows net force, not necessarily velocity.

The SI unit of force is the newton, symbol N. 1 N = 1 kg m s⁻²: one newton gives one kilogram an acceleration of one metre per second squared. Along perpendicular coordinate axes x, y and z, the component equations are Fₓ = maₓ, Fᵧ = maᵧ and F_z = ma_z.

What changes at very high speeds?

When speed approaches the speed of light c, retain the momentum form. In the relativistic-mass convention, m = m₀/√(1 − v²/c²), where m₀ is rest mass and v is speed. Equivalently, momentum is m₀v⃗/√(1 − v²/c²); rest mass itself remains invariant.

Worked example 1. A bullet of mass 0.04 kg enters a heavy wooden block at 90 m s⁻¹ and stops after 60 cm. Find the average resisting force, using constant retardation, meaning acceleration opposing velocity, as the calculation model.

Formula: a = (v² − u²)/(2s); F = ma. Here u and v are initial and final speeds, and s is distance travelled.

Substitute: s = 0.60 m, v = 0; a = −90²/(2 × 0.60) = −6750 m s⁻².

Answer: F = 0.04 × (−6750) = −270 N, so the average resistance has magnitude 270 N. The actual force and retardation may not be uniform.

What is impulse, and why does stopping time matter?

Impulse, denoted J⃗, measures the momentum transferred by a force during an interval. Let Δ mean final value minus initial value, and Δt the duration. For constant force, J⃗ = F⃗Δt = Δp⃗. For a varying force, use its time-average value in this product.

The SI unit of impulse is newton second, N s, equivalent to kg m s⁻¹. Impulse measures total momentum change; force measures its rate.

An impulsive force is a large force acting briefly and producing a finite momentum change. Often force and duration are difficult to determine separately, while momentum change remains measurable.

Why must directions be included?

A ball returning at its original speed has changed momentum because its velocity has reversed. Choose a positive direction before subtracting. Subtracting speed magnitudes alone would incorrectly give zero impulse.

Worked example 2. A batsman returns a 0.15 kg ball straight towards the bowler without changing its speed of 12 m s⁻¹. Find the impulse, assuming linear motion.

Formula: J = m(v − u), where J is the signed impulse along the chosen line.

Substitute: towards the bowler is positive, so u = −12 m s⁻¹ and v = +12 m s⁻¹.

Answer: J = 0.15[12 − (−12)] = 3.6 N s towards the bowler.

A cricketer draws the hands backwards while catching. For the same change of momentum, increasing the stopping time reduces the average force.

How do action, reaction and conservation of momentum fit together?

Newton’s third law states that interacting bodies exert equal and opposite forces on one another. Label two bodies A and B. Write F⃗_AB for the force on A by B and F⃗_BA for the force on B by A: F⃗_AB = −F⃗_BA.

The forces act simultaneously on different bodies. “Action” and “reaction” do not imply a time sequence. Earth attracts a stone, and the stone attracts Earth equally, but Earth’s much greater mass makes its acceleration negligible.

Derivation: conservation of linear momentum

1. Choose an isolated system consisting of interacting bodies A and B. Here isolated means that the net external force is zero.
2. Let p⃗_A and p⃗_B denote their momenta, and let F⃗_A,ext and F⃗_B,ext denote the net external forces on A and B respectively. The second law gives dp⃗_A/dt = F⃗_AB + F⃗_A,ext and dp⃗_B/dt = F⃗_BA + F⃗_B,ext. The assumed zero total external force means F⃗_A,ext + F⃗_B,ext = 0.
3. Add the equations. The third law cancels the internal forces, giving d(p⃗_A + p⃗_B)/dt = 0.
4. Therefore the total momentum remains constant, although either body’s individual momentum can change during the interaction.

p⃗_A + p⃗_B = p⃗′_A + p⃗′_B, where primes identify the final momenta. Conservation applies to the isolated system, not automatically to either member alone.

How can the argument be reversed?

If an isolated pair conserves momentum, its two momentum rates sum to zero. Applying the second law separately gives F⃗_AB + F⃗_BA = 0, obtaining the third law. Setting net force to zero in the constant-mass second law also gives the first law.

This explains the description of the second law as the real law of motion: it quantitatively relates force and motion, includes the first-law result, and connects the third law with momentum conservation. The third law is not derived from the second law alone.

For a gun and bullet initially at rest, final momenta are equal and opposite. If their masses are M and m_b and their signed velocities are V and v_b, then MV + m_bv_b = 0. The negative recoil velocity indicates the gun’s backward motion.

How are concurrent forces and ladder equilibrium analysed?

Concurrent forces have lines of action passing through one point. A line of action is the line along which a force acts; coplanar forces lie in one plane. A particle is in equilibrium when its net external force vanishes.

For three forces, F⃗₁ + F⃗₂ + F⃗₃ = 0. Placing their arrows head to tail gives a closed triangle. Equivalently, the diagonal of a parallelogram gives the resultant of two forces, which must oppose the third.

Why does an extended body also need moment balance?

A rigid body has fixed particle separations; equilibrium requires zero resultant force and torque. Torque, or moment of force, is the turning effect about a chosen point; its magnitude equals force multiplied by perpendicular distance from that point to the force’s line.

Three non-parallel coplanar forces maintaining rigid-body equilibrium must be concurrent and have zero vector sum. Parallel forces also require moment balance. A ladder’s force application points matter.

Weight W = mg is gravitational force, where g is gravitational acceleration. Normal reaction acts perpendicular to contact; static friction prevents relative sliding. A smooth wall is frictionless.

The ladder runs from foot A to wall contact B. Its weight W acts downward at D. The wall force F₁ points horizontally away from the wall; the floor force F₂ is resolved into upward normal force N and friction F towards the wall. These F and N labels denote force magnitudes here.

See Fig. 6.27 in your NCERT textbook

Worked example 3. A uniform ladder of length 3 m and mass 20 kg rests against a frictionless vertical wall, with its foot 1 m from the wall. The rough floor supplies sufficient static friction for equilibrium. Take gravitational acceleration g = 9.8 m s⁻². Find the wall reaction and the two floor-force components.

Formula: W = mg; N = W; F = F₁. The symbol W means weight. The wall-contact height is √(3² − 1²) = 2√2 m; the weight’s horizontal moment arm about the foot is 0.5 m.

Substitute: moment balance about A gives F₁ × 2√2 = (20 × 9.8) × 0.5.

Answer: F₁ = 34.6 N away from the wall; floor friction F = 34.6 N towards the wall; floor normal force N = 196.0 N upwards.

How do free-body diagrams solve connected-body problems?

A free-body diagram shows a chosen system and its external forces. Draw forces exerted on that system. “Free” means isolated for analysis, not free of forces.

Normal reaction is the contact-force component perpendicular to a surface. Tension, written T, is the pulling force transmitted along a taut string. Weight acts vertically downwards. A smooth contact is treated as frictionless; an inextensible string has constant length.

What is a reliable solution sequence?

1. Sketch the bodies, supports and connections, then choose a system boundary.
2. Draw each external force with its direction, leaving unknown magnitudes as symbols.
3. Choose coordinate directions and write the second law separately along them.
4. Use connection constraints to relate accelerations, then solve and check signs and units.

An ideal string of negligible mass over a smooth, negligible-inertia pulley has the same tension on both sides. Its fixed length gives the connected bodies equal acceleration magnitudes in the simple arrangement below. Internal tensions cancel when both bodies form one system.

A 20 kg trolley on a horizontal surface is connected over one pulley to a hanging 3 kg block. The block’s diagram shows tension upwards and 30 N downwards; the trolley’s horizontal diagram shows tension rightwards and kinetic friction leftwards.

See Fig. 4.12 in your NCERT textbook

Worked example 4. For that 20 kg trolley and 3 kg block, take g = 10 m s⁻². The trolley slides rightwards; the kinetic-friction coefficient is 0.04. Assume an inextensible, massless string and an ideal smooth pulley. Find acceleration and tension. Kinetic friction means resistance between sliding surfaces.

Formula: f_k = μ_kN; N = 20g. Here f_k is kinetic-friction magnitude and μ_k is its dimensionless coefficient. For downward block acceleration a, 3g − T = 3a; for the trolley, T − f_k = 20a.

Substitute: f_k = 0.04 × 200 = 8 N; adding the equations gives 30 − 8 = 23a.

Answer: a = 22/23 = 0.96 m s⁻²; T = 30 − 3a = 27.1 N.

How do static and kinetic friction differ?

Friction is the contact-force component parallel to the surfaces, opposing their actual or impending relative motion. Impending motion is the motion that would begin if friction did not prevent it. Friction need not oppose a body’s velocity relative to the ground.

Static friction acts without sliding at the contact and adjusts to what equilibrium or shared acceleration requires, up to a limiting value. Let f_s denote its magnitude and μ_s the coefficient of static friction. The coefficient is a dimensionless force ratio.

f_s ≤ μ_sN, with limiting value f_s,max = μ_sN. The relation concerns the maximum available friction; it does not make every static-friction force equal to that maximum. For the unforced block on a horizontal table, static friction is zero.

What are the empirical laws of friction?

These laws are approximately true empirical relations, not fundamental force laws. Kinetic friction is usually less than maximum static friction. The surfaces determine the coefficients in this model.

Both panels show a block pushed rightwards and friction directed leftwards. Panel (a) labels static friction f_s; panel (b) labels kinetic friction f_k and adds a rightward velocity arrow v.

See Fig. 4.10 in your NCERT textbook

For the usual case of lower kinetic friction, plot applied-force magnitude horizontally and opposing-friction magnitude vertically. Draw a rising equal-value line to f_s,max, then a drop to the lower, approximately constant kinetic-friction level once sliding begins.

Worked example 5. A box must remain stationary relative to a horizontally accelerating train. The coefficient of static friction is 0.15. Take g = 10 m s⁻² and no vertical acceleration. Find the maximum train acceleration.

Formula: ma = f_s ≤ μ_smg; a_max = μ_sg, where a_max denotes the limiting acceleration.

Substitute: a_max = 0.15 × 10.

Answer: a_max = 1.5 m s⁻². Static friction accelerates the box forwards with the train.

How are rough inclines, friction angles and roller forces handled?

For a plane at inclination θ to the horizontal, resolve weight along and perpendicular to it. The symbols sin, cos and tan denote the sine, cosine and tangent trigonometric ratios. With no other perpendicular force, N = mg cos θ; the downward weight component along the plane is mg sin θ.

What changes when the direction of sliding reverses?

For a body sliding down, kinetic friction acts up the plane. Taking down the slope as positive, a = g(sin θ − μ_k cos θ). A negative answer means acceleration is uphill, slowing the downward motion.

For a body sliding up, both gravity’s slope component and kinetic friction point downhill. Taking up the slope as positive, a = −g(sin θ + μ_k cos θ). After the body stops, test static equilibrium before assuming it slides back down.

What are the angles of repose and friction?

The angle of repose, θ_r, is the greatest inclination at which a body can remain at rest without another applied force. At impending downward sliding, mg sin θ_r = μ_smg cos θ_r; therefore tan θ_r = μ_s.

The angle of friction, φ, is the angle between the resultant contact force and the normal at limiting friction. Resolving that contact force gives tan φ = f_s,max/N = μ_s. Hence φ = θ_r for the same pair of surfaces.

Why can pulling reduce resistance compared with pushing?

Let P be an applied-force magnitude and β its angle to the horizontal. Pulling a roller upwards at that angle gives N = mg − P sin β. Pushing downwards gives N = mg + P sin β, provided vertical acceleration is zero and contact remains.

Thus an upward pull reduces the normal load, whereas a downward push increases it. Do not use sliding-friction equations as a complete model of free rolling.

What causes friction, and how can its effects be reduced?

The classical view of friction associates resistance with the interlocking of tiny surface irregularities. The modern microscopic view also involves adhesion and deformation at actual contact regions. These contact interactions ultimately arise from electrical forces between the charged constituents of matter.

How does rolling resistance differ?

Rolling friction is resistance to rolling. In the ideal case of a rigid body rolling uniformly on a perfectly rigid horizontal plane without slipping, there is no relative motion at contact and no dissipative rolling resistance.

Real surfaces deform a little during rolling, creating a finite contact region. For the same weight, rolling friction is much smaller than static or sliding friction. This advantage explains the use of wheels and ball bearings, which reduce sliding between moving machine parts.

What do lubrication and streamlining achieve?

Lubrication introduces material between moving surfaces to reduce kinetic friction. Bearings use rolling contact; a maintained cushion of air can separate moving solid surfaces. Streamlining shapes a body to reduce fluid resistance, the resisting force due to motion through a liquid or gas.

Sliding friction is non-conservative: its work is not recoverable solely from the endpoints of the motion as stored potential energy. Kinetic energy is energy of motion; potential energy depends on position or configuration. Their sum is mechanical energy; friction can convert it into thermal energy.

Friction is also useful. Brakes dissipate mechanical energy; static friction permits walking and supplies traction to accelerating tyres. Reducing friction everywhere would remove these useful interactions along with unwanted resistance.

How are angular displacement, velocity and acceleration related?

Angular displacement θ specifies rotation through an angle. For a point moving on a circle of radius r through arc length s, θ = s/r when θ is measured in radians. A radian, symbol rad, is the angle subtending an arc equal to the radius.

Angular velocity is the rate of angular displacement. For rotation about a fixed axis, use signed angular velocity ω = dθ/dt. Its vector direction is along the axis according to the right-hand rule: curled fingers follow rotation and the thumb gives the direction.

Angular acceleration α is the rate of change of angular velocity: α = dω/dt. The SI unit of angular velocity is radian per second, rad s⁻¹. The SI unit of angular acceleration is radian per second squared, rad s⁻².

Which equations need constant angular acceleration?

Let ω₀ and θ₀ denote initial angular velocity and angular position at t = 0. For a fixed axis and constant α, the angular equations parallel the equations for uniformly accelerated straight-line motion.

For a point at fixed radius r, speed is v = r|ω|, where vertical bars denote magnitude, and signed tangential acceleration is a_t = rα, where a_t is the component along the circular tangent. Uniform circular motion has constant speed and zero tangential acceleration, but still has inward acceleration because velocity changes direction.

Why does uniform circular motion require a centripetal force?

Centripetal acceleration, denoted a_c, is the inward acceleration associated with a circular path. A tangent touches the circle locally and is perpendicular to the radius there. Velocity is tangential, whereas acceleration in uniform circular motion points towards the centre.

Derivation: centripetal acceleration from velocity change

1. Consider two nearby positions separated by time Δt on a circle of radius r, with constant speed v. Let |Δr⃗| mean the magnitude of the displacement between those positions.
2. The two radius vectors and the two velocity vectors turn through the same small angle. Their corresponding triangles are similar, giving |Δv⃗|/v = |Δr⃗|/r.
3. Divide by Δt. As the interval approaches zero, |Δr⃗|/Δt approaches v and |Δv⃗|/Δt approaches a_c.
4. The limiting direction of Δv⃗ is inward. Therefore acceleration has magnitude v²/r and points towards the centre.

a_c = v²/r = rω². The magnitude remains constant in uniform circular motion, while the acceleration vector changes direction continuously.

The panels show successive positions P and P′ on a circle with centre C, tangent velocity arrows, and a separate velocity-difference triangle. As the time interval decreases, the final panel shows acceleration pointing from P towards C.

See Fig. 3.18 in your NCERT textbook

The required inward resultant force is F_c = mv²/r, where F_c denotes centripetal-force magnitude. It is supplied by actual interactions: string tension for a whirled stone, gravity for planetary motion, or tyre-road friction for a level turn.

Note: Centripetal force names the inward resultant required by the motion. Do not add it as another interaction alongside the forces that provide it in a free-body diagram.

How do level and banked roads support circular motion?

On a level circular road, the normal reaction balances weight and static friction supplies the inward force. With r the turn radius, mv²/r ≤ μ_smg. The speed limit is v_max = √(μ_srg), where v_max is the greatest speed allowed by this friction model.

Worked example 6. A cyclist travels at 18 km h⁻¹ round a level turn of radius 3 m. The coefficient of static friction is 0.1. Take g = 9.8 m s⁻². Will the available friction prevent sideways slipping? Here km h⁻¹ means kilometres per hour.

Formula: v² ≤ μ_srg is required for the turn.

Substitute: v = 5 m s⁻¹, so v² = 25 m² s⁻², whereas μ_srg = 0.1 × 3 × 9.8 = 2.94 m² s⁻².

Answer: the limiting speed is √2.94 = 1.71 m s⁻¹, below 5 m s⁻¹, so the cyclist slips.

How does banking reduce reliance on friction?

Banking raises the outer edge of a curved road or railway track. Let θ now be its inclination to the horizontal. Without friction, N cos θ = mg vertically and N sin θ = mv²/r horizontally.

Dividing gives tan θ = v²/(rg). The speed requiring no friction is v₀ = √(rg tan θ), where v₀ denotes this design speed. The same balance explains raising the outer railway rail on a curve.

The level-road panel shows normal force upwards, weight downwards and friction towards the centre. The banked-road panel resolves the inclined normal force and downslope friction into horizontal and vertical components.

See Fig. 4.14 in your NCERT textbook

At the upper limiting speed, friction f acts down the bank: N cos θ − f sin θ = mg and N sin θ + f cos θ = mv²/r. Setting f = μ_sN gives v_max = √[rg(μ_s + tan θ)/(1 − μ_s tan θ)], provided the denominator is positive.

At speeds below v₀, friction acts up the bank. Its direction must therefore be decided from the tendency to slip, not copied unchanged from the upper-speed diagram. A stationary car can remain on the bank only if tan θ ≤ μ_s.

How does motion in a vertical circle differ?

For a particle attached to a light, inextensible string of length r, vertical circular motion generally has changing speed. Gravity has a tangential component during much of the path, so uniform-speed circular equations cannot describe the entire motion without additional forces.

Let v_b and T_b be speed and tension at the bottom; let v_t and T_t be their values at the top. A taut string pulls towards the centre. At the bottom, inward is upwards; at the top, inward is downwards.

What are the radial force equations?

T_b − mg = mv_b²/r at the bottom. At the top, T_t + mg = mv_t²/r. The weight changes its sign in the inward equation because the inward direction reverses between these positions.

A string cannot provide a compressive push. Thus the limiting top condition is T_t = 0, giving v_t,min = √(gr), where the subscript “min” denotes the minimum speed.

How is the minimum bottom speed obtained?

Neglect air resistance and assume a fixed centre. Tension is radial while motion is tangential, so tension does no work. Work is force multiplied by displacement in its direction. Mechanical energy is conserved, with a height increase of 2r between bottom and top.

Hence ½mv_b² = ½mv_t² + mg(2r), or v_b² = v_t² + 4gr. Combining this with the limiting top condition gives v_b,min = √(5gr). These conditions apply to the string model; a rigid rod can also push and has a different constraint.

Glossary

• Inertia — The resistance of a body to changes in rest or uniform straight-line motion.
• Momentum — The product of mass and velocity, directed along the body’s velocity.
• Impulse — The momentum change produced by a force acting over a time interval.
• Isolated system — A system whose net external force is zero, so its total momentum remains constant.
• Normal reaction — The component of a contact force perpendicular to the surfaces in contact.
• Tension — The pulling force transmitted along a taut string or similar connecting member.
• Concurrent forces — Forces whose lines of action pass through one common point.
• Torque — The turning effect of a force about a specified point or axis.
• Static friction — Friction preventing relative sliding, with a magnitude that adjusts up to a limiting value.
• Kinetic friction — The frictional force opposing relative sliding between two surfaces in contact.
• Angle of repose — The greatest incline angle at which a body remains at rest without another applied force.
• Angular acceleration — The rate of change of angular velocity with respect to time.
• Centripetal force — The inward resultant force required to maintain motion along a circular path.
• Banking — Raising the outer edge of a curved road or track above its inner edge.

Common errors and misconceptions

• Misconception: A moving body requires a net forward force. Correct: Constant velocity requires zero net force; resistance may need a balancing applied force.
• Misconception: Equal action and reaction cancel on either body. Correct: They act on different bodies and cancel only as internal forces of the combined system.
• Misconception: An unchanged speed means zero impulse. Correct: Momentum includes direction, so reversal at unchanged speed produces a non-zero impulse.
• Misconception: Static friction is invariably μ_sN. Correct: That expression gives its limiting magnitude; the actual force can be smaller.
• Misconception: Normal reaction equals weight in every problem. Correct: Determine it from perpendicular force balance or acceleration; inclination and other forces change it.
• Misconception: Friction must oppose motion relative to the ground. Correct: It opposes actual or impending relative sliding at the contact and can accelerate a body forwards.
• Misconception: Centripetal force is an extra force to draw. Correct: It is the inward resultant of identified interactions such as tension, friction and gravity.
• Misconception: Zero resultant force is sufficient for ladder equilibrium. Correct: An extended body also requires zero resultant torque.

Exam-style questions with model answers

Q1. State Newton’s first law and explain whether zero net force requires a body to be at rest. [2 marks]

1. A body remains at rest or continues with uniform straight-line motion unless a net external force changes that state.
2. Zero net force gives zero acceleration, so a non-zero constant velocity is also possible; rest is not required.

Q2. A batsman returns a 0.15 kg ball straight towards the bowler at its original speed of 12 m s⁻¹. Assuming linear motion, calculate the impulse and its direction. [3 marks]

1. Choose the direction towards the bowler as positive. The initial velocity is therefore u = −12 m s⁻¹ and the final velocity is v = +12 m s⁻¹.
2. The signed impulse J equals the change in momentum: J = m(v − u), where m is the ball’s mass. Thus J = 0.15[12 − (−12)].
3. The impulse is +3.6 N s, directed towards the bowler. Equal initial and final speeds do not make the momentum change zero because the direction reverses.

Q3. A 0.04 kg bullet enters a heavy wooden block at 90 m s⁻¹ and stops after travelling 60 cm. Using uniform retardation as the calculation model, find the average resisting-force magnitude and explain its qualification as an average. [4 marks]

1. Take the initial direction as positive. Initial speed u = 90 m s⁻¹, final speed v = 0, and stopping distance s = 0.60 m.
2. Use v² − u² = 2as, where a is acceleration. Hence a = −90²/(2 × 0.60) = −6750 m s⁻².
3. The signed force is F = ma = 0.04 × (−6750) = −270 N. Its magnitude is 270 N, opposing the bullet’s initial motion.
4. The calculation treats retardation as constant. The actual resisting force may vary during penetration, so 270 N represents the equivalent average resistance for the stopping distance.

Q4. For two interacting bodies forming an isolated system with zero net external force, derive conservation of total linear momentum from Newton’s laws and explain recoil of an initially stationary gun and bullet. [5 marks]

1. Call the bodies A and B and their momenta p⃗_A and p⃗_B. An isolated system has no net external force, although its members exert internal forces on one another.
2. Let F⃗_AB mean the force on A by B and F⃗_BA the force on B by A. Newton’s third law gives F⃗_AB = −F⃗_BA at the same instant.
3. Let F⃗_A,ext and F⃗_B,ext be the net external forces on A and B respectively, and let t denote time. The second law gives dp⃗_A/dt = F⃗_AB + F⃗_A,ext and dp⃗_B/dt = F⃗_BA + F⃗_B,ext. Adding, using the third law and F⃗_A,ext + F⃗_B,ext = 0, gives d(p⃗_A + p⃗_B)/dt = 0.
4. Consequently, the total momentum remains constant: p⃗_A + p⃗_B = p⃗′_A + p⃗′_B. Primes mean final values; the individual momenta need not remain unchanged.
5. The gun and bullet initially have zero total momentum. After firing, their momenta are equal and opposite, so the gun recoils opposite to the bullet’s motion.

Q5. A 20 kg trolley sliding rightwards on a horizontal surface is connected to a descending 3 kg block by a massless, inextensible string over an ideal smooth pulley. The trolley’s kinetic-friction coefficient is 0.04; take g = 10 m s⁻². Find their acceleration magnitude and string tension. [5 marks]

1. Let a be the common acceleration magnitude, positive rightwards for the trolley and downwards for the hanging block. Let T be the common string tension. The fixed string length supplies the acceleration constraint.
2. The trolley has no vertical acceleration, so its normal reaction is N = 20 × 10 = 200 N. Kinetic friction is f_k = 0.04N = 8 N, directed leftwards.
3. For the hanging block, downward weight minus upward tension gives 30 − T = 3a. For the trolley, rightward tension minus friction gives T − 8 = 20a.
4. Add these equations to eliminate the internal tension: 30 − 8 = 23a. Therefore a = 22/23 = 0.96 m s⁻².
5. Substitute into the block equation: T = 30 − 3(22/23) = 27.1 N. The positive acceleration agrees with the assumed descending block and rightward trolley motion.

Q6. A uniform ladder 3 m long and of mass 20 kg rests in equilibrium against a smooth vertical wall. Its foot is 1 m from the wall on a rough horizontal floor with sufficient static friction. Taking g = 9.8 m s⁻², find the wall reaction, floor normal force and floor friction. [5 marks]

1. The ladder’s weight is W = mg = 20 × 9.8 = 196 N, acting at its midpoint. Here m denotes mass and g gravitational acceleration. The smooth wall exerts a horizontal reaction R.
2. The wall-contact height is h = √(3² − 1²) = 2√2 m. The perpendicular distance from the foot to the weight’s vertical line of action is 0.5 m.
3. Taking moments about the foot removes both floor forces. Rotational equilibrium requires R(2√2) = 196 × 0.5, giving R = 34.6 N away from the wall.
4. Vertical force balance gives floor normal force N = W = 196 N upwards because the wall reaction has no vertical component.
5. Horizontal force balance gives floor friction equal to R, hence 34.6 N towards the wall. It prevents the foot from sliding away and must lie within the available static-friction limit.

Q7. A cyclist takes a level circular turn of radius 3 m at 18 km h⁻¹. The coefficient of static friction between tyres and road is 0.1. Take g = 9.8 m s⁻² and use the sideways-slip model. Determine whether friction is sufficient. [3 marks]

1. Convert the speed: 18 km h⁻¹ = 5 m s⁻¹. On the level road the normal reaction equals weight, and static friction must provide the inward centripetal force.
2. No sideways slipping requires v² ≤ μ_srg, where v is speed, μ_s the static-friction coefficient and r the radius. The right-hand side is 0.1 × 3 × 9.8 = 2.94 m² s⁻².
3. The actual squared speed is 25 m² s⁻², exceeding 2.94 m² s⁻². Available friction is insufficient, so the cyclist slips while taking the turn.

Q8. A particle of mass m moves in a vertical circle on a light, inextensible string of length r about a fixed centre. Neglect air resistance and take gravitational acceleration as g. Derive the minimum top and bottom speeds for a complete circle with the string just taut at the top. [5 marks]

1. At the top, inward is downward. With top tension T_t and speed v_t, radial force balance is T_t + mg = mv_t²/r.
2. A string cannot push, so tension cannot be negative. The limiting condition is T_t = 0; hence the minimum top speed is v_t,min = √(gr).
3. Tension is perpendicular to the instantaneous displacement and does no work. With air resistance neglected, mechanical energy remains constant between the bottom and top.
4. The height increase is 2r. If v_b is bottom speed, energy conservation gives ½mv_b² = ½mv_t² + 2mgr, or v_b² = v_t² + 4gr.
5. Insert the limiting top value v_t² = gr. Then v_b² = 5gr, giving minimum bottom speed v_b,min = √(5gr), under the stated string and energy assumptions.

Key takeaways

• Zero net external force means zero acceleration, allowing either rest or uniform motion in a straight line.
• Force is the rate of momentum change; the constant-mass classical form connects net force directly to acceleration.
• Impulse equals momentum change, so reversing a velocity can produce impulse even when speed stays unchanged.
• Action and reaction act simultaneously on different bodies; isolated systems conserve their total linear momentum.
• Free-body diagrams show forces on the selected system, while rigid-body equilibrium also requires moment balance.
• Static friction adjusts up to a limiting value; kinetic friction opposes relative sliding between contacting surfaces.
• Centripetal force is the inward resultant supplied by real interactions, and banking lets normal reaction contribute to it.
• Vertical-circle string problems require radial force balance, non-negative tension and energy conservation when dissipative effects are neglected.

Test yourself

Does zero net force mean that no individual forces act?

No. Individual forces can act and cancel, as weight and normal reaction do for a resting book on a horizontal table.

Why does drawing the hands backwards help when catching a ball?

It lengthens the stopping time, reducing the average force for the same change in the ball’s momentum.

Why do action and reaction not cancel on a single body?

They act on different bodies. Only the force acting on the selected body belongs in its force equation.

When may static friction be set equal to μ_sN?

At the limiting condition of impending sliding; before that condition, static friction can have a smaller magnitude.

Which way does friction act on a body sliding up a rough incline?

It acts down the incline, opposing the body’s relative sliding over the plane.

Why is acceleration non-zero in uniform circular motion?

The velocity changes direction continuously even though its magnitude stays constant; acceleration points towards the centre.

What supplies centripetal force at the design speed on a frictionless banked road?

The horizontal component of the normal reaction supplies it, while the vertical component balances the vehicle’s weight.

Why is zero top tension allowed as a limiting vertical-circle condition?

A string can pull but cannot push. Zero tension marks the boundary before a lower top speed would require an impossible compressive force.