Laws of Motion Class 11 Notes

All motion is caused by interactions, and in mechanics we call an interaction a force. A force is a push or a pull that can change the state of rest or of uniform motion of a body — it produces acceleration, or deformation, or both.

Forces of two broad kinds act on bodies: contact forces, which arise from direct touch (normal reaction N, tension T in a string, friction f, an applied push/pull F, spring force), and field or non-contact forces, which act at a distance (gravitational, electromagnetic, strong nuclear and weak nuclear forces).

To draw an FBD: (i) select the body and isolate it mentally, (ii) identify all external forces — pulls, pushes, their weights and contact reactions, (iii) draw each force as an arrow from the body, labelling the forces, and (iv) choose an axis system to resolve them into components. Every mechanical problem — lift, pulley, incline, banked curve — begins with a correct FBD.

MCQ Questions

1 Mark Questions

1. 1
   Define force. State its SI unit.
2. 2
   What is a free body diagram (FBD)?
3. 3
   State any two effects that a force can produce on a body.

2 Mark Questions

1. 1
   Draw a free body diagram of a block placed on a rough inclined plane and name all the forces acting on it.

The first law answers Aristotle's old question of why bodies keep moving: it says no force is needed to maintain uniform motion. What force does is change the state of motion, and inertia is the property that resists that change.

Being a response to any change of velocity, inertia shows itself in three forms: inertia of rest (a coin on a card falls into a tumbler when the card is flicked; dust is beaten out of a carpet), inertia of motion (a passenger thrown forward when a bus stops suddenly), and inertia of direction (clothes fly outward when a wet cloth is whirled).

Mass is the measure of the inertia of a body — the larger the mass, the smaller the acceleration produced by a given force. Weight, by contrast, is the gravitational force on the body, W = mg, which varies with g. The second law also defines the newton: 1 N is the force that accelerates a 1 kg mass at 1 m/s².

MCQ Questions

1 Mark Questions

1. 1
   Define inertia.
2. 2
   State Newton's first law of motion (the law of inertia).
3. 3
   Define linear momentum of a body and state its SI unit.
4. 4
   State Newton's second law of motion.
5. 5
   Give one example of inertia of rest.
6. 6
   Why does a body moving with uniform velocity not require any net force?

2 Mark Questions

1. 1
   Explain inertia of rest, inertia of motion and inertia of direction, giving one example of each.
2. 2
   Explain why passengers are thrown forward when a speeding bus stops suddenly.

3 Mark Questions

1. 1
   State and explain Newton's second law of motion and hence deduce the relation F = ma.
2. 2
   Distinguish between the mass and the weight of a body.
3. 3
   A pebble of mass 0.05 kg is thrown vertically upwards. Give the magnitude and direction of the net force on the pebble during its upward motion, its downward motion and at the highest point. Would the answers change if the pebble were thrown at 45° to the horizontal? (Ignore air resistance.)

5 Mark Questions

1. 1
   State and explain Newton's three laws of motion, with suitable examples from everyday life.

Assertion–Reason Questions

Select the correct alternative for each assertion–reason pair from the codes given below: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is NOT the correct explanation of A. (c) A is true but R is false. (d) A is false but R is also false.

1. 1
   Assertion (A): A body moving with constant velocity is in dynamic equilibrium even when no net force acts on it. Reason (R): Newton's first law states that no force is required to maintain uniform motion.
2. 2
   Assertion (A): Linear momentum of a body is a vector quantity. Reason (R): Momentum is the product of the mass (a scalar) and the velocity (a vector) of the body.

Case Study · Newton's First Law and Inertia

According to Newton's first law of motion, a body continues in its state of rest or of uniform motion along a straight line unless it is compelled by a net external force to change that state. This property of a body, by virtue of which it cannot change its own state of rest or of uniform motion, is called inertia. More massive bodies have greater inertia and need a larger force to change their state. Newton's first law is valid only in an inertial frame of reference.

(i) The property of a body by which it resists any change in its state of rest or of uniform motion is called — (a) momentum (b) inertia (c) force (d) impulse.

(ii) When a bus suddenly starts from rest, passengers standing inside are thrown — (a) forward (b) backward (c) sideways (d) upwards.

(iii) A coin placed on a card resting over a glass tumbler falls into the tumbler when the card is flicked sharply. This is because of — (a) the weight of the coin (b) the inertia of motion (c) the inertia of rest of the coin (d) friction.

(iv) Two balls of masses 1 kg and 5 kg are at rest on a table. The ball with more inertia is — (a) the 1 kg ball (b) the 5 kg ball (c) both have equal inertia (d) determined by the surface they rest on.

(v) Newton's first law of motion defines — (a) the unit of force (b) the property of inertia (c) the momentum of a body (d) the energy of a body.

Because the pair acts on different bodies, the two forces never cancel each other when you write the equation of motion of either body alone — each appears in the other body's FBD. Action–reaction pairs always involve both forces together.

Everyday examples of the third law: we walk by pushing the ground backwards and the ground pushes us forward; a horse pulls a cart because the ground pushes the horse's feet forward (which is why a horse cannot pull a cart in empty space); a gun recoils while firing a bullet; a rocket expels gases backwards and moves forwards.

MCQ Questions

1 Mark Questions

1. 1
   State Newton's third law of motion.
2. 2
   Action and reaction are equal and opposite, yet they do not cancel each other. Why?

2 Mark Questions

1. 1
   Explain why a horse can pull a cart on the ground but cannot pull a cart while running in empty space.

Assertion–Reason Questions

Select the correct alternative for each assertion–reason pair from the codes given below: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is NOT the correct explanation of A. (c) A is true but R is false. (d) A is false but R is also false.

1. 1
   Assertion (A): Action and reaction forces cancel each other because they are equal and opposite. Reason (R): Action and reaction always act on two different bodies.

Collisions, impacts and collisions with bats, hammers and air bags all involve very large forces acting for a very short time. It is often impossible to measure the force during the collision, but the total effect — the impulse — can be related to the easily measured change in momentum.

The impulse–momentum theorem follows directly from the second law: since F = dp/dt, integrating over the duration of the force gives J = Δp. Application: a fast ball travelling at 54 km/h deflected through an angle θ without speed change receives an impulse of magnitude J = 2mv sin(θ/2); the change of momentum — not the force — is fixed.

The theorem is the physics behind safe catching and collision protection. Since J = FΔt is fixed by the change in momentum, we reduce the maximum force by increasing the time of contact — a fielder pulls his hands back, air bags and crumple zones stretch the crash time, and sand pits slow a long jumper gradually.

MCQ Questions

1 Mark Questions

1. 1
   Define impulse. State its SI unit.

2 Mark Questions

1. 1
   A 150 g cricket ball moving at 20 m/s is struck and returned along the same line at 15 m/s. If the contact time is 0.02 s, find the impulse and the average force exerted on the ball.
2. 2
   A player catches a fast-moving ball by pulling his hands backwards. Explain, in terms of impulse, why this reduces the force on his hands.
3. 3
   A batsman deflects a ball through an angle of 45° without changing its speed of 54 km/h. Write the expression for the impulse imparted to the 0.15 kg ball and compute its magnitude.

3 Mark Questions

1. 1
   State and prove the impulse–momentum theorem.
2. 2
   A body of mass 0.40 kg moving with a constant speed of 10 m/s to the north is subjected to a constant force of 8.0 N directed towards the south for 30 s. Predict the position of the body at t = −5 s, t = 25 s and t = 100 s.

Assertion–Reason Questions

Select the correct alternative for each assertion–reason pair from the codes given below: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is NOT the correct explanation of A. (c) A is true but R is false. (d) A is false but R is also false.

1. 1
   Assertion (A): Impulse has the same dimensions as linear momentum. Reason (R): Impulse is defined as the change in momentum of a body.
2. 2
   Assertion (A): A fielder pulls his hands backwards while catching a fast-moving cricket ball. Reason (R): Increasing the time of contact reduces the force on the hands, since the impulse J = FΔt is fixed.

Case Study · Impulse in Sports and Safety

When a large force acts for a very short time, its effect is most conveniently described by the impulse J = FΔt, which equals the change in momentum of the body. In a cricket catch or a tennis stroke the change in momentum is fixed, so the force felt can be reduced by increasing the contact time. The same principle is at work in air bags, crumple zones and cushioned packing materials.

(i) Impulse is defined as — (a) F × a (b) F × t (c) m × v (d) F × v.

(ii) The SI unit of impulse is — (a) newton (b) N m (c) N s (d) J s.

(iii) A ball of mass 0.15 kg moving at 10 m/s is struck back along the same line at 10 m/s. The impulse imparted to the ball is — (a) 1.5 N s (b) 3.0 N s (c) 0 N s (d) 0.15 N s.

(iv) An air bag in a car increases the time of impact during a crash so that — (a) the impulse is reduced (b) the force on the occupant is reduced (c) momentum is conserved automatically (d) kinetic energy is increased.

(v) A fielder catches a cricket ball with his hands pulled back because the force on the hands is smaller when — (a) the impulse is smaller (b) the change of momentum is smaller (c) the time of contact is larger (d) the speed of the ball is reduced at once.

It follows from Newton's third law: during an interaction the forces on the two bodies are equal, opposite and act for the same time, so the changes of their momenta balance. It applies to collisions, explosions, recoil of guns, disintegration of a nucleus and rocket propulsion — whenever the external force is zero or negligible.

Recoil of a gun: a gun of mass M shooting a shell of mass m with muzzle speed v acquires the recoil speed V = −mv/M. A nucleus at rest disintegrating into two fragments forces the products to fly in opposite directions with momenta equal in magnitude.

A rocket is an example of a variable-mass system, because it continuously burns and ejects fuel. The thrust on the rocket is F = (dm/dt)·v_rel (ejected mass per second times the exhaust velocity relative to the rocket); the rocket gains velocity forward exactly as hot gases are pushed backwards, so it accelerates even in empty space.

MCQ Questions

1 Mark Questions

1. 1
   State the law of conservation of linear momentum.

2 Mark Questions

1. 1
   A bullet of mass 20 g fired from a gun of mass 2 kg leaves the muzzle at a speed of 80 m/s. Find the recoil velocity of the gun.
2. 2
   A nucleus at rest disintegrates into two smaller nuclei. Show that the two products must move in opposite directions.

3 Mark Questions

1. 1
   State the law of conservation of linear momentum and prove it using Newton's laws of motion.
2. 2
   Explain the working of a rocket on the basis of the conservation of linear momentum. Why can a rocket work even in empty space?

5 Mark Questions

1. 1
   What is a variable-mass system? Derive the equation of motion of a rocket and obtain an expression for its velocity at any instant.

Assertion–Reason Questions

Select the correct alternative for each assertion–reason pair from the codes given below: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is NOT the correct explanation of A. (c) A is true but R is false. (d) A is false but R is also false.

1. 1
   Assertion (A): The linear momentum of the (Earth + stone) system is conserved while a stone falls towards the Earth. Reason (R): The gravitational force between them is an internal force and no external force acts on the system.
2. 2
   Assertion (A): A rocket can accelerate even in empty space. Reason (R): A rocket works on the conservation of linear momentum by ejecting hot gases backwards.

Real problems — blocks in contact, masses on pulleys, people in accelerating lifts, vehicles in equilibrium — are solved by drawing the FBD of each body and applying the second law separately. The procedure is always the same: isolate, draw forces, resolve, set up the equations, solve.

Two blocks in contact on a frictionless table, with a force F pushing m₁ followed by m₂: both move with a = F/(m₁ + m₂), and the contact force that m₁ exerts on m₂ is N = m₂a. For two masses m₁ and m₂ (m₂ > m₁) hung by a string over a frictionless pulley: a = (m₂ − m₁)g/(m₁ + m₂) and T = 2m₁m₂g/(m₁ + m₂).

A person of mass m on a weighing machine in a lift: the machine reads the normal reaction N. Moving up with acceleration a, N = m(g + a); moving down, N = m(g − a); uniform motion gives N = mg; free fall gives N = 0 (apparent weightlessness). The true weight mg never changes.

A body is in equilibrium when the net force on it is zero, ΣF = 0. For concurrent forces this means their vector sum vanishes (the polygon closes), or equivalently the sum of components vanishes in every direction. Whenever strings, pulleys or contacts constrain how bodies move, the constraint relations connect the accelerations (for an inextensible string, the accelerations along the string are equal in magnitude).

In a non-inertial (accelerating) frame, Newton's laws fail unless we invent a pseudo force F = −ma of the observer's own acceleration. A lift accelerating up at a feels like extra gravity; a bus turning left makes passengers feel flung right — that outward push is the pseudo (centrifugal) force, real only in the rotating frame.

MCQ Questions

1 Mark Questions

1. 1
   When a lift moves upward with acceleration, does the weight of a man standing inside it change? Give a reason.

2 Mark Questions

1. 1
   A lift is moving downward with acceleration a. Derive an expression for the apparent weight of a man standing on a weighing machine kept inside the lift.

3 Mark Questions

1. 1
   Two masses of 8 kg and 12 kg are connected by a light inextensible string passing over a frictionless pulley. Find the acceleration of the masses and the tension in the string.
2. 2
   Two blocks of masses 5 kg and 10 kg are kept in contact on a frictionless horizontal table. A force of 60 N acts on the 5 kg block, pushing both blocks. Find the acceleration of the blocks and the contact force between them.
3. 3
   State the conditions for the equilibrium of a body acted upon by concurrent forces.
4. 4
   Distinguish between inertial and non-inertial frames of reference. Explain why a pseudo force F = −ma has to be introduced when analysing motion in a non-inertial frame.

5 Mark Questions

1. 1
   A man of mass m stands on a weighing machine kept in a lift. Derive the expression for the reading of the machine when the lift (i) moves up with uniform velocity, (ii) moves up with acceleration a, (iii) moves down with acceleration a, and (iv) falls freely. Evaluate the readings for m = 70 kg and a = 5 m/s².
2. 2
   Explain the equilibrium of concurrent forces and the constrained motion of masses connected by a string passing over a pulley. Hence solve the two-mass pulley problem for the acceleration and the tension.

Assertion–Reason Questions

Select the correct alternative for each assertion–reason pair from the codes given below: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is NOT the correct explanation of A. (c) A is true but R is false. (d) A is false but R is also false.

1. 1
   Assertion (A): The apparent weight of a man in a lift moving upwards with acceleration a is greater than his true weight. Reason (R): The normal reaction must provide the additional upward acceleration, N = m(g + a).

Case Study · Lift and Apparent Weight

A weighing machine reads the normal reaction exerted on the person standing on it. For a lift moving up with acceleration a, N − mg = ma, giving N = m(g + a); for a lift moving down with acceleration a it gives N = m(g − a). In uniform motion a = 0 and N = mg. If the cable snaps and the lift falls freely, N = 0 and the person appears weightless.

(i) A man of mass 60 kg stands in a lift moving up with acceleration 2 m/s². The reading of the weighing machine is (g = 10 m/s²) — (a) 60 kg (b) 72 kg (c) 48 kg (d) 66 kg.

(ii) If the cable of a lift snaps, the reading of a weighing machine placed inside becomes — (a) mg (b) m(g − a) (c) zero (d) 2mg.

(iii) The apparent weight of a man increases when the lift — (a) moves up with uniform velocity (b) moves down with acceleration a (c) moves up with acceleration a (d) moves down with uniform velocity.

(iv) If the lift moves down with uniform velocity, the reading of the weighing machine is — (a) m(g + a) (b) mg (c) m(g − a) (d) zero.

(v) A man of mass 70 kg stands in a lift accelerating downwards at 5 m/s². The reading of the machine is (g = 10 m/s²) — (a) 35 kg (b) 70 kg (c) 105 kg (d) 0.

Friction is the contact force that opposes relative motion (or its tendency) between two surfaces. It acts along the surfaces of contact, in the direction that opposes the relative sliding. It arises from the roughness and intermolecular adhesion at the surface of contact.

The coefficient of friction μ is dimensionless and depends on the nature of the surfaces, not on the area of contact or (for kinetic friction) appreciably on speed. μs > μk, and μ = tan θ when a body just begins to slide down an incline inclined at θ — this is the angle of repose, which is numerically equal to the angle of friction tan λ = μ.

On a rough inclined plane of inclination θ, a body sliding down has acceleration a = g(sin θ − μ cos θ) (with μ < tan θ for motion to occur); when pulled up the plane the extra acceleration needed is g(sin θ + μ cos θ). When θ reaches the angle of repose (tan θ = μ), the block moves down at constant speed.

MCQ Questions

1 Mark Questions

1. 1
   Define the coefficient of static friction.

2 Mark Questions

1. 1
   Distinguish between limiting friction, static friction and kinetic friction.
2. 2
   Define the angle of friction and the angle of repose and showing that the two are equal.
3. 3
   A block of mass 5 kg rests on a horizontal surface whose coefficient of static friction is 0.2. Find the minimum horizontal force needed to just move the block. (Take g = 10 m/s².)

3 Mark Questions

1. 1
   Obtain an expression for the acceleration of a block sliding down a rough inclined plane.
2. 2
   Show that a body just begins to slide down a rough inclined plane when tan θ = μ, where θ is the angle of inclination and μ is the coefficient of friction.
3. 3
   A block is projected up a rough inclined plane of inclination 30° with a speed of 10 m/s. If the coefficient of friction is 0.2 and g = 10 m/s², find the distance travelled by the block before it comes to rest.

5 Mark Questions

1. 1
   Derive expressions for the acceleration of a body (i) sliding down and (ii) being pulled up a rough inclined plane. Hence show that the body just begins to slide when tan θ = μ.

Assertion–Reason Questions

Select the correct alternative for each assertion–reason pair from the codes given below: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is NOT the correct explanation of A. (c) A is true but R is false. (d) A is false but R is also false.

1. 1
   Assertion (A): Friction opposes the relative motion between two surfaces in contact. Reason (R): Friction always acts along the surfaces and is a self-adjusting contact force.
2. 2
   Assertion (A): The angle of repose of a body on a rough inclined plane is equal to the angle of friction. Reason (R): Both the angles are equal to tan⁻¹(μ).

Case Study · Friction — A Necessary Evil

Friction is the force that opposes relative motion between two surfaces in contact. Static friction is self-adjusting up to a maximum (limiting) value f_s,max = μsN, beyond which the body just begins to slide; kinetic friction f_k = μkN then takes over. Friction is indispensable — walking, writing and driving all need it — yet it also wastes energy, which is why lubricants and ball bearings are used to reduce it.

(i) The maximum static friction between two surfaces is — (a) μkN (b) μsN (c) always equal to the applied force (d) μsmg.

(ii) When a block just begins to slide down a rough inclined plane of inclination θ, the coefficient of friction is — (a) sin θ (b) cos θ (c) tan θ (d) cot θ.

(iii) Which of the following is generally the largest? (a) static friction (b) kinetic friction (c) rolling friction (d) all are equal.

(iv) A block of mass 5 kg rests on a horizontal floor with μs = 0.4 (g = 10 m/s²). The minimum horizontal force needed to move it is — (a) 20 N (b) 10 N (c) 40 N (d) 2 N.

(v) Ball bearings reduce friction by — (a) decreasing μ between the surfaces (b) converting sliding friction into rolling friction (c) removing the surfaces entirely (d) increasing the normal reaction.

Uniform circular motion is one of the simplest cases where velocity changes continuously in direction but not in magnitude. Because the acceleration is directed towards the centre, the net force must also point to the centre — this inward force is the centripetal force.

In circular motion the actual force is provided by whatever else acts: tension in a string, friction between tyres and road, gravity in the case of the Moon, electrostatic attraction for electrons. If the required centripetal force exceeds what is available, the body cannot follow the circle — the vehicle skids, or the string breaks and the stone flies off along the tangent.

For a vehicle on a level road, friction alone supplies mv²/r, which limits the safe speed to v_max = √(μrg). To raise the safe speed, roads are banked: the horizontal component of the normal reaction N sin θ then contributes centripetal force. For optimum banking (no dependence on friction) tan θ = v²/(rg); with friction the maximum safe speed becomes v_max = √(rg(μ + tan θ)/(1 − μ tan θ)).

A conical pendulum is a bob whirled in a horizontal circle with the string sweeping out a cone. Resolving the tension: T cos θ balances mg, while T sin θ provides the centripetal force, giving v = √(rg tan θ). Related applications are the death well (rotor) of an amusement park, where the wall's normal reaction supplies centripetal force, and the effect of the Earth's rotation, which dilutes g at the equator (g' = g − ω²R cos²φ).

MCQ Questions

1 Mark Questions

1. 1
   Define centripetal force and write its expression.

2 Mark Questions

1. 1
   Define banking of roads. Why are roads banked at curves?
2. 2
   A stone of mass 0.25 kg is whirled in a horizontal circle of radius 1.5 m at a speed of 40 revolutions per minute. Find the tension in the string.

3 Mark Questions

1. 1
   Derive the expression for the optimum speed of a vehicle safely taking a circular turn on a banked road, neglecting friction.
2. 2
   A car of mass 1500 kg takes a circular turn of radius 100 m on a level road at 36 km/h. Find the centripetal force required and the minimum coefficient of friction needed between the tyres and the road.
3. 3
   A conical pendulum carries a small bob of mass m suspended by a light string executing uniform circular motion in a horizontal plane. Show that v = √(rg tan θ), where θ is the angle the string makes with the vertical.

5 Mark Questions

1. 1
   What is banking of roads? Derive the expression for the maximum safe speed of a vehicle rounding a banked curve when friction is also taken into account.

Assertion–Reason Questions

Select the correct alternative for each assertion–reason pair from the codes given below: (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is NOT the correct explanation of A. (c) A is true but R is false. (d) A is false but R is also false.

1. 1
   Assertion (A): Roads are banked at curves so that vehicles can take the turn at higher speed. Reason (R): On a banked road the horizontal component of the normal reaction provides the necessary centripetal force.
2. 2
   Assertion (A): Passengers in a bus taking a sharp turn tend to fall outwards. Reason (R): In the frame of the bus a pseudo centrifugal force acts on the passengers in the outward direction.

Case Study · Banking of Roads

On a level circular turn the centripetal force comes entirely from friction, which limits the safe speed. On a banked road the surface is tilted at an angle θ so that the horizontal component of the normal reaction N sin θ contributes centripetal force. For optimum banking with no reliance on friction, tan θ = v²/(rg). When friction is present, a range of safe speeds exists, the maximum being v_max = √(rg(μ + tan θ)/(1 − μ tan θ)).

(i) On a level road, the centripetal force for a vehicle taking a turn is provided by — (a) the engine force (b) friction (c) the normal reaction (d) the weight of the vehicle.

(ii) For optimum banking of a road without depending on friction, the correct relation is — (a) tan θ = rg/v² (b) tan θ = v²/(rg) (c) sin θ = v²/(rg) (d) v = √(2rg).

(iii) A road is banked with tan θ = 0.4 and radius of curvature 250 m (g = 10 m/s²). The optimum safe speed on the road is — (a) 20 m/s (b) 31.6 m/s (c) 100 m/s (d) 10 m/s.

(iv) If the coefficient of friction between the tyres and the road increases, the maximum safe speed on a banked curve — (a) decreases (b) increases (c) remains the same (d) becomes zero.

(v) On a banked road, the horizontal component of the normal reaction on the vehicle provides — (a) the vertical support (b) the centripetal force (c) the frictional force (d) no effect at all.

Exercises (NCERT Q4.1–Q4.23)

• 4.1) Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed, (b) a cork of mass 10 g floating on water, (c) a kite skillfully held stationary in the sky, (d) a car moving with a constant velocity of 30 km/h on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
• 4.2) A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.
• 4.3) Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, (a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h, (c) just after it is dropped from the window of a train accelerating with 1 m s⁻², (d) lying on the floor of a train which is accelerating with 1 m s⁻², the stone being at rest relative to the train. Neglect air resistance throughout.
• 4.4) One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is: (i) T, (ii) T − mv²/l, (iii) T + mv²/l, (iv) 0. T is the tension in the string. [Choose the correct alternative].
• 4.5) A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s⁻¹. How long does the body take to stop?
• 4.6) A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s⁻¹ to 3.5 m s⁻¹ in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
• 4.7) A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.
• 4.8) The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.
• 4.9) A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s⁻². Calculate the initial thrust (force) of the blast.
• 4.10) A body of mass 0.40 kg moving initially with a constant speed of 10 m s⁻¹ to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = −5 s, 25 s, 100 s.
• 4.11) A truck starts from rest and accelerates uniformly at 2.0 m s⁻². At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)
• 4.12) A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s⁻¹. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.
• 4.13) A man of mass 70 kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of 10 m s⁻¹, (b) downwards with a uniform acceleration of 5 m s⁻², (c) upwards with a uniform acceleration of 5 m s⁻². What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
• 4.14) Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only).
• 4.15) Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?
• 4.16) Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
• 4.17) A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
• 4.18) Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s⁻¹ collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
• 4.19) A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 m s⁻¹, what is the recoil speed of the gun?
• 4.20) A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
• 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?
• 4.22) If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks: (a) the stone moves radially outwards, (b) the stone flies off tangentially from the instant the string breaks, (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?
• 4.23) Explain why (a) a horse cannot pull a cart and run in empty space, (b) passengers are thrown forward from their seats when a speeding bus stops suddenly, (c) it is easier to pull a lawn mower than to push it, (d) a cricketer moves his hands backwards while holding a catch.

Answers at a glance

1. 1
   Q4.1 → Zero in every case — each object has zero acceleration, so no net force acts.
2. 2
   Q4.2 → Net force = 0.5 N vertically downward in all three cases; the answers are unchanged for a 45° throw.
3. 3
   Q4.3 → (a) 1 N downward (b) 1 N downward (c) 1 N downward (d) 0.1 N in the direction of the train's motion.
4. 4
   Q4.4 → Option (i): the net force is T, the tension in the string.
5. 5
   Q4.5 → The body stops after 6 s.
6. 6
   Q4.6 → F = 0.18 N in the direction of motion of the body.
7. 7
   Q4.7 → a = 2 m s⁻², at about 37° with the 8 N force.
8. 8
   Q4.8 → Average retarding force = 1162.5 N (rounded 1.2 × 10³ N), against the direction of motion.
9. 9
   Q4.9 → Initial thrust = 3.0 × 10⁵ N.
10. 10
    Q4.10 → Position at t = −5 s: −50 m; at t = 25 s: −6000 m; at t = 100 s: −50000 m.
11. 11
    Q4.11 → (a) v ≈ 22.36 m s⁻¹ at 26.6° with the direction of the truck (b) acceleration = 10 m s⁻² downward.
12. 12
    Q4.12 → (a) Falls vertically downward (b) follows a parabolic (projectile) path.
13. 13
    Q4.13 → (a) 70 kg (b) 35 kg (c) 105 kg (d) 0 kg — weightless in free fall.
14. 14
    Q4.14 → (a) Force = 0 in all three intervals. (b) Impulse = +3 kg m s⁻¹ at t = 0 and −3 kg m s⁻¹ at t = 4 s.
15. 15
    Q4.15 → (i) T = 400 N when the force acts on A (the lighter body) (ii) T = 200 N when it acts on B.
16. 16
    Q4.16 → Acceleration = 2 m s⁻²; tension = 96 N.
17. 17
    Q4.17 → By conservation of momentum v₁ = −(m₂/m₁)v₂ — the products necessarily move in opposite directions.
18. 18
    Q4.18 → Impulse on each ball = 0.6 kg m s⁻¹, opposite in direction for the two balls.
19. 19
    Q4.19 → Recoil speed = 0.016 m s⁻¹ (opposite to the shell).
20. 20
    Q4.20 → Impulse imparted to the ball ≈ 4.16 kg m s⁻¹.
21. 21
    Q4.21 → Tension ≈ 6.57 N; maximum speed ≈ 34.64 m s⁻¹.
22. 22
    Q4.22 → Option (b): the stone flies off tangentially from the instant the string breaks.
23. 23
    Q4.23 → (a) no reaction force in empty space (b) inertia of the upper body (c) pulling reduces, pushing increases effective weight (d) longer impact time ⇒ smaller force.

MCQ Questions

1 Mark Questions

1. 1
   State Newton's third law of motion with an example.
2. 2
   Define impulse and give its SI unit.
3. 3
   State the law of conservation of linear momentum.
4. 4
   What is the angle between two forces of equal magnitude when the resultant equals either of them?

2 Mark Questions

1. 1
   A body of mass 5 kg moving with a speed of 4 m/s is brought to rest by a constant force in 2 s. Find the force. (Ans: F = Δp/Δt = 10 N)
2. 2
   A ball of mass 0.2 kg moving at 10 m/s hits a wall and rebounds with the same speed. Find the impulse. (Ans: J = 2mv = 4 N s)
3. 3
   Two bodies of masses 2 kg and 3 kg move with momenta in the ratio 2 : 3. Find the ratio of their velocities. (Ans: v₁/v₂ = 1 : 1)

3 Mark Questions

1. 1
   Derive the expression for the maximum velocity of a vehicle on a horizontal circular road from the condition that it should not skid. (Ans: vmax = √(μrg))
2. 2
   Derive the expression for the optimum speed of a vehicle moving on a banked road. (Ans: v₀ = √(rg tan θ))
3. 3
   State Newton's second law of motion and hence obtain the relation between force and acceleration. (Ans: F = dp/dt = ma)

5 Mark Questions

1. 1
   State Newton's laws of motion and explain why the first law is a special case of the second. Deduce the law of conservation of linear momentum from Newton's third law.
2. 2
   Explain the centripetal force acting on a vehicle turning on a level road and on a banked road, and derive the condition for no skidding in each case. (Ans: vmax = √(μrg); v₀ = √(rg tan θ))