NEET Physics: Laws of Motion, Work-Energy and Rotation · Chapter 1

Laws of Motion

What to remember

Force is the cause of change in momentum: F = dp/dt, which becomes F = ma for constant mass. Impulse = F × Δt = change in momentum.
Action and reaction are equal and opposite but act on different bodies, so they never cancel each other. The total momentum of an isolated system stays constant.
Friction: f(static) ≤ μs N, f(kinetic) = μk N, with μs > μk. Angle of repose = angle of friction, and tan θ = μ.
Circular motion needs a centripetal force mv²/r towards the centre. On a banked road the ideal speed is v = √(r g tan θ).

Force, inertia and Newton's first law

A force is a push or a pull. It can change the speed, the direction or the shape of a body. Force is a vector. Its SI unit is the newton (N), where 1 N = 1 kg m s⁻². Its dimensions are [M L T⁻²].

Inertia is the property of a body by which it resists any change in its state of rest or of uniform motion in a straight line. The mass of a body is the measure of its inertia. A heavier body has more inertia.

Newton's first law: a body stays at rest, or keeps moving with constant velocity along a straight line, unless an external unbalanced force acts on it. The first law defines force and gives the idea of an inertial frame. Examples of inertia:

Passengers lean forward when a moving bus stops suddenly (inertia of motion).
Passengers lean backward when a bus at rest starts suddenly (inertia of rest).
Dust falls off a carpet when it is beaten with a stick (inertia of rest).
A coin on a card placed over a glass drops into the glass when the card is flicked away.

Momentum, Newton's second law and impulse

Linear momentum p = m v. It is a vector in the direction of the velocity. Unit: kg m s⁻¹. Dimensions: [M L T⁻¹].

Newton's second law: the rate of change of momentum of a body is proportional to the net external force and takes place in the direction of the force.

F = dp/dt. For constant mass, F = m dv/dt = m a.

Impulse J = F × Δt for a constant force. For a variable force, impulse is the area under the force-time graph. Impulse equals the change in momentum: J = Δp = m(v − u). The unit of impulse is N s, which is the same as kg m s⁻¹.

Impulse explains why a cricketer draws the hands back while catching a fast ball. The time of contact becomes longer, so the average force becomes smaller for the same change in momentum.

Worked example 1. A ball of mass 0.2 kg moving at 20 m s⁻¹ is hit back along the same line at 30 m s⁻¹. The contact lasts 0.01 s. Take the original direction as positive. Change in momentum = 0.2 × (−30) − 0.2 × 20 = −10 kg m s⁻¹. Average force = −10 / 0.01 = −1000 N. The magnitude is 1000 N, opposite to the original motion.

Newton's third law and conservation of linear momentum

Newton's third law: to every action there is an equal and opposite reaction. The two forces act on two different bodies at the same time. They are of the same type (both gravitational, or both contact forces). Because they act on different bodies, they do not cancel.

Law of conservation of linear momentum: if the net external force on a system is zero, its total linear momentum remains constant. It follows from the second and third laws.

Applications:

Recoil of a gun: a gun of mass M fires a bullet of mass m with velocity v. Initially the total momentum is zero. So M V + m v = 0 and the recoil velocity V = −m v / M. The gun moves backward. The bullet and gun have equal and opposite momenta, but the lighter bullet has the larger speed and the larger kinetic energy.
Explosion: a body at rest breaks into parts. The vector sum of the momenta of the parts is zero. The kinetic energy increases because chemical energy is converted into kinetic energy.
A rocket moves forward because the ejected gases carry momentum backward.

Worked example 2. A gun of mass 5 kg fires a bullet of mass 20 g at 300 m s⁻¹. Recoil speed = (0.02 × 300) / 5 = 1.2 m s⁻¹.

Equilibrium of concurrent forces and Lami's theorem

A particle is in equilibrium when the vector sum of all forces on it is zero. Then ΣFx = 0 and ΣFy = 0.

Lami's theorem: if three concurrent forces F₁, F₂, F₃ keep a particle in equilibrium, each force is proportional to the sine of the angle between the other two forces.

F₁ / sin α = F₂ / sin β = F₃ / sin γ, where α is the angle between F₂ and F₃, β is the angle between F₁ and F₃, and γ is the angle between F₁ and F₂.

Example of use: a lamp of weight W hangs from two strings. Each string tension can be found from the angles using Lami's theorem. A body of weight W hanging from a single vertical string has tension T = W.

Friction

Friction is the force that opposes the relative motion, or the tendency of relative motion, between two surfaces in contact. It acts along the contact surface.

Types and laws:

Type	Meaning	Value
Static friction	Acts when there is no relative motion. It is self-adjusting.	f ≤ μs N, with maximum value (limiting friction) f(max) = μs N
Kinetic (sliding) friction	Acts when surfaces slide over each other.	fk = μk N
Rolling friction	Acts when a body rolls without slipping.	Much smaller than sliding friction

Laws of friction:

Friction is independent of the apparent area of contact (for a given normal reaction).
Kinetic friction is nearly independent of the speed of sliding.
Friction is proportional to the normal reaction N.
μs is greater than μk for the same pair of surfaces. μ has no unit.

Angle of friction λ: the angle between the normal reaction and the resultant of the normal reaction and the limiting friction. tan λ = μs.

Angle of repose (angle of the rough incline at which a body just begins to slide): tan θ = μs. So the angle of repose equals the angle of friction.

On an incline of angle θ, the component of weight along the plane is m g sin θ and the normal reaction is m g cos θ. A body sliding down with friction has acceleration a = g (sin θ − μk cos θ).

Rolling friction is smaller than sliding friction, which is why wheels and ball bearings are used. Lubrication (oil, grease) puts a thin fluid layer between the surfaces and reduces friction. Friction is also useful: it lets us walk, hold objects and stop vehicles.

Worked example 3. A block of mass 2 kg lies on a rough horizontal table with μs = 0.5 and μk = 0.4. Take g = 10 m s⁻². Limiting friction = 0.5 × 2 × 10 = 10 N. A push of 8 N does not move it, and the friction is 8 N. A push of 14 N moves it. Then friction = 0.4 × 20 = 8 N and acceleration = (14 − 8) / 2 = 3 m s⁻².

Free-body diagrams, connected bodies and apparent weight

A free-body diagram shows one body alone and all the forces on it as arrows: weight mg down, normal reaction N perpendicular to the surface, tension T along a string, friction f along the surface. Then Newton's second law is written along each direction.

Connected bodies:

Atwood machine: masses m₁ and m₂ (m₁ > m₂) hang over a light frictionless pulley. Acceleration a = (m₁ − m₂) g / (m₁ + m₂). Tension T = 2 m₁ m₂ g / (m₁ + m₂).
Mass m₁ on a smooth horizontal table is tied by a string over a pulley at the edge to a hanging mass m₂. Acceleration a = m₂ g / (m₁ + m₂). Tension T = m₁ m₂ g / (m₁ + m₂).
Two blocks in contact, pushed by a force F on a smooth surface: a = F / (m₁ + m₂). The contact force on the second block is m₂ F / (m₁ + m₂).
A light inextensible string has the same tension throughout. A light smooth pulley only changes the direction of the tension.

Apparent weight in a lift (the reading of a spring balance on which a person of mass m stands):

Motion of the lift	Apparent weight
At rest or moving with constant velocity	m g
Accelerating upward with a	m (g + a)
Accelerating downward with a	m (g − a)
Free fall (cable broken)	zero (weightlessness)

Dynamics of uniform circular motion

A body in uniform circular motion has constant speed but changing direction, so it has an acceleration directed towards the centre: a = v²/r = ω² r. The force that provides it is the centripetal force F = m v²/r = m ω² r. It is not a new kind of force. It is supplied by tension, friction, gravity or a normal reaction. It does no work, because it is always perpendicular to the velocity.

Vehicle on a level circular road: friction supplies the centripetal force. The maximum safe speed is v(max) = √(μs r g). It does not depend on the mass of the vehicle.

Banked road of angle θ:

Without friction: N sin θ = m v²/r and N cos θ = m g. So tan θ = v²/(r g) and the ideal speed is v = √(r g tan θ).
With friction, the maximum speed is v(max) = √[ r g (μs + tan θ) / (1 − μs tan θ) ].
At the ideal speed no friction is needed. Banking lets vehicles turn at higher speed safely.

Worked example 4. A road of radius 50 m is banked at 45°. With g = 10 m s⁻², the ideal speed is √(50 × 10 × 1) = √500 ≈ 22.4 m s⁻¹. On a level road with μs = 0.4 and the same radius, v(max) = √(0.4 × 50 × 10) = √200 ≈ 14.1 m s⁻¹.

Formula summary

Quantity	Formula	Unit
Momentum	p = m v	kg m s⁻¹
Second law	F = dp/dt = m a	N
Impulse	J = F Δt = Δp	N s
Recoil velocity	V = − m v / M	m s⁻¹
Limiting friction	f = μs N	N
Angle of friction	tan λ = μs	none
Acceleration down a rough incline	a = g (sin θ − μk cos θ)	m s⁻²
Atwood machine	a = (m₁ − m₂) g / (m₁ + m₂)	m s⁻²
Centripetal force	F = m v²/r	N
Ideal banking speed	v = √(r g tan θ)	m s⁻¹

Constant used: g = 9.8 m s⁻² (taken as 10 m s⁻² in rounded problems when stated).

NEET traps

Action and reaction act on different bodies, so they do not cancel. Two forces that act on the same body and cancel are not an action-reaction pair.
Mass and weight are different. Mass is constant. Weight m g changes with place and in an accelerating lift.
A body in free fall has weight but zero apparent weight.
Friction does not always oppose motion. It opposes relative motion between surfaces. Friction on a walking person acts forward.
Static friction is not always μs N. It equals μs N only at the limit of sliding.
Limiting friction is greater than kinetic friction. A body, once it starts to slide, needs less force to keep moving.
Friction does not depend on the area of contact, but it depends on the nature of the surfaces and on N.
Angle of repose and angle of friction are numerically equal, but they are defined differently.
Centripetal force is not an extra force on the free-body diagram. It is the net inward force of the existing forces.
The speed of a vehicle on a level curve does not depend on its mass, but the safe speed depends on μs and the radius.

One-liners

The SI unit of force is the newton; 1 N = 1 kg m s⁻².
Impulse and momentum have the same dimensions [M L T⁻¹].
Force is the slope of the momentum-time graph.
Impulse is the area under the force-time graph.
A bullet and the gun have equal and opposite momenta after firing.
In an explosion of a body at rest, total momentum after the explosion is zero.
The three forces in Lami's theorem must be concurrent and in equilibrium.
Coefficient of friction has no unit.
tan(angle of repose) = μs.
Rolling friction is less than sliding friction.
In the Atwood machine the tension is less than m₁ g and more than m₂ g (m₁ > m₂).
A person in a lift accelerating upward feels heavier.
Centripetal acceleration = v²/r = ω² r.
Centripetal force does no work.
On a banked road at the ideal speed, friction is not needed.

Practice questions

Impulse has the same SI unit as which of the following?
1. kg m s⁻¹
2. kg m s⁻²
3. kg m⁻¹ s⁻¹
4. kg m² s⁻²
Answer
A. kg m s⁻¹
Impulse = change in momentum, so its unit is N s = kg m s⁻¹.
A constant force of 20 N acts on a body of mass 4 kg initially at rest for 3 s. The speed of the body at the end of 3 s is
1. 30 m s⁻¹
2. 5 m s⁻¹
3. 60 m s⁻¹
4. 15 m s⁻¹
Answer
D. 15 m s⁻¹
a = F/m = 20/4 = 5 m s⁻²; v = at = 5 × 3 = 15 m s⁻¹. Check by impulse: Ft = mv gives 60 = 4v, v = 15.
A gun of mass 4 kg fires a bullet of mass 10 g with a speed of 400 m s⁻¹. The recoil speed of the gun is
1. 0.25 m s⁻¹
2. 4 m s⁻¹
3. 1 m s⁻¹
4. 100 m s⁻¹
Answer
C. 1 m s⁻¹
Momentum of bullet = 0.01 × 400 = 4 kg m s⁻¹. Gun speed = 4/4 = 1 m s⁻¹.
The tangent of the angle of repose of a rough surface is equal to
1. the product of the coefficient of static friction and g
2. the coefficient of limiting static friction
3. the square root of the coefficient of limiting static friction
4. the reciprocal of the coefficient of limiting static friction
Answer
B. the coefficient of limiting static friction
At the angle of repose mg sin θ = μs mg cos θ, so tan θ = μs.
Two masses of 6 kg and 2 kg hang from the two ends of a light string passing over a smooth light pulley. Taking g = 10 m s⁻², the acceleration of the masses is
1. 7.5 m s⁻²
2. 2.5 m s⁻²
3. 10 m s⁻²
4. 5 m s⁻²
Answer
D. 5 m s⁻²
a = (m₁ − m₂)g/(m₁ + m₂) = (4 × 10)/8 = 5 m s⁻². Check: net force 60 − 20 = 40 N on total mass 8 kg gives 5.
A block of mass 3 kg on a smooth horizontal table is tied by a light string over a smooth pulley at the edge to a hanging block of mass 2 kg. Taking g = 10 m s⁻², the tension in the string is
1. 12 N
2. 20 N
3. 8 N
4. 6 N
Answer
A. 12 N
a = m₂g/(m₁ + m₂) = 20/5 = 4 m s⁻². T = m₁a = 3 × 4 = 12 N. Check: m₂g − T = m₂a gives 20 − 12 = 8 = 2 × 4.
A person of mass 60 kg stands on a weighing machine in a lift that accelerates upward at 2 m s⁻². Taking g = 10 m s⁻², the reading of the machine is
1. 600 N
2. 840 N
3. 720 N
4. 480 N
Answer
C. 720 N
N = m(g + a) = 60 × 12 = 720 N. Check: N − mg = ma gives 720 − 600 = 120 = 60 × 2.
A car goes round a level circular road of radius 20 m. The coefficient of friction between the tyres and the road is 0.5. Taking g = 10 m s⁻², the maximum safe speed is
1. 20 m s⁻¹
2. 10 m s⁻¹
3. 50 m s⁻¹
4. 100 m s⁻¹
Answer
B. 10 m s⁻¹
v = √(μ r g) = √(0.5 × 20 × 10) = √100 = 10 m s⁻¹. Check: friction 0.5 mg = m v²/r gives v² = 100.
A road of radius 72 m is banked so that tan θ = 0.2. Taking g = 10 m s⁻², the speed at which a vehicle needs no friction to turn is
1. 24 m s⁻¹
2. 6 m s⁻¹
3. 12 m s⁻¹
4. 144 m s⁻¹
Answer
C. 12 m s⁻¹
v² = r g tan θ = 72 × 10 × 0.2 = 144, so v = 12 m s⁻¹. Check: 12²/(72 × 10) = 0.2.
Which physical quantity is a measure of the inertia of a body?
1. Weight
2. Density
3. Velocity
4. Mass
Answer
D. Mass
Mass is the measure of inertia; it does not change with place.
A block slides down a rough incline of angle 37° for which sin 37° = 0.6 and cos 37° = 0.8. The coefficient of kinetic friction is 0.5. Taking g = 10 m s⁻², its acceleration is
1. 2 m s⁻²
2. 6 m s⁻²
3. 5 m s⁻²
4. 10 m s⁻²
Answer
A. 2 m s⁻²
a = g(sin θ − μ cos θ) = 10(0.6 − 0.4) = 2 m s⁻². Check: net force = 6 − 4 = 2 N per kg.
The area under the force-time graph of a force acting on a body gives the
1. work done by the force
2. impulse of the force
3. power of the force
4. change in kinetic energy
Answer
B. impulse of the force
Area = ∫F dt = impulse = change in momentum.
A ball of mass 0.5 kg changes its speed from 4 m s⁻¹ to 10 m s⁻¹ in the same direction in 0.2 s. The average force on it is
1. 30 N
2. 15 N
3. 7.5 N
4. 35 N
Answer
B. 15 N
F = m Δv/Δt = 0.5 × 6/0.2 = 15 N. Check: impulse = 3 N s, 3/0.2 = 15.
Lami's theorem can be used when a particle is in equilibrium under
1. two parallel forces
2. three forces whose resultant is not zero
3. any three forces on a rigid body
4. three coplanar concurrent forces
Answer
D. three coplanar concurrent forces
Lami's theorem applies to three coplanar, concurrent forces in equilibrium.
The minimum horizontal force needed to just start sliding a 5 kg block on a horizontal floor with μs = 0.3 is (g = 10 m s⁻²)
1. 5 N
2. 50 N
3. 15 N
4. 1.5 N
Answer
C. 15 N
f(max) = μs m g = 0.3 × 5 × 10 = 15 N.
A gun of mass 4 kg fires a bullet of mass 20 g. The ratio of the kinetic energy of the bullet to that of the gun is
1. 1/200
2. 1
3. 20
4. 200
Answer
D. 200
Momenta are equal, so KE = p²/2m. Ratio = M/m = 4/0.02 = 200. Check with numbers: speeds in ratio 200:1 and mass ratio 1:200 gives 200² / 200 = 200.
A 10 kg block rests on a horizontal floor with μs = 0.4 and μk = 0.3. A horizontal push of 35 N is applied. Taking g = 10 m s⁻², the friction force on the block is
1. 35 N
2. 5 N
3. 40 N
4. 30 N
Answer
A. 35 N
Limiting friction = 0.4 × 100 = 40 N, which is more than 35 N, so the block stays at rest and static friction balances the push: 35 N.
A person of mass 50 kg in a lift reads 400 N on a weighing machine. Taking g = 10 m s⁻², the acceleration of the lift is
1. 8 m s⁻² upward
2. 2 m s⁻² downward
3. 2 m s⁻² upward
4. 8 m s⁻² downward
Answer
B. 2 m s⁻² downward
Reading less than mg means a downward acceleration: 500 − 400 = 50a, so a = 2 m s⁻² downward.
A lamp weighing 100 N hangs from two strings. One string is at 30° above the horizontal and the other is at 60° above the horizontal on the other side, so the strings are perpendicular to each other. The tension in the string at 30° is
1. 100/√3 N
2. 50√3 N
3. 50 N
4. 100 N
Answer
C. 50 N
Lami: T₃₀/sin 150° = W/sin 90°, so T₃₀ = 100 × 0.5 = 50 N. Check: horizontal balance 50 cos 30° = 43.3 N = T₆₀ cos 60° with T₆₀ = 86.6 N; vertical 25 + 75 = 100 N.
A car can take a level circular turn safely at a maximum speed v on a dry road with μ = 0.8. On the same turn on a wet road μ falls to 0.2. The maximum safe speed on the wet road is
1. v/2
2. v/√2
3. 2v
4. v/4
Answer
A. v/2
v ∝ √μ, so the speed becomes v√(0.2/0.8) = v/2.
A man of mass 50 kg standing at rest on a boat of mass 200 kg on still water jumps off horizontally with a speed of 4 m s⁻¹ relative to the ground. The speed gained by the boat is
1. 1 m s⁻¹
2. 0.25 m s⁻¹
3. 0.8 m s⁻¹
4. 4 m s⁻¹
Answer
A. 1 m s⁻¹
50 × 4 = 200 × V gives V = 1 m s⁻¹ in the opposite direction.
A car moves on a banked road at a speed lower than the ideal speed for that road. The friction force on its tyres acts
1. horizontally outward, away from the centre
2. not at all, because friction cannot act on a banked road
3. down the incline
4. up the incline
Answer
D. up the incline
Below the ideal speed the car tends to slip down the slope, so friction acts up the incline.
A box just starts to slide when the rough plane it rests on is tilted to an angle θ with tan θ = 0.6. The horizontal force needed to just move a 20 kg box of the same material on a level floor is (g = 10 m s⁻²)
1. 12 N
2. 120 N
3. 200 N
4. 1200 N
Answer
B. 120 N
μs = tan θ = 0.6. Force = μs m g = 0.6 × 200 = 120 N.
Given below are two statements: Statement I: The total momentum of a system is conserved when the net external force on it is zero. Statement II: Internal forces can change the total momentum of a system.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Internal forces occur in equal and opposite pairs and cancel in the sum, so they cannot change the total momentum.
Given below are two statements: Statement I: Static friction is always equal to μs N. Statement II: Kinetic friction is generally less than the limiting static friction.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Static friction equals μs N only at the point of sliding; below that it equals the applied force.
Given below are two statements: Statement I: Action and reaction act on the same body. Statement II: Action and reaction are equal in magnitude and opposite in direction.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
They act on different bodies, so they do not cancel.
Given below are two statements: Statement I: A person in a freely falling lift has zero apparent weight. Statement II: In free fall the gravitational force on the person is zero.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Gravity still acts; only the normal reaction becomes zero.
Given below are two statements: Statement I: The centripetal force does no work in uniform circular motion. Statement II: The centripetal force is always perpendicular to the velocity.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
W = F · d cos 90° = 0, because force is perpendicular to the displacement.
Which of the following statements are correct? 1. Impulse is equal to the change in momentum. 2. Impulse is a scalar quantity. 3. The unit of impulse is N s.
1. 1 and 3 only
2. 1 and 2 only
3. 2 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 3 only
Impulse is a vector, so statement 2 is wrong. Statements 1 and 3 are correct.
Given below are two statements: Statement I: The angle of repose is equal to the angle of friction. Statement II: Rolling friction is greater than sliding friction.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
tan θ = tan λ = μs. Rolling friction is much smaller than sliding friction.
Given below are two statements: Statement I: On a level road a heavier vehicle has a smaller maximum safe speed than a lighter one on the same curve. Statement II: The maximum safe speed is √(μs r g), which does not contain mass.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Mass cancels in μs m g = m v²/r.
Which of the following statements are correct? 1. Friction depends on the apparent area of contact. 2. The coefficient of friction has no unit. 3. Friction is proportional to the normal reaction.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 2 and 3 only
Friction is independent of the apparent area of contact, so statement 1 is wrong.
Given below are two statements: Statement I: The impulse of a force depends only on its magnitude and not on the time for which it acts. Statement II: A rocket moves forward because the exhaust gases push against the surrounding air.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Impulse = F Δt depends on time. A rocket moves by the momentum carried by the exhaust gases, and it works in vacuum.
Assertion (A): Passengers lean forward when a moving bus stops suddenly. Reason (R): A body tends to continue its state of uniform motion unless acted on by an external force.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
This is the inertia of motion, described by the first law.
Assertion (A): Ball bearings are used to reduce friction in machines. Reason (R): Rolling friction is greater than sliding friction.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
Rolling friction is smaller than sliding friction, so R is false.
Assertion (A): A cricketer lowers his hands while catching a fast cricket ball. Reason (R): This increases the time of contact and so reduces the average force for the same change in momentum.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
F = Δp/Δt, so a larger Δt gives a smaller F.
Assertion (A): The centripetal force changes the speed of a body in uniform circular motion. Reason (R): The centripetal force acts towards the centre of the circle.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
Centripetal force is perpendicular to velocity and changes only the direction, not the speed.
Assertion (A): The apparent weight of a person in a lift moving up with constant speed is greater than mg. Reason (R): The apparent weight in a lift accelerating upward is m(g + a).
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
With constant speed a = 0, so the apparent weight is mg. R is a correct statement.
Assertion (A): Static friction is self-adjusting. Reason (R): The coefficient of static friction is greater than the coefficient of kinetic friction.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but the self-adjusting nature comes from f(static) = applied force up to the limit, not from μs > μk.
Match List I with List II. List I: a. Force b. Momentum c. Coefficient of friction d. Centripetal acceleration. List II: i. [L T⁻²] ii. dimensionless iii. [M L T⁻¹] iv. [M L T⁻²]
1. a-iii, b-iv, c-ii, d-i
2. a-iv, b-iii, c-ii, d-i
3. a-i, b-ii, c-iii, d-iv
4. a-iv, b-iii, c-i, d-ii
Answer
B. a-iv, b-iii, c-ii, d-i
Force [M L T⁻²] iv; momentum [M L T⁻¹] iii; μ is a ratio ii; centripetal acceleration [L T⁻²] i.
Match List I with List II for a person of mass m standing in a lift. List I: a. Lift at rest b. Lift accelerating upward at g/2 c. Lift accelerating downward at g/2 d. Lift in free fall. List II: i. mg/2 ii. 3mg/2 iii. zero iv. mg
1. a-iv, b-ii, c-iii, d-i
2. a-ii, b-iv, c-iii, d-i
3. a-iv, b-ii, c-i, d-iii
4. a-iv, b-i, c-ii, d-iii
Answer
C. a-iv, b-ii, c-i, d-iii
Rest: mg (iv). Up: m(g + g/2) = 3mg/2 (ii). Down: m(g − g/2) = mg/2 (i). Free fall: zero (iii).
Match List I with List II. List I: a. Ideal speed on a banked road b. Maximum speed on a level road c. Recoil velocity of a gun d. Acceleration in an Atwood machine. List II: i. √(μ r g) ii. √(r g tan θ) iii. (m₁ − m₂)g/(m₁ + m₂) iv. − m v/M
1. a-i, b-ii, c-iv, d-iii
2. a-ii, b-i, c-iv, d-iii
3. a-ii, b-i, c-iii, d-iv
4. a-iii, b-i, c-iv, d-ii
Answer
B. a-ii, b-i, c-iv, d-iii
Each expression is the standard result for its situation.
Match List I with List II. List I: a. Static friction b. Kinetic friction c. Rolling friction d. Limiting friction. List II: i. μs N at the point of impending slide ii. the smallest of the three common types iii. equals μk N iv. self-adjusting up to a maximum
1. a-iii, b-iv, c-ii, d-i
2. a-i, b-iii, c-ii, d-iv
3. a-iv, b-iii, c-ii, d-i
4. a-iv, b-iii, c-i, d-ii
Answer
C. a-iv, b-iii, c-ii, d-i
Static friction adjusts itself (iv); kinetic friction is μk N (iii); rolling friction is smallest (ii); limiting friction is μs N (i).
Match List I with List II (take g = 10 m s⁻²). List I: a. Weight of a 5 kg body b. Impulse of 10 N acting for 3 s c. Momentum of a 2 kg body moving at 5 m s⁻¹ d. Limiting friction for a 20 kg block with μs = 0.2. List II: i. 40 N ii. 30 N s iii. 50 N iv. 10 kg m s⁻¹
1. a-iii, b-ii, c-iv, d-i
2. a-iii, b-iv, c-ii, d-i
3. a-i, b-ii, c-iv, d-iii
4. a-iii, b-ii, c-i, d-iv
Answer
A. a-iii, b-ii, c-iv, d-i
Weight = 50 N; impulse = 30 N s; momentum = 10 kg m s⁻¹; limiting friction = 0.2 × 200 = 40 N.
Match List I with List II. List I: a. Satellite orbiting the Earth b. Stone whirled on a string c. Car on a level curve d. Car on a frictionless banked road. List II: i. friction ii. gravity iii. tension iv. component of normal reaction. (The List II item is the one that supplies the centripetal force.)
1. a-ii, b-iii, c-iv, d-i
2. a-ii, b-iii, c-i, d-iv
3. a-iii, b-ii, c-i, d-iv
4. a-i, b-iii, c-ii, d-iv
Answer
B. a-ii, b-iii, c-i, d-iv
Gravity, tension, friction and the horizontal component of the normal reaction respectively.

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1.The collision in which the two particles operate simultaneously after collision are called _________.
2.The building up and movement of ships is based on ________.
3.The centripetal force required to keep the moon in its orbit is provided by which force?
4.The following statement applies to which law from the given options?'The line that joins any planet to the sun sweeps equal areas in equal intervals of time.'
5.The force of attraction between two bodies which is directly proportional to the product of masses of the bodies and inversely proportional to the square of the distance of separation is ________.
6.The force of attraction between all masses in the universe, especially the attraction of the earth's mass for bodies near its surface is called ________.
7.The force of attraction applied between molecules of the same substance is called:
8.The force of friction is the speed of some object ______
9.The formula for knowing stress is:
10.The first law of motion is known as the

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