# CBSE Class 11 Physics Chapter 5 Laws of motion Study Materials

### Law of Motion Class 11 Notes Physics Chapter 5

• Dynamics is the branch of physics in which we study the motion of a body by taking into consideration the cause i.e., force which produces the motion.
• Force
Force is an external cause in the form of push or pull, which produces or tries to produce motion in a body at rest, or stops/tries to stop a moving body or changes/tries to change the direction of motion of the body.
• The inherent property, with which a body resists any change in its state of motion is called inertia. Heavier the body, the inertia is more and lighter the body, lesser the inertia.
• Law of inertia states that a body has the inability to change its state of rest or uniform motion (i.e., a motion with constant velocity) or direction of motion by itself.
• Newton’s Laws of Motion
Law 1. A body will remain at rest or continue to move with uniform velocity unless an external force is applied to it.
First law of motion is also referred to as the ‘Law of inertia’. It defines inertia, force and inertial frame of reference.
I here is always a need of ‘frame of reference’ to describe and understand the motion of particle, lhc simplest ‘frame of reference’ used are known as the inertial frames.
A frame of referent, e is known as an inertial frame it, within it, all accelerations of any particle are caused by the action of ‘real forces’ on that particle.
When we talk about accelerations produced by ‘fictitious’ or ‘pseudo’ forces, the frame of reference is a non-inertial one.
Law 2. When an external force is applied to a body of constant mass the force produces an acceleration, which is directly proportional to the force and inversely proportional to the mass of the body.

Law 3. “To every action there is equal and opposite reaction force”. When a body A exerts a force on another body B, B exerts an equal and opposite force on A.
• Linear Momentum
The linear momentum of a body is defined as the product of the mass of the body and its velocity.

• Impulse
Forces acting for short duration are called impulsive forces. Impulse is defined as the product of force and the small time interval for which it acts. It is given by

Impulse of a force is a vector quantity and its SI unit is 1 Nm.
— If force of an impulse is changing with time, then the impulse is measured by finding the area bound by force-time graph for that force.
— Impulse of a force for a given time is equal to the total change in momentum of the body during the given time. Thus, we have

• Law of Conservation of Momentum
The total momentum of an isolated system of particles is conserved.
In other words, when no external force is applied to the system, its total momentum remains constant.

• Recoiling of a gun, flight of rockets and jet planes are some simple applications of the law of conservation of linear momentum.

• Concurrent Forces and Equilibrium
“A group of forces which are acting at one point are called concurrent forces.”
Concurrent forces are said to be in equilibrium if there is no change in the position of rest or the state of uniform motion of the body on which these concurrent forces are acting.
For concurrent forces to be in equilibrium, their resultant force must be zero. In case of three concurrent forces acting in a plane, the body will be in equilibrium if these three forces may be completely represented by three sides of a triangle taken in order. If number of concurrent forces is more than three, then these forces must be represented by sides of a closed polygon in order for equilibrium.

• Commonly Used Forces
(i) Weight of a body. It is the force with which earth attracts a body towards its centre. If M is mass of body and g is acceleration due to gravity, weight of the body is Mg in vertically downward direction.
(ii) Normal Force. If two bodies are in contact a contact force arises, if the surface is smooth the direction of force is normal to the plane of contact. We call this force as Normal force.

Example. Let us consider a book resting on the table. It is acted upon by its weight in vertically downward direction and is at rest. It means there is another force acting on the block in opposite direction, which balances its weight. This force is provided by the table and we call it as normal force.
(iii) Tension in string. Suppose a block is hanging from a string. Weight of the block is acting vertically downward but it is not moving, hence its weight is balanced by a force due to string. This force is called ‘Tension in string’. Tension is a force in a stretched string. Its direction is taken along the string and away from the body under consideration.

• Simple Pulley
Consider two bodies of masses m1 and m2 tied at the ends of an in extensible string, which passes over a light and friction less pulley. Let m1 > m2. The heavier body will move downwards and the lighter will move upwards. Let a be the common acceleration of the system of two bodies, which is given by

• Apparent Weight and Actual Weight
— ‘Apparent weight’ of a body is equal to its ‘actual weight’ if the body is either in a state of rest or in a state of uniform motion.
— Apparent weight of a body for vertically upward accelerated motion is given as
Apparent weight =Actual weight + Ma = M (g + a)
— Apparent weight of a body for vertically downward accelerated motion is given as
Apparent weight = Actual weight Ma = M (g – a).
• Friction
The opposition to any relative motion between two surfaces in contact is referred to as friction. It arises because of the ‘inter meshing’ of the surface irregularities of the two surfaces in contact.
• Static and Dynamic (Kinetic) Friction
The frictional forces between two surfaces in contact (i) before and (ii) after a relative motion between them has started, are referred to as static and dynamic friction respectively. Static friction is always a little more than dynamic friction.
The magnitude of kinetic frictional force is also proportional to normal force.

• Limiting Frictional Force
This frictional force acts when body is about to move. This is the maximum frictional force that can exist at the contact surface. We calculate its value using laws of friction.
Laws of Friction:
(i) The magnitude of limiting frictional force is proportional to the normal force at the contact surface.

(ii) The magnitude of limiting frictional force is independent of area of contact between the surfaces.
• Coefficient of Friction
The coefficient of friction (μ) between two surfaces is the ratio of their limiting frictional force to the normal force between them, i.e.,

• Angle of Friction
It is the angle which the resultant of the force of limiting friction F and the normal reaction R makes with the direction of the normal reaction. If θ is the angle of friction, we have

• Angle of Repose
Angle of repose (α) is the angle of an inclined plane with the horizontal at which a body placed over it just begins to slide down without any acceleration. Angle of repose is given by α = tan-1 (μ)

• Motion on a Rough Inclined Plane
Suppose a motion up the plane takes place under the action of pull P acting parallel to the plane.
• Centripetal Force
Centripetal force is the force required to move a body uniformly in a circle. This force acts along the radius and towards the centre of the circle. It is given by

where, v is the linear velocity, r is the radius of circular path and ω is the angular velocity of the body.
• Centrifugal Force
Centrifugal force is a force that arises when a body is moving actually along a circular path, by virtue of tendency of the body to regain its natural straight line path.
The magnitude of centrifugal force is same as that of centripetal force.

• Motion in a Vertical Circle
The motion of a particle in a horizontal circle is different from the motion in vertical circle. In horizontal circle, the motion is not effected by the acceleration due to gravity (g) whereas in the motion of vertical circle, the motion is not effected by the acceleration due to gravity (g) whereas in the motion of vertical circle, the value of ‘g’ plays an important role, the motion in this case does not remain uniform. When the particle move up from its lowest position P, its speed continuously decreases till it reaches the highest point of its circular path. This is due to the work done against the force of gravity. When the particle moves down the circle, its speed would keep on increasing.

Let us consider a particle moving in a circular vertical path of radius V and centre o tide with a string. L be the instantaneous position of the particle such that

Here the following forces act on the particle of mass ‘m’.
(i) Its weight = mg (verticaly downwards).
(ii) The tension in the string T along LO.

We can take the horizontal direction at the lowest point ‘p’ as the position of zero gravitational potential energy. Now as per the principle of conservation of energy,

From this relation, we can calculate the tension in the string at the lowest point P, mid-way point and at the highest position of the moving particle.
Case (i) : At the lowest point P, θ = 0°

When the particle completes its motion along the vertical circle, it is referred to as “Looping the Loop” for this the minimum speed at the lowest position must be √5gr
• IMPORTANT TABLES

### CBSE Class 11 Physics Chapter-5 Important Questions

1 Marks Questions

1.What is the unit of coefficient of friction?

Ans: It has no unit.

2.Name the factor on which coefficient of friction depends?

Ans:Coefficient of friction  depends on the nature of surfaces in contact and nature of motion.

3.What provides the centripetal force to a car taking a turn on a level road?

Ans: Centripetal force is provided by the force of friction between the tyres and the road.

4.Why is it desired to hold a gun tight to one’s shoulder when it is being fired?

Ans: Since the gun recoils after firing so it must be held lightly against the shoulder because gun and the shoulder constitute one system of greater mass so the back kick will be less.

5.Why does a swimmer push the water backwards?

Ans: A swimmer pushes the water backwards because due to reaction of water he is able to swim in the forward direction

6.Friction is a self adjusting force. Justify.

Ans: Friction is a self adjusting force as its value varies from zero to the maximum value to limiting friction.

7.A thief jumps from the roof of a house with a box of weight W on his head. What will be the weight of the box as experienced by the thief during jump?

Ans: Weight of the box W = m (g – a) = m (g – g) = 0.

8.Which of the following is scalar quantity? Inertia, force and linear momentum.

Ans:Inertia and linear momentum is measured by mass of the body and is a vector quantity and mass is a scalar quantity.

9.Action and reaction forces do not balance each other. Why?

Ans:Action and reaction do not balance each other because a force of action and reaction acts always on two different bodies.

10.If force is acting on a moving body perpendicular to the direction of motion, then what will be its effect on the speed and direction of the body?

Ans:No change in speed, but there can be change in the direction of motion.

11.The two ends of spring – balance are pulled each by a force of 10kg.wt. What will be the reading of the balance?

Ans:The reading of the balance will be 10kgwt.

12.A lift is accelerated upward. Will the apparent weight of a person inside the lift increase, decrease or remain the same relative to its real weight? If the lift is going with uniform speed, then?

Ans:The apparent weight will increase. If the lift is going with uniform speed, then the apparent weight will remain the same as the real weight.

13. 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) ,
(iii) ,

(iv) 0

T is the tension in the string. [Choose the correct alternative].

Ans. (i) When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e.,

F = T =

Where F is the net force acting on the particle.

14. If, in Exercise 5.21, 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 ?

Ans.(b)When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

2 Marks Questions

1.Give the magnitude and direction of the net force acting on

(a) A drop of rain falling down with constant speed.

(b) A kite skillfully held stationary in the sky.

Ans: (1) According to first law of motion F = 0 as a = 0 (particle moves with constant speed)

(2) Since kite is stationary net force on the kite is also zero.

2.Two blocks of masses m1, m2 are connected by light spring on a smooth horizontal surface. The two masses are pulled apart and then released. Prove that the ratio of their acceleration is inversely proportional to their masses.

Ans:The forces F1 and F2 due to masses m1 and m2 acts in opposite directions

Thus F1 + F= 0

m1 a1 + ma2 = 0

m1 a= -ma2

Hence proved.

3.A shell of mass 0.020kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is 80m/s, what is the recoil speed of the gun?

Ans:Momentum before firing = 0

Momentum after firing = momentum of (bullet+gun)

Momentum after firing = mbb – mg

According to law of conservation of linear momentum

0 = mbb – mg

mbb = mg

4.A force is being applied on a body but it causes no acceleration. What possibilities may be considered to explain the observation?

Ans: (1) If the force is deforming force then it does not produce acceleration.

(2) The force is internal force which cannot cause acceleration.

5.Force of 16N and 12N are acting on a mass of 200kg in mutually perpendicular directions. Find the magnitude of the acceleration produced?

Ans:

6.An elevator weighs 3000kg. What is its acceleration when the in the tension supporting cable is 33000N. Given that g = 9.8m/s2.

Ans:Net upward force on the

ElevatorF = T – mg

7.Write two consequences of Newton’s second law of motion?

Ans: (1) It shows that the motion is accelerated only when force is applied.

(2) It gives us the concept of inertial mass of a body.

8.A bird is sitting on the floor of a wire cage and the cage is in the hand of a boy. The bird starts flying in the cage. Will the boy experience any change in the weight of the cage?

Ans:When the bird starts flying inside the cage the weight of bird is no more experienced as air inside is in free contact with atmospheric air hence the cage will appear lighter.

9.Why does a cyclist lean to one side, while going along curve? In what direction does he lean?

Ans:A cyclist leans while going along curve because a component of normal reaction of the ground provides him the centripetal force he requires for turning.

He has to lean inwards from his vertical position i.e. towards the centre of the circular path.

10.How does banking of roads reduce wear and tear of the tyres?

Ans:When a curved road is unbanked force of friction between the tyres and the road provides the necessary centripetal force. Friction has to be increased which will cause wear and tear. But when the curved road is banked, a component of normal reaction of the ground provides the necessary centripetal force which reduces the wear and tear of the tyres.

11.A monkey of mass 40 kg climbs on a rope which can stand a maximum tension 600 N. In which of the following cases will the rope break? The monkey (a) climbs up with an acceleration of 6m/s2 (b) climbs down with an acceleration of 4m/s2 (c) climbs up with a uniform seed of 5m/s (d) falls down the rope freely under gravity. Take g = 10m/s2 and ignore the mass of the rope.

Ans:m = 40kg, T = 600N (max tension rope can hold)

Rope will break if reaction (R) exceeds Tension (T)

(a) a = 6m/s2

R = m (g + a) = 40 (10 + 6) = 640 N (Rope will break)

(b) a = 4m/s2

R = m (g – a) = 40 (10 – 6) = 240 N (Rope will not break)

(c) = 5m/s (constant) a = 0

R = mg = 40 10 = 400 N (Rope will not break)

(d) a = g;  R = m (g – a) = m (g – g)

R = zero (Rope will not break)

12.A soda water bottle is falling freely. Will the bubbles of the gas rise in the water of the bottle?

Ans:bubbles will not rise in water because water in freely falling bottle is in the state of weight – lessens hence no up thrust force acts on the bubbles.

13.Two billiard balls each of mass 0.05kg moving in opposite directions with speed 6m/s collide and rebound with the same speed. What is the impulse imparted to each ball due to other.

Ans:Initial momentum to the ball A = 0.05(6) = 0.3 kg m/s

As the speed is reversed on collision,

final momentum of ball A = 0.05(-6) = -0.3 kg m/s

Impulse imparted to ball A = change in momentum of ball A = final momentum – initial momentum = -0.3 -0.3 = -0.6 kg m/s.

14.A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must be emitted in opposite directions.

Ans:According to the principle of conservation of linear momentum, total momentum remains constant.

Before disintegration linear momentum = zero

After disintegration linear momentum =

15.Explain why passengers are thrown forward form their seats when a speeding bus stops suddenly.

Ans:When the speeding bus stops suddenly, lower part of the body in contact with the seat comes to rest but the upper part of the body of the passengers tends to maintain its uniform motion. Hence the passengers are thrown forward.

16. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m. Calculate the initial thrust (force) of the blast.

Ans. Mass of the rocket, m= 20,000 kg

Initial acceleration, a= 5 m/s2

Acceleration due to gravity, g

Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation:

F– mg = ma

Fm (g + a)

17. 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. 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.

Ans. (a) Vertically downward

(b) Parabolic path

(a) At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.

(b)At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.

18. Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms-1collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Ans.Mass of each ball = 0.05 kg

Initial velocity of each ball = 6 m/s

Magnitude of the initial momentum of each ball, = 0.3 kg m/s

After collision, the balls change their directions of motion without changing the magnitudes of their velocity.

Final momentum of each ball, = -0.3 kg m/s

Impulse imparted to each ball = Change in the momentum of the system

=

= -0.3 – 0.3 = -0.6 kg m/s

The negative sign indicates that the impulses imparted to the balls are opposite in direction.

19. A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is  kg. What provides the centripetal force required for this purpose – The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Ans.Radius of the circular track, r= 30 m

Speed of the train, v= 54 km/h = 15 m/s

Mass of the train, mkg

The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail

The angle of banking θ, is related to the radius (r) and speed (v) by the relation:

Therefore, the angle of banking is about 36.87°.

20. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of. How long does the body take to stop?

Ans. Retarding force, F =

Mass of the body, m= 20 kg

Initial velocity of the body, u= 15 m/s

Final velocity of the body, v= 0

Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:

Fma

= 20 × a

Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:

vu + at

= 6 s

21. 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.

Ans. Let m, and  be the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest.

Initial momentum of the system (parent nucleus) = 0

Let and  be the respective velocities of the daughter nuclei having masses m1and m2.

Total linear momentum of the system after disintegration =

According to the law of conservation of momentum:

Total initial momentum = Total final momentum

Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

22. 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, what is the recoil speed of the gun?

Ans. Mass of the gun, M= 100 kg

Mass of the shell, m= 0.020 kg

Muzzle speed of the shell, v= 80 m/s

Recoil speed of the gun = V

Both the gun and the shell are at rest initially.

Initial momentum of the system = 0

Final momentum of the system =

Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.

According to the law of conservation of momentum:

Final momentum = Initial momentum

3 Marks Questions

1.A train runs along an unbanked circular bend of radius 30m at a speed of 54km/hr. The mass of the train is 106kg. What provides the necessary centripetal force required for this purpose? The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Ans:  (1) The centripetal force is provided by the lateral force acting due to rails on the wheels of the train.

(2) Outer rails

(3)

2.   A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.

Ans.

(a) Mass of the block, m = 15 kg

Coefficient of static friction,  = 0.18

Acceleration of the trolley, a =

As per Newton’s second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:

F = ma =   = 7.5 N

This force is acted in the direction of motion of the trolley.

Force of static friction between the block and the trolley:

f =

= 0.18 x 15 x 10 = 27 N

The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest.

When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.

(b) An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.

3.What is the acceleration of the blocks? What is the net force on the block P? What force does P apply on Q. What force does Q apply on R?

Ans:If a is the acceleration

Then F = (3m)a

(1) Net force on P

(2) Force applied on Q

F= (m + m)a

(3) Force applied on R by Q

4.How is centripetal force provided in case of the following?

(i) Motion of planet around the sun,

(ii) Motion of moon around the earth.

(iii) Motion of an electron around the nucleus in an atom.

Ans:(i) Gravitational force acting on the planet and the sun provides the necessary centripetal force.

(ii) Force of gravity due to earth on the moon provides centripetal force.

(iii) Electrostatic force attraction between the electron and the proton provides  the necessary centripetal force.

5.State Newton’s second, law of motion. Express it mathematically and hence obtain a relation between force and acceleration.

Ans:  According to Newton’s second law the rate of change of momentum is directly proportional   to the force.

i.e. F rate of change of momentum

(In S.I. unit K = 1)

6.A railway car of mass 20 tonnes moves with an initial speed of 54km/hr. On applying brakes, a constant negative acceleration of 0.3m/s2 is produced.

(i) What is the breaking force acting on the car?

(ii) In what time it will stop?

(iii) What distance will be covered by the car before if finally stops?

Ans:

(a) F = ma

F = -6000N

(b)

t = 50s

(c)

S = 375m

7.What is meant by coefficient of friction and angel of friction? Establish the relation between the two?   OR

A block of mass 10kg is sliding on a surface inclined at a angle of 30with the horizontal. Calculate the acceleration of the block. The coefficient of kinetic friction between the block and the surface is 0.5

Ans:Angle of friction is the contact between the resultant of limiting friction and normal reaction

with the normal reaction

Coefficient of static friction

The limiting value of static frictional force is proportion to the normal reaction is

Or

From (1) & (2)

OR

A block of mass 10kg is sliding on a surface inclined at a angle of 30o with the horizontal. Calculate the acceleration of the block. The coefficient of kinetic friction between the block and the surface is 0.5

a = 0.657m/s2

8.State and prove the principle of law of conservation of linear momentum?

Ans: The law of conservation of linear momentum states that if no external force acts on the system. The total momentum of the system remains unchanged.

i.e. if

Impulse experienced by

Impulse experienced by

According to Newton’s third law

Thus momentum gained by one ball is lost by the other ball. Hence linear momentum remains conserved.

9.A particle of mass 0.40 kg moving initially with constant speed of 10m/s to the north is subject to a constant force of 8.0 N directed towards south for 30s. Take at that instant, the force is applied to be t = 0, and the position of the particle at that time to be x = 0, predict its position at t = -5s, 25s, 30s?

Ans. m = 0.40kg

u = l0m/s due North

F = – 8.0N

(1) At t = -5s

x = -50m

(2) At t = 25s

x = -6000m

(3) At t = 30s

(4) At t = 30s

Motion from 30s to 100s

Total distance x = x1 + x2

x = -50000m

10.   A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

Ans.

750 N and 250 N in the respective cases; Method (b)

Mass of the block, m = 25 kg

Mass of the man, M = 50 kg

Acceleration due to gravity, g =

Force applied on the block, F = 25 x 10 = 250 N

Weight of the man, W = 50 x 10 = 500 N

Case (a): When the man lifts the block directly

In this case, the man applies a force in the upward direction. This increases his apparent weight.

∴Action on the floor by the man = 250 + 500 = 750 N

Case (b): When the man lifts the block using a pulley

In this case, the man applies a force in the downward direction. This decreases his apparent weight.

∴Action on the floor by the man = 500 – 250 = 250 N

If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.

11.(a) State impulse – momentum theorem?

(b) A ball of mass 0.1kg is thrown against a wall. It strikes the wall normally with a velocity of 30m/s and rebounds with a velocity of 20m/s. calculate the impulse of the force exerted by the ball on the wall.

Ans: (a) It states that impulse is measured by the total change in linear momentum is

Impulse =

(b) m = 0.1kg v = 30m/s

Impulse =

Impulse = s

Impulse = m (-20 – 30) = -5Ns

12.Ten one rupee coins are put on top of one another on a table. Each coin has a mass m kg. Give the magnitude and direction of

(a) The force on the 7th coin (counted from the bottom) due to all coins above it.

(b) The force on the 7th coin by the eighth coin and

(c) The reaction of the sixth coin on the seventh coin.

Ans.(a) The force on 7th coin is due to weight of the three coins lying above it.

Therefore, F = (3 m) kgf = (3 mg) N

Where g is acceleration due to gravity. This force acts vertically downwards.

(b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e.

F = 2m + m = (3m) kg f = (3 mg) N

The force acts vertically downwards.

(c) The sixth coins is under the weight of four coins above it

Reaction, R = -F = -4 m (kgf) = – (4 mg) N

13.   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.

Ans (a) Zero net force

The rain drop is falling with a constant speed. Hence, it acceleration is zero. As per Newton’s second law of motion, the net force acting on the rain drop is zero.

(b) Zero net force

The weight of the cork is acting downward. It is balanced by the buoyant force exerted by the water in the upward direction. Hence, no net force is acting on the floating cork.

(c) Zero net force

The kite is stationary in the sky, i.e., it is not moving at all. Hence, as per Newton’s first law of motion, no net force is acting on the kite.

(d) Zero net force

The car is moving on a rough road with a constant velocity. Hence, its acceleration is zero. As per Newton’s second law of motion, no net force is acting on the car.

(e) Zero net force

The high speed electron is free from the influence of all fields. Hence, no net force is acting on the electron.

14.   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.

Ans. 0.5 N, in vertically downward direction, in all cases

Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as:

Where,

F= Net force

m= Mass of the pebble = 0.05 kg

a = g =

F = 0.5 N

The net force on the pebble in all three cases is 0.5 N and this force acts in the downward direction.

If the pebble is thrown at an angle of 45° with the horizontal, it will have both the horizontal and vertical components of velocity. At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble.

15.   A constant force acting on a body of mass 3.0 kg changes its speed from  to  in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Ans. 0.18 N; in the direction of motion of the body

Mass of the body, m= 3 kg

Initial speed of the body, u= 2 m/s

Final speed of the body, v= 3.5 m/s

Time, t = 25 s

Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:

vu + at

As per Newton’s second law of motion, force is given as:

Fma

= 0.18 N

Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.

16.   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.

Ans. , at an angle of 37° with a force of 8 N

Mass of the body, m= 5 kg

The given situation can be represented as follows:

The resultant of two forces is given as:

is the angle made by with the force of 8 N

The negative sign indicates that is in the clockwise direction with respect to the force of magnitude 8 N.

As per Newton’s second law of motion, the acceleration (a) of the body is given as:

F = ma

17.   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?

Ans.

Mass of the stone, m = 0.25 kg

Radius of the circle, r = 1.5 m

Number of revolution per second,

Angular velocity,

The centripetal force for the stone is provided by the tension T, in the string, i.e.,

Maximum tension in the string, = 200 N

Therefore, the maximum speed of the stone is 34.64 m/s.

18.   Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg.)

Figure 5.18

Ans. Mass of the man, m= 65 kg

Acceleration of the belt,

Coefficient of static friction, = 0.2

The net force F, acting on the man is given by Newton’s second law of motion as:

= 65 × 1 = 65 N

The man will continue to be stationary with respect to the conveyor belt until the net force on the man is less than or equal to the frictional force, exerted by the belt, i.e.,

Therefore, the maximum acceleration of the belt up to which the man can stand stationary is.

19.   A stream of water flowing horizontally with a speed of gushes out of a tube of cross-sectional area, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

Ans.  Speed of the water stream, v= 15 m/s
Cross-sectional area of the tube, A=
Volume of water coming out from the pipe per second,
VAv =
Density of water,
Mass of water flowing out through the pipe per second = = 150 kg/s
The water strikes the wall and does not rebound. Therefore, the force exerted by the water on the wall is given by Newton’s second law of motion as:
F= Rate of change of momentum

20.   An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?

Ans.  Speed of the aircraft, v = 720 km/h
Acceleration due to gravity, g =
Angle of banking, θ = 15°
For radius r, of the loop, we have the relation:

= 14925.37 m
= 14.92 km

4 Marks Questions

1. 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.

Ans. (a) 1 N; vertically downward

Mass of the stone, m= 0.1 kg

Acceleration of the stone, a= g =

As per Newton’s second law of motion, the net force acting on the stone,

Fma = mg

= 0.1 x 10 = 1 N

Acceleration due to gravity always acts in the downward direction.

(b) 1 N; vertically downward

The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, no force is acting on the stone in the horizontal direction.

The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. The magnitude of this force is 1 N.

(c) 1 N; vertically downward

It is given that the train is accelerating at the rate of.

Therefore, the net force acting on the stone, F‘ = ma = 0.1 x 1 = 0.1 N

This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F,’ stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations.

Therefore, the net force acting on the stone is given only by acceleration due to gravity.

Fmg = 1 N

This force acts vertically downward.

(d) 0.1 N; in the direction of motion of the train

The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train.

Acceleration of the train,

The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by:

Fma

= 0.1 x 1 = 0.1 N

2. 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.

Ans. Initial speed of the three-wheeler, u = 36 km/h = 10 m/s

Final speed of the three-wheeler, v = 0 m/s

Time, t = 4 s

Mass of the three-wheeler, m = 400 kg

Mass of the driver, m‘ = 65 kg

Total mass of the system, M = 400 + 65 = 465 kg

Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as:

v = u + at

The negative sign indicates that the velocity of the three-wheeler is decreasing with time.

Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as:

F = Ma

The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.

3.  A body of mass 0.40 kg moving initially with a constant speed of 10 ms-1to 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.

Ans. Mass of the body, m= 0.40 kg

Initial speed of the body, u= 10 m/s due north

Force acting on the body, F

Acceleration produced in the body,

(i) At t= -5 s

Acceleration, a‘ = 0 and u = 10 m/s

(ii) At t= 25 s

Acceleration, a” = and u = 10 m/s

(iii) At t= 100 s

For

a

u= 10 m/s

For

As per the first equation of motion, for t= 30 s, final velocity is given as:

vu + at

Velocity of the body after 30 s =

For motion between 30 s to 100 s, i.e., in 70 s:

∴Total distance,

4. A truck starts from rest and accelerates uniformly at. 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.)

Ans. (a) 22.36 m/s, at an angle of 26.57° with the motion of the truck

(b)

(a) Initial velocity of the truck, u = 0

Acceleration, a =

Time, t = 10 s

As per the first equation of motion, final velocity is given as:

v = u + at

= 0 + 2 × 10 = 20 m/s

The final velocity of the truck and hence, of the stone is 20 m/s.

At t = 11 s, the horizontal component () of velocity, in the absence of air resistance, remains unchanged, i.e.,

= 20 m/s

The vertical component () of velocity of the stone is given by the first equation of motion as:

Where, = = 1 s and

The resultant velocity (v) of the stone is given as:

Let θ be the angle made by the resultant velocity with the horizontal component of velocity,

= 26.57°

(b) When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. Hence, the acceleration of the stone is and it acts vertically downward.

5.  Figure 5.16 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).

Figure 5.16

Ans.

(a) For t < 0

It can be observed from the given graph that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. Hence, the force acting on the particle is zero.

For t > 4 s

It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of
3 m from the origin. Hence, no force is acting on the particle.

For 0 < t < 4

It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.

(b) At t = 0

Impulse = Change in momentum

=

Mass of the particle, m = 4 kg

Initial velocity of the particle, u = 0

Final velocity of the particle,

∴Impulse

At t = 4 s

Initial velocity of the particle,

Final velocity of the particle, v = 0

∴ Impulse

6.  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?

Ans.

Horizontal force, F = 600 N

Mass of body A, = 10 kg

Mass of body B, = 20 kg

Total mass of the system, = 30 kg

Using Newton’s second law of motion, the acceleration (a) produced in the system can be calculated as:

Fma

When force is applied on body A:

The equation of motion can be written as:

= 400 N … (i)

When force is applied on body B:

The equation of motion can be written as:

… (ii)

7.  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.)

Ans. The given situation can be represented as shown in the following figure.

Where,

AO = Incident path of the ball

OB = Path followed by the ball after deflection

∠AOB = Angle between the incident and deflected paths of the ball = 45°

∠AOP = ∠BOP = 22.5° = θ

Initial and final velocities of the ball =v

Horizontal component of the initial velocity = vcos along RO

Vertical component of the initial velocity = vsin  along PO

Horizontal component of the final velocity = vcos θ along OS

Vertical component of the final velocity = vsin  along OP

The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.

∴Impulse imparted to the ball = Change in the linear momentum of the ball

Mass of the ball, m= 0.15 kg

Velocity of the ball, v= 54 km/h = 15 m/s

∴Impulse =  = 4.16 kg m/s

8.  Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

Figure 5.17

Ans.

A ball rebounding between two walls located between at x= 0 and x= 2 cm; after every 2 s, the ball receives an impulse of magnitude from the walls

The given graph shows that a body changes its direction of motion after every 2 s. Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions x= 0 and x= 2 cm. Since the slope of the xgraph reverses after every 2 s, the ball collides with a wall after every 2 s. Therefore, ball receives an impulse after every 2 s.

Mass of the ball, m= 0.04 kg

The slope of the graph gives the velocity of the ball. Using the graph, we can calculate initial velocity (u) as:

Velocity of the ball before collision,

Velocity of the ball after collision,

(Here, the negative sign arises as the ball reverses its direction of motion.)

Magnitude of impulse = Change in momentum

9.  A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

 Lowest Point Highest Point (a) (b) (c) (d)

Ans. (a)The free body diagram of the stone at the lowest point is shown in the following figure.

According to Newton’s second law of motion, the net force acting on the stone at this point is equal to the centripetal force, i.e.,

… (i)

Where, = Velocity at the lowest point

The free body diagram of the stone at the highest point is shown in the following figure.

Using Newton’s second law of motion, we have:

… (ii)

Where, = Velocity at the highest point

It is clear from equations (i) and (ii) that the net force acting at the lowest and the highest points are respectively and (T + mg).

10.  A disc revolves with a speed of  rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

Ans. Coin placed at 4 cm from the centre
Mass of each coin = m
Radius of the disc, r= 15 cm = 0.15 m
Frequency of revolution,   rev/min
Coefficient of friction,  = 0.15

In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.

Coin placed at 4 cm:

Radius of revolution, r‘ = 4 cm = 0.04 m
Angular frequency,
Frictional force, f = mg = 0.15 × m× 10 = 1.5mN
Centripetal force on the coin:

= 0.49m N
Since f > ,the coin will revolve along with the record.
Coin placed at 14 cm:
Radius, = 14 cm = 0.14 m
Angular frequency,
Frictional force, f‘ = 1.5m N
Centripetal force is given as:

= 1.7m N
Since f < , the coin will slip from the surface of the record.

11. A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Ans. Mass of the man, m= 70 kg
Radius of the drum, r= 3 m
Coefficient of friction,  = 0.15
Frequency of rotation, v= 200 rev/min
The necessary centripetal force required for the rotation of the man is provided by the normal force (FN).
When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man
(mg) acting downward is balanced by the frictional force (f = ) acting upward.
Hence, the man will not fall until:
mg < f

The minimum angular speed is given as:

5 Marks Questions

1.(a) Define impulse. State its S.I. unit?

(b) State and prove impulse momentum theorem?

Ans:(a) Force which are exerted over a short time intervals are called impulsive forces.

Impulse

Unit – NS

Impulse is a vector quantity directed along the average force

(b) Impulse of a force is equal to the change in momentum of the body.

According to Newton’s second law

2.  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,

(b) downwards with a uniform acceleration of,

(c) upwards with a uniform acceleration of.

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?

Ans. (a) Mass of the man, m = 70 kg

Acceleration, a = 0

Using Newton’s second law of motion, we can write the equation of motion as:

= ma

Where, ma is the net force acting on the man.

As the lift is moving at a uniform speed, acceleration a = 0

R = mg

= 70 × 10 = 700 N

∴Reading on the weighing scale =

(b) Mass of the man, m = 70 kg

Acceleration, a = downward

Using Newton’s second law of motion, we can write the equation of motion as:

R + m= ma

=

= 350 N

∴Reading on the weighing scale =

(c) Mass of the man, m = 70 kg

Acceleration, upward

Using Newton’s second law of motion, we can write the equation of motion as:

R = m(g + a)

= 70 (10 + 5) = 70 × 15

= 1050 N

∴Reading on the weighing scale =

(d) When the lift moves freely under gravity, acceleration a = g

Using Newton’s second law of motion, we can write the equation of motion as:

R + m= ma

= 0

∴Reading on the weighing scale =

The man will be in a state of weightlessness.

3.  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.

Ans.

The given system of two masses and a pulley can be represented as shown in the following figure:

Smaller mass, = 8 kg

Larger mass, = 12 kg

Tension in the string = T

Mass, owing to its weight, moves downward with acceleration a, and mass moves upward.

Applying Newton’s second law of motion to the system of each mass:

For mass:

The equation of motion can be written as:

For mass:

The equation of motion can be written as:

Adding equations (i) and (ii), we get:

Therefore, the acceleration of the masses is.

Substituting the value of in equation (ii), we get:

Therefore, the tension in the string is 96 N.

4.  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.

Ans.(a) In order to pull a cart, a horse pushes the ground backward with some force. The ground  in turn exerts an equal and opposite reaction force upon the feet of the horse. This reaction force causes the horse to move forward.

An empty space is devoid of any such reaction force. Therefore, a horse cannot pull a cart and run in empty space.

(b) When a speeding bus stops suddenly, the lower portion of a passenger’s body, which is in contact with the seat, suddenly comes to rest. However, the upper portion tends to remain in motion (as per the first law of motion). As a result, the passenger’s upper body is thrown forward in the direction in which the bus was moving.

(c) While pulling a lawn mower, a force at an angle θ is applied on it, as shown in the following figure.

The vertical component of this applied force acts upward. This reduces the effective weight of the mower.

On the other hand, while pushing a lawn mower, a force at an angle θ is applied on it, as shown in the following figure.

In this case, the vertical component of the applied force acts in the direction of the weight of the mower. This increases the effective weight of the mower.

Since the effective weight of the lawn mower is lesser in the first case, pulling the lawn mower is easier than pushing it.

(d) According to Newton’s second law of motion, we have the equation of motion:

Where,

F = Stopping force experienced by the cricketer as he catches the ball

m = Mass of the ball

Δt = Time of impact of the ball with the hand

It can be inferred from equation (i) that the impact force is inversely proportional to the impact time, i.e.,

Equation (ii) shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa.

While taking a catch, a cricketer moves his hand backward so as to increase the time of impact (Δt). This is turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.

5.  A helicopter of mass 1000 kg rises with a vertical acceleration of. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

(c) force on the helicopter due to the surrounding air.

Ans.(a) Mass of the helicopter, = 1000 kg

Mass of the crew and passengers, = 300 kg

Total mass of the system, m = 1300 kg

Acceleration of the helicopter,

Using Newton’s second law of motion, the reaction force R, on the system by the floor can be calculated as:

= ma

= 300 (10 + 15) = 300 × 25

= 7500 N

Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is 7500 N, directed downward.

(b) Using Newton’s second law of motion, the reaction force R’, experienced by the helicopter can be calculated as:

m(g + a)

= 1300 (10 + 15) = 1300 × 25

= 32500 N

The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.

(c) The force on the helicopter due to the surrounding air is 32500 N, directed upward.

6.  Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of

(a) the force on the  coin (counted from the bottom) due to all the coins on its top,

(b) the force on the  coin by the eighth coin,

(c) the reaction of the  coin on the  coin.

Ans.

(a) Force on the seventh coin is exerted by the weight of the three coins on its top.

Weight of one coin = mg

Weight of three coins = 3mg

Hence, the force exerted on the  coin by the three coins on its top is 3mg. This force acts vertically downward.

(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.

Weight of the eighth coin = mg

Weight of the ninth coin = mg

Weight of the tenth coin = mg

Total weight of these three coins = 3mg

Hence, the force exerted on the  coin by the eighth coin is 3mg. This force acts vertically downward.

(c) The  coin experiences a downward force because of the weight of the four coins () on its top.

Therefore, the total downward force experienced by the  coin is 4mg.

As per Newton’s third law of motion, the  coin will produce an equal reaction force on the  coin, but in the opposite direction. Hence, the reaction force of the 6th coin on the  coin is of magnitude 4mg. This force acts in the upward direction.

7.   A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(a) climbs up with an acceleration of

(b) climbs down with an acceleration of

(c) climbs up with a uniform speed of

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

Fig. 5.20

Ans.

Case (a)

Mass of the monkey, m= 40 kg

Acceleration due to gravity, g= 10 m/s

Maximum tension that the rope can bear, = 600 N

Acceleration of the monkey, aupward

Using Newton’s second law of motion, we can write the equation of motion as:

T– mg= ma

Tm(g+ a)

= 40 (10 + 6)

= 640 N

Since T >, the rope will break in this case.

Case (b)

Acceleration of the monkey, a=downward

Using Newton’s second law of motion, we can write the equation of motion as:

mg – T = ma

Tm (g– a)

= 40(10 – 4)

= 240 N

Since T <, the rope will not break in this case.

Case (c)

The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a= 0.

Using Newton’s second law of motion, we can write the equation of motion as:

T– m= ma

T– mg = 0

T = mg

= 400 N

Since T < the rope will not break in this case.

Case (d)

When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., a = g

Using Newton’s second law of motion, we can write the equation of motion as:

mg – Tmg

Tm(g g) = 0

Since T <, the rope will not break in this case.

8.   Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between.

Fig. 5.21

Ans.

(a) Mass of body A, = 5 kg

Mass of body B, = 10 kg

Applied force, F = 200 N

Coefficient of friction, = 0.15

The force of friction is given by the relation:

= 0.15 (5 + 10) × 10

= 1.5 × 15 = 22.5 N leftward

Net force acting on the partition = = 177.5 N rightward

As per Newton’s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force.

Hence, the reaction of the partition will be 177.5 N, in the leftward direction.

(b) Force of friction on mass A:

= 0.15 × 5 × 10 = 7.5 N leftward

Net force exerted by mass A on mass B == 192.5 N rightward

As per Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A, i.e., 192.5 N acting leftward.

When the wall is removed, the two bodies will move in the direction of the applied force.

Net force acting on the moving system = 177.5 N

The equation of motion for the system of acceleration a, can be written as:

Net force causing mass A to move:

= 5 × 11.83 = 59.15 N

Net force exerted by mass A on mass B =  = 133.35 N

This force will act in the direction of motion. As per Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3 N, acting opposite to the direction of motion.

9.   The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Fig. 5.22

Ans. Mass of the box, m= 40 kg

Coefficient of friction,  = 0.15

Initial velocity, u= 0

Acceleration, a =

Distance of the box from the end of the truck, s‘ = 5 m

As per Newton’s second law of motion, the force on the box caused by the accelerated motion of the truck is given by:

Fma

= 40 × 2 = 80 N

As per Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction f, acting between the box and the floor of the truck. This force is given by:

f

∴Net force acting on the block:

The backward acceleration produced in the box is given by:

Using the second equation of motion, time t can be calculated as:

Hence, the box will fall from the truck after  from start.

The distance s, travelled by the truck in  is given by the relation:

= 20m

10.  You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?

Ans.

In a death-well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure.

The net force acting on the motorcyclist is the sum of the normal force (FN) and the force due to gravity ( ).

The equation of motion for the centripetal acceleration ac, can be written as:

mac

Normal reaction is provided by the speed of the motorcyclist. At the minimum speed (

11. A thin circular loop of radius R rotates about its vertical diameter with an angular frequency  . Show that a small bead on the wire loop remains at its lowermost point for .What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for  ?Neglect friction.

Ans.
Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.

OP = R = Radius of the circle
N= Normal reaction
The respective vertical and horizontal equations of forces can be written as:
… (i)
… (ii)
In ΔOPQ, we have:

l =  … (iii)
Substituting equation (iii) in equation (ii), we get:

Substituting equation (iv) in equation (i), we get:

… (v)
Since cosθ ≤ 1, the bead will remain at its lowermost point for, i.e., for
For or
On equating equations (v) and (vi), we get:

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