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Important Questions for CBSE Class 11 Physics Chapter 4 - Law of Motion

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CBSE Class 11 Physics Chapter 4 Important Questions - Free PDF Download

CBSE Class 11 Physics Chapter 4 - Law of Motion delves into the fundamental principles that govern the motion of objects in our everyday lives. This chapter introduces you to the renowned Newton's laws of motion, which form the backbone of classical physics. Understanding these laws is vital for comprehending the behavior of objects under the influence of various forces. To help you grasp these concepts thoroughly, this chapter presents a set of important questions carefully curated to test your understanding and problem-solving skills. The questions are picked up by the experts at Vedantu and are surely beneficial for your final examination. We prepare all types of questions based on the guidelines provided by the CBSE board. The following Class 11 Physics Important Questions can give you sufficient ideas. Some Class 11 Physics Chapter 4 Extra Questions (short and long) are also provided here.


Important Questions for CBSE Class 11 Physics Chapter 4 (Law of Motion) - Salient Features  

The salient features of the Important Questions for CBSE Class 11 Physics Chapter 4 (Law of Motion) are as follows: 

  1. The important questions and answers consist of all the important topic coverage of Chapter Law of Motion.

  2. The important questions and answers are prepared in such a way that it will help the students to study the chapter keeping in mind the important questions and answers. 

  3. The study material is prepared to take into account the types of questions asked in the CBSE Class 11 examination

  4. Laws of Motion consist of many important concepts and formulas which are explained in a lucid manner for better understanding purposes. 

  5. The important questions and answers are in PDF format. The students can download this PDF at no cost and avail advantage of the same. 

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CBSE Class 11 Physics Chapter 4 Important Questions - Free PDF Download

1. What is the unit of coefficient of friction?

Ans: Since, Coefficientpffriction=FrictionalForce×[NormalForce]1 ……(1)

Also, we know that

Force(F) = Mass × acceleration   = Mass × Velocity ×  [ Time ]  - 1

And

Velocity = Displacement×[ Time ]1

The dimensions of Force =[M]×[LT1]×[T]1=[M1L1T2]             ……(2)

On putting the equation (2) in equation (1) we get,

μ=[M1L1T2]×[M1L1T2]1=[M0L0T0].

Since mass, velocity and time all having zero dimensions. Hence, the coefficient of friction has no units.

 

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

Ans: The coefficient of friction will mainly depends upon two factor, they are as following:

i. The materials of the surfaces in contact.

ii. The characteristics of the surfaces.


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

Ans: The frictional contact between the tyres and the road provides centripetal force.

 

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

Ans: As the gun recoils after shooting, it must be held softly on the shoulder. Here the gun and the shoulder are one mass system, due to this the back kick will be reduced. A gunman must keep his weapon securely against his shoulder when shooting.

 

5. Why does a swimmer push the water backwards?

Ans: From the Newton's 3rd law of motion, we know that "when one body exerts a force on the other body, the first body experiences a force equivalent in magnitude in the opposite direction of the force exerted". As a result, in order to swim ahead, the swimmer pushes water backward with his hands.

 

6. Friction is a self-adjusting force. Justify.

Ans: Friction is a self-adjusting force that changes in magnitude from zero to maximum to limit 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: The thief is in free fall during the jump. Both he/she and the box will be weightless during that time. So, the weight of the box experience by the thief during the jump will be zero.

So, mathematically it can be written as:

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: The resistance of a body to its own acceleration is measured by its inertia. As a result, mass becomes a qualitative indicator of inertia. Because mass is a scalar quantity, linear inertia is also a scalar quantity. Hence, inertia will be the scalar quantity among them.

 

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

Ans: Because a force of action and response always operates on two separate bodies, action and reaction do not balance each other.

 

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: When a force acts in a perpendicular direction on a moving body, the work done by the force is zero.

Since 

W=FScosθ, where S=90 and cos90=0, therefore W=0.

As a result, the magnitude of the body's velocity (or speed) will remain unchanged. The direction of motion of the body, however, will be altered.

 

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: As the spring balancing is based on the tension in the spring, it gauges weight. Now, if both ends are pulled by a 10kg weight, the tension is 10kg, and the reading will be 10kg.

 

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: There will be an increase in perceived weight. The apparent weight will stay the same as the true weight if the lift moves at a constant pace.

 

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.

ii. Tmv2l

iii. T+mv2l

iv. 0

T is the tension in the string. (Choose the correct alternative).

Ans: (i) The tension created in the string provides the centripetal force when a particle attached to a string spins in a circular motion around a centre. As a result, in the given situation, the particle's net force is tension. T, i.e.

F=T=mv2l

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) The stone will go in the direction of the velocity at the time the string breaks. The direction of the velocity vector is tangential to the path of the stone at that time, according to Newton's first law of motion. As a result, as soon as the string snaps, the stone will fly off in a tangential direction.

 

 Solve the Following Questions                            2 Mark

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: 

(a) As the raindrop is falling with a constant speed, so its acceleration will be 0. As the force acting on a particle is given by, so the net force acting on the rain drop will be 0.

(b) As the kite is held stationary, so by Newton's first law of motion, the algebraic sum of forces acting on the kite is 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: Due to inertia, the mass of the two bodies tries to expand, and the acceleration will act in the opposite direction as it shrinks.

So let us assume that the F1 and F2 be the forces acting in opposite directions due to masses m1 and m2

Thus F1+F2=0

m1a1+m2a2=0

m1a1=m2a2

a1a2=m1m2

Hence the above is proved.

 

3. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80ms1 , what is the recoil speed of the gun?

Ans: From the question, we have the shell having mass 0.020 kg and is fired by a gun having the mass 100 kg. We need to find the recoil speed of the gun, when the muzzle speed is given as 80ms1.

Since, the momentum before firing =0

As the momentum after firing = momentum of (bullet + gun). 

Therefore, momentum after firing will be =mbvbmgvg

As per the law of conservation of linear momentum:

0=mbvbmgvg         mbvb=mgvg     vg=mbvbmgvg=mbvbmg   vg   =0.02×80100   vg=0.016 ms1

 

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 a deforming force, no acceleration is produced.

(2) Internal force is incapable of causing acceleration.

 

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

Ans: In the given question, we have the force of 16N and 12N given and they are acting on a mass of 200kg in mutually perpendicular directions. We need to find the magnitude of the acceleration produced.

 F=F12+F22+2F1F2cosθ

Since, the forces are in mutually perpendicular directions. Therefore, θ=90. Hence, the force will become:

F=F12+F22

Now substituting the values, we get

                                                F=(16)2+(12)2)   =20N

Hence, the magnitude of the acceleration will be:

a=Fm   a=20200   a=0.1ms2

 

6. An elevator weighs 3000 kg. What is its acceleration when the tension supporting cable is 33000 N. Given that g=9.8m2 .

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Ans: From the question, we have an elevator having the weighs given as 3000 kg. We need to find the acceleration, if the tension in the supporting cable is given as 33000 N.

Net upward force on the Elevator F is equal to =T mgbecauseF=ma

ma=Tmg

T=m(a+g)

T=33000N=3000(a+9.8)

a=330003000×9.83000     =1.2 ms2

 

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

Ans: 

1. It demonstrates that only when force is added to the motion is it accelerated. 

2. It introduces the notion of a body's inertial mass.

 

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 begins to fly within the cage, the weight of the bird is no longer felt since the air inside is in direct touch with ambient air, making the cage look 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 riding along a curve because a component of the ground's natural response supplies him with the centripetal force he needs to turn.

He must lean inwards from his vertical posture, towards the circular path's centre.

 

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

Ans: When a curving road is unbanked, the centripetal force is provided by friction between the tyres and the road. Friction must be increased, resulting in wear and tear. When the curving road is banked, however, a component of the ground's natural response supplies the necessary centripetal force, reducing tyre wear and tear.

 

11. A monkey of mass 4 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  6ms2 (b) climbs down with an acceleration of 4ms2 (c) climbs up with a uniform seed of 5 ms (d) falls down the rope freely under gravity. Take g=10ms2 and ignore the mass of the rope.

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Ans: Here, from the question we have:

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

If the response (R) is greater than the tension, the rope will break (T) 

(a) a=6ms2

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

(b) a=4ms2

R=m(ga)=40(106)=240N (Rope will not break)

 

(c) v=5ms1 (constant a=0)

R=mg=40×10=400N (Rope will not break)

(d) a=g;R=m(ga)=m(gg)

R=0 (Rope will not break)

Since, the rope only breaks when the monkey climbs up with an acceleration of 6ms2.

Hence, option (a) will be correct.

 

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

Ans: As the water in a freely falling bottle is in a state of weight – lessening. So no up thrust force occurs on the bubbles and bubbles will not ascend in the water.

 

13. Two billiard balls each of mass 0.05 kg 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: From the question, we have the initial momentum to the ball A=0.05(6)=0.3kgms1

Because the speed of ball A is reversed when it collides, its ultimate momentum is: A=0.05(6)=0.3kgms1

Impulse imparted to ball A= change in momentum of ball A= final momentum – initial momentum =0.30.3=0.6kgms1.

 

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: Total momentum is conserved according to the conservation of linear momentum concept.

Before disintegration linear momentum will be equal to zero.

After disintegration linear momentum =m1v1+m2v2

m1v1+m2v2=0v2=m1v1m2

 

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

Ans: When a fast bus comes to a complete stop, the bottom half of the body in touch with the seat comes to a complete halt, while the upper section of the passengers' bodies prefer to retain their uniform motion. As a result, the passengers are pushed forward.

 

16. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0ms2. Calculate the initial thrust (force) of the blast.

Ans: From the question, we have:

The mass of the rocket given as m=20,000kg

Initial acceleration given as a=5ms2

Acceleration due to gravity given as, g=10ms2

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

Fmg=ma

F=m(g+a)

F=20000×(10+5)

F=20000×15 

F=3×105N

 

17. A bob of mass 0.1kg hung from the ceiling of a room by a string 2m long is set into oscillation. The speed of the bob at its mean position is 1ms1. 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) The bob's velocity is 0 at its most extreme position. The bob will fall vertically to the ground if the string is severed at this point.

(b) The bob's velocity is 1ms1 at its average location. This velocity runs perpendicular to the arc produced by the oscillating bob. The bob will follow a projectile path with only the horizontal component of velocity if it is severed at the mean location. As a result, it will take a parabolic course.

 

18. Two billiard balls each of mass 0.05kg 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: As from the question, we have the mass of each ball given as 0.05kg

Initial velocity of each ball will be =6ms1

Magnitude of the initial momentum of each ball given as, pi=0.3kgms1

The balls alter their directions of motion after colliding, but their velocity magnitudes do not change.

Final momentum of each ball given as, pf=0.3kgms1

Each ball's impulse equals a change in the system's momentum.

=pfpi

=0.30.3=0.6kgms1

The negative indication implies that the balls are receiving opposite-direction shocks.

 

19. A train runs along an unbanked circular track of radius 30m at a speed of 54kmh1. The mass of the train is 106kg. 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: In the given question, we have the radius of the circular track given as, r=30m

Speed of the train is given as, v=54kmh1=15ms1

Mass of the train is given as, m=106kg

The lateral thrust of the rail on the wheel provides the centripetal force. The wheel exerts an equal and opposite force on the rail, according to Newton's third law of motion. The wear and damage of the rail are caused by this reaction force.

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

tanθ=v2rg

=(15)230×10=225300

θ=tan1(0.75)=36.87

Therefore, the angle of banking is about 36.87.

 

20. A constant retarding force of 50N is applied to a body of mass 20kg moving initially with a speed of 15ms1. How long does the body take to stop?

Ans: As in the given question, we have the retarding force given as, F=50N

Mass of the body is given as, m=20kg

Initial velocity of the body is given as, u=15ms1

Final velocity of the body is given as, v=0

The acceleration created in the body may be estimated using Newton's second rule of motion:F=ma

50=20×a

a=5020=2.5ms2

The time it takes for the body to come to rest may be estimated using the first equation of motion:

i.e. v=u+at

t=ua=152.5

t=6s

 

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 us assume that m,m1, and m2 represent the parent nucleus's and the two daughter nuclei's respective masses. The parent nucleus is in a state of dormancy.

Therefore, the initial momentum of the system (parent nucleus) will be = 0

Also let us assume that v1 and v2 will be the particular velocities of the daughter nuclei having masses m and 2m.

After disintegration, the system's total linear momentum =m1v1+m2v2

As per the law of conservation of momentum the total initial momentum will be equal to total final momentum.

0=m1v1+m2+v2

v1=m2v2m1

The negative indication here implies that the parent nucleus pieces are moving in opposing directions.

 

22. A shell of mass 0.020kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is 80ms1, what is the recoil speed of the gun?

Ans: In the given question, we have the mass of the gun given as, M=100kg

Mass of the shell is given as, m=0.020kg

Muzzle speed of the shell is given as, v=80ms1

Recoil speed of the gun will become V

Since the gun and the shell, both are at rest initially.

Therefore, initial momentum of the system will be and

Final momentum of the system will be =mvMV

Because the shell and the gun are pointing in opposing directions, the negative sign emerges.

As per the law of conservation of momentum the total initial momentum will be equal to total final momentum.

mvMV=0

V=mvM

=0.020×80100×1000=0.016ms1


Solve the Following Questions                                 3 Mark

1. A train runs along an unbanked circular bend of radius 30 m at a speed of 54 kmhr1. The mass of the train is 106 kg. 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: From the question, we have the radius of circular bend given as, r=30 m

Speed of train =v=54kmh1=54×518=15 ms1

Mass of train given as, m=106 kg 

Then we need to find the angle of banking θ.

(1) The centripetal force is generated by the lateral force exerted by rails on the train's wheels. 

(2) The centripetal force is provided by the lateral thrust by the outer rail.

(3) According to Newton's third law of motion, the train exerts (i.e., causes) an equal and opposite thrust on the outer rail causing its wear and tear.

Therefore, the angle of baking:

tanθ=v2rg=(15)230×9.8

θ=37.4

 

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 ms2 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 is given as, m=15 kg

Coefficient of static friction is given as, μ=0.18

Acceleration of the trolley is given as, a=0.5 ms2

According to Newton's second law of motion, the force F exerted on the block by the trolley's motion is given by the relationship:

F=ma=15×0.5=7.5 N

This force is applied in the trolley's forward motion.

The block and the trolley have a static friction force of:

f=μmg

f=0.18×15×10=27 N

The applied external force is larger than the static friction force between the block and the trolley. As a result, the block will appear to be at rest to a ground observer.

There will be no applied external force while the trolley moves at a constant speed. In this case, the only force acting on the block is friction.  

(b) When moving with the trolley, a spectator experiences some acceleration. This is a non-inertial frame of reference situation. A pseudo force of equal size opposes the frictional force pushing on the trolley backward. This force, on the other hand, works in the opposite direction. For the spectator moving with the trolley, the trolley will appear to be at rest.

 

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 ? (Given: tanθ=0.7653)

Ans: If a is the acceleration

Then F=(3 m)a

a=F/3m

(1) Net force on P

F1=ma=m×F3m

F1=F/3

(2) Force applied on Q

F2=(m+m)a

F2=2m×a=2m×F3m

F2=2F3

(3) Force applied on R by Q

F3=m×a=m×F3m

F3=F3


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) The centripetal force is provided by the gravitational force acting on the earth and the sun.

(ii) Centripetal force is provided by the earth's gravitational attraction on the moon.

(iii) The centripetal force is provided by the electrostatic attraction between the electron and the proton.

 

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 precisely proportional to the force. i.e. Fα rate of change of momentum (dpdt)

F=kdpdt (P=mv)

F=kmdvdt

F=kma(In S.I. unit k=1)

F=kma

 

6. A railway car of mass 20 tonnes moves with an initial speed of 54 km/hr. .On applying brakes, a constant negative acceleration of 0.3 m/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: m=20 tonnes =20×1000 kg, u=54kmhr1=15 ms1

a=0.3 ms2,ϑ=0

(a) F=ma

F=20000×(0.3)

F=6000 N

(b) v=u+at

vu=at

t=vua=0150.3

t=50 s

(c) v2v2=2 as

(0)2(15)2=2(0.3)s

S=375 m

 

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

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

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Ans: The contact between the outcome of limiting friction and normal reaction is known as the angle of friction.

tanα=FsR , will be with the normal reaction.

Coefficient of static friction

The limiting value of static frictional force is proportion to the normal reaction is given by:FsRFs=μsR

Or μs=FsR......(2)

From (1) & (2)

μs=tanα=FsR

OR

A block of mass 10 kg is sliding on a surface inclined at an angle of 30 with the horizontal.

Calculate the acceleration of the block. The coefficient of kinetic friction between the block

and the surface is 0.5.

m=10 kg,θ=30,μk=0.5

a=g(sinθμkcosθ)

a=9.8(sin300.5cos30)

a=9.8(.50.5×0.866)

a=0.657 m/s2

 

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

Ans: Law of conservation of momentum states that unless an external force is applied, the two or more objects acting upon each other in an isolated system, the total momentum of the system remains constant. This also means that the total momentum of an isolated system before is equal to the total momentum of the isolated system after.

The law of conservation of linear momentum applies if no external force occurs on the system. The entire momentum of the system remains unchanged.

i.e. if Fext=0then P= constant

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Impulse experienced by m1=F12Δt=m1u1m1v1

Impulse experienced by m2=F21Δt=m2u2m2v2

According to Newton's third law

F12=F21

m1u1m1v1=(m2u2m2v2)

m1u1+m2v2=m1u1+m2v2

As a result, the momentum gained by one ball is lost by the other. As a result, linear momentum is conserved.

 

9. A particle of mass 0.40 kg moving initially with constant speed of 10 m/s to the north is subject to a constant force of 8.0 N directed towards south for 30 s. 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=0.40 kg

u=l0 ms1 due North

F=8.0 N

a=Fm=8.00.40

a=20 ms2 

(1) At t=5 s

x=ut=10×(5)

x=50m

(2) At t=25 s

x=ut+12at2

x=10×25+(20)(25)2

x=6000 m

(3) At t=30 s

x1=ut+12at2

x1=10×30+12(20)(30)2

x1=8700 m

(4) At t=30 s

v=u+at

v=10+(20)(30)

v=590 ms1

Motion from 30 s to 100 s

x2=ut=590×70

x2=41300 m

Total distance x=x1+x2

x=50000 m

 

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?

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Ans:

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

From the question, we have the mass of the block I given as m=25 kg

Mass of the man given as M=50 kg

Acceleration due to gravity is given as, g=10 ms2

Force applied on the block is given as, F=25×10=250 N

Weight of the man is given as, W=50×10=500 N 

Case (a): When the man directly raises the block.

In this situation, the man exerts an upward push. This makes him appear heavier.

Action on the floor by the manman=250+500=750 N

Case (b): The man uses a pulley to lift the block.

In this situation, the man exerts a downward force. His apparent weight is reduced as a result of this.

Action on the floor by the manman=500250=250 N

If the floor can withstand a normal force of 700N, the man should use the second approach to lift the block more readily while using less force.

 

11. 

(a) State Impulse-momentum theorem?

(b) A ball of mass 0.1 kg is thrown against a wall. It strikes the wall normally with a velocity of 30 ms1 and rebounds with a velocity of 20 ms1. Calculate the impulse of the force exerted by the ball on the wall.

Ans: 

(a) The impulse-momentum theorem generally states that the impulse applied to a body is equal to the change in momentum of that body.

Impulse m(uv)=P2P1

(b) m=0.1 kg, v=30 ms1u=20 ms1

Impulse, 

P2P1=mvmu=m(vu)=m(30(20))=m(30+20)=(0.1×50)Ns=5Ns

 

12. Ten one rupee coins are put on top of one another on a table. Each coin has a mass mkg. 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=(3mg)N

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

(b) The eighth coin is already weighed down by the weight of the two coins above it, as well as its own. As a result, the force on the 7th  coin owing to the 8th  coin is equal to the sum of the two forces, i.e.

F=2 m+m=(3 m)kgF=(3mg)N

The force acts in a vertical downward direction. 

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

Reaction, 

R=F=4 m(kg)F=(4mg)N

The -ve symbol denotes that the reaction is vertically upwards.

 

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 kmh1 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 raindrops are falling at a steady rate. As a result, the acceleration is zero. The net force acting on the rain drop is zero, according to Newton's second law of motion. 

(b) Zero net force

The cork's weight is acting downward. The buoyant force exerted by the water in an upward direction balances it. As a result, there is no net force acting on the floating cork. 

(c) Zero net force

In the sky, the kite is stationary, i.e. it is not moving at all. As a result, according to Newton's first rule of motion, there is no net force acting on the kite.  

(d) Zero net force

The car is going at a consistent speed down a bumpy route. As a result, it has no acceleration. There is no net force operating on the car, according to Newton's second law of motion. 

(e) Zero net force

All fields have no affect on the high-speed electron. As a result, there is no net force 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

Gravitational acceleration always operates downward, regardless of the direction of motion of an item. In all three scenarios, the gravitational force is the only force acting on the stone. Newton's second law of motion gives its magnitude as: F=m×a

Where,

F= Net force

m= Mass of the pebble =0.05 kg

a=g=10 ms2

F=0.05×10=0.5 N

In all three circumstances, the net force acting on the stone is 0.5N, and this force acts downward. 

The horizontal and vertical components of velocity will be present if the stone is thrown at an angle 45to the horizontal. Only the vertical component of velocity becomes zero at the highest point. Throughout its travel, however, the stone will have a horizontal component of velocity. The net force applied on the stone is unaffected by this component of velocity. 

 

15. A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s1 to 3.5 m s1 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 is given as, m=3 kg

Initial speed of the body is given as, u=2 ms1

Final speed of the body is given as, v=3.5 ms1

Time is given as, t=25 s

The acceleration (a) produced in the body can be estimated using the first equation of motion:v=u+at

a=vut

=3.5225=1.525=0..06 ms1

Newton's second law of motion states that force is equal to:

F=ma

=3×0.06=0.18 N

The net force acting on the body is in the direction of its motion since the application of force does not affect the direction of the body.

 

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: In the given question, we have the mass of the body given as, m=5 kg.

It is acting on two perpendicular forces, given as 8 N and 6 N.

The following is a representation of the situation:

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The resultant of two forces is given as:

R=(8)2+(6)2=64+36=10N

θ is the angle made by R with the force of 8 N

θ=tan1(68)=36.87

The –ve sign indicates that θ will be in the clockwise direction with respect to the force having the magnitude 8 N.

The acceleration (a) of a body is provided by Newton's second law of motion:

F=ma

a=Fm=105=2 ms2

 

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 is given as, m=0.25 kg

Radius of the circle is given as, r=1.5mt

Number of revolution per second is given as, n=4060=23rps

Since, Angular velocity is given by the formula, ω=vr=2πn(i)

The tension T in the string provides the centripetal force for the stone, i.e.,

T=FCentripetal

=mv2r=mrω2=mr(2πn)2

=0.25×1.5×(2×3.14×23)2

=6.57 N

Since, the maximum tension in the string, Tmax=200 N

Tmax=mv2maxr

vmax=Tmax×rm

=200×1.50.25

=1200=34.64 ms1

Therefore, the maximum speed of the stone will be equal to 34.64 ms1.

 

18. Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s2. 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? (Massoftheman=65kg)

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Ans: From the question, we have the mass of the man is given as, m=65kg

Acceleration of the belt is given as, a=1 ms2

Coefficient of static friction is given as, μ=0.2

Newton's second law of motion gives the net force F operating on the man as:Fnet=ma=65×1=65 N

The individual will remain motionless in relation to the conveyor belt until his net force is equal to or less than the frictional force fz, exerted by the belt, i.e.,

Fnet=fz

ma=μmg

a=0.2×10=2 ms2

As a result, the maximum belt acceleration at which the guy may remain immobile is 2 ms2.

 

19. A stream of water flowing horizontally with a speed of 15 ms1 gushes out of a tube of cross-sectional area 102 m2, 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: Since, the speed of the water stream is given as, v=15 ms1 

Cross-sectional area of the tube is given as, A=102 m2 

Therefore, the volume of water coming out from the pipe per second will be: V=Av=15×102 m3s1

As the density of water is, p=103 kgm3

Flow rate of water via the pipe in gallons per second =p×V=150 kgs1 

When the water hits the wall, it does not bounce back. As a result, Newton's second rule of motion gives the force exerted by the water on the wall as:

F= Rate of change of momentum =ΔPΔt =mvt

=150×15=2250 N

 

20. An aircraft executes a horizontal loop at a speed of 720 kmh1 with its wings banked at 15. What is the radius of the loop?

Ans: 

From the question, it is given that the peed of the aircraft, v=720 kmh1=720×518=200 ms1

Also the acceleration due to gravity, g=10 m/s2 and the angle of banking, θ=15 

Therefore, for radius r, of the loop, we have the relation:

tanθ=v2rg

r=v2gtanθ

=200×20010×tan15=40000.268

=14925.37 m

=14.92 km


Solve the Following Questions                            4 Mark

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 s2,

(d) lying on the floor of a train which is accelerating with 1 m s2, the stone being at rest relative to the train. Neglect air resistance throughout.

Ans: 

(a) 1 N; vertically downward

From the question, we have the mass of the stone given as, m=0.1 kg

Acceleration of the stone is given as, a=g=10 m/s2

The net force exerted on the stone, according to Newton's second law of motion, is

F=ma=m g

=0.1×10=1 N

Gravitational acceleration always works in the downward direction. 

(b) 1 N; vertically downward

The train is travelling at a constant speed. As a result, its acceleration in the horizontal direction, where it is moving, is zero. As a result, there is no horizontal force acting on the stone. 

The net force acting on the stone is due to gravity's acceleration, and it is always vertically downward. This force has a magnitude of 1 N.

(c) 1 N; vertically downward

It is given that the train is accelerating at the rate of 1 m/s2.

Hence, the net force acting on the stone will be equal to, F=ma=0.1×1=0.1 N

This force has a horizontal component to it. The horizontal force F, no longer acts on the stone when it is dropped. This is due to the fact that the force acting on a body at any one time is determined by the current circumstance rather than previous ones.

As a result, the net force acting on the stone is determined only by gravity's acceleration.F=mg=1 N

This force acts vertically downward.

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

The typical reaction of the floor balances the weight of the stone. The train's horizontal motion is the only source of acceleration.

Acceleration of the train, a=0.1 m/s2

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

F=ma 

=0.1×1=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: From the question, it is given that the initial speed of the three-wheeler, u=36 km/h=10 m/s

Also the final speed of the three-wheeler is given as, v=0 m/s

Time, t=4 s

Mass of the three-wheeler is given as, m=400 kg

Mass of the driver is given as, m=65 kg

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

So, with the use of first law of motion, the acceleration (a) of the three-wheeler can be calculated by using the formula: v=u+at

a=vut=0104=2.5 m/s2

The negative indication implies that the three-velocity wheeler's is decreasing over time.

The net force operating on the three-wheeler may be estimated using Newton's second law of motion:

F=Ma

=465×(2.5)=1162.5 N

The minus symbol shows that the force is acting in the opposite 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: From the question, we have the value given as the mass of the body, m=0.40 kg

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

Force acting on the body is given as, F=8.0 N

Therefore, the acceleration produced in the body, a=Fm=8.00.40=20 m/s2

(i) At t=5 s

Acceleration, a=0 and u=10 m/s

s=ut+12at2

=10×(5)=50 m

(ii) At t=25 s

Acceleration, a=20 m/s2 and u=10 m/s

s=ut+12at2

=10×25+12×(20)×(25)2

=250+6250=6000 m

(iii) At t=100 s

For 0t30 s

a=20 m/s2

u=10 m/s

s1=ut+12ant2

=10×30+12×(20)×(30)2

=3009000

=8700 m

For 30 < t  100 s

According to the first equation of motion, for t=30 s, final velocity will be calculated by the formula:

v=u+at

=10+(20)×30=590 m/s

Velocity of the body after 30 s=590 m/s

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

s2=vt+12ant2

=590×70=41300 m

=590×70=41300 m

Total distance, s=s1+s2=870041300=50000m

 

4. A truck starts from rest and accelerates uniformly at 2.0 ms2. 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.)

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

(b) 10 m/s2

Ans: 

(a) In the given question, we have the initial velocity of the truck as, u=0

Acceleration, a=2 m/s2

Time, t=10 s

The final velocity is provided by the first equation of motion:

v=u+at

=0+2×10=20 m/s

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

At t=11 s, the horizontal component (vx) of velocity, in the absence of air resistance, the rate of change of velocity remains constant, i.e.vx=20 m/s

The first equation of motion gives the vertical component of the stone's velocity as: 

vy=u+ayt

Where, t=1110=1 s and ay=g=10 m/s2

vy=0+10×1=10 m/s

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

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v=vx2+vy2

=202+102=400+100

=500=22.36 m/s

Assume that the resultant velocity makes an angle θ with the horizontal component of velocity.vx

tanθ=(vyvz)

θ=tan1(1020)

=tan1(0.5)

=26.57

(b) The horizontal force acting on the stone is zero when it is dropped from the truck. The stone, however, continues to move due to gravity's pull. As a result, the stone's acceleration is 10m/s2 and it operates vertically downward.

 

5. Figure 5.16 shows the position-time graph of a particle of mass 4kg. What is the (a) force on the particle for t<0, 0<t<4s, 0<t<4s ? (b) impulse at t=0 and t=4s (Consider one-dimensional motion only).

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Ans:

(a) For t<0

The location of the particle is coincident with the time axis, as can be seen in the graph. It means that the particle's movement in this time interval is zero. As a result, there is no force acting on the particle.

For t>4 s

The location of the particle in the provided graph is parallel to the time axis, as can be seen. It means the particle is resting at a distance of from the origin. As a result, there is no force acting on the particle.

For 0 < t < 4

The provided position-time graph exhibits a constant slope, as can be seen. As a result, the particle's acceleration is zero. As a result, there is no force acting on the particle. 

(b) At t=0

Impulse = Change in momentum

=mvmu

Mass of the particle is given as, m=4 kg

Initial velocity of the particle, u=0

Final velocity of the particle will be, v=34 m/s

Impulse =4(340)=3 kg m/s

At t=4 s

Initial velocity of the particle, u=34 m/s

Final velocity of the particle, v=0

Impulse 4(340)=3 kg m/s

 

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:

From the question, we have the horizontal force given as, F=600 N

Mass of body A given as, m1=10 kg

Mass of body B given as, m2=20 kg

Therefore, the total mass of the system will become, m=m1+m2=30 kg

The acceleration produced in the system may be estimated using Newton's second rule of motion:F=ma

a=Fm=60030=20 m/s2

When force F is applied on body A:

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The motion equation may be expressed as follows:

FT=m1a

T=Fm1a

=60010×20=400 N (i)

When force F is applied on body B :

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The motion equation may be expressed as follows:

FT=m2a

T=Fm2a

T=60020×20=200 N

 

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 0.15kg.)

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

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Where,

AO will be the incident path of the ball and OB will be the path followed by the ball after deflection.

AOB will be the angle between the incident and the deflected paths of the ball and is equal to 45.

AOP=BOP=22.5=θ

Let us assume that the initial and final velocities of the ball =v

Horizontal component of the initial velocity will be =vcosθ along RO

Vertical component of the initial velocity will be=vsinθ along PO

Horizontal component of the final velocity will be=vcosθ along OS

Vertical component of the final velocity = vsinθ along OP

There is no change in the horizontal components of velocities. The vertical components of velocities are in a clockwise orientation.

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

=mvcosθ(mvcosθ)

=2mvcosθ

Mass of the ball, m=0.15 kg

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

Impulse =2×0.15×15cos22.5=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?

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Ans:

A ball bounces back and forth between at x=0 and x=2 cm; after every 2 s, the ball receives an impulse of magnitude 0.08×102 kg m/s from the walls.

A body changes its direction of motion every 2 s, as seen in the graph. In terms of physical representation, imagine a ball rebounding Aback and forth between two fixed walls located at x=0 and x=2 cm. The ball collides with a wall every 2 s because the slope of the xt graph flips after every 2 s. As a result, every2 s, the ball receives an impulse.

Mass of the ball, m=0.04 kg

The ball's velocity is determined by the graph's slope. We may compute initial velocity using the graph as follows: 

u=(20)×102(20)=102 m/s

Velocity of the ball before collision will be, u=102 m/s

Velocity of the ball after collision will be, v=102 m/s

(As the ball reverses its direction of motion, the negative sign appears.)

Since, magnitude of impulse is equal to change in momentum.

=|mvmu|

=|0.04(vu)|

=∣0.04(102=102

=0.08×102 kg m/s

 

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)

mg - T1

mg + T2

(b)

mg + T1

mg - T2

(c)

mg + T1 - (mv12) / R

mg - T1 + (mv12) / R

(d)

mz + T1 - (mv12) / R

mz + T1 + (mv12) / R


Ans: (a) The following graphic depicts the free body diagram of the stone at its lowest position. 

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The net force exerted on the stone at this moment, according to Newton's second law of motion, is equal to the centripetal force, i.e.

Fnet=T=mg=mv12R(i)

Here, v1= Velocity at the lowest point

The following graphic depicts the stone's free body diagram at its highest position.

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We may calculate the following using Newton's second rule of motion:

T+mg=mv22R (ii)

Where, v2= Velocity at the highest point

The net force operating at the lowest and highest locations is (Tmg) and (T+mg), as shown by equations (i) and (ii), respectively.

 

10. A disc revolves with a speed of 3313 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, v=3313rev/min=1003×60=59rev/s

Coefficient of friction, μ=0.15

The coin with a friction force higher than or equal to the centripetal force supplied by the rotation of the disc will rotate with the disc in the present circumstance. If this isn't the case, the coin will fall out of the disc.

When the coin placed at 4 cm :

Radius of revolution, r=4 cm=0.04 m 

Angular frequency, ω=2πv=2×227×59=3.49 s1

Frictional force, f=μmg=0.15×m×10=1.5 mN

Centripetal force on the coin:

=0.49 m N

Asf>Fcent., along with the record, the coin will spin.

Coin placed at 14 cm :

Radius, r=14 cm=0.14 m

Angular frequency, ω=3.49 s1 

Frictional force, f=1.5 m N 

Centripetal force is given as:

Fcent=mrω2

=m×0.14×(3.49)2=1.7 m N

Since $f < {F_{\text {cert }}$, 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 3m rotating about its vertical axis with200 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: In the given question, we have the mass of the man given as,m=70 kg

Radius of the drum given as, r=3 m

Coefficient of friction, μ=0.15 

Frequency of rotation, v=200rev/min=20060=103rev/s

The normal force provides the essential centripetal force for the rotation of the man(FN)

The man adheres to the drum's wall while the floor spins. As a result, the frictional force (f=μFN) acting upward balances the weight of the man mg acting downward.

As a result, the man won't fall until: 

mg<f

mg<μFN=μmrω2

g<μrω2

ω>gμr

The minimum angular speed is calculated as follows:

ωmin=gμr

=100.15×3=4.71rads1


Solve the Following Questions                            5 Mark

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

(b) State and prove impulse momentum theorem?

Ans: (a) Impulsive forces are those that are applied repeatedly over a short period of time.

Impulse I=F×t

Unit =NS

The term "impulse" refers to a vector quantity that is oriented along the average forceFav.

(b) The change in momentum of the body is equal to the force's impulse. Newton's second law is that:

F=dpdt

or dp=Fdt

At t=0P=P1 and at t=tP=P2

P1P2dP=vtFdt

P2P1=Ft

P2P1=I

[Ft=I(Impulse)]

 

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 m s1,

(b) downwards with a uniform acceleration of 5 m s2,

(c) upwards with a uniform acceleration of 5 m s2.

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) In the question, we have the mass of the man given as, m=70 kg

Acceleration given as, a=0

We may express the equation of motion as: using Newton's second law of motion.

Rmg=ma

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

As the lift moves at a constant pace, acceleration a=0

R=m g

=70×10=700 N

Taking a weighing scale reading =700g=70010=70 kg

(b) Mass of the man, m=70 kg

Acceleration, a=5 m/s2 downward

We may express the equation of motion as: using Newton's second law of motion.

R+m g=ma

R=m(ga)

=70(105)=70×5

=350 N

Reading on the weighing scale =350g=35010=35 kg

(c) Mass of the man, m=70 kg

Acceleration, a=5 m/s2 upward

We may express the equation of motion as: using Newton's second law of motion.

Rmg=ma

R=m( g+a)

=70(10+5)=70×15

=1050 N

Reading on the weighing scale =1050g=105010=105 kg

(d) When the lift is allowed to move due to gravity, acceleration a=g

We may express the equation of motion as: using Newton's second law of motion.

R+m g=ma

R=m(ga)

=m( gg)=0

Reading on the weighing scale =0g=0 kg

The man will experience 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 following diagram shows how the given system with two masses and a pulley may be represented:

seo images

So, the smaller mass is given as, m1=8 kg

Larger mass is given as, m2=12 kg

Therefore, tension in the string =T

Mass m2, owing to its weight, moves downward with acceleration a, and mass m1 moves upward.

The system of each mass is subjected to Newton's second law of motion:

For mass m1:

The motion equation may be expressed as follows:

Tm1 g=ma(i)

For mass m2:

The motion equation may be expressed as follows:

m2gT=m2a(ii)

Now on adding equations (i) and (ii), we get:

(m2m1)g=(m1+m2)a

a=(m2m1m1+m2)g

=(12812+8)×10=420×10=2 m/s2

As a result, the acceleration of the masses is 2 m/s2.

When we replace the value a in equation (ii), we get:

m2gT=m2(m2m1m1+m2)g

T=(m2m22m1m2m1+m2)g

=(2m1m2m1+m2)g

=(2×12×812+8)×10

=2×12×820×10=96 N

As a result, the string tension 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) A horse pushes the earth backward with some force in order to drive a waggon. The horse's feet, in turn, are subjected to an equal and opposite reaction force from the ground. The horse moves forward as a result of this response force.

There is no response force in an empty area. As a result, a horse can't pull a cart and run in the open. 

(b) When a fast bus comes to a quick stop, the lower section of a passenger's body in touch with the seat comes to an abrupt stop. The top part, on the other hand, prefers to stay in motion (as per the first law of motion). As a result, the upper body of the passenger is pushed forward in the direction of the bus's movement. 

(c) A force at an angle is applied to a lawn mower as it is being pulled, as indicated in the diagram below.

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This applied force's vertical component acts upward. This decreases the mower's effective weight.

When pulling a lawn mower, on the other hand, a force is delivered at an angle, as indicated in the diagram below.

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The vertical component of the applied force operates in this situation in the direction of the mower's weight. This raises the mower's effective weight.

Because the lawn mower's effective weight is lower in the first scenario, pulling the lawn mower is easier than pushing it. 

(d) The equation of motion, according to Newton's second law of motion, is:

F=ma=mΔvΔt(i)

Where,

F= As he catches the ball, the cricketer feels a braking force.

m= Mass of the ball

Δt= Time of impact of the ball with the hand

From equation I it can be shown that the impact force is inversely proportional to the impact time, i.e.

F1Δt(ii)

Equation (ii) demonstrates that as the time of impact rises, the force received by the cricketer reduces, and vice versa.

A cricketer extends his hand backward while taking a catch to lengthen the time of impactt. As a result, the stopping force decreases, protecting the cricketer's hands from being injured.

 

5. A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms2. 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 is given as, mh=1000 kg

Mass of the crew and passengers is given as, mp=300 kg

Therefore, the total mass of the system, m=1300 kg

As the acceleration of the helicopter is given as, a=15 m/s2

The reaction force R exerted on the system by the floor may be computed using Newton's second equation of motion:

Rmpg=ma

=mp(g+a)

=300(10+15)=300×25

=7500 N

The response force will likewise be directed upward because the chopper is accelerating vertically upward. As a result, the force exerted on the floor by the crew and passengers is 7500 N, directed downward, according to Newton's third law of motion. 

(b) The reaction force R experienced by the helicopter may be computed using Newton's second equation of motion as follows:

=32500 N

The helicopter is being pushed higher by the reaction force of the surrounding air. As a result, the rotor's action on the surrounding air will be 32500N, directed downward, according to Newton's third law of motion. 

(c) The surrounding air exerts a force of 32500Non the helicopter, which is 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 7th  coin (counted from the bottom) due to all the coins on its top,

(b) the force on the 7th  coin by the eighth coin,

(c) the reaction of the 6th  coin on the 7th  coin.

Ans:

(a) The weight of the three coins on top of the seventh coin exerts force on it.

So, let us assume that the weight of one =3mg

Therefore, the weight of three coins =3mg

As a result, the three coins on top of the 7th  coin exert 3mg of force on it. This force works in a vertical downward direction. 

(b) Because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top, the eighth coin exerts force on the seventh coin.

Therefore, the weight of the eighth coin will be equal to coin=m g.

Weight of the ninth and tenth coin will also be the same i.e. coin=m g

Therefore, the total weight of these three coins =3mg

Hence, the force exerted on the 7th  coin by the 8th coin is 3mg. This force acts vertically downward.

(c) Because of the weight of the four coins (7th ,8th ,9th ,10th ) on top, the 6th coin suffers a downward pull.

Therefore, 4mgwill be the total downward force experienced by 6th coin

As per Newton's third law of motion, the 6th  coin will produce an equal reaction force on the 7th  coin, but in the opposite direction. Hence, the reaction force of the 6th coin on the 7th  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 6 m s2

(b) climbs down with an acceleration of 4 m s2

(c) climbs up with a uniform speed of 5 m s1

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

(Ignore the mass of the rope).

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Ans:

Case (a)

From the given question, we have the information given as mass of the monkey, m=40 kg

Acceleration due to gravity is given as, g=10 m/s

Maximum tension that the rope can bear is given as, Tmax=600 N

Acceleration of the monkey is given as, a=6 m/s2 upward

So, the equation of motion can be written by using the Newton’s second law of motion:

Tm g=ma

T=m(g+a)

=40(10+6)

=640 N

As T>Tmax, in this situation, the rope will break.

Case (b)

Now in this case, the acceleration of the monkey, a=4 m/s2 acting downward.

Here we will use the Newton’s second law of motion to write the equation of motion. So it can be written as:

mgT=ma

T=m( ga)

=40(104)

=240 N

As T < Tmax , in this situation, the rope will not break.

Case (c)

In this case the monkey is climbing with a uniform speed of 5 m/s. Hence, the acceleration will become zero, i.e.,

a=0.

Here we will use the Newton’s second law of motion to write the equation of motion. So it can be written as:

Tmg=ma

Tmg=0

T=mg

=40×10=400 N

As T<T in this situation, the rope will not break.

Case (d)

When a monkey will fall freely under the presence of the gravity, the acceleration will become equal to the gravitational acceleration. 

i.e., a=g

Here we will use the Newton’s second law of motion to write the equation of motion. So it can be written as:

mgT=mg

T=m(gg)=0

As T<Tmax , in this situation, the rope will not break.

 

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 μs and μk.

seo images

Ans:

(a) In the given question, the mass of body A is given as, mA=5 kg

Mass of body B is given as, mB=10 kg

Applied force, F=200 N

Coefficient of friction, μs=0.15

The friction force is calculated using the following formula:fs=μ(mA+mB)g

=0.15(5+10)×10

=1.5×15=22.5 N in leftward

Net force acting on the partition =20022.5=177.5 N rightward

The reaction force of the partition will be in the opposite direction as the net applied force, according to Newton's third law of motion.

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

(b) Force of friction on mass A:

fA=μmAg

=0.15×5×10=7.5 N leftward

Net force exerted by mass A on mass B=2007.5=192.5 N rightward

According to Newton's third rule of motion, mass B will exert an equivalent amount of reaction force on mass A, i.e., 192.5 N acting leftward.

The two bodies will move in the direction of the applied force after the wall is removed.

On the moving system, there is a net force=177.5 N

As a result, the system of acceleration's equation of motion may be represented as:

Net force =(mA+mB)a

a= Net force mA+mB

=177.55+10=177.515=11.83 m/s2

Net force causing mass A to move:

FA=mAa=5×11.83=59.15 N

Net force exerted by mass A on mass B=192.559.15=133.35 N

The direction of motion will be affected by this force. According to Newton's third law of motion, mass B will exert an equal amount of force on mass A, i.e., 133.3 N, acting in the opposite 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 s2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

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Ans: In the give question, we have the mass of the box given as, m=40 kg

Coefficient of friction is given as, μ=0.15

Initial velocity is given as, u=0

Acceleration is given as, a=2 m/s2

Distance of the box from the end of the truck is given as, s=5 m

The force on the box generated by the truck's accelerated speed is described by Newton's second law of motion:

F=ma

=40×2=80 N

A reaction force of 80 N is operating on the box in the rearward direction, according to Newton's third law of motion. The force of friction f occurring between the box and the truck's floor opposes the box's backward motion. The following factors contribute to this force:

f=μmg

=0.15×40×10=60 N

On the block, the net force is:

Fmat =8060=20 N acting in backward.

The box's rearward acceleration is calculated as follows:

ahxk=Fmtm=2040=0.5 m/s2

Time t may be computed using the second equation of motion as follows:

s=ut+12akxcit2

5=0+12×0.5×t2

t=20 s

As a result, after 20s, the box will fall from the truck.

The below formula will give the distance s travelled by the vehicle in20 s:

s=ut+12at2

=0+12×2×(20)2

=20 m

 

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:

Because both the force of normal response and the weight of the biker act downward and are balanced by the centripetal force, a motorcyclist does not fall at the top point of a vertical loop in a death-well. The scenario is depicted in the diagram below.

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The total of the normal force (FN) and the force due to gravity(Fg=mg) acting on the biker is the net force.

The centripetal acceleration (a) equation of motion may be expressed as:

Fnet =mac

FN+Fg=mac

FN+mg=mv2r

The motorcyclist's speed provides a normal reaction. At the slowest possible speed (vmin),FN=0

mg=mvmin2r

vmin=rg

=25×10=15.8 m/s

 

11. A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for ωg/R. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω=2g/R ? Neglect friction.

Ans:

Let us assume that the radius vector joining the bead with the centre is making an angle θ, in the vertically downward direction.

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OP =R=will be the radius of the circle

N= Normal reaction

The vertical and horizontal force equations can be expressed as follows:

mg=Ncosθ(i)

mlω2=Nsinθ(ii)

In ΔOPQ, we have:

sinθ=lR

l=Rsinθ (iii)

Putting equation (iii) in equation (ii), we get:

m(Rsinθ)ω2=Nsinθ

mRω2=N

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

mg=mRω2cosθ

cosθ=gRω2(v)

Since cosθ1, the bead will remain at its lowermost point for gRω21, i.e., for ωgR 

For ω=2gR or ω2=2gR(vi)

On equating equations (v) and (vi), we get:

2gR=gRcosθ

cosθ=12

θ=cos1(0.5)=60


Important Questions for CBSE Class 11 Physics Chapter 4 (Law of Motion) - Benefits of Studying the Important Questions and Answers

The CBSE Class 11 students are advised to include this study material of important questions and answers to get the following benefits:

  1. Important questions and answers make the students aware of the important topics in the chapter.

  2. The chapter might seem lengthy for some students, in that case, they can always avail themselves of these important questions and answers. 

  3. Students can also use this study material at the time of revising this chapter. This will help them to cover the important points right before their exam.

  4. Laws of Motion can seem complicated for some students, in that case, they can only study the important questions and answers before the Physics exam.


Conclusion  

The CBSE Class 11 Physics Chapter 4 - Law of Motion covers fundamental concepts that form the basis of classical mechanics. Understanding these principles is crucial for building a strong foundation in physics. The chapter introduces Newton's laws of motion, emphasizing their application in real-world scenarios. By mastering these concepts, students can analyze the motion of objects, calculate forces, and solve practical problems. Important questions in this chapter provide invaluable practice and reinforce students' understanding of the topic. Regular practice of these questions enhances problem-solving skills, critical thinking, and conceptual clarity, ultimately preparing students for higher-level physics topics and various competitive exams.


Related Study Materials for Class 11 Physics Chapter 4

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Important Other Links for Class 11 Physics Chapter 4

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CBSE Class 11 Law of Motion Notes

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CBSE Class 11 Law of Motion Solutions



CBSE Class 11 Physics Chapter-wise Important Questions

CBSE Class 11 Physics Chapter-wise Important Questions and Answers cover topics from other chapters, helping students prepare thoroughly by focusing on key topics for easier revision.



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FAQs on Important Questions for CBSE Class 11 Physics Chapter 4 - Law of Motion

1. How to use Vedantu’s Important Questions for Class 11 Physics Chapter 4?

Vedantu's Important Questions for Class 11 Physics Chapter 4 "Laws of Motion" should be used in conjunction with the NCERT textbook and its solutions. Once you are thorough with the concepts and explanations of the NCERT textbook, try solving the NCERT exercise questions. Following that refer to Vedantu’s important questions which are available free of cost on the Vedantu website (vedantu.com). This way you can get scrupulous practice for "Laws of Motion". Practicing the important questions will give you an idea about exam questions and how to answer them well. 

2. Who discovered the three laws of motion?

Isaac  Newton who was a brilliant mathematician and physicist discovered the three laws of motion. These three laws are as follows:

  • Objects at rest stay at rest, and objects in motion stay in motion unless acted upon by an external force.

  • The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

  • For every action, there is an equal and opposite reaction.

3. Why are the laws of motion important?

Newton's laws of motion form a very important part of Physics. They explain almost all types of movements that we experience in our daily lives. The laws pose an explanation for simple things like why we do not fall out of chairs when sitting, how does a car move, what role does friction play in a moving bus. All of these movements can be easily explained using Newton's laws of motion.

4. How is centripetal force provided for an electron moving around the nucleus in an atom?

Centripetal force is the force required to make an object move in a circular path. In the case of an electron moving around the nucleus in an atom, the required centripetal force is provided by the electrostatic force of attraction between electron and proton. Many such important questions are carefully selected by our expert teachers. These questions and their answers for this chapter can be downloaded from the page Important Questions for Class 11 Physics or from Vedantu's mobile app for free.

5. How many questions from Chapter 4 “Laws of Motion” appear in NEET?

NEET exam will have 45 questions from Physics. Of these questions 3% questions come from the topic of Laws of Motion. Hence it is very important that students prepare for this chapter well for CBSE exams and if they aspire for NEET.  Vedantu provides many important questions from this chapter in one place which may be beneficial for students preparing for competitive exams as well. You can access these questions and other study material from the Vedantu mobile app.