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:
The important questions and answers consist of all the important topic coverage of Chapter Law of Motion.
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.
The study material is prepared to take into account the types of questions asked in the CBSE Class 11 examination.
Laws of Motion consist of many important concepts and formulas which are explained in a lucid manner for better understanding purposes.
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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,
Also, we know that
And
On putting the equation
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
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
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
i.
ii.
iii.
iv. 0
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.
Where,
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
(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
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
Thus
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
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
Since, the momentum before firing
As the momentum after firing = momentum of (bullet + gun).
Therefore, momentum after firing will be
As per the law of conservation of linear momentum:
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
Since, the forces are in mutually perpendicular directions. Therefore,
Now substituting the values, we get
Hence, the magnitude of the acceleration will be:
6. An elevator weighs 3000 kg. What is its acceleration when the tension supporting cable is 33000 N. Given that
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
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
Ans: Here, from the question we have:
If the response (R) is greater than the tension, the rope will break
(a)
(b)
(c)
(d)
Since, the rope only breaks when the monkey climbs up with an acceleration of
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
Because the speed of ball
Impulse imparted to ball
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
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
Ans: From the question, we have:
The mass of the rocket given as
Initial acceleration given as
Acceleration due to gravity given as,
Using Newton's second law of motion, the net force (thrust) acting on the rocket is given by the relation:
17. A bob of mass
Ans: (a) Vertically downward
(b) Parabolic path
(a) The bob's velocity is
(b) The bob's velocity is
18. Two billiard balls each of mass
Ans: As from the question, we have the mass of each ball given as
Initial velocity of each ball will be
Magnitude of the initial momentum of each ball given as,
The balls alter their directions of motion after colliding, but their velocity magnitudes do not change.
Final momentum of each ball given as,
Each ball's impulse equals a change in the system's momentum.
The negative indication implies that the balls are receiving opposite-direction shocks.
19. A train runs along an unbanked circular track of radius
Ans: In the given question, we have the radius of the circular track given as,
Speed of the train is given as,
Mass of the train is given as,
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
Therefore, the angle of banking is about
20. A constant retarding force of
Ans: As in the given question, we have the retarding force given as,
Mass of the body is given as,
Initial velocity of the body is given as,
Final velocity of the body is given as,
The acceleration created in the body may be estimated using Newton's second rule of motion:
The time it takes for the body to come to rest may be estimated using the first equation of motion:
i.e.
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
Therefore, the initial momentum of the system (parent nucleus) will be = 0
Also let us assume that
After disintegration, the system's total linear momentum
As per the law of conservation of momentum the total initial momentum will be equal to total final momentum.
The negative indication here implies that the parent nucleus pieces are moving in opposing directions.
22. A shell of mass
Ans: In the given question, we have the mass of the gun given as,
Mass of the shell is given as,
Muzzle speed of the shell is given as,
Recoil speed of the gun will become
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
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.
Solve the Following Questions 3 Mark
1. A train runs along an unbanked circular bend of radius
Ans: From the question, we have the radius of circular bend given as,
Mass of train given as,
Then we need to find the angle of banking
(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:
2. A block of mass
Ans:
(a) Mass of the block is given as,
Coefficient of static friction is given as,
Acceleration of the trolley is given as,
According to Newton's second law of motion, the force
This force is applied in the trolley's forward motion.
The block and the trolley have a static friction force of:
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
Ans: If
Then
(1) Net force on
(2) Force applied on
(3) Force applied on
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.
6. A railway car of mass
(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)
(b)
(c)
7. What is meant by coefficient of friction and angle of friction? Establish the relation between the two? OR
A block of mass
Ans: The contact between the outcome of limiting friction and normal reaction is known as the angle of friction.
Coefficient of static friction
The limiting value of static frictional force is proportion to the normal reaction is given by:
Or
From (1) & (2)
OR
A block of mass
Calculate the acceleration of the block. The coefficient of kinetic friction between the block
and the surface is
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
Impulse experienced by
Impulse experienced by
According to Newton's third law
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
Ans:
(1) At
(2) At
(3) At
(4) At
10. A block of mass
Ans:
From the question, we have the mass of the block I given as
Mass of the man given as
Acceleration due to gravity is given as,
Force applied on the block is given as,
Weight of the man is given as,
Case (a): When the man directly raises the block.
In this situation, the man exerts an upward push. This makes him appear heavier.
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.
If the floor can withstand a normal force of
11.
(a) State Impulse-momentum theorem?
(b) A ball of mass
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
(b)
Impulse,
12. Ten one rupee coins are put on top of one another on a table. Each coin has a mass
(a) The force on the
(b) The force on the
(c) The reaction of the sixth coin on the seventh coin.
Ans:
(a) The force on
Therefore,
Here
(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
The force acts in a vertical downward direction.
(c) The sixth coins is under the weight of four coins above it
Reaction,
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
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of
(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
(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
Ignore air resistance.
Ans:
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:
Where,
In all three circumstances, the net force acting on the stone is
The horizontal and vertical components of velocity will be present if the stone is thrown at an angle
15. A constant force acting on a body of mass
Ans:
Mass of the body is given as,
Initial speed of the body is given as,
Final speed of the body is given as,
Time is given as,
The acceleration (a) produced in the body can be estimated using the first equation of motion:
Newton's second law of motion states that force is equal to:
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
Ans: In the given question, we have the mass of the body given as,
It is acting on two perpendicular forces, given as
The following is a representation of the situation:
The resultant of two forces is given as:
The –ve sign indicates that
The acceleration (a) of a body is provided by Newton's second law of motion:
17. A stone of mass
Ans:
Mass of the stone is given as,
Radius of the circle is given as,
Number of revolution per second is given as,
Since, Angular velocity is given by the formula,
The tension
Since, the maximum tension in the string,
Therefore, the maximum speed of the stone will be equal to
18. Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with
Ans: From the question, we have the mass of the man is given as,
Acceleration of the belt is given as,
Coefficient of static friction is given as,
Newton's second law of motion gives the net force
The individual will remain motionless in relation to the conveyor belt until his net force is equal to or less than the frictional force
As a result, the maximum belt acceleration at which the guy may remain immobile is
19. A stream of water flowing horizontally with a speed of
Ans: Since, the speed of the water stream is given as,
Cross-sectional area of the tube is given as,
Therefore, the volume of water coming out from the pipe per second will be:
As the density of water is,
Flow rate of water via the pipe in gallons per second
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:
20. An aircraft executes a horizontal loop at a speed of
Ans:
From the question, it is given that the peed of the aircraft,
Also the acceleration due to gravity,
Therefore, for radius
Solve the Following Questions 4 Mark
1. Give the magnitude and direction of the net force acting on a stone of mass
(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
(c) just after it is dropped from the window of a train accelerating with
(d) lying on the floor of a train which is accelerating with
Ans:
(a)
From the question, we have the mass of the stone given as,
Acceleration of the stone is given as,
The net force exerted on the stone, according to Newton's second law of motion, is
Gravitational acceleration always works in the downward direction.
(b)
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
(c)
It is given that the train is accelerating at the rate of
Hence, the net force acting on the stone will be equal to,
This force has a horizontal component to it. The horizontal force
As a result, the net force acting on the stone is determined only by gravity's acceleration.
This force acts vertically downward.
(d)
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,
The net force acting on the stone will be directed in the train's direction of travel. Its magnitude is given by:
2. The driver of a three-wheeler moving with a speed of
Ans: From the question, it is given that the initial speed of the three-wheeler,
Also the final speed of the three-wheeler is given as,
Time,
Mass of the three-wheeler is given as,
Mass of the driver is given as,
Therefore, the total mass of the system,
So, with the use of first law of motion, the acceleration
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:
The minus symbol shows that the force is acting in the opposite direction of motion of the three-wheeler.
3. A body of mass
Ans: From the question, we have the value given as the mass of the body,
Initial speed of the body is given as,
Force acting on the body is given as,
Therefore, the acceleration produced in the body,
(i) At
Acceleration,
(ii) At
Acceleration,
(iii) At
For
For 30 < t
According to the first equation of motion, for
Velocity of the body after
For motion between
4. A truck starts from rest and accelerates uniformly at
(a)
(b)
Ans:
(a) In the given question, we have the initial velocity of the truck as,
Acceleration,
Time,
The final velocity is provided by the first equation of motion:
The final velocity of the truck and the stone is
At
The first equation of motion gives the vertical component of the stone's velocity as:
Where,
The resultant velocity
Assume that the resultant velocity makes an angle
(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
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).
Ans:
(a) For
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
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
Impulse = Change in momentum
Mass of the particle is given as,
Initial velocity of the particle,
Final velocity of the particle will be,
At
Initial velocity of the particle,
Final velocity of the particle,
6. Two bodies of masses
Ans:
From the question, we have the horizontal force given as,
Mass of body A given as,
Mass of body B given as,
Therefore, the total mass of the system will become,
The acceleration produced in the system may be estimated using Newton's second rule of motion:
When force
The motion equation may be expressed as follows:
When force
The motion equation may be expressed as follows:
7. A batsman deflects a ball by an angle of
Ans: The given situation can be represented as shown in the following figure.
Where,
AO will be the incident path of the ball and OB will be the path followed by the ball after deflection.
Let us assume that the initial and final velocities of the ball
Horizontal component of the initial velocity will be
Vertical component of the initial velocity will be
Horizontal component of the final velocity will be
Vertical component of the final velocity =
There is no change in the horizontal components of velocities. The vertical components of velocities are in a clockwise orientation.
Mass of the ball,
Velocity of the ball,
8. Figure 5.17 shows the position-time graph of a body of mass
Ans:
A ball bounces back and forth between at
A body changes its direction of motion every
Mass of the ball,
The ball's velocity is determined by the graph's slope. We may compute initial velocity using the graph as follows:
Velocity of the ball before collision will be,
Velocity of the ball after collision will be,
(As the ball reverses its direction of motion, the negative sign appears.)
Since, magnitude of impulse is equal to 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) | m | m |
(b) | m | m |
(c) | m | m |
(d) | m | m |
Ans: (a) The following graphic depicts the free body diagram of the stone at its lowest position.
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.
Here,
The following graphic depicts the stone's free body diagram at its highest position.
We may calculate the following using Newton's second rule of motion:
Where,
The net force operating at the lowest and highest locations is
10. A disc revolves with a speed of
Ans: Coin placed at
Mass of each coin
Radius of the disc,
Frequency of revolution,
Coefficient of friction,
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
Radius of revolution,
Angular frequency,
Frictional force,
Centripetal force on the coin:
As
Coin placed at
Radius,
Angular frequency,
Frictional force,
Centripetal force is given as:
Since $f < {F_{\text {cert }}$, the coin will slip from the surface of the record.
11. A
Ans: In the given question, we have the mass of the man given as,
Radius of the drum given as,
Coefficient of friction,
Frequency of rotation,
The normal force provides the essential centripetal force for the rotation of the man
The man adheres to the drum's wall while the floor spins. As a result, the frictional force
As a result, the man won't fall until:
mg<f
The minimum angular speed is calculated as follows:
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
Unit
The term "impulse" refers to a vector quantity that is oriented along the average force
(b) The change in momentum of the body is equal to the force's impulse. Newton's second law is that:
or
At
2. A man of mass
(a) upwards with a uniform speed of
(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) In the question, we have the mass of the man given as,
Acceleration given as,
We may express the equation of motion as: using Newton's second law of motion.
Where,
As the lift moves at a constant pace, acceleration
(b) Mass of the man,
Acceleration,
We may express the equation of motion as: using Newton's second law of motion.
(c) Mass of the man,
Acceleration,
We may express the equation of motion as: using Newton's second law of motion.
(d) When the lift is allowed to move due to gravity, acceleration
We may express the equation of motion as: using Newton's second law of motion.
The man will experience weightlessness.
3. Two masses
Ans:
The following diagram shows how the given system with two masses and a pulley may be represented:
So, the smaller mass is given as,
Larger mass is given as,
Therefore, tension in the string
Mass
The system of each mass is subjected to Newton's second law of motion:
For mass
The motion equation may be expressed as follows:
For mass
The motion equation may be expressed as follows:
Now on adding equations (i) and (ii), we get:
As a result, the acceleration of the masses is
When we replace the value
As a result, the string tension is
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.
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.
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:
Where,
From equation I it can be shown that the impact force is inversely proportional to the impact time, i.e.
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 impact
5. A helicopter of mass
(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,
Mass of the crew and passengers is given as,
Therefore, the total mass of the system,
As the acceleration of the helicopter is given as,
The reaction force
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
(b) The reaction force
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
(c) The surrounding air exerts a force of
6. Ten one-rupee coins are put on top of each other on a table. Each coin has a mass
(a) the force on the
(b) the force on the
(c) the reaction of the
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
Therefore, the weight of three coins
As a result, the three coins on top of the
(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
Weight of the ninth and tenth coin will also be the same i.e.
Therefore, the total weight of these three coins
Hence, the force exerted on the
(c) Because of the weight of the four coins
Therefore,
As per Newton's third law of motion, the
7. A monkey of mass
(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).
Ans:
Case (a)
From the given question, we have the information given as mass of the monkey,
Acceleration due to gravity is given as,
Maximum tension that the rope can bear is given as,
Acceleration of the monkey is given as,
So, the equation of motion can be written by using the Newton’s second law of motion:
As
Case (b)
Now in this case, the acceleration of the monkey,
Here we will use the Newton’s second law of motion to write the equation of motion. So it can be written as:
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
Here we will use the Newton’s second law of motion to write the equation of motion. So it can be written as:
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.,
Here we will use the Newton’s second law of motion to write the equation of motion. So it can be written as:
As T<Tmax , in this situation, the rope will not break.
8. Two bodies
Ans:
(a) In the given question, the mass of body A is given as,
Mass of body B is given as,
Applied force,
Coefficient of friction,
The friction force is calculated using the following formula:
Net force acting on the partition
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
(b) Force of friction on mass A:
Net force exerted by mass
According to Newton's third rule of motion, mass B will exert an equivalent amount of reaction force on mass A, i.e.,
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
As a result, the system of acceleration's equation of motion may be represented as:
Net force
Net force causing mass
Net force exerted by mass
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.,
9. The rear side of a truck is open and a box of
Ans: In the give question, we have the mass of the box given as,
Coefficient of friction is given as,
Initial velocity is given as,
Acceleration is given as,
Distance of the box from the end of the truck is given as,
The force on the box generated by the truck's accelerated speed is described by Newton's second law of motion:
A reaction force of
The box's rearward acceleration is calculated as follows:
Time
As a result, after
The below formula will give the distance
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
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.
The total of the normal force
The centripetal acceleration
The motorcyclist's speed provides a normal reaction. At the slowest possible speed
11. A thin circular loop of radius
Ans:
Let us assume that the radius vector joining the bead with the centre is making an angle
OP
The vertical and horizontal force equations can be expressed as follows:
In
Putting equation (iii) in equation (ii), we get:
Putting equation (iv) in equation (i), we get:
Since
For
On equating equations
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:
Important questions and answers make the students aware of the important topics in the chapter.
The chapter might seem lengthy for some students, in that case, they can always avail themselves of these important questions and answers.
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.
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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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.
S.No | Chapter-wise Important Questions for Class 11 Physics |
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5 | Chapter 6 - Systems of Particles and Rotational Motion Questions |
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Additional Study Materials for Class 11 Physics
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.

















