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Ever wondered how rockets launch or why objects move the way they do? Class 11 Physics Chapter 4 Laws of Motion gives all the answers with NCERT solutions made by our subject matter experts, you will easily understand concepts from friction to free-body diagrams. These Class 11 Physics Chapter 4 Laws of Motion Exercise Solutions follow the NCERT syllabus and make even tough problems easy to understand.
The NCERT concepts and exercises are the source of many of the problems in the CBSE board exams, JEE Main and also NEET; therefore, NCERT Solutions for Class 11 Physics Laws of Motion are very important. These answers provide a direct explanation of each exercise question, which helps students to learn faster. Understanding how things move around us, such as why you feel pushed back as a car speeds up or how objects react when moving, is all made easier by the laws of motion.
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You can download the Motion in a Plane Class 11 solved problems PDF for free. The detailed explanation for all the questions in this chapter will help strengthen both your board exam and other competitive exams like JEE. It's a great resource for grasping the core concepts and improving problem-solving skills.
Q4.1 (a) Give the magnitude and direction of the net force acting ona drop of rain falling down at a constant speed
Answer:
Since the drop is falling with constant speed, so the net acceleration of the drop is zero. This means net force on the drop is zero.
Q4.1 (b) Give the magnitude and direction of the net force acting on a cork of mass 10 g floating on water,
Answer:
Since the cork is floating on the water, that implies the gravitational force of cork is balanced by the buoyant force. Thus net force acting on cork is zero.
Answer:
Since the kite is stationary in the sky, thus according to Newton's law, the net force on the kite is zero.
Answer:
Since the car is moving with constant velocity thus the net acceleration of the car is zero. Thus net force acting on the car is zero.
Q4.1 (e) Give the magnitude and direction of the net force acting ona high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Answer:
Since the electron is free from the electric and magnetic field so net force acting on the electron is zero.
Answer:
During upward motion, the force acting on the pebble is only the gravitational force which is acting in a downward direction .
The gravitational force :
=mg=0.05(10)N=0.5N
Q4.2 (b) A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, during its downward motion
Answer:
Since the pebble is moving in the downward direction the net force acting is also in the downward direction .
The net force is the gravitational force.
Gravitational force:-
=mg=0.(05)10=0.5N
Answer:
At the highest point velocity becomes zero for a moment. At this point also, the net force acting is the gravitational force which acts in the downward direction .
Gravitational force:-
When a pebble is thrown at 450 with the horizontal direction then also net force will be the same gravitational force.
Q4.3 (a) Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,just after it is dropped from the window of a stationary train, Neglect air resistance throughout.
Answer:
Since the train is stationary so the net force acting on the stone is the gravitational force which acts in the downward direction .
Gravitational force :
or
or
Q4.3 (b) Give the magnitude and direction of the net force acting on a stone of mass 0.1 kgjust after it is dropped from the window of a train running at a constant velocity of 36 km/h, Neglect air resistance throughout.
Answer:
Since the train is travelling with the constant speed so the acceleration of the train is zero. Thus there is no force on stone due to train.
The net force acting on the stone will be the gravitational force which acts in a downward direction.
Gravitational force:-
or
or
Answer:
Just after the stone is dropped, the stone is free from the acceleration of the train. Thus the force acting on the stone will be just the gravitational force which acts in the downward direction.
The gravitational force is given by:-
or
or
Q4.3 (d) Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,lying on the floor of a train which is accelerating with 1 m s-2 , the stone being at rest relative to the train. Neglect air resistance throughout.
Answer:
As the stone is in contact with the train thus the acceleration of stone is the same as that of train i.e. 1 m/s 2 .
Thus force acting on the stone is given by :
or
This force is acting in the horizontal direction.
(i)
(ii)
(iii)
(iv) 0
Answer:
When the particle is moving in a circular path, the centripetal force will be :
This centripetal force will be balanced by the tension in the string.
So, the net force acting is :
Answer:
We are given the retarding force. So we can find the deacceleration this force is causing.
By Newton's second law of motion, we get :
or
or
Now we will use the first equation of motion,
The final velocity, in this case, will be zero (Since the vehicle stops).
Thus
Thus the time taken to stop the vehicle is 6 sec.
Answer:
Since the velocity of the body is increased by applying the force. This is possible only when the force is applied in the direction of the motion.
For finding the magnitude of the force, we need to calculate acceleration.
By using first equation of the motion,
or
or
Thus force can be written as :
Answer:
The magnitude of the resulting force can be found by :
or
or
Now force direction,
or
The acceleration of the body is given by :
or
Hence acceleration of the body is
Answer:
Total mass of the system = 400 + 65 = 425 kg
Using first law of motion, we get
Since the car comes to rest, thus final velocity will be zero.
or
So the force required :
or
or
So the magnitude of the force is 1162.5 N and it is retarding force.
Answer:
Let the initial thrust be F Newton.
Using Newton's second law of motion, we get
or
Substituting values in this equation, we get :
or
Q4.10 A body of mass 0.40 kg moving initially with a constant speed of
Answer:
The acceleration of force is given by :
At t = - 5 s :
There is no force acting so acceleration is zero and u = 10 m/s
or
or
At t = 25 s :
Acceleration is - 20 m/s 2 and u = 10 m/s
or
or
At t = 100 s
We have acceleration for first 30 sec, and then it will move with constant speed.
So for 0 < t < 30 :
or
or
Now for t > 30 s :
We need to calculate velocity at t = 30 sec which will be used as the initial velocity for 30 < t < 100.
or
Now
or
or
Hence total displacement is : - 8700 + (- 41300 ) = - 50000 m.
Answer:
The initial velocity of the truck is given as zero.
We need to find the final velocity (at t = 10 s), so we will use the equation of motion :
or
This is the velocity imparted to stone by the truck that's why it is a horizontal component of velocity.
The stone is dropped at t = 10 sec. so it has travelled 1 sec in the air (11 - 10 = 1 s). We need to find the final vertical velocity.
Thus resultant of both the component is the required velocity.
or
or
Direction :
or
Answer:
When the stone is dropped, the stone comes only in effect of gravity.
So the acceleration of stone is 10 m/s 2 and it acts in the downward direction.
Answer:
At the extreme positions, the velocity of the bob will become zero for a moment. So if we cut the string at this time, then Bob will fall vertically downward due to gravity.
Answer:
At the mean position, the bob will have velocity tangential to the circular path (it will be completely horizontal). If the bob is cut at this place then it will follow a parabolic path having only horizontal velocity.
Answer:
Since lift is moving with a constant speed or in this case constant velocity so the acceleration provided to man by lift is zero.
The net force on the man will be zero. So a = 0.
Using Newton's law of motion, we can write :
or
Thus
So, reading on weighing scale will be :
Answer:
Using Newton&##39;s law of motion, we have :
or
or
Thus the reading on the weighing scale will be :
Answer:
The acceleration of lift is given to be 5 m/s 2 . Let us assume the upward direction to be positive.
Using Newton's law of motion, we can write :
or
or
Thus the reading of the weighing scale will be :
Answer:
If the lift falls freely then the acceleration of lift will be acceleration due to gravity.
Using Newton's law of motion we can write :
or
or
Thus the reading of weighing scale will also be zero since there is no normal force.
This state is called the state of weightlessness.
Answer:
(i) For t < 0 :
In this range the position of particle coincides with time, that implies no motion takes place. Hence net force on the particle is zero.
(ii) For t > 4 :
In this range displacement of the particle is not changed so the net force is zero.
(iii) For 0 < t < 4 :
In this range the slope of the position-time graph is constant that means the particle is moving with constant speed. And hence net force, in this case, is zero.
Q4.14 (b) Figure shows the position-time graph of a particle of mass 4 kg. What is the impulse at t = 0 and t = 4 s ? (Consider one-dimensional motion only).
Answer:
Impulse is defined by :
At t = 0 s :
or
or
At t = 4 s :
Thus
or
Answer:
We will consider A and B in a system. So total mass in the system is = 10 + 20 = 30 Kg.
Thus acceleration of the system is given by :
or
When force is applied at block A :
Using Newton's law of motion :
or
Thus
Answer:
We will consider A and B in a system. So total mass in the system is = 10 + 20 = 30 Kg.
Thus acceleration of the system is given by :
or
When force is applied at block B :
Using Newton's law of motion, we can write
or
Thus
Answer:
Since both the masses are connected with string so they will have the same acceleration, let say 'a'.
We will apply Newton's law for each block individually.
For smaller block (8 Kg) :
For larger block (12 Kg) :
The equation of motion is given by :
Adding both the equations we get :
The acceleration is given by :
or
or
Now put the value of acceleration in any of the equations to get the value of T.
or
or
Thus the tension in the string is 96 N.
Q4.17 Anucleus 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.
Answer:
Let the mass of parent nuclei be m.
And the mass of daughter nuclei be m 1 and m 2 .
Initial momentum is zero since the nuclei is at rest.
But after dissociation, the momentum becomes :
Using conservation of momentum,
or
Thus both velocities have opposite direction.
Answer:
Impulse imparted can be calculated by knowing the change in momentum.
The initial momentum of each ball is :
The final momentum is given by :
So the impulse is :
or
or
Answer:
In this question, we will use the conservation of momentum.
Final momentum = Initial momentum
For initial momentum :
Both gun and shell are at rest initially so momentum is zero.
For final momentum :
The direction of the velocity of the shell is opposite to that of the gun.
So,
The recoil speed of the gun :
or
Answer:
The situation is shown below :
The horizontal components of velocity are to be considered for imparting impulse as vertical components are in the same direction thus impulse in the vertical direction is zero.
The impulse is given by a change in momentum.
Initial momentum =
Final momentum =
Thus impulse is, J
or
or
Answer:
We are given, frequency :
So, the angular velocity becomes :
By Newton's law of motion, we can write :
or
or
or
Now, we are given maximum tension and we need to find the maximum velocity for that :
Thus,
or
or
(a) the stone moves radially outwards,
Answer:
The stone should move in the direction of velocity at that instant. Since velocity is tangential at that moment the stone will not move radially outward.
(b) the stone flies off tangentially from the instant the string breaks,
Answer:
(b) This statement is correct as the direction of velocity at the instant of the breaking of the string is tangential thus stone will move tangentially.
(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle.
Answer:
The direction of velocity at the instant of breakage of the string is tangential so the stone will fly tangentially. Hence given statement is false.
Q4.23 (a) Explain whya horse cannot pull a cart and run in empty space
Answer:
The horse moves forward by pushing ground backwards. The ground will then give the normal force to the horse (action-reaction pair), which is responsible for the movement of the horse. In an empty space, no such force is present so the horse cannot get the push to run forward. Thus, a horse cannot pull a cart and run in empty space.
Q4.23 (b) Explain whypassengers are thrown forward from their seats when a speeding bus stops suddenly
Answer:
This is because of inertia. When the bus is moving our body has the same speed. But when the bus comes to rest, the inertia of our body opposes to stop and continues its motion. That's why passengers are thrown forward from their seats when a speeding bus stops suddenly.
Q4.23 (c) Explain whyit is easier to pull a lawnmower than to push it
Answer:
Because when you pull the lawnmower at some angle, one component of force is in the upward direction and one in horizontal (to move). The vertical force reduces the effective weight of the motor which makes it easier. But in case of a push, the vertical force is directed downward which makes its effective weight even greater than before. That's why it is said 'pull is easier than push'.
Q4.23 (d) Explain whya cricketer moves his hands backwards while holding a catch.
Answer:
According to Newton's law, we can write :
or
Thus
It can be seen from the equation that if we increase the impact time then the experienced force will be lesser.
So a cricketer increases the impact time by taking his hands backward while holding a catch, resulting in less force on their hand.
Q1. Consider a block and trolley system as shown in the figure. If the coefficient of kinetic friction between the trolley and the surface is 0.4, the acceleration of the system
Answer:
Hence, The answer is 2/3.
Q.2 The system shown in the figure is in equilibrium and at rest. The spring and string are massless. Now the string between B and C is cut. Find the acceleration of A, B and C just after the string is cut.
Answer:
At equilibrium, the free body diagrams are-
For block A,
For block B,
For block C,
When string is cut
Block A
Block B
Block C
Q.3 The reading of the spring balance (mass-stress) in the figure given below is
Answer:
FBD of the block of mass m1
FBD of block mass m2
For massless spring balance reading
From the equation (1) and (2), we get-
Spring balance reading
Q.4 The arrangement is shown in the given figure. If the coefficient of friction between the 2kg block and the table is 0.2 . What would be the maximum value of m (in kg) such that two blocks do not move?(10=m/s2) ?
Answer:
Given-
mass of block on table
coefficient of friction,
As we keep on increasing the value of
Let the tension in the string be
F.B.D of a block of mass
For equilibrium-
F.B.D of a block of mass 2 kg -
Limiting friction-
For equilibrium-
From equation (1) and (2) -
Q.5 A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in the figure. The other end of the string is pulled with constant acceleration ' a ' vertically downward.The tension in the string is equal to :
Answer:
As we learned
Newton's 2nd Law -
Apply the concept of pseudo force as shown in the below figure
Applying Newton's law perpendicular to string
Applying Newton's law along string,
The topics of class 11 physics chapter 4 is listed below.
4.1 Introduction
4.2 Aristotle’s Fallacy
4.3 The Law Of Inertia
4.4 Newton’s First Law Of Motion
4.5 Newton’s Second Law Of Motion
4.6 Newton’s Third Law Of Motion
4.7 Conservation Of Momentum
4.8 Equilibrium Of A Particle
4.9 Common Forces In Mechanics
4.9.1 Friction
4.10 Circular Motion
4.11 Solving Problems In Mechanics
These formulas form the foundation of classical mechanics and are critical for solving both conceptual and numerical problems in exams like CBSE, JEE, and NEET.
F = ma
Force is the product of mass and acceleration.
Momentum is the product of mass and velocity.
To understand the solutions of NCERT Class 11 Physics Chapter 4 Laws of Motion, it is necessary to know how to draw a free-body diagram. To familiarise with free body diagrams practice as many problems as possible. NCERT solutions for Class 11 are important for competitive exams like NEET and JEE Main and final exams of Class 11. NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion helps to do well in these exams.
In order to learn how to determine the behaviour of objects in various forces, as well as to resolve real-life problems in mechanics.
No, it can only apply in the frames that are inertial and break down at very high velocities or quantum levels.
Practical examples are rocket propulsion, jumping, swimming and walking.
Yes, it will move at a constant velocity according to the first law of Newton.
Unbalanced external forces, such as friction or air resistance, tend to bring motion to a halt.
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