## NCERT Solutions Class 11 Physics Chapter 5 - Laws of MotionOne of the essential resources for effectively preparing for the Physics exams is the NCERT Solutions for Class 11 Physics Chapter 5 Laws of Motion. Students can learn more about the format of the exam's question papers and its grading system by using these NCERT Solutions. Below are the answers to Chapter 5, Laws of Motion's problems, to help students easily understand the chapter and new ideas that it contains. We largely concentrated on a particle's motion in the previous chapter. In contrast to non - uniform motion, which depends on acceleration, uniform motion depends on velocity. What controls the movement of bodies has received little attention from us thus far. This question now emerges as a fundamental one for the chapter. NCERT Solutions are written with precise and original information based solely on the most recent CBSE Syllabus for 2023-2024 and its curriculum, making them the best resource for learning all the fundamental ideas presented in this chapter. Let's look at the Class 11 solutions for NCERT laws of motion. ## NCERT Solutions Class 11 Physics Chapter 5
Give the magnitude and direction of the net force acting on - a drop of rain falling down with a constant speed
- a cork of mass 10 g floating on water
- a kite skillfully held stationary in the sky
- a car moving with a constant velocity of 30 km/h on a rough road
- a high - speed electron in space far from all material objects, and free of electric and magnetic fields.
- The raindrops are falling with a constant speed. As a result, acceleration will become zero. The force acting on the drop will become zero when the acceleration is zero since F = ma.
- Because the cork is floating in the water, the upthrust balances the cork's weight. The net force on the cork will be zero as a result.
- The acceleration is zero because the car is moving at a constant speed. The force will therefore be zero.
- Because the high - speed electron is far from the physical objects and is free of electric and magnetic fields, the net forc
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 upward motion
- during its downward motion
- at the highest point where it is momentarily at rest. Do your Solutions change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.
(a) The acceleration caused by gravity acts downward during the pebble's upward motion, hence the force acting on the stone is given by, F = mg = 0.05 kg × 10 ms = 0.5 N Also the force is acting in downward direction. (b) The force applies downward and has a magnitude of 0.5 N when the object is moving downward. (c) The pebble's velocity will have both horizontal and vertical components if it is thrown at a 45° angle with respect to the horizontal axis. The vertical component of velocity will be 0 at the pebble's highest point, while the horizontal component of velocity will always be present. The force acting on the pebble won't be impacted by this element in any way. Due to the fact that the pebble is only subject to acceleration, the force acting on it will be downward and have a value of 0.5 N.
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
- just after it is dropped from the window of a train running at a constant velocity of 36 km/h
- just after it is dropped from the window of a train accelerating with 1 m s
^{ - 2} - 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.
(a) Given that, mass of stone = 0.1 kg acceleration = 10 m s F = mg = 0.1 × 10 = 1.0 N Also the force is acting in vertically downward direction. (b) The train travels at a constant velocity. The acceleration will therefore be equal to zero. The motion of the train is therefore not exerting any force on the stone. As a result, the stone will continue to be subject to the same force (1.0 N) (c) A force of F' = ma = 0.1 x 1 = 0.1 N is applied to the stone while the train accelerates at a speed of 1 m/s The net force acting on the stone, however, is F = mg = 0.1 x 10 = 1.0 N because the force F' ceases to exist once the stone is dropped (vertically downwards). (d) The stone's acceleration will be the same as the train's when it rests on the train's floor. F = ma = 0.1 x 1 = 0.1 N is therefore the force's magnitude acting on the stone. It moves in the direction of motion of the train.
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 : T is the tension in the string. [Choose the correct alternative].
(i) T T is the net force exerted on the particle, and it is pointed in the direction of the centre. It gives the particle the centripetal force needed to move in a circular motion.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m s
Here, force = - 50 N (negative sign indicates retarding force) mass m = 20 kg final velocity v = 0 initial velocity u = 15 ms
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms
Here, mass m = 3.0 kg final velocity v = 3.5 m/s initial velocity u = 2 ms t = 25 s Also the force is acting in the direction of motion of the body.
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.
Given,
The driver of a three - wheeler moving at 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.
Given, initial velocity,u = 36 km/h final velocity,v = 0 mass of the three wheeler,m mass of the driver,m time taken to bring the vehicle to rest = 4.0 s Now, F = ma here,m = m m = 400 + 65 = 465 kg F = 465 × - 2.5 F = - 1162.5 N = - 1.6 × 10 The negative sign indicates that the force is retarding.
A rocket with a lift - off mass of 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms
Given, mass of the rocket,m = 20000 kg = 2 × 10 initial acceleration = 5 ms acceleration due to gravity,g = 9.8 ms The initial thrust should have a 5 ms A net acceleration of 9.8 + 5.0 = 14.8 ms Newton's second law of motion can be used to calculate the rocket's initial thrust. Thrust = force = mass × acceleration F = 20000 × 14.8 = 2.96 × 10
A body of mass 0.40 kg moving initially with a constant speed of 10 ms
Given, mass of the body,m = 0.40 kg initial velocity,u = 5 m/s force,f = - 8 N (retarding force) Using equation of motion we get, a.) Position when time is - 5 s The force acts on the body from starting point i.e. when t is 0 Therefore, we can say that the acceleration of the body is 0 when t is - 5 s. b.) Position when time is 25 s (c) Position at the time t = 100 s The body will move slowly for the first 30 seconds of the retarded force's application, after which the speed will remain constant.
A truck starts from rest and accelerates uniformly at 2.0 ms
Given that, initial velocity,u = 0 acceleration,a = 2 ms time,t = 10 seconds Using the equation of motion we get, v = u + at v = 0 + 2 × 10 = 20 m/s The horizontal component of the velocity in the absence of air resistance is unaltered at time t = 11 seconds. V The vertical component of the velocity is given by the equation V here, t = 11 - 10 = 1 sec and a Thus, the resultant velocity V will be given as,
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m/s. 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?
a.) The velocity is zero when the bob is in one of its extreme positions. The bob will fall vertically downward under the force of its weight F = mg if the string is cut. b.) The bob moves horizontally at its mean location. The bob will behave like a bullet and fall to the earth after travelling a parabolic route if the string is cut.
A man of mass 70 kg, stands on a weighing machine in a lift, which is moving (a) upwards with a uniform speed of 10 ms (b) downwards with a uniform acceleration of 5 ms (c) upwards with a uniform acceleration of 5 ms 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?
Given that, mass of the man,m = 70 kg acceleration due to gravity,g = 10 ms In each instance, the weighing machine determines the apparent weight by measuring the reaction R. a.) The acceleration is 0 when the lift moves upwards at a uniform speed of 10 m/s. R = mg = 70 × 10 = 700 N b.) When the lift is moving in a downward direction with an acceleration of 5 ms The equation can be written as R = m(g - a) By using Neton's second law of motion = 70(10 - 5) = 350 N c.) When the lift is moving in a upward direction with an acceleration of 5 ms The equation can be written as R = m (g + a) By using Neton's second law of motion = 70 (10 + 5) = 1050 N d.) If the lift is having a free fall under gravity in downward direction. Therefore g = a. ∴ R = m (g - a) = m (g - g) = 0 The person will experience weightlessness.
The figure shows the position - time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one - dimensional motion only).
a.) When t< 0, the particle travels a distance of zero. The force acting on the particle is therefore zero. The particle is travelling at a constant velocity when 0<t<4s. The force will therefore be zero. The particle maintains a constant distance for t>4s. As a result, the particle will have no force. b.) impulse at t = 0 initial velocity,u = 0 mass,m = 4 kg impulse = total change in momentum = mv - mu = m(u - v) = 4(0 - 0.75) = - 3 kg m/s impulse at t = 4 seconds Initial velocity, u = 0.75 m/s Final velocity, v = 0 m/s impulse = m(v - u) = 4(0 - 0.75) = - 3 kg m/s
Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a tight 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?
Given, mass of the body A,m mass of the body A,m horizontal force = 600 N The total mass of the system is, m = m m = 30 kg By Newton's Second Law of Motion we get, F = ma i.) When the force is applied on body A which is weighted at 10 kg. F - T = m T = F - m T = 600 - (10 × 20) = 600 - 200 = 400 N ii.) When the force is applied on body B which is weighted at 20 kg. F - T = m T = F - m T = 600 - (20 × 20) = 600 - 400 = 200 N
Two masses 8 kg and 12 kg are connected at the two ends of a light in extensible string that goes over a friction less pulley. Find the acceleration of the masses and the tension in the string when the masses are released.
Given, smaller mass m larger mass m Tension in the String = T The smaller mass m By Newton's Second Law of Motion we get, Adding Equations (1) and (2) we get, As a result, the mass accelerates at a rate of 2 m/s By putting the value of acceleration in equation (2) we gwt, As a result, the string is under 96 N of tension.
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.
Let m1, m2 be the masses of the two daughter nuclei and v1,v2 be their respective velocities of the daughter nuclei. Let m be the mass of the parent nucleus. After disintegration, the total linear momentum is, m The nucleus is at rest before disintegration. Therefore, it has zero linear momentum before disintegrating. By law of conservation of momentum, Total linear momentum before disintegration = Total linear momentum after disintegration The negative sign implies that v
Two billiard balls, each of mass 0.05 kg, moving in opposite directions with speed 6 ms
given mass of each ball is = 0.05 kg initial velocity,u of each ball is = 6 m/s Therefore, the initial momentum of the ball before the collision is, = 0.05 × 6 = 0.3 kg m/s Following the collision, the balls' motion direction changes without their speed changing significantly. Final momentum following the first ball's collision is equal to 0.05 x 6 = - 0.3 kg m/s. The second ball's collision produces a final momentum of 0.3 kg/s. The first ball received an impulse of ( - 0.3) − (0.3) = - 0.6 kg m/s. The second ball received an impulse of (0.3) - ( - 0.3) = 0.6 kg m/s. The directions of the two impulses are reversed.
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms
mass of the shell,m = 0.020 kg mass of the gun,M = 100 kg speed of the shell = 80 m/s The shell's and the gun's starting velocities are both zero. As a result, the system's initial momentum is zero. The initial and final momentum are equal according to the law of conservation of momentum.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
velocity of the ball = 54 km/h the ball is deflected in such a way that it makes an angle of = 45° Initial momentum of the ball is,
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?
Given that,mass of the stone = 0.25 kg radius,r = 1.5 m In 1 second, the number of revolution is Thus, the maximum speed of the stone is 36.64 m/s
If in problem 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 stoneflies off at an angle with the tangent whose magnitude depends on the speed of the particle
(b) The velocity will be tangential at all of the circular motion's points. Following Newton's first law of motion if the string breaks suddenly, the stone will move in a tangential path.
Explain why - a horse cannot pull a cart and run in empty space
- passengers are thrown forward from their seats when a speeding bus stops suddenly
- it is easier to pull a lawnmower than to push it
- a cricketer moves his hands backwards while holding a catch
a.) When pulling the cart, the horse pushes the ground with a particular amount of force. The third rule of motion states that the ground will place an equal and opposite reaction force on the horse's feet. The horse advances as a result of this. The horse won't feel a response force in an empty space. As a result, in an open area the horse cannot move the cart. b.) Because of the inertia of motion, when a bus abruptly stops, the part of a person's body that is in touch with the seat abruptly comes to rest while the upper part of their body is still in motion. The person's upper torso is thrust forward in the direction of the bus' travel as a result. c.) The lawnmower is dragged upwards by the vertical component of the applied force. This lowers the lawnmower's actual weight. The vertical part of the lawn mower pushes in the direction of the weight of the mower. Because of this, the mower's weight has increased. It is therefore simpler to pull a lawnmower than to push one. d.) The ball has a lot of momentum when the batsman smacks it. The force is decreased when he moves his hands backward because the contact duration is prolonged.
Figure shows the position - time graph of a particle of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the particle? What is the magnitude of each impulse?
A ball could be rebounding between two walls that are 2 cm apart in this graph. Every two seconds, the ball bounces uniformly off of the walls. The time between two consecutive impulses is 2 sec, so the ball receives an impulse every 2 seconds.
Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 ms
Here,the acceleration of conveyor belt a = 1ms Coefficient of static friction,μ Mass of the man,m = 65 kg Net Force = ma = 65 × 1 = 65 N
A stone of mass m tied to the end of a string is revolving in a vertical circle of radius R. The net force at the lowest and highest points of the circle directed vertically downwards are: (choose the correct alternative). T1 and v1 denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point.
(a) The net force at the lowest point is (mg - T Since mg and T
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms - force on the floor by the crew and passengers
- the action of the rotor of the helicopter on surrounding air
- force on the helicopter due to the surrounding air
Mass of the helicopter,m = 1000 kg Crew and passengers weight = 300 kg Vertical acceleration,a = 15 ms Acceleration due to gravity,g = 10 ms Total mass of the system, m (a) Force on the floor of the helicopter by the crew and passengers R - mg = ma = m(g + a) = 300(10 + 15) = 7500 N (b) Action of the rotor of the helicopter on surrounding air is due to the mass of the helicopter and the passengers. R', - m R', = m = 1300 × (10 + 15) = 32500 N (c) The reaction of the force applied by the rotor on the air is the force on the helicopter owing to the surrounding air. Since action and reaction are complementary and equal, the force of reaction is F = 32500 N. This force has an upward, vertical motion.
A stream of water flowing horizontally with a speed of 15 m/s pushes out of a tube of cross - sectional area 10
Speed of flowing water,v = 15m/s Cross - sectional area of the tube,A = 10 The distance travelled will be equal to the velocity in 1 second.
Ten one rupee coins are put on top of one another on a table. Each coin has a mass of m kg. Give the magnitude and direction of: (a) the force on the 7th coin (counted from the bottom) due to all coins above it (b) the force on the 7th coin by the eighth coin (c) the reaction of the sixth coin on the seventh coin
(A) The weight of the three coins held above the seventh coin is what is exerting pressure on it. One coin has a mass of milligrammes. Consequently, three coins weigh 3mg. The force acting vertically downward on the seventh coin is (3mg)N. (b) The eighth coin has its own weight in addition to the weight of the two coins above it. As a result, the force exerted on the seventh coin by the eighth coin will be equal to the force exerted on the seventh coin by the three coins above it. As a result, the force exerted by the eighth coin on the seventh coin is (3mg)N and acts vertically downward. (c) The fourth, fifth, and sixth coins are pressing downward on the sixth coin as a result of the weight of the four coins above it. The sixth coin has a total downward force of 4mg. In accordance with Newton's third law of motion, the sixth coin will generate an upward response force. The force applied by the sixth coin to the seventh coin is therefore 4mg and acts upward.
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
Speed of the aircraft executing the horizontal loop = 720 km/h
A train runs along an unbanked circular track of radius of 30 m at a speed of 54 km/h. The mass of the train is 106 kg. What provides the centripetal force required for this purpose the engine or the rails? What is the angle of banking required to prevent wearing out of the rail?
The required centripetal force is provided by the force of lateral friction due to the rails on the wheels of the train The angle of banking required to prevent wearing out of the rails
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 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?
Mass of the block = 25 kg Massof the man = 50 kg Acceleration due to gravity,g = 10 ms Weight of the block,F = 25 × 10 F = 250 N Weight of the man,W = 50 × 10 W = 500 N In the first instance, the man uses a 250 N upward force to lift the block directly (same as the weight of the block) The third law of motion by Newton predicts that there would be a downward reaction on the ground. The man's movement on the ground. = 500 N + 250 N = 750 N. In the second instance, the man pushes down with a 25 kg weight. The third law of motion of Newton states that the response will move upward. In this instance, the man's movement on the floor = 500 N - 250 N = 250 N. The man should therefore choose the second strategy.
A monkey of mass 40 kg climbs on a rope (Fig.) which can stand a maximum tension of 600 N. In which of the following cases will the rope break, when the monkey: - climbs up with an acceleration of 6 ms
^{ - 2} - climbs down with an acceleration of 4 ms
^{ - 2} - climbs up with a uniform speed of 5 ms
^{ - 2} - falls down the rope nearly freely under gravity
(Ignore the mass of the rope).
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.). 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 Solution to (b) change, when the bodies are in motion? Ignore the difference between μ
Mass of the body,m Mass of the body,m Applied force = 200 N Coefficient of friction between the bodies and the table μs = 0.15 (a) The force of friction is given by the relation: Therefore, the net force on the partition is 200 - 22.5 = 177. 5 N rightward According to Newton's third law, action and reaction are in the opposite direction Therefore, the reaction of the partition will be 177.5 N, in the leftward direction (b) Force of friction on mass A The net force exerted by mass A on mass B = 200 - 7.5 = 192.5 N rightwards An equal amount of reaction force will be applied on mass A by B, i.e., 192.5 N acting leftward When the wall is removed, the two bodies move in the direction of the applied force The net force acting on the moving system = 177. 5 N The equation of motion for the system of acceleration a, can be written as Net force exerted by the mass A on mass B = 192.5 - 59.15 = 133. 35 N This force acts in the direction of motion. As per Newton's third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3N, acting opposite to the direction of motion.
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 ms
Mass of the block = 15 kg Coefficient of static friction between the block and the trolley, p = 0.18 Acceleration of the trolley = 0.5 m/s (a) Force experienced by block, F = ma = 15 x 0.5 = 7.5 N, this force acts in the direction of motion of the trolley Force of friction, Force experienced by the block will be less than the friction. So the block will not move. So, for a stationary observer on the ground, the block will be stationary. (b) The observer moving with the trolley has an accelerated motion. He forms a non - inertial frame in which Newton's laws of motion are not applicable. The trolley will be at rest for the observer moving with the trolley
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. 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 ms
Force experienced by box, F = ma = 40 x 2 = 80 N Frictional force, If t is time taken by the box to travel s = 5 metre and fall off the truck then from
A disc revolves with a speed of 33⅓ rpm 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 coefficient of friction between the coins and record is 0.15, which of the coins will revolve with the record?
For coin A, r = 4 cm The condition (r≤ 12) is satisfied for coin placed at 4cm, so coin A will revolve with the disc For coin B, r = 14 cm The condition (r≤ 12) is not satisfied for the coin placed at 14cm, so coin B will not revolve with the disc.
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?
When the motorcyclist is at the uppermost point of the death - well, the normal reaction R on the motorcyclist by the ceiling of the chamber acts downwards. His weight mg also acts downwards. The outward centrifugal force acting on the motorcyclist is balanced by these two forces. Here, v is the velocity of the motorcyclist m is the mass of the motorcyclist and the motorcycle Because of the balance between the forces, the motorcyclist does not fall The minimum speed required at the uppermost position to perform a vertical loop is given by the equation (i) when R = 0
A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?
Mass of the man,m = 70 kg Radius of the drum,r = 3 m Coefficient of friction between the wall and his clothing, μ = 0.15 Number of revolution of hollow cylindrical drum, The centripetal force required is provided by the normal N of the wall on the man When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force acting vertically upwards. The man will not fall if mg ≤ limiting frictional force f mg ≤ μN mg ≤ μ (mω ω2 ≥ g/Rμ Therefore, for minimum rotational speed of the cylinder ω ω = √22.2 = 4.7 rad/s
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 ω ≤ √g / R . What is the angle made by the radius vector joining the centre to the bead with the vertically downward direction for ω = √2g/ R? Neglect friction.
Let θ be the angle made by the radius vector joining the bead and the centre of the wire with the downward direction. Let, N be the normal reaction mg = N cosθ —-(1) Mrω (or) m (Rsinθ) ω mRω Substituting the value of N in (1) mg = mRω (or) cos θ = g/Rω As l cos θ I ≤ 1 the bead will remain at the lowermost point g/Rω For ω = √2g/ R , equation (3) becomes cos θ = g/Rω cos θ = (g/R) (R/2g) = 1/2 θ = 60 |