NCERT Solutions for Class 11 Physics Physics Chapter 5 Laws of motion includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Physics exams must go through NCERT Solutions for Class 11 Physics Physics Chapter 5 Laws of motion. Going through the solutions provided on this page will help you to know how to approach and solve the problems.
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NCERT Solutions for Class 11 Physics Physics Chapter 5 Laws of motion
NCERT Solutions for Class 11 Physics Physics Chapter 5 Laws of motion are been solved by expert teachers of CBSETuts.com. All the solutions given in this page are solved based on CBSE Syllabus and NCERT guidelines.
NCERT Exercises
Question 1.
[For simplicity in numerical calculations, take g= 10 m s^{2}]
Give the magnitude and direction of the net force acting on
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with a constant velocity of 30 km h^{_1} on a rough road,
(e) a highspeed electron in space far from all material objects, and free of electric and magnetic fields.
Answer:
(a) Since the drop of rain is falling downward with a constant speed, so the acceleration of the rain drop is zero. Hence according to Newton’s first law of motion, net force on the drop is zero i.e. a = 0,
∴ From F = ma, F = 0.
(b) As the cork is floating in water, so the weight of the cork is balanced by upthrust (equal to weight of water displaced). Therefore net force on the cork is zero.
(c) Since the kite is held stationary, so its acceleration is zero. Hence according to Newton’s first law of motion, the net force acting on the kite is zero.
(d) Since the car is moving with a constant velocity, so its acceleration is zero. Hence according to Newton’s first law of motion, the net force acting on the car is zero.
(e) As there are no electric and magnetic fields and no material (gravitating) objects around, so no force (gravitational/electric/ magnetic) is acting on the electron, so net force on it is zero.
Question 2.
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 directions? Ignore air resistance.
Answer:
Whenever an object is thrown vertically upwards or it moves vertically downwards, the gravitational pull of Earth gives uniform acceleration a = + g = 10 m s^{2} in the vertically downward direction. If m = mass of the object, then net force on it = mg Here m = 0.05 kg = mass of the pebble.
(a) ∴ Net force on the pebble = mg ( ∴ a = g)
= 0.05 x 10 = 0.5 N (acts vertically downwards)
(b) Net force on the pebble = mg (∴ a = g)
= 0.05 x 10 = 0.5 N (acts vertically downwards)
(c) When the stone is at the highest point then also the net force (= mg) acts in vertically downward direction
∴ Net force on the pebble = mg
= 0.05 x 10 = 0.5 N
If the pebble was thrown at an angle of 45° with the horizontal direction, then it will have horizontal and vertical components of velocity which will not affect the force on the pebble. Hence our answer will not.change in any cases. However in case (c), the pebble will not be at rest at the highest point. It will have a horizontal component of velocity at this point.
Question 3.
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^{1},
(c) just after it is dropped from the window of a train accelerating with 1 m s^{2},
(d) lying on the floor of a train which is accelerating with 1 m s^{2}, the stone being at rest relative to the train. Neglect air resistance throughout.
Answer:
(a) Here, m = 0.1 kg, a = +g= 10ms^{2
}Net force, F = ma = 0.1 x 10 = 1.0 N
This force acts vertically downwards.
(b) When the train is running at a constant velocity, its acceleration = 0.
No force acts on the stone due to this motion. Therefore, force on the stone
F = weight of stone = mg = 0.1 x 10 = 1.0 N This force also acts vertically downwards.
(c) When the train is accelerating with 1 m s^{2}, an additional force
F’ = ma = 0.1 x 1 = 0.1 N acts on the stone in the horizontal direction. But once the stone is dropped from the train, F’ becomes zero and the net force on the stone is
F = mg = 0.1 x 10 = 1.0 N, acting vertically downwards.
(d) As the stone is lying on the floor of the train, its acceleration is same as that of the train.
.’. Force acting on stone,
F = ma = 0.1 x 1 = 0.1 N
This force is along the horizontal direction of motion of the train. Note that weight of the stone in this case is being balanced by the normal reaction.
Question 4.
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 center) is
(a) T
(b) Tmv^{2}/l
(c) T+mv^{2}/l
(d) 0
T is the tension in the string. [Choose the correct alternative].
Answer:
The net force on the particle directed towards the center is T. This provides the necessary centripetal force to the particle moving in the circle.
Question 5.
A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms^{_1}. How long does the body take to stop?
Answer:
Here, M = 20 kg;
F = – 50 N (retarding force)
Now, F = Ma
\( a=\cfrac { F }{ M } =\frac { 50 }{ 20 } =2.5\)
Also, υ= u + at
Here, u = 15m s^{1}; v = 0
∴ 0 = 15 + (2.5)t
or t = 6 s
Question 6.
A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s^{_1} to m s^{_1} in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?
Answer:
Here, m = 3.0 kg, u = 2.0 m s^{1}; v = 3.5 m s^{1} t = 25 s,
Now, υ= u + at
3.5 = 2 + a x 25
\(a=\cfrac { 3.52 }{ 25 } =0.06ms\)
Therefore, force acting on the body,
F = ma = 3 x 0.06 = 0.18 N in the direction of motion.
Question 7.
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.
Answer:
This is the direction of resultant force and hence the direction of acceleration of the body.
Question 8.
The driver of a threewheeler moving with a speed of 36 km h^{1} 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 threewheeler is 400 kg and the mass of the driver is 65 kg.
Answer:
Question 9.
A rocket with a liftoff mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s ^{2}. Calculate the initial thrust (force) of the blast.
Answer:
Here, M = 20,000 kg;
Initial acceleration = 5 m s^{2
}The rocket moves up against gravity.
Therefore, the blast has to produce a total acceleration given by
a = 10 + 5 = 15 m s^{2
}Hence, the initial thrust of the blast,
F = ma = 20,000 x 15 = 3 x 10^{5} N
Question 10.
A body of mass 0.40 kg moving initially with a constant speed of 10 m s^{_1} to the north is subjected 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.
Answer:
Question 11.
A truck starts from rest and accelerates uniformly at 2.0 m s ^{2}. At f = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are
(a) velocity, and
(b) acceleration of the stone at t = 11 s? (Neglect air resistance.)
Answer:
Initial velocity, u = 0, a =0 m s ^{2}, t = 10 s Let v be the velocity of truck when the stone is dropped from it after
t = 10 s.
Using the relation, v = u + at, we get
υ = 0 + 2.0 x 10 = 20 m s^{1}
(a) Horizontal velocity of the stone when it is dropped from the truck is
υ_{x} = v = 20 m s^{_1}.
As air resistance is neglected, so υ_{x} = constant.
Motion in the vertical direction :
Initial velocity of the stone, υ_{y} = 0 at t = 10 s
acceleration, υ_{y} = g = 10 m s^{2}, time t = 1110 = 1 s
If v_{y} be velocity of the stone after 1 s of drop (i.e. at t = 11 s,) then
υ_{y} = uy + a_{y}t = 0 + 10 x 1 = 10 m s^{1
}If v be the velocity of the stone after 11s, then
horizontal direction OA i.e.. with υ_{x} Then from ΔOAC
(b) At the moment, the stone is dropped from the truck, the horizontal force on the stone is zero,so,
a_{x} = 0 and a_{y} = acceleration along vertical direction = +g = 10 m s^{2} which acts in downward direction.
If a = resultant acceleration of the stone, then
Question 12.
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^{1}. 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.
Answer:
(a) At each extreme position, the instantaneous velocity of the bob is zero. If the string is cut at the extreme position, the bob is under the action of ‘g’ only, hence the bob will fall vertically downwards.
(b) When the bob is at the mean position, it is affected by gravity. At mean position the bob is having a velocity of 1 m s^{1} along the tangent to the arc which is in the horizontal direction. If the string is cut at the mean position, the bob will behave as a horizontal projectile. Hence it will follow a parabolic path.
Question 13.
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 ms^{1},
(b) downwards with a uniform acceleration of 5 ms ^{2},
(c) upwards with a uniform acceleration of 5 ms^{2}.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?
Answer:
Here, m = 70 kg, g = 10 m s^{2
}The weighing machine in each case measures the reaction R i.e. the apparent weight.
(a) When the lift moves upwards with a uniform speed, its acceleration is zero.
∴ R = mg = 70 x 10 = 700 N
(b) When the lift moves downwards with a = 5 ms^{2 }R – m(g a) = 70(10 – 5) = 350 N
(c) When the lift moves upwards with
a = 5ms^{2
}R = m(g + a) = 70(10 + 5) = 1050 N
(d) If the lift was to come down freely under gravity, downward acceleration
a = g :. R = m(g a) = m(g g) = zero.
Question 14.
Figure shows the positiontime graph of a particle of mass 4 kg. What is the
(a) force on the particle for t < 0, t > 4 s, 0 < t < 4s?
(b) impulse at t = 0 and t = 4 s? (Consider onedimensional motion only).
Answer:
(a)
.

 For t < 0, the position time graph is AO which means displacement of the particle is zero, i.e. particle is at rest at the origin. Hence force on the particle must be zero.
For t > 4 s, the position time graph BC is parallel to time axis. Therefore, the particle remains at a distance of 3 m from the origin, i.e., it is at rest. Hence force on the particle is zero.
For 0 < t < 4 s, the position time graph OB has a constant slope. Therefore, velocity of the particle is constant in this interval e. particle has zero acceleration. Hence force on the particle must be zero.
(b)
Impulse at t = 0
Impulse = change in linear momentum. Before t = 0, particle is at rest i.e. u = 0. After t = 0, particle has a constant velocity
∴ Impulse = m(υ u)
= 4(0.750) = 3 kg ms^{1
}Impulse at t = 4 s
Before t = 4 s, particle has a constant velocity u = 0.75 ms^{1}^{ }After t = 4s, particle is at rest i.e. v = 0
Impulse = m(υ u)
= 4 (0 – 0.75) = – 3 kg ms^{1}
Question 15.
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
(1) A,
(2) B along the direction of string. What is the tension in the string in each case?
Answer:
Here, F = 600 N; m_{1} = 10 kg; m_{2} = 20 kg. Let T be tension in the string and a be the acceleration of the system, in the direction of force applied.
Question 16.
Two masses 8 kg and 12 kg are connected at the two ends of a light in extensible string that passes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.
Answer:
Let m_{1} and m_{2} be the masses suspended at the ends of a light in extensible string passing over the pulley.
m_{1}=8 kg, m_{2} = 12 kg,
Let T be the tension in the string and a be the common acceleration with which m_{1} moves upwards and m_{2} moves downward = ?
The equation of motion of m_{1} and m_{2} are given by
Question 17.
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.
Answer:
Let m = mass of the nucleus at rest.
\(\bar {u} \) = its initial velocity = 0 as it is at rest.
Let m_{1} ,m_{2} be the masses of the two smaller nuclei also called product nuclei and be their respective velocities.
If \(\bar {p}\)_{i} and \(\bar {p}\)_{f} be the initial and final momentum of the nucleus and the two nuclei respectively, then
The negative sign in equation (iii) shows that \(\bar {v}\)_{1} and \(\bar {v}\)_{2} are in opposite directions i.e. the two smaller nuclei are moved in opposite directions.
Question 18.
Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 m s^{_1} collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Answer:
Let the two balls A and B moving in opposite directions collide and rebound with same speed.
Initial momentum of the ball A
= 0.05 x (6) = 0.3 kg m s^{1}^{
}Final momentum of the ball A
= 0.05(6) = 0.3 kg m s^{1
}Impulse imparted to ball A = change in momentum of ball
A = final momentum – initial momentum = 0.3 – 0.3 = 0.6 kg m s^{1}^{
}Initial momentum of the ball B = 0.05 x (6) = 0.3 kg m s^{1
}Final momentum of the ball B = 0.05 x (6) = 0.3 kg m s^{1
}Impulse imparted due to B = 0.3 – (0.3) = 0.6 kg m s^{1}^{
}Impulse on each ball is 0.6 kg m s^{1} in magnitude but these two impulses are opposite in direction.
Question 19.
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 m s^{1}
what is the recoil speed of the gun?
Answer:
Here, mass of the shell, m_{1} = 0.02 kg ; mass of the gun, m_{2} = 100 kg;
Initial velocities of both the shell and the gun are zero i.e. u_{1}= u_{2} = 0
After firing, speed of the shell, υ = 80 m s^{1
}Let v_{2} be the recoil speed of the gun.
According to the principle of conservation of momentum,
Question 20.
A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal
to 54 km h^{1}. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)
Answer:
Let the ball of mass m moving along AO with initial speed u hits the bat PQ and is defected by the batsman along OB (with out change in the speed of the ball), such that
∠AOB = 45°.
The initial momentum of the ball can be resolved into the following two components :
 mucos22.5° along NO and
 mucos22.5° along PQ
Also, the final momentum of the ball can be resolved into the following components :
 mucos22.5° along ON and
 mucosin22.5° along PQ
The component of the momentum of the ball along PQ remains unchanged (both in magnitude and direction). However, the components of the momentum of the ball along ON are equal in magnitude but opposite in direction. Since the impulse imparted by the batsman to the ball is equal to the change , in momentum of the ball along ON,
impulse = mucos22.5° – (mucos22.5°)
= 2 mucos22.5°
Here, m = 0.15 kg; u = 54 km h^{1} = 15 m s^{1}^{
}Therefore, impulse = 2×0.15 x 15 x cos22.5°
= 2 x 0.15 x 15 x 0.9239
= 4.16 kg m s^{1}
Question 21.
A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5m 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?
Answer:
Question 22.
If, in question number 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?
Answer:
Option (b) correctly describes the trajectory of the stone after the string breaks i.e. the stone flies off tangentially from the instant the string breaks.
The velocity always acts tangentially to the circle at each point in the circular motion. At the time, the string breaks, the particle continues to move in the tangential direction according to Newton’s first law of motion.
Question 23.
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.
Answer:
(a) While trying to pull a cart, a horse pushes the ground backwards with a certain force at an angle. The ground offers an equal reaction in opposite direction, on the feet of the horse. The forward component of this reaction is responsible for the motion of the cart.In empty space, there is no reaction and hence horse cannot pull the cart and run.
(b) This is due to inertia of motion possessed by the passengers in a speeding bus.
(c) While pulling a lawn mower, force is applied upwards along the handle. The vertical component of this force is upwards and reduces the effective weight of the mower as shown in figure
 While pushing a lawn mower, force is applied downwards along the handle. The vertical component of this force is downwards and increases the effective weight of the mower, as shown in figure
 As the effective weight is lesser in case of pulling than in case of pushing, therefore, pulling is easier than pushing.
(d) While holding a catch, the impulse received by the hands, F x t = change in momentum of the ball is constant. By moving his hands backwards, the cricketer increases the time of impact (t) to complete the catch. As t increases, F decreases and as a result, the ball hurts his hands lesser.
Question 24.
Figure shows the positiontime 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?
Answer:
Here, m = 0.04 kg. Position time graph shows that the particle moves from x = 0 at O to x = 2 cm at A in 2 s. As xt graph is a straight line, the motion is with a constant velocity
We can visualise a ball moving between two walls located at x = 0 and x = 2 cm, getting rebounded repeatedly on striking against each wall. On every collision with a wall, linear momentum of the ball changes. Therefore, the ball receives impulse after every two seconds.
Magnitude of impulse = total change in linear momentum
= mu – mυ = m(u – υ )
= 0.04 [10^{2} – (10^{2})]
= 0.04 (10^{2} + 10^{2}) = 0.08 x 10^{2
}= 8 x 10^{4} kg m s^{_1}
Question 25.
Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s^{2}. What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg)
Answer:
Here, acceleration of conveyor belt, a = 1 m s^{2
}As the man is standing stationary w.r.t. the belt, so
acceleration of the man = acceleration of belt a = 1 m s^{2
}As m = 65 kg
Net force on the man, F = ma = 65 x 1 = 65 N
Now μ = 0.2
Force of limiting friction;f= μR = μmg
If the man remains stationary upto maximum
acceleration a’ of the belt, then
f= ma’ =μ mg
a’ = mg = 0.2 x 10 = 2 m s^{2
}
Question 26.
A stone of mass m tied to the end of a string revolves in a vertical circle of radius The net force at the lowest and highest points of the circle directed vertically downwards are :
[Choose the correct alternative]
T_{1 }and v_{1} denote the tension and speed at the lowest point. T_{2} and v_{2} denote corresponding values at the highest point.
Answer:
In the figure shown here L and H shows the lowest and highest points respectively.
At point L: T_{1} acts towards the center of the circle and mg acts vertically downward.
.’. Net force on the stone at the lowest point in the downward direction = mg – T,
At point H : Both T_{2} and mg act vertically downward towards the center of the vertical circle.
.’. Net force on the stone at the highest point in the downward direction = T, + mg
So option (a) is the correct alternative.
Question 27.
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s^{2}. 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.
Answer:
Here, mass of the helicopter, M = 1000 kg
Mass of the crew and the passengers, m = 300 kg
Acceleration, a = 15 m s^{2} (vertically upwards) g = 10 m s^{2}
(a) Force on the floor by the crew and the passengers will be equal to their apparent weight. If the helicopter is rising up with an acceleration a, then the apparent weight and hence the required force,
F = m(g + a)
= 300(10 +15)
= 7500 N (vertically downwards)
(b) Action of the rotor of the helicopter on the surrounding air,
F = (M + m) (g + a)
= (1000 + 300) (10 +15) = 1300 x 25
= 32500 N (vertically downwards)
(c) Force on the helicopter due to surrounding air will be equal and opposite to the action of the rotor of the helicopter on the surrounding air (third law of motion). Therefore, the required force is given by
F = 32500 N (vertically upwards)
Question 28.
A stream of water flowing horizontally with a speed of 15 m s^{1} gushes out of a tube of crosssectional area 10^{2} m^{2}, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
Answer:
Here, υ = 15 m s^{1
}Area of cross section, A = 10^{2} m^{2}, F = ?
Volume of water pushing out per second
= A x υ = 10^{2} x 15 m^{3} s^{1
}As density of water is 10^{3} kgm^{3} therefore, mass of water striking the wall per second is
m = (15 x 10^{2}) x 10^{3} = 150 kg s^{1
}
Question 29.
Ten onerupee 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 7^{th} coin (counted from the bottom) due to all the coins on its top,
(b) the force on the 7^{th} coin by the eighth coin,
(c) the reaction of the 6^{th} coin on the 7^{th
}Answer:
Mass of each coin = m
(a) If F_{7} be the force on 7^{th} coin (counted from the bottom) experienced due to all coins above it, then
F_{7} = weight of three coins above it = 3 mg N (downwards)
(b) F_{87 }= force on 7^{th} coin by 8^{th} coin, then the 8^{th} coin has to support the weight of the two coins above it. So, the 8^{th} coin shall exert the force T_{87 }such that F_{87} = weight of 8^{th} coin + weight of two coins above the 8^{th} coin = mg + 2mg = 3mg (N) and it acts downwards.
(c) The sixth coin experiences force equal to weight of the four coins above it. Hence reaction due to 6^{th} coin on 7^{th} coin = 4mg N and it acts vertically upwards.
Question 30.
An aircraft executes a horizontal loop at a speed of 720 km h^{1} with its wings banked at 15°. What is the radius of the loop?
Answer:
Question 31.
A train runs along an unbanked circular track of radius 30 m at a speed of 54 km h^{1}. The mass of the train is 10^{6} 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?
Answer:
Radius of circular bend, r = 30 m,
speed of train = v = 54 km h^{1
}\(=54X\quad \cfrac { 5 }{ 18 } =15ms\)–^{1
}mass of train, m = 10^{6} kg
angle of banking = θ= ?
The centripetal force is provided by the lateral thrust by the rails on the wheels. According to Newton’s third law of motion, the train exerts an equal and opposite thrust on the rails causing its wear and tear.
The angle of banking is given by
Question 32.
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in figure. 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?
Answer:
Here, mass of block, m = 25 kg
Mass of man, M = 50 kg
Force applied to lift the block,
F = mg = 25 x 10 = 250 N.
Weight of man, W = Mg = 50 x 10 = 500 N
1. When block is raised by man as shown in figure (a), force is applied by the man in the upward direction. Action on the floor by the man = WF = 500 + 250 = 750 N
2. When block is raised by man as shown in figure (b), force is applied by the man in the downward direction.
Action on the floor by the man
= WF = 500 – 250 = 250N
As the floor yields to a normal force of 700 N, the mode (b) has to adopted by the man to lift the block.
Question 33.
A monkey of mass 40 kg climbs on a rope as shown in the figure 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 s^{2
}(b) climbs down with an acceleration of 4 ms^{2}^{
}(c) climbs up with a uniform speed of 5 m s^{1}^{
}(d) falls down the rope nearly freely under gravity? (Ignore the mass of the rope).
Answer:
Here, mass of monkey m = 40 kg
Maximum tension the rope can stand, T = 600 N. In each case, actual tension in the rope will be equal to apparent weight of monkey (R).
The rope will break when R exceeds T.
(a) When monkey climbs up with o = 6m s^{2}, R = m(g + a) = 40(10 + 6) = 640 N (which is greater than T) Hence the rope will break.
(b) When monkey climbs down with a = 4 m s^{2}, R = m(ga) = 40(10 – 4) = 240 N (which is less than T) :. The rope will not break.
(c) When monkey climbs up with a uniform speed v = m s^{1}, its acceleration a = 0
:. R = mg = 40 x 10 = 400 N (which is less than T)
:. The rope will not break.
(d) When monkey falls down the rope nearly freely under gravity, a = g
:. R = m(g a) = m(g g) = zero
Hence the rope will not break.
Question 34.
Two bodies A and 8 of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall as shown in figure.
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 actionreaction 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 μ_{s} and μ_{k}.
Answer:
Here, mass of body A, m, = 5 kg
mass of body B, m_{2} = 10 kg
Coefficient of friction between the bodies and the table, m = 0.15.
Horizontal force applied on body A, P = 200 N
(a) Reaction of partition = ?
Let f = force of limiting friction acting to the left, then
f= μR = μ(m_{1} + m_{2})g [∴R = (m, + m_{2})g]
= 0.15 (5 + 10) x 10 = 22.5 N
∴ According to Newton’s 3^{rd} law of motion,
Reaction of B on A = 192.5 N towards left.
Note : If we assume perfect contact between bodies A and B and the rigid partition, then the self adjusting normal force on B by the partition (reaction) equals the applied force i.e. 200 N. There is no impending motion and no friction. The actionreaction forces between A and B are also 200 N.
When the partition is removed : When the partition is removed, the kinetic friction comes into play, the masses move together as a system of two bodies under the action of net force F given by
Does the answer to (b) change if m_{1} and m_{2} are in motion?
Yes, when the bodies are in motion, then the answer to (b) changes and can be proved as follows :
When the bodies are moving, the force exerted by A on B is given by
P_{BA} = F – force required to produce an acceleration of 11.83 m s^{2} in body A alone
= P f_{1} – m_{1}a = 200 – 7.5 – 5 x 11.83
= 200 – 7.5 – 59.30 = 192.5 – 59.15 = 133.35 N
Action of A on B when partition is removed = 133.35 N
reaction of B on A, when partition is removed = 133.35 N to the left.
Question 35.
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 m s^{2} 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.
Answer:
Here, m = 15 kg; a = 0.5 m s^{2}, t = 20 s and m = 0.18
Force on the block due to the motion of the trolley,
F = ma = 15 x 0.5 = 7.5 N Force of limiting friction on the block,
f’ = μR = μmg = 0.18 x 15 x 10 = 27 N
(a) To a stationary observer op the ground, force F on the block acts so as to cause the motion and the force f opposes the motion of the block. Since f> F, the block will continue to remain stationary. In fact, the force of limiting friction/adjusts itself to be equal to the force F.
(b) The motion of the trolley is an accelerated one. Therefore, the trolley is a noninertial frame of reference and the observer moving with the trolley is in a noninertial frame. The laws of mechanics are no longer valid in such a frame.
Question 36.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m from the open end as shown in figure. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s^{2}. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).
Answer:
Here, mass of the box, m = 40 kg; acceleration of the truck, a = 2 m s^{2};
distance of the box from the open end, S = 5 m;
and the coefficient of friction between the box and the surface below it, p = 0.15 As the truck moves in forward direction with the acceleration a = 2 m s^{2}, the box experiences a force F in the opposite (backward) direction given by
F = ma = 40 x 2 = 80 N
Under the action of this force, the box will tend to move to the open side of the truck. As it does so, its motion will be opposed by the force of friction. The limiting friction acting between the box and the truck,
f = μmg = 0.15 x 40 x 10 = 60 N
The net force on the box in the backward direction,
Question 37.
A disc revolves with a speed of 33 \( \frac {1}{3} \) rev / min, and has a radius of 15 cm.Two coins are placed at 4 cm and 14 cm away from the center of the record. If the coefficient of friction between the coins and the records is 0.15, which of the coins will revolve with the record?
Answer:
The coin revolves with the record in the case when the force of friction is enough to provide the necessary centripetal force. If this force is not sufficient to provide centripetal force, the coin slips on the record. Now, the frictional force μR where R is the normal reaction, and R = mg
Hence force of friction = μmg and centripetal force required is mυ^{2}/r or mrω^{2} μω are same for both the coins and we have different values of r for the two coins. So to prevent slipping i.e. causing coins to rotate
Here, μg > rω^{2} is not satisfied, so this coin will not revolve with record.
Note : We have nothing to do with the radius of the record (= 15 cm).
Question 38.
You may have seen in a circus a motorcyclist driving in a vertical loops inside a ‘deathwell'(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 ^{1} chamber is 25 m?
Answer:
When the motorcyclist is at the highest point of the death well, the normal reaction R on him by the ceiling of the chamber acts downwards. His weight mg also acts downwards. These two forces are balanced by the outward centrifugal force acting on him.
where v = speed of the motorcyclist, m = mass of the motorcyclist (mass of motorcycle + driver)Because of the balancing of two forces, the motorcyclist does not fall down.The minimum speed required to perform a vertical loop is given by equation (i) when R = 0
Question 39.
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?
Answer:
The cylinder being vertical, the normal reaction of the wall on the man acts horizontally and provides the necessary centripetal force
Question 40.
A thin circular loop of radius R rotates about its vertical diameter with an angular frequency co. 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 center to the bead with the vertical downward direction for ω=√2g/r ?
Neglected friction.
Answer:
We have shown that radius vector joining the bead to the center of the wire makes an angle 0 with the vertical downward direction.If N is normal reaction, then as is clear from the figure
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