Laws of Motion
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5.31 A train runs along an unbanked circular track of radius 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 ?
5.31 A train runs along an unbanked circular track of radius 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 ?
Given
Radius, r = 30m
Speed, v = 54 km/h = 15 m/s
The mass of the train, m = 106 kg
(a) The required centripetal force is provided by the rails to the wheels of the train
(b) The angle of banking required to prevent wearing out of the rails
= / rg = / (30
= 37 
5.30 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 ?
5.30 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 ?
The speed of the aircraft, v = 720 km/h = 200 m/s
The angle of banking = 15
From the relation = /rg we get r = / ( g = 14928 m
5.29 Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of
(a) The force on the 7th coin (counted from the bottom) due to all the coins on its top
(b) The force on the 7th coin by the eighth coin
(c) The reaction of the 6th coin on the 7th coin
5.29 Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of
(a) The force on the 7th coin (counted from the bottom) due to all the coins on its top
(b) The force on the 7th coin by the eighth coin
(c) The reaction of the 6th coin on the 7th coin
(a) The force on the 7th coin is due to the 3 coins kept above it The weight of the 3 coins = 3m Force exerted on the 7th coin is (3mg) N and the force is acting vertically downwards (b) The force on the 7th coin by the 8th coin will be the force exerted by the 3 coins above it = (3mg) N, acting downwards (c) The 6th coin will experience the force of 4 coins above it, acting downwards = 4mg |
5.28 A stream of water flowing horizontally with a speed of 15 m s-1 gushes out of a tube of cross-sectional area 10-2 m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?
5.28 A stream of water flowing horizontally with a speed of 15 m s-1 gushes out of a tube of cross-sectional area 10-2 m2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound ?
Speed of water, v = 15 m/s
Cross-sectional area of the tube, A = 10-2 m2
Density of water = 1000 kg/m3
Mass of the water hitting the wall = = 1000 10-2 = 150 kg/s
Force exerted on the wall because of impact of water = mass velocity = 150 N = 2250 N
5.27 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
5.27 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
Given
Mass of the helicopter m1 = 1000 kg
Mass of the crew and passenger m2 = 300 kg
The vertical acceleration of the helicopter, a = 15 m/s2 and acceleration due to gravity g = 10 m/s2
The total mass of the system, m = m1 + m2 = 1000 + 300 = 1300 kg
(a) Force on the floor by the crew and passengers :
R
5.26 A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : [Choose the correct alternative]
Lowest Point Highest Point
(a) mg – T1 mg + T2
(b) mg + T1 mg – T2
(c) mg +T1 – (m ) / R mg – T2+ (m ) / R
(d) mg – T1 – (m ) / R mg + T2 + (m ) / R
T1 and v1 denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point.
5.26 A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are : [Choose the correct alternative]
Lowest Point Highest Point
(a) mg – T1 mg + T2
(b) mg + T1 mg – T2
(c) mg +T1 – (m ) / R mg – T2+ (m ) / R
(d) mg – T1 – (m ) / R mg + T2 + (m ) / R
T1 and v1 denote the tension and speed at the lowest point. T2 and v2 denote corresponding values at the highest point.
The net force at the lowest point is denoted by (mg – T1)and the net force at the highest point is denoted by (mg + T2 ), hence option (a) is correct. The forces mg and T1 are in mutually opposite direction at the lowest point and mg and T2 are in the same direction at the highest point.
5.25 Figure 5.18 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.)

5.25 Figure 5.18 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.)

The acceleration of the conveyer belt, a = 1 m/s2
Coefficient of static friction, = 0.2
Mass of the man, m = 65 kg
The net force experienced by the man, MA = 1 N = 65 N
This net force is due to the friction between the belt and the man
At maximum static friction
= =
5.24 Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body ? What is the magnitude of each impulse?

5.24 Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body ? What is the magnitude of each impulse?

This graph could be of a ball rebounding between two walls separated by a distance of 2 cm.
The ball rebounds every 2 secs between the walls with uniform velocity.
Velocity of rebounding = displacement / time = ( 2 )/2 = 0.01 m/s
Initial momentum = mu = 0.04 4 kgm/s
Fina
5.23 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
5.23 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
(a) A horse can pull a cart by the reaction force generated by the ground to its feet. In an empty space, no reaction force will be generated and it cannot pull a cart.
(b) During the movement of the bus, passenger's whole body experiences the inertia of motion of the bus. When a bus brakes,
5.22 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 ?
5.22 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 ?
(a) & (c) solution not possible as the stone will fly off tangentially. The answer is (b)