NCERT Solution (Class 11) > Physics Chapter 5 Laws of Motion
NCERT Solution Class 11 Physics Chapter 5 Laws of Motion: National Council of Educational Research and Training Class 11 Physics Chapter 5 Solution – Laws of Motion. Free PDF Download facility available at our website.
Board |
NCERT |
Class |
11 |
Subject |
Physics |
Chapter |
5 |
Chapter Name |
Laws of Motion |
Topic |
Exercise Solution |
Laws of Motion Chapter all Questions and Numericals Solution
5.1) 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 on a rough road,
(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
ANSWER-
- The rain drop is falling with a constant speed. Hence, it acceleration is zero. According to Newton’s second law of motion, the net force acting on the rain drop is zero if acceleration is zero.
- Out of total forces acting on cork one is The weight of the cork which is acting downward and other is the buoyant force exerted by the water in the upward direction. Both balances each other hence net force will be zero.
- The kite is not moving . Hence according to Newton’s first law of motion, no net force is acting on the kite.
- The car is moving on a rough road with a constant velocity. Hence, its acceleration zero. According to Newton’s second law of motion, no net force is acting on the car.
- The high speed electron is far from all material objects, and free of electric and magnetic fields. Hence, no net force is acting on the electron.
5.2) 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 direction ? Ignore air resistance.
ANSWER-
Acceleration due to gravity always acts downward. The gravitation force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as:
F = m × a
Where,
F = Net force
m = Mass of the pebble = 0.05 kg
a = acceleration due to gravity= 10 m/s2
∴ F = 0.05 × 10 = 0.5 N
The net force on the pebble is 0.5 N in the downward direction.
5.3) 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,
(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)a stone of mass 0.1 kg just after it is dropped from the window of a stationary train
Mass of the stone, m = 0.1 kg
Acceleration of the stone, a = 10 m/s2
As per Newton’s second law of motion, the net force acting on the stone,
F = ma
F = 0.1 × 10 = 1 N
This acts in vertically downward direction.
(b)The train is moving with a constant velocityof 36 km/h. Hence, its acceleration is zero in the horizontal directions velocity is in horizontal direction. Hence, no force is acting on the
stone in the horizontal direction.
The net force acting on the stone is because of acceleration due to gravity and it always
acts vertically downward. The magnitude of this force is 1 N.
(c)as soon as stone is dropped from the accelerating train it is acted by only one force which is gravitational force hence net force is 1N vertically downward.
(d)) as stone is lying on the floor of a train ,The only acceleration is in horizontal direction due to acceleration of train a = 0.1 m/s.
The net force acting on the stone will be in the direction of motion of the train and given by:
F = ma
F = 0.1 × 1 = 0.1 N
5.4) 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 centre) is
(i) T
(II) T- m v2/l
(iii) T- m v2/l
(iv) 0
ANSWER-
When a particle connected to a string revolves in a circular path around a centre, required centripetal force is given by the tension in the string. Hence the net force on the particle is the tension T only. Hence option (i).
5.5)
5.7)
5.8)
5.10)
5.14)
5.15)
5.16)
5.19)
5.20)
5.21)
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5.22)
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.
ANSWER-
(a)To pull a cart, a horse pushes the ground backward with some force. The groundin turn exerts an equal and opposite reaction force on the feet of the horse. this reaction force causes the horse to move forward but in case of An empty space reaction force will be absent. Therefore, a horse cannot pull a cart and run in empty space.
(b)When bus is moving passengers are having speed of bus but When a speeding bus stops suddenly, the lower portion of a passengers body, which is in contact with the seat, suddenly comes to rest. On the other hand the upper portion remains in motion. Hence the passengers’ upper body is thrown forward in the direction in which the bus was moving.
(c)While pulling a lawn mower, the vertical component of this applied force acts upward. This reduces the effective weight of the mower. on the other hand when pushing the lawn mover one component of force is in downward direction which in turn increases the weight of lawn mover hence it is easier to pull a lawn mower than to push it.
(d) According to Newton’s second law of motion, we have the equation of motion
F = ma = m x dv/dt
Where,
F = force experienced by the cricketer
m = Mass of the ball
dt = Time of impact of the ball with the hand
As we can see force is inversely proportional to time of impact so if time of impact increases which decreases the force experienced by the cricketer hence to get that relief a cricketer moves his hands backwards while holding a catch.
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 v2/R)mg – T2 + (m v2 / R)
(d) mg- T1 – (m v2/R)mg + T2 + (m v2 / R)
ANSWER-
At lowest point weight mg acting downward while tension in the string acts upward towards the centre of the circle. So net force is T1 – mg.
At highest point, weight mg acting downward and tension in the string also acts downward towards the centre hence the net force is downward and given by addition. Hence net force = T2 + mg.
5.27)
5.28)
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.
ANSWER-
(a)Force on the seventh coin counted from the bottom is exerted by the weight of the three coins on its top.
Weight of one coin = mg
Weight of three coins = 3mg
Hence, the force exerted on the 7th coin is 3mg downward.
(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin
and the other two coins (ninth and tenth) on its top.
Weight of the eighth coin = mg
Weight of the ninth coin = mg
Weight of the tenth coin = mg
Total weight of these three coins = 3mg
Hence, the force exerted on the 7th coin due to 8th coin is 3mg downward.
(c)The 6th coin experiences a downward force because of 7th ,8th, 9th, and 10thon its top.
Therefore, the total downward force on the 6th coin is 4mg . so 6th coin will exert equal and opposite reaction on 7th which is of magnitude 4mg and acting upward.
5.30)
5.32) A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. 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-
Mass of the block, m1 = 25 kg
Mass of the man, m2 = 50 kg
Weight of the man, W = 50 × 10 = 500 N
Acceleration due to gravity, g = 10 m/s
When the man lifts the block directly
Here, the man applies a force in the upward direction directly. This increases the force on the floor.
∴Action on the floor by the man = 250 + 500 = 750 N
When the man lifts the block using a pulley
Here, the man applies a force in the downward direction. This decreases his apparent weight.
∴Action on the floor by the man = 500 N
If the floor can yield to a normal force of 700 N, then the man should use pulley to easily lift the block.
5.33)
5.35)
5.38) You may have seen in a circus a motorcyclist driving in 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 chamber is 25 m?
ANSWER-
In a death-well, when a motorcyclist is at uppermost point with no support from below, the centripetal force due to circular motion is balancing the weight of motorcyclist and normal reaction hence motorcyclist does not drop down.
Let FN be the normal reaction at top and v be the velocity of motor cyclist then for balancing condition;
Normal reaction + weight of motorcyclist = centripetal force
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