# Chapter 5 – Laws of motion Questions and Answers: NCERT Solutions for Class 11 Physics

Class 11 Physics NCERT book solutions for Chapter 5 - Laws of motion Questions and Answers.

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Class 11 Physics NCERT book solutions for Chapter 5 - Laws of motion Questions and Answers.

(a) a drop of rain falling down with a constant speed,

(b) a cork of mass 10 g floating on water,

(c) a kite skilfully 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.

(b) As the cork is floating on water, its weight is being balanced by the upthrust (equal to.weight of water displaced). Hence net force on the cork is zero.

(d) Since car is moving with a constant velocity, the net force on the car is zero.

(e) Since electron is far away from all material agencies producing electromagnetic and gravitational forces, the net force on electron is zero.

(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 1 Ignore air resistance.

(b) In this case also F = mg = 0.05 x 10 = 0.5 N. (downwards).

(c) The pebble is not at rest at highest point but has horizontal component of velocity. The direction and magnitude of the net force on the pebble will not alter even if it is thrown at 45° because no other acceleration except ‘g’ is acting on pebble.

(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 ms-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.

Net force, F = mg = 0.1 x 10 = 1.0 N. (vertically downwards).

(b) When the train is running at a constant velocity, its acceleration is zero. No force acts on the stone due to this motion. Therefore, the force on the stone is the same (1.0 N.).

(c) The stone will experience an additional force F’ (along horizontal) i.e.,F = ma = 0.1 x l = 0.1 N

As the stone is dropped, the force F’ no longer acts and the net force acting on the stone F = mg = 0.1 x 10 = 1.0 N. (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 the stone, F = ma = 0.1 x 1 = 0.1 N.

It acts along the direction of motion of the train.

(i) T, (ii) T – mv2/l, (iii) T +mv2/l, (iv) 0

T is the tension in the string. [Choose the correct alternative].

The net force T on the particle is directed towards the centre. It provides the centripetal force required by the particle to move along a circle.

Answer:Here m = 20 kg, F = – 50 N (retardation force)

As F = ma

u = 36 km/h = 36 x — m/s 10 ms-1, final speed, v – 0 and t = 4s.

Initial acceleration = 5 ms-2

Clearly, the thrust should be such that it overcomes the force of gravity besides giving it an upward acceleration of 5 ms-2.

Thus the force should produce a net acceleration of 9.8 + 5.0 = 14.8 ms-2.

Since, thrust = force = mass x acceleration

F = 2 x 104x 14.8 = 2.96 x 105N.

Using equation, v = u + at, we get

v = 0 + 2 x 10 = 20 ms-1

(a) Let us first consider horizontal motion. The only force acting on the stone is force of gravity which acts vertically downwards.

Its horizontal component is zero. Moreover, air resistance is to be neglected. So, horizontal motion is uniform motion.

.-. vx = v = 20 ms-1

Let us now consider vertical motion which is controlled by force of gravity.

u=0, a = g = 10 ms-2, t = (11 — 10) s = 1 s

The speed of the bob at its mean position is 1 ms-1. What is the trajectory of the bob if the stringis cut when the bob is (a) at one of its extreme positions, (b) at oits mean position ?

(a) When the bob is at its extreme position (say B), then its velocity is zero. Hence on cutting the string the bob will fall vertically downward under the force of its weight F = mg.

(b) When the bob is at its mean position (say A), it has a horizontal velocity of v = 1 ms-1and on cutting the string it will experience an acceleration a = g = 10 ms-2in vertical downward direction. Consequently, the bob will behave like a projectile and will fall on ground after describing a parabolic path.

(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?

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 witha = 5 ms-2

R = m (g + a) = 70 (10 + 5) = 1050 N

(d) If the lift were to come down freely under gravity, downward acc. a = g

:. R = m(g -a) = m(g-g) = Zero.

Therefore, its linear momentum before disintegration is zero.

According to the principle of conservation of linear momentum,

= 0.05 x 6 kg ms-1= 0.3 kg ms-1

Final momentum of each ball after collision

= – 0.05 x 6 kg ms-1= – 0.3 kgms-1Impulse imparted to each ball due to the other

= final momentum – initial momentum = 0.3 kg m s-1 – 0.3 kg ms-1

= – 0.6 kg ms-1= 0.6 kg ms-1(in magnitude)

The two impulses are opposite in direction.

(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?

(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.

(b) The passengers in a speeding bus have inertia of motion. When the bus is suddenly stopped the passengers are thrown forward due to this inertia of motion.

(c) In the case of pull, the effective weight is reduced due to the vertical component of the pull. In the case of push, the vertical component increases the effective weight.

(d) The ball comes with large momentum after being hit by the batsman. When the player takes catch it causes large impulse on his palms which may hurt the cricketer. When he moves his hands backward the time of contact of ball and hand is increased so the force is reduced.

in fig. (a). Hence to maintain his equilibrium, he exerts a force F = – Fs= ma = 65 x 1 = 65 N in forward direction i.e., direction of motion of belt.

.’. Net force acting on man = 65 N (forward)

As shown in fig. (b), the man can continue to be stationary with respect to belt, if force of friction

Since mg and T1are in mutually opposite directions at lowest point and mg and T2are in same direction at the highest point.

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on surrounding air,

(c) force on the helicopter due to the surrounding air,

Mass of the crew and passengers, m2= 300 kg upward acceleration, a = 15 ms-2and g = 10 ms-2

(a)Force on the floor of helicopter by the crew and passengers = apparent weight of crew and passengers

= m2(g + a) = 300 (10 + 15) N = 7500 N

(b)Action of rotor of helicopter on surrounding air is obviously vertically downwards, because helicopter rises on account of reaction to this force. Thus, force of action

F = (m1+ m2) (g + a) = (1000 + 300) (10 + 15) = 1300 x 25 = 32500 N

(c)Force on the helicopter due to surrounding air is the reaction. As action and reaction are equal and opposite, therefore, force of reaction, F = 32500 N, vertically upwards.

Volume of water hitting the wall per second, V = av where a is the cross-sectional area of the tube and v is the speed of water coming out of the tube.

V = 10-2m2x 15 ms-1= 15 x10-2m3s-1

Mass of water hitting the wall per second

= 15 x 10-2x 103kg s-1= 150 kg s-1[v density of water = 1000 kg m-3] Initial momentum of water hitting the wall per second

= 150 kg s-1x 15 ms-1= 2250 kg ms-2or 2250 N Final momentum per second = 0 Force exerted by the wall = 0 – 2250 N = – 2250 N Force exerted on the wall = – (- 2250) N = 2250 N.

(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 and

(c) the reaction of the sixth coin on the seventh coin.

F = (3 m) kgf = (3 mg) N

where g is acceleration due to gravity. This force acts vertically downwards.

(b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7th coin due to 8th coin is sum of the two forces i.e.

F = 2 m + m = (3 m) kg f = (3 mg) N The force acts vertically downwards.

(c) The sixth coin is under the weight of four coins above it.

Reaction, R = – F = – 4 m (kg) = – (4 mgf) N Minus sign indicates that the reaction acts vertically upwards, opposite to the weight.

The action on the floor by the man.

= 50 kg wt. + 25 kg wt. = 75 kg wt = 75 kg x 10 m/s2= 750 N.

In case II, the man applies a downward force of 25 kg wt. According to Newton’s third law, the reaction is in the upward direction.

In this case, action on the floor by the man

= 50 kg wt – 25 kg wt. = 25 kg wt. = 25 kg x 10 m/s2= 250 N.

Therefore, the man should adopt the second method.

(a) climbs up with an acceleration of 6 ms-2

(b) climbs down with an acceleration of 4 ms-2

(c) climbs up with a uniform speed of 5 ms-1

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

where T represents the tension (figure a).

T mg = 40 kg x 10 ms-2= 400 N The rope will hot break.

(d) When the monkey is falling freely, it would be a state of weightlessness. So, tension will be zero and the rope will not break.

(a) reaction of the partition = – (force applied on A) = 200 N towards left.

(b) action-reaction forces between A and B are 200 N each. A presses B towards right with an action force 200 N and B exerts a reaction force on A towards left having magnitude 200 N.

(ii) When the wall is removed, motion can take place such that net pushing force provides the acceleration to the block system. Hence, taking kinetic friction into account, we have

(b) The observer moving with trolley has an accelerated motion i.e., he forms non-inertial frame in which Newton’s laws of motion are not applicable. The box will be at rest relative to the observer.

The minimum speed required to perform a vertical loop is given by equation (1) when R = 0.

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