NCERT SOLUTION OF CLASS 11TH PHYSICS LAW OF MOTION

QUESTIONS FROM TEXTBOOK ( law of motion )

Question 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 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.
Answer: (a) As the drop of rain is falling with constant speed, in accordance with first law of motion, the net force on the drop of rain is zero.
(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.
(c) Net force on a kite skilfully held stationary in sky is zero because it is at rest.
(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.

Question 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 1 Ignore air resistance.
Answer: (a) When the pebble is moving upward, the acceleration g is acting downward, so the force is acting downward is equal to F = mg = 0.05 kg x 10 ms^-2 = 0.5 N.
(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.

Question 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 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.
Answer: (a) Mass of stone = 0.1 kg
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.

Question 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 – mv²/l, (iii) T +mv²/l, (iv) 0
T is the tension in the string. [Choose the correct alternative].
Answer: (i) T
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.

Question 5. 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^-3. How long does the body take to stop?
Answer:  Here m = 20 kg, F = – 50 N (retardation force)
As F = ma

Question 5. 9. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 ms^-2. Calculate the initial thrust (force) of the blast.
Answer: Here, m = 20000 kg = 2 x 10⁴ kg 
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 10⁴ x 14.8 = 2.96 x 10⁵ N.

Question 5. 10. A body of mass 0.40 kg moving initially with a constant speed of 10 ms^-1 to the north is subject 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 5. 11.  A truck starts from rest and accelerates uniformly at 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11s? (Neglect air resistance.)
Answer: u = 0, a = 2 ms^-2, t 10 s
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

Answer:  Let the bob be oscillating as shown in the figure.
(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^-1 and on cutting the string it will experience an acceleration a = g = 10 ms^-2 in vertical downward direction. Consequently, the bob will behave like a projectile and will fall on ground after describing a parabolic path.

Question 5. 13. 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^-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/s²
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 = 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.

Question 5. 14. 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).

Question 5. 18. Two billiard balls, each of mass 0.05 kg, moving in opposite directions with speed 6 ms^-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?
Answer: Initial momentum of each ball before collision
= 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 kg ms^-1 Impulse 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.

Answer: Here acceleration of conveyor belt a = 1 ms^-2, μ_s= 0.2 and mass of man m = 65 kg. t As the man is in an accelerating frame, he experiences a pseudo force F_s = ma as shown
in fig. (a). Hence to maintain his equilibrium, he exerts a force F = – F_s = 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

Question 5. 27. A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms^-2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of
(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,
Answer: Here, mass of helicopter, m₁= 1000 kg
Mass of the crew and passengers, m₂ = 300 kg upward acceleration, a = 15 ms^-2 and 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.

5.28.A stream of water flowing horizontally with a speed of 15 ms^-1 pushes out of a tube of cross sectional area 10^-2 m², and hits at a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming that it does not rebound?
Ans.In one second, the distance travelled is equal to the velocity v.
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^-2 m² x 15 ms^-1 = 15 x 10^-2 m³ s^-1
Mass of water hitting the wall per second
= 15 x 10^-2 x 10³ kg 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^-1 x 15 ms^-1 = 2250 kg ms^-2 or 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.

Question 5. 29. Ten one rupee coins are put on top of one another on a table. Each coin has a mass 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 and
(c) the reaction of the sixth coin on the seventh coin.
Answer: (a) The force on 7th coin is due to weight of the three coins lying above it. Therefore,
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.

Answer: In 1st case, man applies an upward force of 25 kg wt.r (same as the weight of the block). According to Newton’s third law of motion, there will be a downward reaction on the floor.
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.

Question 5. 33. 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: the monkey
(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).

Question 5. 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 ms^-1 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:  (a) Force experienced by block, F = ma = 15 x 0.5 = 7.5 N Force of friction,F_f= p mg = 0.18 x 15 x 10 = 27 N. i.e., force experienced by block will be less than the friction.So the block will not move. It will remain stationary w.r.t. trolley for a stationary observer on ground.
(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.

Question 5. 36. 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^-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).