At Saralstudy, we are providing you with the solution of Class 11 Physics Laws of Motion according to the latest NCERT (CBSE) Book guidelines prepared by expert teachers. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. We are trying our best to give you detailed answers to all the questions of all the topics of Class 11 Physics Laws of Motion so that you can prepare for the exam according to your own pace and your speed.
(a) Zero net force
The rain drop is falling with a constant speed. Hence, it acceleration is zero. As per
(b) Zero net force
The weight of the cork is acting downward. It is balanced by the buoyant force exerted by the water in the upward direction. Hence, no net force is acting on the floating cork.
(c) Zero net force
The kite is stationary in the sky, i.e., it is not moving at all. Hence, as per
(d) Zero net force
The car is moving on a rough road with a constant velocity. Hence, its acceleration is zero. As per
(e) Zero net force
The high speed electron is free from the influence of all fields. Hence, no net force is acting on the electron.
0.5 N, in vertically downward direction, in all cases
Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational 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 x a
Where,
F = Net force
m= Mass of the pebble = 0.05 kg
a = g = 10 m/s2
∴F = 0.05 x 10 = 0.5 N
The net force on the pebble in all three cases is 0.5 N and this force acts in the downward direction.
If the pebble is thrown at an angle of 45° with the horizontal, it will have both the horizontal and vertical components of velocity. At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble.
(a)1 N; vertically downward
Mass of the stone, m = 0.1 kg
Acceleration of the stone, a = g = 10 m/s2
As per
F = ma = mg
= 0.1 × 10 = 1 N
Acceleration due to gravity always acts in the downward direction.
(b)1 N; vertically downward
The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the 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)1 N; vertically downward
It is given that the train is accelerating at the rate of 1 m/s2.
Therefore, the net force acting on the stone, F' = ma = 0.1 × 1 = 0.1 N
This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F,' stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations.
Therefore, the net force acting on the stone is given only by acceleration due to gravity.
F = mg = 1 N
This force acts vertically downward.
(d)0.1 N; in the direction of motion of the train
The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train.
Acceleration of the train, a = 0.1 m/s2
The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by:
F = ma
= 0.1 × 1 = 0.1 N
(i)
When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e.,
F = T = mv2 / l
Where F is the net force acting on the particle.
Retarding force, F = -50 N
Mass of the body, m= 20 kg
Initial velocity of the body, u= 15 m/s
Final velocity of the body, v= 0
UsingNewton's second law of motion, the acceleration (a) produced in the body can be calculated as:
F= ma
-50 = 20 × a
∴ a = -50 / 20 = -2.5 m/s2
Usingthe first equation of motion, the time (t) taken by the body to come to rest can be calculated as:
v= u + at
∴ t = -u /a = -15 / -2.5 = 6s
0.18 N; in the direction of motion of the body
Mass of the body, m= 3 kg
Initial speed of the body, u= 2 m/s
Final speed of the body, v= 3.5 m/s
Time, t = 25 s
Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:
v= u + at
∴ a = v-u / t
= (3.5 - 2) / 25 = 1.5 / 25 = 0.06 m/s2
As per Newton's second law of motion, force is given as:
F= ma
= 3 × 0.06 = 0.18 N
Since the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion.
Initial speed of the three-wheeler, u = 36 km/h = 10 m/s
Final speed of the three-wheeler, v = 0 m/s
Time, t = 4 s Mass of the three-wheeler, m = 400 kg
Mass of the driver, m' = 65 kg
Total mass of the system, M = 400 + 65 = 465 kg
Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as:
v = u + at
∴ a = v - u / t = 0-10/4 = -2.5 m/s2
The negative sign indicates that the velocity of the three-wheeler is decreasing with time.
Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as:
F = Ma = 465 × (–2.5)
= –1162.5 N
The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.
Mass of the rocket, m = 20,000 kg
Initial acceleration, a = 5 m/s2
Acceleration due to gravity, g = 10 m/s2
Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation:
F – mg = ma
F = m (g + a)
= 20000 × (10 + 5)
= 20000 × 15 = 3 × 105 N
(a) Vertically downward
(b) Parabolic path
(a) At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.
(b) At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.
Mass of each ball = 0.05 kg
Initial velocity of each ball = 6 m/s
Magnitude of the initial momentum of each ball, pi= 0.3 kg m/s
After collision, the balls change their directions of motion without changing the magnitudes of their velocity.
Final momentum of each ball, pf= -0.3 kg m/s
Impulse imparted to each ball = Change in the momentum of the system
= pf- pi = -0.3 - 0.3 = -0.6 kg m/s
The negative sign indicates that the impulses imparted to the balls are opposite in direction.
Mass of the gun, M= 100 kg
Mass of the shell, m= 0.020 kg
Muzzle speed of the shell, v = 80 m/s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = 0
Final momentum of the system = mv - MV
Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
According to the law of conservation of momentum:
Final momentum = Initial momentum
mv - MV= 0
∴ V = mv / M
= 0.020 x 80 / 100 x 1000
= 0.016 m/s
(b)
When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.
(a) Force on the seventh coin 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 by the three coins on its top is 3mg. This force acts vertically 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 by the eighth coin is 3mg. This force acts vertically downward.
(c) The 6th coin experiences a downward force because of the weight of the four coins (7th, 8th, 9th, and 10th) on its top.
Therefore, the total downward force experienced by the 6th coin is 4mg.
As per Newton's third law of motion, the 6th coin will produce an equal reaction force on the 7th coin, but in the opposite direction. Hence, the reaction force of the 6th coin on the 7th coin is of magnitude 4mg. This force acts in the upward direction.
Speed of the aircraft, v = 720 km/h = 720 x 5/18 =200 m/s
Acceleration due to gravity, g = 10 m/s2
Angle of banking, θ = 15°
For radius r, of the loop, we have the relation:
tanθ = v2 / rg
r = v2 / gtanθ
= 200 x 200 / 10 tan15
= 4000 / 0.268
= 14925.37 m
= 14.92 km
Radius of the circular track, r = 30 m
Speed of the train, v = 54 km/h = 15 m/s
Mass of the train, m= 106 kg
The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton's third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail.
The angle of banking θ, is related to the radius (r) and speed (v) by the relation:
tanθ = v2 / rg
= 152 / 30 x 10 = 225 / 300
θ = tan-1 (0.75) = 36.87°
Therefore,the angle of banking is about 36.87°.
750 N and 250 N in the respective cases; Method (b)
Mass of the block, m = 25 kg
Mass of the man, M = 50 kg
Acceleration due to gravity, g = 10 m/s2
Force applied on the block, F = 25 x 10 = 250 N
Weight of the man, W = 50 x 10 = 500 N
Case (a): When the man lifts the block directly
In this case, the man applies a force in the upward direction. This increases his apparent weight.
∴Action on the floor by the man = 250 + 500 = 750 N
Case (b): When the man lifts the block using a pulley
In this case, the man applies a force in the downward direction. This decreases his apparent weight.
∴Action on the floor by the man = 500 - 250 = 250 N
If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.
Case (a)
Mass of the monkey, m= 40 kg
Acceleration due to gravity, g= 10 m/s
Maximum tension that the rope can bear, Tmax= 600 N
Acceleration of the monkey, a= 6 m/s2upward
UsingNewton's second law of motion, we can write the equation of motion as:
T- mg= ma
∴T= m(g+ a)
= 40 (10 + 6)
= 640 N
Since T > Tmax, the rope will break in this case.
Case (b)
Acceleration of the monkey, a= 4 m/s2downward
UsingNewton's second law of motion, we can write the equation of motion as:
mg - T = ma
∴T= m (g- a)
= 40(10 - 4)
= 240 N
Since T < Tmax, the rope will not break in this case.
Case (c)
The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a= 0.
UsingNewton's second law of motion, we can write the equation of motion as:
T- mg = ma
T- mg = 0
∴T = mg
= 40 x 10
= 400 N
Since T < Tmax, the rope will not break in this case.
Case (d)
When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., a = g
UsingNewton's second law of motion, we can write the equation of motion as:
mg - T= mg
∴T= m(g- g) = 0
Since T < Tmax, the rope will not break in this case.
(a) Mass of the block, m = 15 kg
Coefficient of static friction, μ = 0.18
Acceleration of the trolley, a = 0.5 m/s2
As per Newton's second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation:
F = ma = 15 x 0.5 = 7.5 N
This force is acted in the direction of motion of the trolley.
Force of static friction between the block and the trolley:
f = μmg = 0.18 x 15 x 10 = 27 N
The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest.
When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.
(b) An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.