1st PUC Physics Question Bank Chapter 5 Laws of Motion

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Karnataka 1st PUC Physics Question Bank Chapter 5 Laws of Motion

Question 1.
Define Inertia and mention the types.
Answer:
Inertia is the property of the body due to which it resists to change its state of rest or uniform motion, when external force is applied on it.
Types of Inertia :

1. Inertia of rest
2. Inertia of motion .
3. Inertia of direction

Question 2.
State Newton’s first law motion. Give examples.
Answer:
Every body continues in its state of rest or uniform motion along a straight line unless it is compelled to change that state by an external force applied on it. In absence of external force body will be in equilibrium [both static and dynamic equilibrium].
Examples.

• A book on the table continues to be at rest.
• A passenger in moving vehicle tends to fall forward when vehicle suddenly stops.
• A passenger of a train moving at constant velocity throws a ball vertically upwards it returning to the passenger’s hand due to inertia of motion.

Question 3.
Define linear momentum of a body and mention its SI unit.
Answer:
The momentum of a body is defined as the product of mass and velocity of the body.

Question 4.
Is linear momentum scalar or vector quantity?
Answer:
Vector quantity

Question 5.
State Newton’s Second law of motion and hence derive the relation F = ma.
Answer:
The rate of change of momentum of a body is directly proportional to the impressed force on it and takes place in the direction of the force.

To derive F – ma :
Consider a body of mass ‘m’ moving with a initial velocity u.
∴ Initial momentum of body = mu
Let a force ‘F’ is acted on the body for time interval ‘t’ second to change its velocity from u to v.
∴ Final momentum of body = mv
According to Newton’s II-law of motion
F ∝ rate of change of momentum
F ∝ \(\frac{m v-m u}{t}\)
F ∝ \(m\left(\frac{v-u}{t}\right)\)
F ∝ ma
f = k ma
if k = 1
f = ma

Question 6.
Mention the SI unit of Force.
Answer:
SI unit of Force is newton (N).

Question 7.
Define one newton [unit Force].
Answer:
One newton:
Consider F=ma
if m = 1 kg,a = 1 ms^-2
then IF =1 newton
The force is said to be one newton if it causes an acceleration of 1 ms ^-2 in a body of mass one kg. Note: 1 kg weight: It is the force experience by a body of mass one kg, by earth on its surface.
∴ 1 kg weight = m g, where g = 9.8 ms^-2, m = 1 kg
1 kg weight = 9.8 N

Question 8.
Briefly explain the application of Newton’s II law applied to variable mass system.
Answer:
If mass of a body is not constant but changes continuously with time, the Newton’s II law cannot be applied
directly.
Examples:

• For a moving rocket its mass continuously decreases.
• The truck loaded with sand moving in rain it mass increases.

During ejection or addition of mass, the main mass is subjected to thrust force.
Thrust is the force exerted by mass leaving or entering the system.
Magnitude of thrust force depends on relative velocity v_r, and rate of change of mass.
Now Newton’s II law takes form
F + F_t =ma
Where F_t → Thrust force
And F_t = \(v_{r} \frac{d m}{d t}\)
∴ F + v_r\(\frac{d m}{d t}\) = ma
This equation represents Newton’s II law applied to variable mass system.

Question 9.
Define Impulse and obtain the expression for Impulse.
Answer:
A large force acting on a body for a short interval is called an impulsive force.
Examples: (1) Kicking of a foot ball.
(2) Hitting a nail by a hammer.
During impulse we can measure the change in momentum
F = ma = \(m\left(\frac{v-v_{0}}{t}\right)\)
Ft = mv – mv₀
∴ Impulse = change in momentum,
i.e. Impulse = Force x time interval

Question 10.
Write the graph representing variation of Force and time for Impulse. Also mention SI unit of impulse.
Answer:
The variation of force with respect to time is given as follows:
SI unit of Impulse : newton second (Ns)
Dimensional formula : [M¹L¹T^-1]

Question 11.
State and explain Newton’s III law of motion.
Answer:
“For every action there is an equal and opposite reaction”.
When a body exerts a force on another body at same time the second body exerts an equal and opposite force on the first.
Explanation:
According to III law, when ‘A’ exerts force on ‘B’, at same time ‘B’ exerts equal and opposite force on ‘A’
∴ F_AB = F_BA
Where F_AB – Force ob B by A
F_BA – Force ob A by B

Examples:

1. Recoiling of gun when a bullet is fired.
2. Swimmer pushes water backward and moves forward.

Note:

1. Time interval of action is equal to that of reaction.
2. Action and reaction acts on two different bodies.
3. Centripetal force and Centrifugal force do not constitute action and reaction pair.

Question 12.
State and prove the law of conservation of Linear momentum.
Answer:
Statement:
The total momentum of an isolated system of interacting particles is conserved.
Proof: Consider anisolated system of particles A and B are moving with initial momenta P_A and P_B respectively. Let they collide and get apart with final momenta P’_A and P’_B respectively.

According to II law of Newton:

According to III law of Newton :
F_AB x ∆t = -(F_BA x ∆t)
i.e P’_A-P_A = -(P’_B-P_B)
P’_A – P_A = -P’_B + PB
P’_A + P’_b = P_A +P_B
i.e., momentum after collision = momentum before collision.

Question 13.
Give examples for law of conservation of linear momentum.
Answer:

1. Recoiling of gun when bullet is fired.
2. When a body explodes in to two or more pieces, then total momentum of pieces is equal to momentum of body.
3. Principle of rocket propulsion is based on conservation of momentum.

Question 14.
Briefly explain common forces in mechanics.
Answer:
Weight: Weight of a body is the gravitational force exerted on it near surface of earth.
W = mg
Tension: The force exerted by a string is called Tension.
It is denoted by T.
It acts in a direction away from body.
Spring force: A compressed or stretched spring develops a force to retain its original position called spring force i.e. F ∝ -x
F = -kx
where, k → spring constant
x → displacement

Contact force: When two bodies are in contact the molecules at interface interact and results in net force called contact force.
This contact force is divided (components) in to Normal reaction (N) and friction (f).

Question 15.
What is friction?
Answer:
Friction: Friction is the force that arise when surfaces of two bodies in contact or when they move or tend to move relative to each other.

Question 16.
Explain the origin of Friction.
Answer: Origin of friction: When two bodies are in contact due to molecular forces of attraction the net contact force arise.
The vertical component of contact force is Normal reaction (AO and Horizontal component is called friction (f).

Note:

1. At microscopic level due to irregularities in surfaces “high points” interlock each other. It is called “cold welding”.
2. Friction always acts in the direction against the applied external force.
3. Friction do not oppose motion of body. It oppose relative motion between bodies in contact.
4. Friction is self adjusting force with respect to both magnitude and direction.

Question 17.
Mention the types of friction.
Answer:
There are three types of friction.

1. Static friction.
2. Kinetic fiction.
3. Rolling friction

Question 18.
What is static friction? Mention expression.
Answer:
Static friction (/j): It is the friction between two bodies when external force is applied but body is at rest.
It is given by
f_s ∝ N
f_s = μ_sN
where N = mg
μ_s→ coefficient of static friction

Question 19.
What is Kinetic friction? Mention expression for it.
Answer:
Kinetic friction (fg): The friction that arises between two bodies when the body is set into motion.
It is given by
f_d ∝ N
f_k = μ_kN
where N = mg
μ_k→ coefficient of static friction

Question 20.
What do you mean by Limiting friction?
Answer:
Limiting friction (f_s max):
When magnitude of external force is increased, the magnitude of static friction also increases and reach maximum value. This maximum value of static friction is known as limiting friction.

Question 21.
State laws of static friction.
Answer:
Laws of static friction:
(i) f_s ≤ f_s ^max
(2) friction is independent of area of contact and nearly independent of velocity of body.
(3) Static friction depend on nature of surfaces and normal reaction. f_s = μ_sN

Question 22.
What is rolling friction?
Answer:
Rolling friction:
The friction between a rolling body and surface is known as rolling friction.
Rolling friction very less than static friction because of negligible area of contact between rolling body and surface. Example: We use ball bearings in automobile wheels to avoid friction between axel and wheel.

Question 23.
Mention advantages and disadvantages of friction.
Answer:
Advantages:

1. Brakes work due to friction
2. Friction helps in walking
3. Friction helps to hold objects

Disadvantages:

1. It causes wear and tear of machinery
2. Energy is wasted as heat due to friction

Question 24.
Define angle of friction and obtain expression for it
Answer:
It is the angle between the normal reaction and the resultant of limiting friction and normal is called angle of friction (θ)
In the diagram

Question 25.
Name different methods of reducing friction.
Answer:
Friction between two surfaces can be reduced by (i) Polishing (ii) use of lubricants (iii) stream lining (iv) use of ball bearings.

Question 26.
Define angled repose and obtain expression for it.
Answer:
It is the angle between the inclined plane and the horizontal surface such that a body placed on the plane just begins to slide down is called angle of repose (α).
In the diagram, let ‘α’ be the angle of repose.
Now Mg sin α = Friction (f) ….. (1)
Mg cosα = R …….. (2)
∴ 1/2
tanα = f/R
tanα = f/R = μ
∴ tanα = μ
Since tan θ = μ
∴ θ = α

Question 27.
What is centripetal force? Mention the expression for it.
Answer:
An external force required to keep an object in a circular path with uniform velocity is known as centripetal force.
F = \(\frac{m v^{2}}{R}\) = mω²R
R = Radius of circle
ω = angular velocity
v = velocity

Question 28.
What is centrifugal force? Mention expression for it.
Answer:
Centrifugal force is the force experienced by a body when it moves in a circular path. It is also called pseudo force.
mv2
It is directed along the radius and away from centre of circle F = \(\frac{m v^{2}}{R}\) = mω²R
Note: (a) Centripetal force and centrifugal force do not constitute action – reaction pair.

Question 29.
Define angle of Banking of roads.
Answer:
It is the angle by which the outer edge of road is raised with respect to inner edge. θ = angle of banking of road.

Question 30.
Derive an expression for maximum velocity for an automobile on a level circular road.
Answer:
Consider a car is moving on a level circular road of radius of curvature ‘r’.
Now N = mg and f_s^max is providing necessary centripetal acceleration

Question 31.
Derive an expression for maximum velocity of an automobile on a circular banked road
Answer:
Consider a car of mass ‘m’ is moving round a curved path of radius ‘r’ with speed ‘v’.
If ‘θ’ is the angle of banking. Let μ be the coefficient of friction between road and Lyres of car.
If ‘f is the force of friction between banked road and tyres then

Numerical Problems

Question 1.
A car having a mass of 1000 kg is moving at a speed of 30 ms^-1. Brakes are applied to bring the car to rest. If the frictional force between the tyres and the road surface is 5000 N, the car will come to rest in what time?
Answer:
m = 1000 kg, u = 30 ms^-1, v = 0
F = -5000 N
Now F = ma
or -5000 = 1000\(\left(\frac{v-u}{t}\right)\)
or -5 = \(\frac{(0-30)}{t}\) ⇒ t = 6s

Question 2.
A 30 kg block rests on a rough horizontal surface. A force of 200 N is applied on the block. The block acquires a speed of 4 ms^-1 starting from rest in 2s. What is the value of coefficient of friction?
Answer:
m = 30 kg, u = 0,
F = 200 N, v = 4 ms^-1
t = 2 s
F_app= 200 N
R = mg = 30 x 10 = 300 N
Now, force of friction
F = μR = μ mg = μ x 300
Also, f = F _spp – F
ma = 200 – μ x 300
30 x \(\left(\frac{v-u}{t}\right)\) = 200 – μ x 300

Question 3.
A body of mass 10 kg is acted upon by two perpendicular forces 6 N and 8 N. The resultant acceleration of the body?
Answer:

A bullet of mass 0.04 kg moving with a speed of 90 ms^-1 enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?
(N.C.E.R.T Solved example)
Answer:
m = 0.04 kg
u = 90 ms^-1, v = 0 s = 60 cm = 0.6 m, F = ?
Since v² – u² =2 as
∴ 0 – (90)² =2 x a x 0.6
or a = \(\frac{-90 \times 90}{2 \times 0.6}\) = -6750ms^-2
Now, F = ma
∴ F = -6750 x 0.04 = -270 N
Here negative sign indicates resistive force.

Question 4.
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 ms-1. If the mass of the balls is 0.15 kg, determine the impulse imparted to the ball. (N.C.E.R.T Solved example)
Answer:
u = 12 ms^-1, v = -12 ms^-1
m = 0.15 kg
I = m(v-u)
= 0.15 x (-12-12)
= -3.6 Ns.

Question 5.
Calculate the maximum velocity with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding.
Answer:

Question 6.
A rope of mass 0.1 kg is connected at the same height of two opposite walls. It is allowed to hang under its own weight. At the contact point between the rope and the wall, the rope makes an angle of loo with respect to horizontal. Calculate the tension in the rope at its mid point between the wall.
Answer:

Question 7.
A hammer of mass 1 kg moving with a speed of 6 ms^-1 strikes a wall and comes to rest in 0.1 s. Calculate (i) impulse of the force (ii) the retardation of hammer and (iii) the retarding force that stops the hammer.
Answer:
m = 1 kg,
u = 6 ms^-1, v=0, t = 0.1 s
(i) f = m(v-u) = 1(0-6) =-6 Ns
(ii) F = ma => 1 x t = ma
or -6 x 0.1 = 1 x a
∴ Retardation = 0.6 ms^-2
(iii) Retardation force F = l x t .
= +0.6 N

Question 8.
A shell of mass 0.020 kg is fired by the gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun?
Answer:
m₁ = 0.020 kg, m₂ = 100 kg
v₁ = 80 ms^-11, v₂ = ?
By law of conservation of linear momentum
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0 = (0.02 x 80) +100 x v₂
100 v₂ = -16
v₂ = \(\frac{-16}{100}\) = -0.16 ms^-1

Question 9.
In Fig. a mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the mid point P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium? Take g = 10 ms^-2 Neglect mass of the rope.
Answer:
In equilibrium,
T₂ = mg = 6×10 = 60 N
Along horizontal direction
T₁sin θ = 50N ….. (1)
and along vertical direction
T₁cosθ = T₂ =60N …. (2)
Dividing eq. (1) by eq. (2)
tanθ = \(\frac{5}{6}\)
or θ = tan^-1( \(\frac{5}{6}\) )
θ = 40°

Question 10.
What is the acceleration of the block and trolley system shown in Fig. if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? Take g = 10 ms^-2. Neglect
the mass of the string

Answer:
As the string is inextensible, and the pulley is smooth, the 3 kg block and the 20 kg trolley, both have same magnitude of acceleration (a). Applying Newton’s second law of motion to free body diagram of W = 20 kg,
Fig. 3(b). 26(b), we get
T – f_k = 20 a …(i)
Now, f_k =μ_kR = μ_kmg =0.04 x 20 x 10 = 8 N
∴ T – 8 = 20 a …(ii)
Again applying Newton’s second law of motion to free body diagram of w = 3 kg, Fig. 3(b). 26(c),
we get 30-T -3a …… (iii)
Adding (ii) and (iii), we get
22 = 23 a
a = \(\frac{22}{23}\) = 0.96 ms^-2
From (ii) T = 20 a + 8
T = 20×0.96 + 8 = 19.2 + 8 = 27.2 N

Question 11.
A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. For sliding block A on B, a horizontal force of 12 N is required to be applied on A. How much maximum force can be applied on B so that both A and B move together? Also, find out acceleration produced by this force.
Answer:
Here, m₁= 4 kg, m₂ = 5 kg
Force applied on A,f= 12 N, Fig. 3(b). 27.
This must atleast be equal to force kinetic friction applied on A by B
i-e- f = f_k = μ_kR = μ_km₁g
12 = f_k x 4g
∴μ_k = \(\frac{12}{4 g}=\frac{3}{g}\)

As block B is on smooth surface, therefore to move A and B together, (maximum) force F required to be applied on B = frictional force applied on A by B plus frictional force applied on B by A
F = μ_km₁g + μ_km₂g = μ_k(m₁ + m₂)g
F = \(\frac{3}{8}\) (4 + 5 ) = 27 N
As both the blocks are moving on a smooth table under the action of this force,
a = \(\frac{F}{m_{1}+m_{2}}=\frac{27}{4+5}\) = 3ms^-2

Question 12.
A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until an angle θ = 15° with the horizontal and the mass just begins to slide. What is the coefficient of static friction between the block and the surface? NCERT Solved Example.
Answer:
Here, m = 4 kg, θ = 15°, μ_s= ?
As the mass just begins to slide at this angle of inclination, therefore, θ = 15° = α, the angle of repose.
μ_s = tan α = tan 15° = 0.27

Question 13.
A mass of 200 kg is placed on a rough inclined plane of angle 30°. If coefficient of limiting friction is 1/ √3 , find the least forces in netwon, acting parallel to the plane (i) to keep the mass from sliding down, (ii) to move the mass up the plane.
Answer:
Here, m – 200 kg, θ = 30°; μ = \(\frac{1}{\sqrt{3}}\)
As is clear from Fig. a and b R = mg cosθ
∴ Force of friction F – μ R – μ mg cos θ = \(\frac{1}{\sqrt{3}}\) x 200 x 9.8 cos 30°
F = \(\frac{1}{\sqrt{3}}\) x 200 x 9.8\(\frac{1}{\sqrt{3}}\) = 980 N

Also, component of weight acting down the plane = mg sin θ
= 200 x 9.8 sin 30° = 200 x 9.8 x \(\frac{1}{2}\) = 980 N
(i) Hence the least force required to keep the body from sliding down, Fig. (a).
f₁ = mg sin θ – F = 980 – 980 = Zero.
(ii) To move the body up the plane, the least force required, Fig. (b)
f₂ = mg sinθ + F = 980 + 980 = 1960 N

Question 14.
A particle of mass 150 g is attached to one end of a massless inextensible string. It is made to describe a vertical circle of radius 1 m. When the string is making an angle of 48.2° with the vertical, its instantaneous speed is 2 m/s. What is the tension in the string in this position? Would this particle be able to complete its circular path? NCERT Solved Example
Answer:
Here,

As instantaneous velocity is 2 ms^-1, it cannot increase (to 3.16 ms^-1) on reaching the highest point as gravity is opposing the motion. Therefore, the particle cannot complete its circular path.

Question 15.
A bucket containing 4 kg of water, is tied to a rope of length 2.5 m and rotated in a vertical circle in such a way that the water in it just does not spill over when the bucket is in upside down position. What is the speed of the bucket at (a) highest point and (b) lowest point of its circular path? Take g = 10 m/s²
Answer:
Here, m = 4 kg, l = r = 2.5 m
In the upside down position, water does not spill. Therefore,

Question 16.
A cyclist speeding at 18 km/h on a level road takes a sharp circular turn of radius 3 m without reducing the speed. The coefficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?
Answer:
Here, v = 18 km / h = \(\frac{18 \times 1000}{60 \times 60}\)= 5 ms^-1
r = 3m, μ_s=0.1
On an unbanked road, frictional force alone can provide the centripetal force. Therefore, condition for the cyclist not to slip is that
\(\frac{m v^{2}}{r}\) ≤ F_s,{= μ_sR = μ_smg)
v² ≤ μ_srg
As v² = 5² =25, and μ_srg = 0.1 x 3 x 10 = 3
∴ the condition is not satisfied. Hence the cyclist will slip.

Question 17.
A body moving on the ground with a velocity of 16 ms^-1 comes to rest after covering a distance of 25 m. If the acceleration due to gravity is 10 ms^-1, find the coefficient of friction between the ground and the body.
Answer:
u = 15 ms^-1, v = 0, S = 25 m, g = 10 ms^-2
Now, v² – u² = 2aS
=> 0-(15)² = 2 x a x 25
a = \(\frac{-225}{2 \times 25}\)
or a = -4.5 ms^-2
∴ F = ma
∴μ = \(\frac{F}{R}=\frac{m a}{m g}=\frac{4.5}{10}\) = 0.45

Question 18.
A train weighing 1000 quintals is running on a level road with a uniform speed of 72 kmh^-1 If the frictional resistance amounts to 50 gwt. per quintal, find power in Watt, take g = 9.8 ms^-2.
Answer:
m = 1000 quintal = 10⁵ kg
v = 72 kmh^-1 = 72 x \(\frac{5}{18}\)
= 20 ms^-1
Total frictional force
= \(\frac{50}{1000}\) x 1000 x 9.8 N
= 50 x 9.8 N
Power, P = F x v
=50 x 9.8 x 20
= 9800 watt.

Question 19.
What is the acceleration of the block and trolley system shown in Fig. if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? Take g = 10 ms^-2. Neglect the mass of the string.
Answer:
Applying Newton’s 2^nd law of motion to W = 20 kg.
T-f_k=20a
Since, f_k = μ_kR
= μ_kmg
= 0.04 x 20 x 10 = 8 N
∴ T – 8 = 20a …(1)
Again applying Newton’s 2^nd law of motion to W= 3 kg
∴ 30 -T = 3a …(2)
Adding equation (1) and (2)
22 = 23a
or a = \(\frac{22}{23}\) = 0.96 ms^-2 23 .
Also, T = 20 a + 8
= 20 x 0.96 + 8 = 27.2 N

Question 20.
A block of mass 2 kg rests on a plane inclined at 30° with the horizontal. The coefficient of friction between the block and the surface is 0.7. What will be the frictional force acting on the block?
Answer:
As the block is stationary,
∴ F = mg sin 30°
= 2 x 9.8 x \(\frac{1}{2}\)
= 9.8 N
This is the force of friction.

Question 21.
A 20 kg box is gently placed on a rough inclined plane of inclination 30° with horizontal. The coefficient of sliding friction between the box and the plane is 0.4. Find the acceleration of the box down the incline.
Answer:
m = 20 kg
μ = 0.4
From Fig.
f = mg sin30°-F
F = μR = μ mg cos 30°
f = ma
∴ ma = mg sin 30° – μ mg cos 30°
a = \(\frac{g}{2}-0.4 \times g \frac{\sqrt{3}}{2}\)
= \(\frac{g}{2}\) (1 – 0.4 x 1.73)
= 1.505 ms^-2