Class 11 Physics Dynamics Notes

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Dynamics, Newton’s First Law and Force

Dynamics: The branch of mechanics which deals with the study of the motion of the objects with taking into account of the cause of the motion in the objects.

First Law of Newton’s

It states that, “Every body continue in its state of rest or of uniform motion in a straight line, unless it is compelled (act) by some external force.”

Force

Force is an external agency which changes or tends to change the state of rest or uniform motion of a body. It is denoted by F and given by:

F = ma

Where, m = mass and a = acceleration. Force is a vector quantity. Its unit is kg m s⁻² (N).

Effect of Force

It changes the state of body.
It changes the speed of body.
It changes the direction of body.
It changes the shape of body.

First Law Gives Definition of Force

According to the first law, force is an external agent that changes or tends to change the state of rest or uniform motion of a body, which is the definition of force.

PDF p. 2

First Law as the Law of Inertia

First law is also given or called law of inertia.

Inertia: The inability of a body to change the state of rest or of uniform motion by itself is called inertia.

According to Newton’s first law of motion, a body is unable to change its state of rest or of uniform motion without any external force. So, first law and law of inertia are equivalent to each other.

Short Question: The dust fall down when cloth are beaten with stick?

Before the cloth is beaten, both dust and cloth are in rest. After cloth is beaten, cloth is in motion and dust wants to be in rest. So dust fall down when cloth are beaten.

PDF pp. 3–4

Second Law of Newton’s

It states that, “The force acting on any body is directly proportional to its rate of change of momentum.”

F ∝ dpdt

or, F = K dpdt

Where, K is proportionality constant whose value is equal to 1.

∴ F = dpdt

Since, p = mv

or, F = d(mv)dt

or, F = m dvdt

Since, a = dvdt

∴ F = ma

The supplied scan repeats this same second-law statement and derivation on the next scanned page; it is represented once here without adding new material.

PDF p. 5

Third Law of Newton’s

It states that, “Every action there will be equal and opposite reaction.”

Short Question: If action and reaction are equal and opposite, then why can they cancel each other?

For example, the gun is fired when equal and opposite in bullet takes place action and to gun takes place reaction. Here action and reaction act in different two bodies. Hence, the supplied note concludes that they sometimes cancel each other.

Short Question: When a balloon filled with air and its mouth downwards is released, it moves upwards. Why?

According to Newton’s third law, every action has an equal and opposite reaction. When the mouth of an air-filled balloon is released, air moves downward as action and the balloon moves upward as reaction. So, when the air-filled balloon’s mouth is released, it moves upwards.

PDF pp. 6–7

Momentum and Principle of Conservation of Linear Momentum

Momentum: The momentum is defined as the total quantity of motion contained in a body and it is equal to the product of mass and velocity.

p⃗ = m v⃗

It is a vector quantity and its unit is kg m s⁻¹.

Principle of Conservation of Linear Momentum

It states that, “If no external force acts on a system of interacting particle then total linear momentum of the system remains constant.”

From Newton’s second law, F = dpdt

If F = 0, then dpdt = 0

⇒ p = constant

Collision between two spheres as shown in the source.

Proof

Let us consider two bodies A and B with masses m₁ and m₂, moving with initial velocities u₁ and u₂ (u₁ > u₂). Suppose they collide with each other for small time Δt, then after collision their velocities become v₁ and v₂ respectively.

Initial momentum of body A = m₁u₁

Final momentum of body A = m₁v₁

Initial momentum of body B = m₂u₂

Final momentum of body B = m₂v₂

Change in momentum of body A = m₁v₁ − m₁u₁

Change in momentum of body B = m₂v₂ − m₂u₂

According to Newton’s second law:

F_A = m₁v₁ − m₁u₁Δt

Similarly, F_B = m₂v₂ − m₂u₂Δt

Also, from Newton’s third law: F_A = −F_B

m₁v₁ − m₁u₁Δt = − m₂v₂ − m₂u₂Δt

m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂

i.e., total final momentum = total initial momentum.

PDF pp. 8–9

Problem of Mass and Pulley

Mass and pulley arrangement, with m₂ > m₁ as drawn in the scan.

For body m₁

F = m₁g − T

Since, F = m₁a

m₁a = m₁g − T

For body m₂

F = T − m₂g

Since, F = m₂a

m₂a = T − m₂g

L.Q. 5(A): Atwood’s machine with 15 kg and 10 kg masses

In the Atwood’s machine in the given figure, the system starts from rest. What is the acceleration of the given mass?

For body m₁ = 15 kg:

F = m₁g − T

15a = 150 − T

T = 150 − 15a …(i)

For body m₂ = 10 kg:

F = T − m₂g

10a = T − 100

T = 10a + 100 …(ii)

From (i) and (ii):

150 − 15a = 10a + 100

50 = 25a

a = 2 m s⁻²

L.Q. 5(B): Two masses 7 kg and 12 kg over a frictional pulley

Two masses 7 kg and 12 kg are connected at the two ends of light inextensible string that passes over a frictional pulley. Using free diagram method, find the acceleration of masses and the tension in the string when the masses are released.

For body m₁ = 7 kg: 7a = T − 70 ⇒ T = 7a + 70 …(i)

For body m₂ = 12 kg: 12a = 120 − T ⇒ T = 120 − 12a …(ii)

From (i) and (ii): 7a + 70 = 120 − 12a

19a = 50 ⇒ a = 2.63 m s⁻²

Again, T = 7(2.63) + 70 = 88.41 N

Acceleration = 2.63 m s⁻², tension = 88.41 N.

PDF p. 10

Collision Numerical

L.Q. 5(C)

A ball A of mass 0.1 kg moving with a velocity of 6 m s⁻¹ collides directly with a ball B of mass 0.2 kg at rest. Calculate their common velocity if both balls move off together. If A had rebounded with a velocity of 2 m s⁻¹ in the opposite direction after collision, what would be the new velocity of B?

m_A = 0.1 kg, u_A = 6 m s⁻¹, m_B = 0.2 kg, u_B = 0.

(I) Common velocity

m_Au_A + m_Bu_B = (m_A + m_B)V

(0.1 × 6) + (0.2 × 0) = (0.1 + 0.2)V

0.6 = 0.3V

V = 2 m s⁻¹

(II) If A rebounds with v_A = −2 m s⁻¹

m_Au_A + m_Bu_B = m_Av_A + m_Bv′_B

(0.1 × 6) + (0.2 × 0) = (0.1 × −2) + (0.2 × v′_B)

0.6 + 0.2 = 0.2v′_B

v′_B = 4 m s⁻¹

PDF pp. 11–13

Apparent Weight

The reaction of machine on a person is called apparent weight.

Case I: Lift moving upwards

F = R − mg

ma = R − mg

R = mg + ma

Apparent weight increases.

Case II: Lift moving downward

F = mg − R

ma = mg − R

R = mg − ma

Apparent weight decreases.

Case III: At rest / constant velocity

a = 0

R = mg

Apparent weight = actual weight.

Lift cases from the source: upward acceleration, downward acceleration, and zero acceleration.

L.Q. 5(D): 50 kg man in a lift accelerating at 2 m s⁻²

A lift moves (1) up and (2) down with an acceleration of 2 m s⁻². In each case, calculate the reaction of the floor on a man of mass 50 kg standing in the lift.

Upward: 50(2) = R − 50(10) ⇒ R = 600 N.

Downward: 100 = 500 − R ⇒ R = 400 N.

Upward reaction = 600 N; downward reaction = 400 N.

L.Q. 5(E): Scale reading falls from 550 N to 450 N

A 550 N physics student stands on a bathroom scale in an elevator. As the elevator starts moving, the scale reads 450 N. Draw free body diagram and find the magnitude and direction of the acceleration of the elevator.

mg = 550 N ⇒ m × 10 = 550 ⇒ m = 55 kg.

R = 450 N. In this case, acceleration is downward:

ma = mg − R

55a = 550 − 450

55a = 100

a = 1.8 m s⁻² downward.

PDF p. 14

Moment of Force (Torque)

The product of force and moment arm (r) is called torque. It measures the turning effect of force.

τ = Fr

Its unit is N m.

Torque: force F acting at a moment arm r from pivot O.

Clockwise and Anticlockwise Moment

The moment of force which rotates the body in clockwise direction is called clockwise moment. The moment of force which rotates the body in anticlockwise direction is called anticlockwise moment.

F₂r₂ gives clockwise moment; F₁r₁ gives anticlockwise moment in the source figure.

PDF pp. 15–16

Principle of Moment

It states that, “For a body to be in a rotational equilibrium, the sum of anticlockwise moment is equal to the sum of clockwise moment.”

Experimental arrangement for verification of the principle of moment.

To Verify Principle of Moment in a Lab

A scale is taken and pivoted at point O. Masses m₁, m₂, m₃, m₄ and m₅ are suspended on both sides as shown so that the scale becomes equilibrium. Let r₁, r₂, r₃, r₄ and r₅ be distances on both sides of point O at which respective masses are placed, whose weights act vertically downward.

The weights m₁g and m₂g tend to rotate the scale in anticlockwise direction. So, anticlockwise moment is:

m₁gr₁ + m₂gr₂

Similarly, the weights m₃g, m₄g and m₅g tend to rotate the scale in clockwise direction. So, clockwise moment is:

m₃gr₃ + m₄gr₄ + m₅gr₅

If the scale remains in equilibrium:

m₁gr₁ + m₂gr₂ = m₃gr₃ + m₄gr₄ + m₅gr₅

i.e., anticlockwise moment = clockwise moment, which verifies principle of moment.

PDF pp. 16–17

Parallel Forces and Couple

Parallel forces: The forces whose line of action are parallel to each other are called parallel forces.

Types:

Like parallel force
Unlike parallel force

Like Parallel Force

The parallel forces acting in the same direction are called like parallel force.

Unlike Parallel Force

The parallel forces acting in opposite direction are called unlike parallel force.

Like and unlike parallel forces as represented in the scan.

Couple

Two equal and unlike parallel force with different line of action produce couple.

Couple of forces acting on a body pivoted at O.

Consider a couple of force acting on a body pivoted at point O. The torque due to couple is the sum of torque due to individual force.

Torque due to force at point A: τ_A = F · AO

Similarly, torque due to force at point B: τ_B = F · BO

Therefore, torque due to couple:

τ_c = τ_A + τ_B

τ = F·AO + F·BO

τ = F(AB)

Hence, torque due to couple is product of magnitude of force and perpendicular distance between two forces.

PDF pp. 18–19

Equilibrium

A body is said to be equilibrium if net force or net torque acting on a body is equal to zero.

1. Translation Equilibrium

A body is said to be in translation equilibrium if net force acting on a body is equal to zero.

ΣF = 0

a = 0

Body is at rest or moving with constant velocity.

If ΣF = 0 and body is at rest, it is said to be static translation equilibrium.
If ΣF = 0 and body is moving with constant velocity, it is said to be dynamic translation equilibrium.

2. Rotational Equilibrium

A body is said to be in rotational equilibrium if net torque acting on a body is equal to zero.

Στ = 0

Angular acceleration (α) = 0

Body is at rest or moving with constant angular velocity.

If Στ = 0 and body is at rest, it is said to be static rotational equilibrium.
If Στ = 0 and body is moving with constant angular velocity, it is said to be dynamic rotational equilibrium.

Stable Equilibrium

The equilibrium in which body returns to its original position by itself after being slightly displaced from initial position is called stable equilibrium.

Unstable Equilibrium

The equilibrium in which body doesn’t return to its original position by itself after being slightly displaced from initial position is called unstable equilibrium.

Neutral Equilibrium

The equilibrium in which body always stay in the displaced position after it has been slightly displaced from initial position is called neutral equilibrium.

PDF p. 19

Centre of Mass and Centre of Gravity

Centre of mass: The point of the body at which whole mass of the body can be considered to be concentrated is called centre of mass. It always lies in the body.

Centre of gravity: The centre of gravity is the point at which whole weight of the body acts in downward direction. It may or may not lie within the body.

For example: centre of gravity of hollow sphere and ring lies at its centre where there is no mass.

PDF pp. 20–21

Numericals on Moments

L.Q. 6(A): Roller over a brick

A roller whose diameter is 1 m weighs 360 N. What horizontal force is necessary to pull the roller over a brick 0.1 m high when the force is applied at the centre?

Source geometry for pulling the roller over a 0.1 m brick.

OB = AC = (0.5 − 0.1) m = 0.4 m.

BC = √(OC² − OB²) = √((0.5)² − (0.4)²) = 0.3 m.

F × AC = W × BC

F × 0.4 = 360 × 0.3

F = 270 N.

L.Q. 6(B): Two forces on a metre scale

Two forces of 1.5 N and 2 N act vertically at the two ends of a metre scale. Where and in which direction should a force be applied so that the scale remains horizontally stable?

From the figure, F₁r₁ = F₂r₂.

1.5x = 2(1 − x)

3.5x = 2

x = 0.57 m

r₂ = 1 − x = 0.43 m.

Force should act 0.57 m from the 1.5 N end and 0.43 m from the 2 N end.

L.Q. 6(C): Carrying a 3 m wooden board

Two people are carrying a uniform wooden board that is 3 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what position does the other person lift?

The other upward force = 160 − 60 = 100 N.

Taking moment as in the source: 60(1.5) = 100x.

90 = 100x ⇒ x = 0.9 m.

Position of the other person’s lift = 1.5 + 0.9 = 2.4 m from the 60 N end.

PDF pp. 21–22

Body Moving on a Smooth Horizontal Surface

Block on a smooth horizontal surface connected to a hanging mass.

For m₂

F = m₂g − T

m₂a = m₂g − T

T = m₂g − m₂a

For m₁

F = T

Since, F = m₁a

T = m₁a

L.Q. 7(A): 4 kg block, tension 10 N

A light rope is attached to a block with mass 4 kg that rests on a frictionless horizontal surface. The horizontal rope passes over a frictionless pulley and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 10 N. Draw free body diagrams and calculate the acceleration of either block and the mass of the hanging block.

For the 4 kg block: F = T = 10 N.

4a = 10 ⇒ a = 2.5 m s⁻².

For hanging mass: ma = m(10) − 10.

m(2.5) = 10m − 10

10 = 7.5m

a = 2.5 m s⁻², m = 1.33 kg.

PDF p. 23

Friction and Terms in Friction

Friction: The force which comes between two surfaces in contact and tends to oppose the relative motion between them is called frictional force.

Types of Friction

Static friction: The force of friction acting between two surfaces when they are at rest is called static friction. The maximum value of static friction is called limiting friction.
Kinetic or dynamic friction: The force of friction acting between two surfaces when they are in motion is called kinetic or dynamic friction.

Terms in Friction

Normal reaction (R): The reaction acting perpendicular to the surface of contact of two bodies is called normal reaction.

Normal reaction acts perpendicular to the surface of contact.

Coefficient of friction (μ): The ratio between force of friction (frictional force) to the normal reaction is called coefficient of friction.

μ = F_f / R

PDF pp. 24–25

Angle of Friction and Angle of Repose

Angle of Friction

The angle made by resultant of normal reaction and friction force to normal reaction is called angle of friction.

Angle θ between resultant reaction and normal reaction.

tan θ = F_f / R

Since, μ = F_f / R

tan θ = μ …(A)

Angle of Repose

The minimum angle of inclined surface with horizontal such that a body placed on it just begins to slide down by itself is called angle of repose.

Forces on a body at the angle of repose.

Let an object of mass m be kept on an inclined plane whose inclination is slowly increased such that the object slides downward. Its weight mg acts vertically downward and is resolved into components mg cos α and mg sin α.

At equilibrium:

mg sin α = F_f …(i)

mg cos α = R …(ii)

Dividing (i) by (ii): tan α = F_f/R

Since F_f/R = μ

tan α = μ …(B)

Comparing equation (A) and (B):

α = θ

i.e., angle of repose = angle of friction.

PDF p. 25

Motion on an Inclined Plane

For Downward Motion

F = mg sin θ − F_f

F = mg sin θ − μR

F = mg sin θ − μmg cos θ

ma = mg sin θ − μmg cos θ

a = g sin θ − μg cos θ

For Upward Motion

F = mg sin θ + F_f

a = g sin θ + μg cos θ

Direction of friction reverses between downward and upward motion.

PDF pp. 26–29

Numericals on Friction and Force

L.Q. 8(A): Block sliding down a 45° incline

What would be the acceleration of a block sliding down an inclined plane that makes an angle of 45° with the horizontal if the coefficient of sliding friction between two surfaces is 0.3?

F = mg sin 45° − F_f

F = mg sin45° − μR

F = mg sin45° − μmg cos45°

ma = mg sin45° − μmg cos45°

a = g sin45° − μg cos45°

a = 10(1/√2) − (0.3)(10)(1/√2)

a = 7.07 − 2.12

a = 4.9 m s⁻²

L.Q. 8(B): Initial acceleration of a toy rocket

The mass of gas emitted from the rear of toy rocket is initially 0.2 kg s⁻¹. If the speed of the gas relative to the rocket is 40 m s⁻¹ and the mass of rocket is 4 kg, what is the initial acceleration of the rocket?

Given in the scan: m/t = 0.2 kg s⁻¹, v = 40 m s⁻¹, m = 4 kg, a = ?, u = 0 m s⁻¹.

F = ma and F = m(v − u)/t.

⇒ (v − u)/t × m = 40(0.2)

4a = 40(0.2)

a = 2 m s⁻²

L.Q. 8(C): Ball strikes a wall four times

A ball of mass 0.05 kg strikes a smooth wall normally four times in 2 seconds with a velocity of 10 m s⁻¹. Each time the ball rebounds with the same speed of 10 m s⁻¹. Calculate the average force on the wall.

Given: m = 0.05 kg, t = 2 s, u = 10 m s⁻¹, v = −10 m s⁻¹.

For each time, F = m[(v − u)/t]

F = 0.05[(-10 − 10)/2] = −0.5 N

|F| = 0.5 N

For 4 strikes: average force = 0.5 × 4

Average force = 2 N.

L.Q. 8(D): Pulling a crate at 30°

Suppose you try to move a crate by tying a rope around it and pulling on the rope at an angle of 30° above the horizontal. What is the tension required to keep the crate moving with constant velocity? Assume weight of the crate W = 500 N and coefficient of dynamic friction μ_k = 0.40.

Source force diagram for a crate pulled at 30°.

F = T cos30° − F_f

ma = T cos30° − μR

ma = T cos30° − μ(W − T sin30°)

For constant velocity, a = 0:

0 = T(√3/2) − (0.4)[500 − T(1/2)]

200 − 0.2T = 0.86T

200 = 1.06T

T = 187.6 N

L.Q. 8(E): Iron block on a 30° wooden plane

An iron block of mass 10 kg rests on a wooden plane inclined at 30° to the horizontal. It is found that the least force parallel to the plane which causes the block to slide up is 100 N. Calculate the coefficient of friction between the two surfaces.

F = mg sin30° + F_f

F = mg sin30° + μR

F = mg sin30° + μmg cos30°

The source then uses F = ma: 100 = 10a ⇒ a = 10 m s⁻².

Therefore, a = g sin30° + μg cos30°

10 = 10(1/2) + μ(10)(√3/2)

5 = μ(8.66)

μ = 0.577

L.Q. 8(F): Force required on a 6 kg box

In a physics lab experiment, a 6 kg box is pushed across a flat table by a horizontal force F. If the box is moving at a constant speed of 0.35 m s⁻¹ and the coefficient of kinetic friction is 0.12, find the magnitude of force F. What is the magnitude of force F if the box is moving with a constant acceleration 0.18 m s⁻²?

Case I: Constant speed, a = 0

F = F_f = μR = (0.12)(mg) = (0.12)(6)(10) = 7.2 N.

Case II: Constant acceleration, a = 0.18 m s⁻²

F = ma + F_f

= 6(0.18) + μR

= 6(0.18) + (0.12)(6)(10)

= 1.08 + 7.2

F = 8.28 N.

PDF pp. 30–31

Law of Friction and Verification

Law of Friction

The force of friction between two surfaces depends upon the nature of surface.
The force of limiting friction is directly proportional to the normal reaction, i.e., F_f ∝ R.
The force of limiting friction is independent to the area of contact between two surface.
The kinetic friction is independent to the relative velocity of two surface.

Verification of Law of Friction

(I) To verify F_f ∝ R

Let us take two blocks A and B of masses m₁ and m₂ placed on horizontal table with a pulley at one end. A rope/string is attached to the block and runs over the pulley with another end attached to a scale pan on which load can be placed. The load on the scale pan required to just slide the block along the table is noted. Let W₁ and W₂ be the weights on the scale pan required to just slide blocks A and B of weights m₁g and m₂g respectively, such that normal reactions R₁ = m₁g and R₂ = m₂g.

W₁ / R₁ = W₂ / R₂

W / R = constant

Since at limiting friction applied force must be equal to frictional force, W = F_f.

F_f / R = constant

or, F_f = constant × R

⇒ F_f ∝ R

Experimental arrangement used in the source to verify limiting-friction laws.

(II) To verify force of limiting friction is independent of area of contact between two surfaces

Similarly, to verify force of limiting friction is independent of area of contact, two blocks A and B are first placed separately on a table. Here area of contact is equal to sum of area of contact of block A and B. The weight required to just slide the blocks A and B is noted as W. In the second case block B is placed over block A. In this case the area of contact is less than in the first case. The weight required to just slide them is again noted and is the same weight W. The force of friction is equal to applied load in case of limiting friction. Thus, although area of contact in both cases are different, force of limiting friction is same.