Unit 2: Force and Motion — Long Questions

Scalars Physical quantities which are described only by magnitude without any mention of direction.
e.g. mass, distance, speed, time, energy, temperature etc.
(i) Mass: The amount of matter in an object. For example, 2kg.
(ii) Distance: The total length of path travelled by an object irrespective of the direction. For example, 50m.
(iii) Speed: The rate at which an object covers distance. For example, 40 km h-1.
(iv) Time: The duration between two events taking place. For examples, 20 seconds.
(v) Energy: The capacity to do work. For example, 25 J.
(vi) Temperature: A measure of the average kinetic energy of particles in a substance. For example, 20°C.

Vectors Those physical quantities which have magnitude as well as direction. e.g:
(i) Displacement: The change in position of an object. It has length, a distance (magnitude) and a direction (e.g 10 m towards west).
(ii) Velocity: The speed of an object in a particular direction (e.g., 50 km h-1 towards west).
(iii) Acceleration: The rate of change of velocity, which includes changes in speed or direction (e.g. 10 ms-2 upward).
(iv) Force: A push or pull acting on an object, determined by its magnitude and direction (e.g. 20N to the right).

The vectors are denoted by bold face characters such as, A, d, r, v, the magnitude of a vector d is 'd' (light face).

Graphically, a vector is represented by a directed line segment with an arrow head at its one end. While length of the line segment represents the magnitude of the vector according to the chosen scale.

Rectangular Components of a Vector:
Such components of a vector which are at right angle to each other are called rectangular components of vector. A component of a vector is its effective value in a given direction.

Explanation Consider a vector A is represented by a line of making an angle θ with x-axis. If we draw projections from point P on x-axis and on y-axis, then it will meet the x-axis at point M and y-axis at point N. As OM is along x-axis named as Ax and ON is along y-axis named as Ay. Also ON and MP are equal in magnitude and direction. So MP is also named as Ay.

OMP is a right-angled triangle in which by head to tail rule,
A is the resultant vector of Ax and Ay. So, Ax and Ay are components of A. Because these components are at right angle to each other are called rectangular components of A. In right angle triangle OMP

Cosθ = OM/OP Sinθ = OM/OP

Cosθ = Ax/A Sinθ = Ay/A

A Cosθ = Ax ........ (1) Ay = A Sinθ ........ (2)

From equations (1) and (2) if we know the vector A and angle θ we can determine its two rectangular components.

Determination of a vector by rectangular components:
If Ax and Ay are two rectangular components of a vector A then by Pythagorean theorem:

hyp² = base² + per²

OP² = OM² + MP2

A² = Ax² + Ay²

A = √(Ax² + Ay²)

From this equation if we know two rectangular components of vector then magnitude of the resultant vector can be determined.

For direction (θ) Using ΔOMP

tanθ = MP/OM

tanθ = Ay/Ax

θ=tan⁻¹(Ay/Ax)

From this equation we can find the direction of resultant vector by putting values of Ax and Ay.

Definition "When two vectors are multiplied and their product results into scalar quantity, such a product is called a scalar product."
Scalar product is also called dot product because a dot (.) is placed between the vectors.

Mathematical Expression The scalar product of two vectors A and B is written as A.B.

A.B= AB Cosθ

Where A and B rare magnitudes of vectors A and B and 'θ' is angle between them.

Explanation Consider two vectors A & B making an angle θ as shown in Fig. Draw a perpendicular from Q on OP which represents projection of vector B along A.

Thus
A.B= A (Projection of vector B onto vector A)
A.B= A (BCosθ)
A.B= AB Cosθ ........---- (1)

Similarly, if we change the order of two vectors. The perpendicular from point P is drawn on OQ for representation of projection of A.

Thus;
B.A = B (Projection of vector A onto vector B)
B.A = B (ACosθ)
B.A = BACosθ ........---- (2)

By comparing eq. (1) & (2) we conclude that
A.B = B.A

Examples 1. Work:
Work done by a force F is an example of scalar product.
When a force F is applied to a body in which result it covers a displacement d making an angle θ then work done is
Work done = (effective component of force × Distance moved in the direction of motion)
W = (FCosθ)d
W = FdCosθ = F.d

2. Power:
Another example of scalar product is power. As we known that
P=F(vCosθ)
P=FvCosθ
P= F.v

Characteristics of Scalar Product

1. Scalar Product is Commutative:
As in Scalar Product
A.B = AB Cosθ
B.A = BACosθ
B.A = ABCosθ
∴ B.A = A.B

Thus
A.B = B.A
The order of multiplication is irrelevant i.e. scalar product is commutative.

2. The scalar product of two mutually perpendicular vectors:
The scalar product between two vectors A & B when are perpendicular to each other is:
A.B = ABCos90° (∵ Cos90° = 0)
A.B = 0 (Minimum value)

3. Scalar Product of Two Parallel or anti-parallel Vectors:
The scalar product of two parallel vectors is equal to the product of their magnitudes
i.e. for parallel vectors (θ = 0°).
A.B = ABCos0° [∵ Cos0° = 1]
A.B = AB (Maximum value)

For antiparallel vectors θ = 180°
A.B = ABCos180° = -AB [∵ Cos180° = -1]

4. The Dot product with itself:
The self dot product of a vector A is equal to square of its magnitude i.e.
A.A = AA Cos0° = A²
∵ Cos0° = 1

5. Scalar product of two vectors A and B in terms of their rectangular components:
A = Ax i + Ay j + Az k
B = Bx i + By j + Bz k
A.B = AxBx + AyBy + AzBz

6. To find the angle between two vectors in terms of their components.
A.B=AB Cosθ = AxBx + AyBy + AzBz

Cosθ= (AxBx + AyBy + AzBz) / AB

θ = Cos⁻¹((A.B) / AB)

By definition of scalar product
A.B=AB Cosθ

(A.B / AB) = Cosθ

or

θ = Cos⁻¹((A.B) / AB)

Definition "When two vectors are multiplied and their product results in a vector quantity then their product is said to be vector product or cross product."
(OR)
"If the product of two vectors is a vector quantity, the product is known as vector product."

The vector product is also called cross product because a symbol of "×" is placed in between the vectors.

Magnitude of Vector Product Consider two vectors A and B, making an angle θ with each other as shown in Fig. (a) then, magnitude of vector product is defined as:

|A × B| = AB Sinθ

The direction of A × B is determined by right hand rule.

Right Hand Rule According to this rule, the vector A is rotated toward vector B through small angle. Curling the fingers of right hand in the direction of smaller rotation, the extended thumb indicates the direction of A × B. The direction of A × B is perpendicular to the plane containing vectors A and B as shown in Fig. (b).

If n is a unit vector in the direction of A × B the vector product A × B is completely represented by.

A × B = AB sinθ n̂ -------- (1)

Examples

1. Torque:
Definition: "The turning effect of force on body is called torque."

Mathematical Expression / Formula Mathematically, it can be defined as the vector product of position vector and force.

τ = r × F = rFsinθ n̂

2. Angular Momentum:
"Angular Momentum can be defined as "the cross product of position vector and linear momentum." Mathematically

L = r × P = rPsinθ n̂

Characteristics of vector product

1. Vector product is non commutative:
A × B = ABSinθ n̂ --------- (1)

By right hand rule, the direction of vector n̂ is upward.

B× A = - ABSinθ n̂

Negative sign shows that the direction of n̂ is downward while the magnitude is same Fig.(c).

B× A = - ABSinθ n̂ ------- (2)

By comparing eq. (i) & (ii):

A × B ≠ B × A or (A × B ) = - (B× A ) or |A × B| = |B× A|

Thus, it shows that vector product is not commutative.

2. Cross product of perpendicular vectors:
The cross product of two perpendicular vectors has maximum magnitude i.e.

A × B = AB sin 90 n̂ = AB n̂ [∵ Sin90° = 1]

3. Cross Product of Parallel or Anti-parallel Vectors:
The cross product of two parallel vectors is null vector i.e.

A × B = ABSin0°n̂ = AB(0)n̂ = 0 [∵ Sin0° = 0]

Also If θ = 180° then

A × B = AB sin 180° n̂ = 0[∵ Sin180° = 0]

A × B = AB (0)n
A × B = 0

4. Self-Cross Product of a vector A :
Self-cross product is also equal to null vector i.e.

A × A = AA Sin0° n̂ = AA (0) n = 0

A × A = 0

5. If A and B are two adjacent sides of a parallelogram then the magnitude of their vector product will be area of parallelogram.

Area of Parallelogram = |A × B|

A × B = ABSinθ

Equations of motion can be used to describe the motion of an object in terms of its three kinematic variables: velocity (v), position (s) and time (t). There are three ways to pair these variables up: velocity-time, position-time and velocity-position. In this order they are called first equation of motion, second equation of motion and third equation of motion, respectively.

Limitation These equation of motion can only be applied to those objects which are moving in a straight line with constant acceleration.

Derivation of First Equation of Motion:
Suppose a body is moving with uniform acceleration along a straight line with an initial velocity vi. Let velocity changes from initial value vi to a final value vf in time interval t. Then the acceleration produced in the body during this time interval is given as

a = (vf - vi) / t

at = vf - vi

vf = vi + at

Above equation is the first equation of motion. It correlates the final velocity attained by a body with initial velocity and the time interval t, when moving with constant acceleration.

Derivation of First Equation of Motion by Graphical Method:
First equation of motion can be derived using velocity time graph for an object moving with initial velocity vi final velocity vf and constant acceleration a.

Let the velocity of a body at point A is vi which changes to vf at point B in time interval t as shown in Fig. A perpendicular BD is drawn from point B to x-axis and another perpendicular BE from B to y-axis, such that

OA = vi =Initial velocity of the body
BD = vf =final velocity of the body

From the graph it can observed that
BD = BC + CD
BD = BC + OA (As OA=CD)
Therefore
vf = BC + vi ............ (1)

The value of BC in above equation can be determined by taking the slope of line AB, Which is equal to acceleration a.

a = BC / AC

as
AC = t

so
a = BC / t

or
BC = at

Put in eq. (1)
vf = at + vi
vf = vi + at

This is the first equation of motion.

Derivation of Second Equation of Motion:
Suppose a body is moving with uniform acceleration a along 'a' straight line with an initial velocity vi, which become vf after time interval t. let it covers a distance S in a particular direction during time t, then using the definition of velocity as rate of change of displacement, we can write

Velocity = Displacement / time

Or
Displacement = Velocity × time

If velocity of the body is not constant, we can use average velocity instead of velocity. Thus

Displacement = Average velocity × time

Displacement = ((Initial velocity + Final velocity) / 2) × time

S = ((vi + vf) / 2) × t

Put vf = vi + at from first eq. of motion

S = ((vi + vi + at) / 2) × t ⇒ ((2vi + at) / 2) × t

S = (vi + at/2) × t

S= vi t +1/2 at²

This is the second equation of motion.

Derivation of Second Equation of Motion by Graphical Method:
Second equation of motion can be derived using velocity-time graph for a body moving with initial velocity vi which attains a final value vf in time interval t. While moving with constant acceleration a it covers a displacement 'S' in time t. it can be seen from the graphs that distance travelled by the body is, S = v × t.

Also
S = Area of the figure OABD
S = (Area of the rectangle OACD) + (Area of the triangle ABC)
S = (OA × OD) + (1/2 (AC × BC))

As
OA = vi and OD = AC =t. So, above equation becomes:
S = (vi t) + 1/2 (t × BC) --------(1)
∴ slop of AB = a

a = BC/AC = BC/t

BC = at put in Eq. (1)

S = (vi t) + (1/2 × t × at)

S = vi t + (1/2 at²)

S = vi t + 1/2 at²

Derivation of third equation of motion:
Consider a body moving along a straight line with an initial velocity vi which attains a final value vf in time t. let the displacement of the body be S during this time interval. Then we can write

Displacement = ((Initial Velocity + Final Velocity) / 2) × time

S = ((vi + vf) / 2) × time

S = ((vi + vf) / 2) × t --------(1)

Using the first equation of motion
vf = vi + at

Or
t = (vf - vi) / a

Putting the value of t in Eq. (1)

S = ((vf + vi) / 2) × ((vf - vi) / a) = (vf² - vi²) / 2a

2aS = vf² - vi²

This is the third equation of motion.

Derivation of third Equation of Motion by Graphical method:
In the speed time graph shown in the figure, the total distance S travelled by a body is given by the area OABD under the graph, such that

S = Area of trapezium
S = 1/2 (Sum of parallel sides) × height
S = 1/2 (OA + BD) × OD

Since
OA = vi, AD = vf and OD = t

S = 1/2 (vi + vf) × t

Put t = ((vf - vi) / a) from 1st eq. of motion

S = 1/2 (vi + vf) × ((vf - vi) / a) = 1/2 ((vf² - vi²) / a)

vf² - vi² =2aS---------- (2)

This is third equation of motion.

In the absence of air resistance, all objects in free fall near the surface of the Earth, move toward the Earth with a uniform acceleration. This acceleration is known as acceleration due to gravity, denoted by g and its average value near the Earth surface is taken as 9.8 ms-2 in the downward direction. The equation for uniformly accelerated motion can also be applied to free fall motion of the object by replacing a by g.

i. vf = vi + gt

ii. h = vi t + 1/2 gt²

iii. 2gh = vf² - vi²

According to Galileo, all bodies falls freely (in vacuum) under the acceleration due to gravity denoted by 'g'. its experimental value is 9.8 ms-2 in SI units. This means that different bodies, when allowed to fall from the same height, strike the ground with the same velocity. As regards the sign of g, it is taken positive for a falling body (when initial velocity is zero) and negative for a body projected vertically upward (when initial velocity is not zero).

Projectile Motion

Definition "The two dimensional motion of the body along the curved path under the action of gravity alone is known as projectile motion".
The curved path followed by the body is known as trajectory and body itself is known as projectile.

Examples 1. Motion of the football when it is kicked. 2. A missile fired from a launching pad.
3. Motion of the bomb dropped by the bomber. 4. Jumping animals.

The projectile motion is two dimensional motion in which body covers distance along horizontal and vertical direction simultaneously.

In this topic we will discuss two cases separately for better understanding.

Case 1 Linear Projectile
When an object is thrown horizontally from a certain height.

Case 2 Oblique Projectile
When an object is projected from ground making angle θ with the horizontal axes.

Explanation

Horizontal Acceleration of Projectile The horizontal direction in projectile motion is taken along x-axis. Suppose a ball is thrown horizontally with velocity 'v'. it will experience only one force which is its weight acting along y-axis, if the friction of air is neglected. Its acceleration along x-axis will be

ax = Fx / m

As no force is acting along x-direction.
i.e. Fx = 0

ax = 0 / m

ax = 0

There is no acceleration, therefore, the velocity of the projectile along x-axis will remain same.

Acceleration along y-direction The force acting on the projectile in the absence of air friction is only its weight "w". Therefore, its acceleration along y-direction is given by:

ay = Fy / m

As Fy = w

ay = w / m (∵ w = mg)

ay = mg / m

ay = g

Hence acceleration of projectile will be equal to acceleration due to gravity. The motion of body along vertical direction is just like free fall motion.

Distance covered Along Vertical Direction:
Consider the vertical motion of the body thrown with horizontal velocity 'vi' from a height 'y'.

As S= vi t + 1/2 at²

For the motion of Projectile along y-direction in which;
S=y, vi = viy and a = ay

∴ y = viy t + 1/2 ay t²

As viy = 0 and ay = g

∴ y = 1/2 gt²

Distance Covered Along Horizontal Direction:
Suppose the body covers a distance 'x' in the horizontal direction in 't' second, then

S = vi t + 1/2 at²

Where S = x, vi = vix = vx

And a = ax = 0

x = vix t + 1/2 ax t²

x = vix t + 1/2 (0)t²

x = vix t

Expression for the Resultant Velocity of Projectile:
Suppose a projectile is launched in a direction at angle θ with horizontal with velocity vi. Components of velocity vi are vi cosθ and vi sinθ.

Now along horizontal
vfx = vix + ax t

As ax = 0 so

Horizontal component of final velocity vfx = vx = vix = vi cosθ

Along vertical

Vertical component of final velocity vfy = viy + ay t

ay = -g and viy = vi sinθ

∴ vfy = vy = vi sinθ - gt

Thus during the motion of projectile it has two components of velocity vx and vy. The horizontal component of velocity remains constant whereas its vertical component of velocity is variable.

Magnitude of resultant velocity of projectile at any instant can be determined by putting the value of vx and vy as

v² = vx² + vy²

Therefore
v = √(vx² + vy²) ----------(1)

Eq. (1) shows that the velocity of projectile. The velocity becomes maximum just before hitting the ground.

Direction of Resultant Velocity The resultant velocity 'v' making an angle 'φ' with x-axis which represents its direction. From Fig (b):-

Tanφ= vy / vx

Or φ = tan⁻¹(vy / vx)

As vy changes with time, therefore, the direction of resultant velocity will also change.

Time of Flight of Projectile:
Time taken by the projectile to complete its journey is known as time of flight.

Consider a projectile launched with an initial velocity vi at an angle θ with the horizontal. The components of the initial velocity are:

vix = vi cosθ
viy = vi sinθ

Let 't' is the time taken by the projectile to complete its journey. Using 2nd eq. of motion along vertical direction i.e. along y-axis.

S = vi t + 1/2 at²

y = viy t + 1/2 ay t²

Substituting viy = vi sin θ, ay = -g, s = y = 0

0 = vi sin θ t - 1/2 gt²

1/2 gt² = vi sin θ t

t = 2vi Sinθ / g

Range of Projectile "The horizontal distance covered by the projectile during its total flight is known as its horizontal range".

From diagram, it is clear that, the horizontal range is OQ =R

∴ The horizontal distance covered by projectile due to its motion along x-axis can be written as

R = S = vix × t ----------(2)

Here vix = vi cosθ and

and t = 2vi sinθ / g.

Put in equation (2)

R = (vi cosθ)(2vi sinθ) / g

R = (2v²cosθsinθ) / g

R = vi² (2cosθ sinθ) / g

As sin 2θ=2 sinθ cos θ

∴ R = vi² Sin2θ / g

For Maximum Range The range will become maximum when sin2θ = 1, therefore above equation will become:

Rmax = vi² × 1 / g

Rmax = vi² / g

As sin2θ =1

2θ = sin⁻¹ (1)

2θ = 90°

θ = 90° / 2

θ=45°

Therefore for maximum range projectile should be launched at θ = 45°.

Minimum Range It has minimum range when angle of projection is 90°.

i.e. |R = vi² Sin2θ| = vi² Sin2(90°)

R = vi² Sin180°

R = vi² × 0

R = 0

Height of the Projectile "Maximum vertical distance covered by the projectile in vertical direction is called height of projectile.

For vertical motion we have:
ay = -g
viy = vi sin θ and vfy = 0
s = y = h = ?

Using 3rd eq. of motion along y – axis i.e.

2ay = vfy² - viy²

Substituting the above values we get

2 (-g)h = 0 - (vi sin θ)²

Air resistance will slow down projectile forward motion, reducing its velocity vi. The reduction in vi will result in a decrease in the range of projectile.

Furthermore, air resistance is not constant throughout the flight of object. As the object slows down, the air resistance experienced by it also decreases. This means that the object retards more slowly and accelerates more slowly as it falls down. This results in a trajectory that is not perfectly parabolic, but is skewed, with steeper descent than ascent.

Momentum

Definition "The quantity of motion possessed by a body is known as momentum."
"It is defined as the product of mass and velocity".

Mathematically,
P = m v

It is a vector quantity. Its direction is same as that of velocity.

Unit Its SI units are kg ms⁻¹ or N s. its dimensions are [MLT⁻¹].

Newton's second law of motion in terms of linear momentum:

Consider a body of mass m moving with an initial velocity vi. Suppose an external force F acts upon it for time t after which the velocity becomes vf.

The acceleration a produced by this force is given by

a = (vf - vi) / t

By Newton's second law, the acceleration is given as

a = F / m

Equating the two expressions, we have

F / m = (vf - vi) / t

F = (mvf - mvi) / t

Therefore

F = (Pf - Pi) / t

F = ΔP / t

This shows that force is always equal to the rate of change of linear momentum.

Impulse

Definition "The product of force and time interval (very short) for which it acts on the body is known as impulse." It is represented by 'I'.

Mathematically Impulse = F×Δt
Where F is the average force that acts on the body during the time t.

Unit and Direction Impulse is a vector quantity. The direction of impulse is the same as that of the force.
Its S.I units are Ns or kgm s⁻¹ and dimensions is [MLT⁻¹].

Relation between impulse and momentum:
We know that
F = Δp / Δt

F × Δt = Δp

∴ Impulse = F × Δt
∴ Impulse = Δp
∴ Impulse = pf - pi

Impulse = I = mvf - mvi

Impulse is always equal to change in linear momentum.

Law of Conservation of Linear Momentum:

The total linear momentum of an isolated system of interacting bodies remain constant.

Isolated System

Definition "The system, which is not subjected to an external force, is known as isolated system."
e.g. gas molecules enclosed in a closed vessel form an isolated system.

Proof

Consider an isolated system of two interacting bodies of masses m₁ and m₂ such that their respective velocities are v₁ and v₂ in the same direction. After collision they are moving with velocities v₁' and v₂' as shown in Fig.

Let at the time of collision they exert forces F and F' on each other.

By 3ʳᵈ law of Newton
F = −F'

Or F + F' = 0 ..................(1)

Where F is the force acting on m₂ and F' is the force exerted on m₁.

Then according to 2ⁿᵈ law of motion in terms of momentum, force acting on m₂ can be written as:

F = (m₂v₂' - m₂v₂) / t

⇒ F×t = m₂v₂' - m₂v₂ -------(2)

Similarly
Force acting on m₁ is:

F' = (m₁v₁' - m₁v₁) / t

F't = m₁v₁' - m₁v₁ ------- (3)

Adding eqs. (2) & (3)

F×t + F'×t = (m₂v₂' - m₂v₂) + (m₁v₁' - m₁v₁)

(F + F')t = (m₂v₂' - m₂v₂) + (m₁v₁' - m₁v₁) -------- (4)

From Newton's 3ʳᵈ law

F = - F or F + F = 0

Put value from Eq. (1) in Eq. (4), we have.

0×t = (m₂v₂' - m₂v₂) + (m₁v₁' - m₁v₁)

0 = (m₂v₂' - m₂v₂) + (m₁v₁' - m₁v₁)

Hence m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Total linear momentum before and after collision is equal which proves law of conservation of linear momentum.

Collision

Definition "When two bodies come close to such an extent that there is some sort of interaction takes place between them, such an interaction is called collisions."

Types There are two types of collision.
(i) Inelastic collision (ii) Elastic collision

(i) Inelastic Collision:

Definition "When the total linear momentum is conserved but K.E is not conserved, the collision is known as inelastic collision."

It should be noted that during inelastic collision a portion of K.E is lost partially due to friction, heat and sound energies.

Examples (i) Car crashes
(ii) Two clay balls colliding and sticking together

(ii) Elastic Collision:

Definition "When the total K.E. energy and total linear momentum of the system is conserved, then the collision is known as elastic collision." Collision are considered nearly elastic under certain conditions.

Elastic Collision in One Dimension:

The collision is said to be one dimensional if before and after collision the direction of motion of colliding bodies will remain same. Let v₁ and v₂ are the initial velocities of two spherical bodies of mass m₁ and m₂ respectively v₁' and v₂' are the velocities after collision. Both the bodies are moving along the line joining their centres before and after the collision, and they are non-rotating, therefore, the collision is one dimensional. By applying the conditions of elastic collision:

Conservation of momentum Initial momentum = final momentum

Rearranging,
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
m₁v₁ - m₁v₁' = m₂v₂' - m₂v₂
m₁(v₁ - v₁') = m₂(v₂' - v₂) ..................(1)

Conservation of K.E.

According to second condition of elastic collision
1/2 m₁v₁² + 1/2 m₂v₂² = 1/2 m₁v₁'² + 1/2 m₂v₂'²

Multiply by (2) on both sides

Rearranging, m₁v₁² + m₂v₂² = m₁v₁'² + m₂v₂'²
m₁v₁² - m₁v₁'² = m₂v₂'² - m₂v₂²
m₁(v₁² - v₁'²) = m₂(v₂'² - v₂²) ...........................(2)

Dividing eq. (2) by (1)
m₁(v₁² - v₁'²) / m₁(v₁ - v₁') = m₂(v₂'² - v₂²) / m₂(v₂' - v₁)

(v₁ - v₁')(v₁ + v₁') / (v₁ - v₁') = (v₂' - v₂)(v₂' + v₂) / (v₂' - v₂)

v₁ + v₁' = v₂' + v₂ ...........(3)

OR
v₁ - v₂ = v₂' - v₁'

(v₁ - v₂) = -(v₁' - v₂') ................... (4)

(v₁ - v₂) is the relative speed of two bodies with which they are approaching each other before collision. It is also called relative speed of approach. (v₂' - v₁') is the relative speed of separation after collision. The negative sign shows that direction of these velocities is opposite to each other. Therefore, Speed of approach = Speed of recession

Calculation of v₁' and v₂':

For v₁' From eq. (3) v₂' = v₁ + v₁' - v₂ ............. (a)

Substituting value of v₂' in eq. (1)
m₁(v₁ - v₁') = m₂(v₁ + v₁' - v₂ - v₂)
m₁(v₁ - v₁') = m₂(v₁ + v₁' - 2v₂)
m₁v₁ - m₁v₁' = m₂v₁ + m₂v₁' - 2m₂v₂
m₁v₁ - m₂v₁ + 2m₂v₂ = m₁v₁' + m₂v₁'
(m₁ - m₂)v₁ + 2m₂v₂ = v₁'(m₁ + m₂)

Dividing both sides by (m₁ + m₂)

[(m₁ - m₂) / (m₁ + m₂)]v₁ + [2m₂v₂ / (m₁ + m₂)] = v₁' ..................(5)

For v₂' By putting the value of v₁' in eq. (a) we get,

v₂' = v₁ - v₂ + v₁'

v₂' = v₁ - v₂ + [(m₁ - m₂) / (m₁ + m₂)]v₁ + [(2m₂) / (m₁ + m₂)]v₂

v₂' = v₁ + [(m₁ - m₂) / (m₁ + m₂)]v₁ + [(2m₂) / (m₁ + m₂)]v₂ - v₂

v₂' = [1 + (m₁ - m₂) / (m₁ + m₂)]v₁ + [(2m₂) / (m₁ + m₂) - 1]v₂

v₂' = [(m₁ + m₂ + m₁ - m₂) / (m₁ + m₂)]v₁ + [(2m₂ - m₁ - m₂) / (m₁ + m₂)]v₂

v₂' = [(2m₁) / (m₁ + m₂)]v₁ + [(m₂ - m₁) / (m₁ + m₂)]v₂

v₂' = [(2m₁) / (m₁ + m₂)]v₁ + [(m₂ - m₁) / (m₁ + m₂)]v₂ ................(6)

Special Cases

Case I When m₁ = m₂ = m

Eq. (5) and (6) will become

v₁' = [(m - m) / (m + m)]v₁ + [(2m)v₂ / (m + m)]
v₁' = 0(v₁) + [2mv₂ / 2m]
v₁' = 0 + v₂
v₁' = v₂

v₂' = [(2m) / (m + m)]v₁ - [(m - m) / (m + m)]v₂
v₂' = [2m / 2m]v₁ - (0)v₂
v₂' = v₁

Hence, after collision the bodies will interchange their velocities.

Case II When masses of two bodies are equal and second body is at rest i.e. m₁ = m₂ = m and v₂ = 0

Eq. (5) and (6) can be written as:

v₁' = [(m - m) / (m + m)]v₁ + [(2m)0 / (m + m)]
v₁' = 0

v₂' = [(2m) / (m + m)]v₁ - [(m - m) / (m + m)] × 0
v₂' = [2m / 2m]v₁ + 0
v₂' = v₁

In this case bodies will again interchange their velocities.

Case III When a light body collides with the massive body at rest. In this case m₁ is very small as compared to m₂
i.e. m₁ <<< m₂ i.e. m₂ ≈ 0

Neglecting m₁ in eq. 5 and 6 and putting v₂ = 0:

v₁' = [-m₂ / m₂]×v₁ + [2m₂ / m₂](0) ⇒ v₁' = -m₂/m₂ v₁ + (0)
⇒ v₁' = -v₁

Similarly v₂' = [2(0)v₁ + m₂ / m₂] × (0) v₂' = 0 + 0
v₂' = 0

Hence heavier body will remain at rest and lighter body will rebound with its own velocity.

Case IV In this case m₁ is very very large than m₂
i.e. m₁ >>> m₂ and v₂ = 0 i.e. m₂ ≈ 0

Neglecting m₂ in eq. (5) & (6) and putting v₂ = 0

v₁' = [m₁ / m₁]×v₁ + (0 / m₁)|0

v₁' = v₁

v₂' = [2m₁v₁ + (-m₁) / m₁]0 ⇒ v₂' = [2m₁v₁ + 0] / m₁

v₂' = 2v₁

Hence lighter body will move with double of the velocity of m₁, while the velocity of heavy body will remain same.

Elastic Collision in Two Dimensions

Consider the motion of two balls of mass m₁ and m₂ in a straight line with velocities v₁ and v₂ respectively undergoing an elastic collision with each other as shown in Fig.

Assume the bodies move off in different directions after collision with velocities v₁' and v₂' making angles θ₁ and θ₂ respectively with x-axis.

As, the collision is elastic, so we apply both the laws of conservation of momentum and law of conservation of kinetic energy. Momentum is a vector quantity, we resolve it into its rectangular components and apply the law of conservation of momentum along both axes.

Momentum conservation along x-axis is:

Momentum before collision = momentum after collision
m₁v₁+m₂v₂ = m₁v₁' cosθ₁+m₂v₂' cosθ₂ ............(1)

Momentum conservation along y-axis

Momentum before collision = Momentum after collision
0 = m₁v₁' sinθ₁ - m₂v₂' sinθ₂ .....................(2)

Conservation of Energy

Kinetic Energy before collision = Kinetic Energy after collision
1/2 m₁v₁² + 1/2 m₁v₂² = 1/2 m₁v₁'² + 1/2 m₂v₂'² ...(3)

These equations (1 to 3) are useful in solving problems about Elastic collision in two dimensions.

Consider two bodies having masses m and m₁ and m₂, moving with velocities v₁ and v₂ along the same line shown in Fig. such that v₁ > v₂. In such a case m₁ is regarded as projectile and m₂ as target. After time t both the bodies make inelastic collision and stick together. Let their combined mass become m₁ + m₂ which moves with final velocity vf after collision.

Since the collision is perfectly inelastic, the total momentum of balls is conserved. Using law of conservation of momentum.

Total momentum of system before collision = Total momentum of the system after collision
m₁v₁ + m₂v₂ = (m₁ + m₂)vf

vf = [m₁ / (m₁ + m₂)]v₁ + [m₂ / (m₁ + m₂)]v₂ ............. (1)

Which gives the common velocity of the body after inelastic collision.

In a special case when the target m₂ is at rest, v₂ = 0, the above equation becomes:

vf = [m₁ / (m₁ + m₂)]v₁

It shows that velocity of m₁ is reduced by the mass ratio i.e., m₁ / (m₁ + m₂).

Inelastic Collision in Two Dimensions

The macroscopic collisions are generally inelastic and do not conserve Kinetic energy. The perfect inelastic collision is one in which the colliding objects stick together to make a single mass after collision. Its analysis can be carried out as follows:

Let us take two balls having masses m₁ and m₂ moving with velocities, v₁ and v₂, respectively, in a two-dimensional xy-plane. Assume that the first body is moving along the x-axis while the second body moves in a direction, making an angle θ with x-axis. Both the bodies collide at the origin as shown in the Fig.

After collision, bodies stick together, having combined mass M = m₁ + m₂, which moves with velocity vf, making an angle φ with x-axis.

Momentum in the x-direction m₁v₁ + m₂v₂ cosθ = Mvf cosφ ............(1)

Momentum in the y-direction 0 + m₂v₂ sinθ = MVf sinφ ......... (2)

Equation (1) and (2) can be used to find the final velocity.

Kinetic Energy

Since collision is inelastic, the kinetic energy of colliding system is not conserved. The loss of kinetic energy is computed as follows:

Initial Kinetic Energy

The total initial kinetic energy K.Ei of the system before the collision is:

(K.E)i = 1/2 m₁v₁² + 1/2 m₂v₂² ---- (3)

Since K.E. is a scalar quantity, so velocities involving in the formula of K.E. does not require to break velocities into their components.

Final Kinetic Energy

(K.E)f = 1/2 Mvf² .............(4)

Where vf is magnitude of the final velocity which can be calculated from Eq. (2.41)

Energy Loss in the Collision

Since the collision is inelastic, there is a loss in kinetic energy, represented by ΔK.E.

ΔK.E. = (K.E)i - (K.E)f

This lost kinetic energy is transformed into other forms of energy, such as heat, sound, or in deformation.

Some Examples of inelastic collision:

i. When a karate chop breaks a pile of bricks, it's an examples of an inelastic collision. In this type of collision, the objects involved don't bounce back after impact. Instead, some of the energy from the strike is absorbed by the bricks, turning into heat, sound, and the force needed to break them. This means the energy goes into braking the bricks rather than causing the hand to rebound. If the karate chop is not perfectly vertical and involves some horizontal motion, the momentum transfer and the resulting forces will have both horizontal and vertical components.

ii. In a car crash, the collision is inelastic in nature. When the vehicles collide and absorb the impact energy, causing them to crumple and deform. This energy absorption slows down the cars, stopping them from bouncing back. Most of the kinetic energy is lost, turning into heat, sound, and damage to the vehicles.

iii. In real-world collisions, a ball and bat show inelastic behaviour. When the bat hits the ball, some of the kinetic energy is lost because the ball deforms, and energy is also converted into heat and sound. Even though the bat is rigid, it doesn't transfer energy perfectly and absorbs some energy itself. The ball compresses upon impact, which leads to further energy loss. Consequently, not all of the initial kinetic energy is conserved, making the collision inelastic overall.

Rocket Propulsion

The motion of rocket is according to the principle of the law of conservation of linear momentum. The Rocket and fuel makes an isolated system. Due to combustion of fuel by the engine of the rocket it is converted into hot gases. These gases are expelled at high velocity from the tail of rocket. As a result, rocket gains momentum in the upward direction equals to momentum of the gas expelled from engine but in opposite direction. Such gases are expelled continuously, so rocket goes on moving faster and faster as long as engines are operating.

Expression of Acceleration of the Rocket

Let 'm' is the mass of the hot gases ejected per second from the tail of rocket with velocity 'v'. the momentum of the hot gases = - mv

The negative sign shows the direction of velocity of hot gases vertically downward. Initial momentum of the fuel = 0

Change in momentum during combustion of fuel.
Δp = pf - pi
Δp = -mv - 0
Δp = -mv

Where Δp is the change in momentum per second or rate of change of momentum is always equal to the force, therefore, the force on hot gases. F = -mv

By third law of Newton force of same magnitude will act on rocket in the upward direction.

Force on rocket is
F = mv ....................(1)

The acceleration of the rocket is

a = F / M

As F = mv

Therefore a = mv / M

a = (m / M)v ...(2)

The equation (2) shows the acceleration of the rocket is inversely proportional to its mass. Since mass of the rocket decreases due to consumption of fuel, its acceleration will increase.

It should be noted that:
i. Typical rocket consumes fuel at the rate of 10,000 kg/s
ii. The velocity of hot gases is 4000 m/s
iii. More than 80% mass of rocket at launching paid is due to its fuel.
iv. To overcome the problem of fuel several rocket are linked together. It is called multistage rocket.