The gravitational potential energy U(r) of a system of two masses m₁ and m₂ separated by a distance r as the amount of work done to bring the mass m₂ from infinity to a distance r assuming m₁ to be fixed in its position and is written as
\(\mathrm{U}(\mathrm{r})=-\frac{G m_{1} m_{2}}{r}\)

Mass spring = m
Force constant = k
Un-stretched length = l 
Angular velocity = ω
Let 'x' be the increase in the length of the spring.
The new length = (l +x) = r
When the spring is rotated in a horizontal circle, Spring force = centripetal force.
kx = mω² (l + x)
x=\(\frac{mω^2l}{k-mω^2}\)

m₁ = 9000 kg, u₁ = 36 km/h = 10 m/s
m₂ = 9000 kg, u₂ = 0, v = v₁= v₂ =?
By conservation of momentum:
m₁u₁ + m₂u₂ = (m₁ + m₂)v
\(\therefore\) v = 5 m/s
Total K.E. before collision = \(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }{ u }_{ 2 }^{ 2 }=45000J\)
Total K.E. after collision = \(\frac { 1 }{ 2 } \left( { m }_{ 1 }+{ m }_{ 2 } \right) { v }^{ 2 }=225000J\)
As total K.E. after collision < Total K.E. before the collision
\(\therefore\) Collision is inelastic.

F = K.x so x = \(\frac { F }{ K } \)
For same F, \({ W }_{ A }=\frac { 1 }{ 2 } { K }_{ A }{ x }^{ 2 }=\frac { 1 }{ 2 } \frac { { F }^{ 2 } }{ { K }_{ A } } \)
and \({ W }_{ B }=\frac { { F }^{ 2 } }{ { 2K }_{ B } } \)
\(\therefore\) \(\frac { { W }_{ A } }{ { W }_{ B } } =\frac { { K }_{ B } }{ { K }_{ A } } \)
As K_A > K_B so W_A < W_B

No acceleration is towards the centre only in case of uniform circular motion.

Average velocity
\(\vec v_{avg}=\frac{\vec r_p-\vec r_O}{\Delta t}\)
Here \(\Delta t=5s\)
 \(\vec r_0-i ;\vec r_p=10\hat i\)
\(\vec v_{avg}=\frac{10\hat i}{5 sec}=2\hat icms^{-1}\)
The average velocity is in the positive x-direction.
The average speed total path length/time taken (the path is semi-circular)
= \(\frac{5\pi cm}{5s}=\pi cms^{-1}=3.14 cms^{-1}\)
Note that the average speed is greater than the magnitude of the average velocity.

M.I. of the disc about its own axis.
\(I={1\over2}mr^2\)
As v = \(rω\therefore ω^2={v^2 \over r^2}\)
Rotational K.E. \(={1\over 2}Iω^2={1\over2}\times{1\over 2}mr^2\times{v^2 \over r^2}={1 \over 4}mv^2\)
Translational K.E \(={1\over 2}mv^2\)
Translational K.E = Rotational K.E. + Translational K.E

As shown in figure, resolve the tension T₁ along horizontal and vertical directions. As the body is in equilibrium,
T₁ sin60⁰ = 4kg wt =4 x 9.8N        ....(1)
T₁ cos60° = T₂                           ......(2)
From (1), T₁ = \(\frac { 4\times 9.8 }{ sin60^{ 0 } } =\frac { 4\times 9.8\times 2 }{ \sqrt { 3 } } \) = 45.26 N
From (2), T₂ = T₁ cos60⁰ = 45.26 x 0.5 = 22.63N

According to conservation of linear momentum.
mv + MV = 0
v = \(\frac { -mv }{ M } =\frac { -0.1\ \times\ 2000 }{ 10 } \)
v = -20 m/s

Given:
t=3s
Distance s = Area of \(\triangle OAB\)

\(=\frac{1}{2}\times OA\times BA\)
\(=\frac{1}{2}\times 3\times 6=9m\)
If the speed varies with the time: Then \(v=\frac{ds}{dt}\Rightarrow ds=v\ dt\)
\(\Rightarrow \int { ds } =\int { v }\ dt\)
\(or\ s=\int { v }\ dt\)

In the diagram \(\angle\) LOR is called the parallax angle or parallactic angle

Mass of ball(m) = \(\frac { 35 }{ 100 } =0.035kg\)
A ball hits by a stumps at an angle (\(\theta\)₁) = 30°
Ball rebounds at an .angle (\(\theta\)₂) = 60°
Change in momentum along horizontal direction
= -mu cos30° - (mu cos30⁰)
= -2 mu cos30°
= -2\(​​\times\)0.035\(​​\times\)20\(​​\times\)cos30°
= -2 \(​​\times\)0.035\(​​\times\)20\(​​\times\)\(\sqrt { \frac { 3 }{ 2 } } \)  \(\quad \left[ \because cos{ 30 }^{ 0 }=\sqrt { \frac { 3 }{ 2 } } \right] \)
= \(-1.4\times \frac { \sqrt { 3 } }{ 2 } =1.21\quad kgms^{ -1 }\)
The impulse received by a ball j = 1.2 kg ms^-1

(i) Unit vector:
A vector divided by its magnitude is a unit vector. The unit vector for\(\overrightarrow { A } \)is denoted by\(\hat { A } \). It has a magnitude equal to unity or one.
Since, \(\hat { A } =\frac { \overrightarrow { A } }{ A } we\ van\ write\ \overrightarrow { A } =A\hat { A } \)
Thus, we can say that the unit vector specifies only the direction of the vector quantity.
(ii) Orthogonal unit vectors:
Let \(\hat { i } ,\hat { j } \)  and \(\hat { k } \) be three unit vectors which specify the directions along positive x-axis, positive y-axis and positive z-axis respectively. These three unit vectors are directed perpendicular to each other, the angle between any two of them is 90°. \(\hat { i } \hat { j } \) and \(\hat { k } \) are examples of orthogonal vectors. Two vectors which are perpendicular to each other are called orthogonal vectors.

Gross Error
(i) The error caused due to the shear carelessness of an observer is called gross error.
For example:
(ii) Reading an instrument without setting it properly.
(iii) Taking observations in a wrong manner without bothering about the sources of errors and the precautions.
(iv) Recording wrong observations. These errors can be minimized only when an observer is careful and mentally alert.

Newton's third law states that for every action there is an equal and opposite reaction.

Limitations of Dimensional analysis: 
(i) This method gives no information about the dimensionless constants in the formula like 1, 2,................ \(\pi\), e, etc.
(ii) This method cannot decide whether the given quantity is a vector or a scalar.
(iii) This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.
(iv) It cannot be applied to an equation involving more than three physical quantities.
(v) It can only check on whether a physical relation is dimensionally correct but not the correctness of the relation.
For example, using dimensional analysis, s = ut + 1/3 at² is dimensionally correct whereas the correct relation is s = ut+1/2 at².

Escape speed is the minimum speed of an object thrown vertically up such that it escapes the Earth's gravity and would never come back.

The line of action of normal force N exerted by the table on the block does not depend on h because the reactionary force N exerted by the table which is directed vertically upward and passes through its centre of gravity. Since the block is in equilibrium, N=mg.
Friction is always perpendicular to the normal force N acting between the surfaces. It acts tangential to the surface of contact.

(i) Bottle filled with water rolls, faster than the empty bottle. Due to M.I. I = mr².
(ii) When it rolls, down it possesses translational K.E. and rotational K.E.
(iii) For the empty bottle 100% of the mass of the bottle spins as the bottle rolls.
(iv) But for full bottle, much of the water in the bottle is effectively sliding down without spinning.
(v) Thus 100% of the mass of the sliding water goes into translational K.E. and full bottle have a greater speed

Law of conservation of angular momentum states that, when no external torque acts on the body, the net angular momentum of a rotating body remains constant.

Let the height of the pole be h
Solution \(\frac { x+45 }{ h } \)  = cot 30^o ⇒ h =\(\frac { x+45 }{ cot\quad { 30 }^{ o } } \)
\(\frac { x }{ h } \) = cot 30^o ⇒ x = h cot 60^o
Substituting the values of x in the above equation
h = \(\frac { h\quad cot \ { 60 }^{ o }+45 }{ cot \ { 30 }^{ o } } \)
\(h \cot 30^{\circ} =h \cot 60^{\circ}+45  \)
\(h\left(\cot 30^{\circ}-\cot 60^{\circ}\right) =45  \)
\(h =\frac{45}{\cot 30^{\circ}-\cot 60^{\circ}}=\frac{45}{\sqrt{3}-\frac{1}{\sqrt{3}}}=38.97 \mathrm{~m}\)