In figure

Elastic region : O to E.
Plastic region: E to B.
Upto the point E, the steel wire will regain its status immediately on the removal of stress and the ratio of \(\frac{\text { Stress }}{\text { Strain }}\) will be a constant.
(i) Strain increases in proportion to the load upto P. But beyond P, it increases by an increasingly greater amount for a given increase in the load. Beyond the elastic limit E, it does not retrace the curve backward. The wire is unloaded but returns to 'O' along the dotted line AO'. Point O corresponding to zero load which implies a permanent strain in wire.
(ii) Prom C to B, strain increases even if the wire is being unloaded and at B it fractures. Stress upto that corresponding to C can be applied without causing fracture.

(a) Let a body with constant velocity v, between time t₁ and t₂ as shown in graph.
Area below = area ABCD = v (t₂ - t₁)
Also, the displacement
= velocity x time = v x (t₂ - t₁)
Thus proved.

(b) Slope of v- t graph is constant. So there is uniform acceleration. Area below the graph gives the displacement (x).

Displacement = Area of trapezium
\(=\mathrm{ON} \times \mathrm{OP}+\frac{1}{2} \mathrm{NA} \times \mathrm{MA}\)
\(=v_{0} \times t+\frac{1}{2}(t)\left(v-v_{0}\right)\)
Since \(a=\frac{v-v_{0}}{t}\)
We have, \(x=v_{0} t+\frac{1}{2} t \text { at }=v_{0} t+\frac{1}{2} a t^{2}\)
(c) u = 126 km/hr = 35 m/sec, v = 0, S = 200 m
using, v² = u ² + 2as
\(a=\frac{v^{2}-u^{2}}{2 s}=\frac{35^{2}}{2 \times 200}\)
= - 3.00 ms^-2
v = u + at
0 = 35 - 3 x t
\(t=\frac{35}{3}=11.66 \mathrm{sec}\) 

Elastic collision is one in which both momentum and energy are conserved. Consider two bodies A and B of masses m₁ and m₂ moving in a straight line with velocities u₁ and u₂ respectively (u₁ > u₂). After collision, let their velocities change to v₁ and v₂ respectively. If the centres of two colliding bodies are in one line, then the collision is said to be head-on-collision.

Total linear momentum of the system before collision = m₁ u₂ + m₂u₂
Total linear momentum of the system after collision = m₁ v₁ + m₂v₂
As collision is elastic, so the linear momentum and kinetic energy of the system are conserved.
According to the law of conservation of linear momentum,
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂        ..........(1)
or m₁ (u₁ - v₁) = m₂ (v₂ - u₂)       ..........(2)
According to the law of conservation of kinetic energy,
\(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }{ u }_{ 2 }^{ 2 }=\frac { 1 }{ 2 } { m }_{ 1 }{ v }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }{ u }_{ 2 }^{ 2 }\)  ......(3)
or \({ m }_{ 1 }({ u }_{ 1 }^{ 2 }-{ u }_{ 1 }^{ 2 })={ m }_{ 2 }({ v }_{ 2 }^{ 2 }-{ u }_{ 2 }^{ 2 })\)
or m₁ (u₁ + v₁) (u₁ - v₁)= m₂ (v₂ + u₂) (v₂ - u₂)  .......(4)
Dividing eqn. (4) by eqn. (2), we get
u₁ + v₁ = v₂ + u₂                        ............(5)
u₁ + u₂ = v₂ + v₁                    ............(6) 
From eqn. (6) v₂ = u₁ - u₂ + v₁  ........(7)
Substituting the value of v₂ in eqn. (2), we get
m₁(u₁-v₁) = m₂ [u₁-2u₂+v₁]
or \({ v }_{ 1 }=\frac { 2{ m }_{ 2 }{ u }_{ 2 }+{ u }_{ 1 }({ m }_{ 1 }-{ m }_{ 2 }) }{ ({ m }_{ 1 }+{ m }_{ 2 }) } \)
Substituting value of v₁ in eqn. (7), we get
\({ v }_{ 2 }=\frac { 2{ m }_{ 1 }{ u }_{ 1 }+{ u }_{ 2 }({ m }_{ 2 }-{ m }_{ 1 }) }{ ({ m }_{ 1 }+{ m }_{ 2 }) } \)

Let the field be zero at a point at distance x from M₁.
\(\therefore \frac{GM_1}{x^2}=\frac{GM_2}{(d-x)^2}\)
\(\therefore \frac{x}{d-x}=\sqrt {\frac{M_1}{M_2}}\Rightarrow x\sqrt {M_2}=\sqrt {M_1}.d-x\sqrt {M_1}\)
\(x[\sqrt {M_!}+\sqrt {M_2}]=\sqrt {M_1}.d\)
\(x=\frac{d\sqrt {M_2}}{\sqrt {M_1}+\sqrt {M_2}}\)
\(d-x=\frac{d\sqrt {M_2}}{\sqrt {M_1}+\sqrt {M_2}}\)
Potential at this point due to both the masses will be
\(=-\frac{GM_1}{x}-\frac{GM_2}{(d-x)}=-G[\frac{M_1(\sqrt {M_1}+\sqrt {M_2})}{d\sqrt {M_1}}+\frac{M_2(\sqrt {M_1}+\sqrt {M_2})}{d\sqrt {M_2}}]\)
\(=-\frac{G}{d}(\sqrt {M_1}+\sqrt {M_2})^2=\frac{G}{d}(M_1+M_2+2\sqrt {M_1}\sqrt {M_2}).\)

The angle between two vectors is included in the expression of dot product or scalar product.
A.B = ABcos Q
The magnitude of \(\overrightarrow { A } \) is given by
 \(A=\quad \sqrt { \left( 1 \right) ^{ 2 }+\left( 2 \right) ^{ 2 }+\left( -1 \right) ^{ 2 } } =\sqrt { 6 } \)
The magnitude of \(\overrightarrow { B } \) is given by
\(B=\quad \sqrt { \left( -1 \right) ^{ 2 }+\left( 1 \right) ^{ 2 }+\left( -2 \right) ^{ 2 } } =\sqrt { 6 } \)
We can separately evaluate the left side of Eq.(i) by writing the vectors in unit vector notation and using the distributive law.
\(A.B=\left( \hat { i } +\hat { 2j } -\hat { k } \right) .\left( \hat { -i } +\hat { j } -\hat { 2k } \right)\)
\(=\left( 1.-1 \right) \left( \hat { i } .\hat { i } \right) +\left( 1.1 \right) \left( \hat { i } .\hat { j } \right) +\left( 1.-2 \right) \left( \hat { i } .\hat { k } \right) +\left( 1.-1 \right) \)
\(\\ \left( \hat { i } .\hat { j } \right) +\left( 2.1 \right) \left( \hat { j } .\hat { j } \right) +\left( 2.-2 \right) \left( \hat { j } .\hat { k } \right) +\left( -1.-1 \right)\)
\( \\ \left( \hat { k } .\hat { i } \right) +\left( -1.1 \right) \left( \hat { k } .\hat { j } \right) +\left( -1.-2 \right) \left( \hat { k } .\hat { k } \right) \)
\(\left[ \begin{matrix} \hat { i } .\hat { i } =1,\hat { i } .\hat { j } =1,\hat { i } .\hat { k } =1 \\ \hat { j } .\hat { j } =1,\hat { j } .\hat { i } =1,\hat { j } .\hat { k } =1 \\ \hat { k } .\hat { k } =1,\hat { k } .\hat { i } =1,\hat { k } .\hat { j } =1 \end{matrix} \right] \)
Substituting the values of A from Eq.(ii) and B from Eq.(iii) and A.B=6 in Eq.(i)we get
\(3=\left( \sqrt { 6 } \right) \left( \sqrt { 6 } \right) cos\theta \)
\(\\ \Rightarrow cos\theta =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \Rightarrow \theta =cos^{ -1 }\left( \frac { 1 }{ 2 } \right) =60^{ 0 }\) 

Power. The rate of doing work is called power.
\(\mathrm{P}=\underset{\Delta t \rightarrow 0}{L t} \frac{\Delta \mathrm{W}}{\Delta t}=\frac{d \mathrm{~W}}{d t}\)
\(=\frac{d}{d t}(\vec{F} \cdot \vec{s})=\vec{F} \cdot \frac{d \vec{s}}{d t}\)
\(\mathrm{P}=\vec{F} \cdot \vec{v}\)
⇒ P = Fv

Kinetic energy and speed

Yes, it will be negative if the angle between the two vectors lies between 90⁰ to 270 ^o.

Curved roads are generally banked so as to help in providing centripetal force needed to balance the centrifugal force, arising due to circular motion on the curved road.

The change in the position co-ordinates of a particle over a given period of time is called the displacement of the particle.

The ratio of lateral strain to the longitudinal strain is called Poisson's ratio.

The universe would collapse. We would be thrown away because of the centrifugal force. Eating, drinking and infact all activities would become impossible.

Angular momentum L is the product of twice the mass and aerial velocity.
\(\therefore \ \mathrm{L}=2 \mathrm{M} \frac{d \mathrm{~A}}{d t}=2 \mathrm{M}_{1} \times \frac{\pi r^{2}}{\mathrm{~T}}\)
\(=2 \mathrm{M}_{1} \frac{\pi r^{2} v}{2 \pi r}\)
\(=\mathbf{M}_{1} r \sqrt{\frac{\mathrm{GM}_{2}}{r}}\)
\(=\mathrm{M}_{1} \sqrt{\mathrm{GM}_{2} r}\)

Work done on a body is reflected as change in kinetic energy, according to Work-Energy theorem.
\(\mathrm{W}=\int \mathrm{F} d x=\int m \frac{d v}{d f} d x\)
\(=\int m v d v=\left|\frac{1}{2} m v^{2}\right|_{v i}^{v_{f}}\)
Work done \(=\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)\)
\(=\frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}\)
Work done = Change in kinetic energy.

When no external force acts on a body, there is no change in linear momentum or the total momentum of an isolated system of interacting particles remain conserved.
Let us consider an isolated system (which is free from the influence of any external forces) comprising of two bodies A and B with initial momentum  P A and P B. Let them collide for a small time Δt and separate with final momentum \(\overrightarrow{p_{A}}^{\prime} \text { ând } \overrightarrow{p_{B}}^{\prime} \mathrm{r}\) respectively. During collision, If \(\overrightarrow{F_{A B}}\) is force on A exerted by B, and \(\overrightarrow{F_{B A}}\) is force on B exerted by A, then according to Newton's second law.
\(\overrightarrow{F_{A B}} \times \Delta t\) = Change in momentum of A
\(=\overrightarrow{p_{A}^{\prime}}-\overrightarrow{p_{A}}\)                ...........(i)
\(\overrightarrow{F_{B A}} \times \Delta t\) = Change in momentum of B
\(=\overrightarrow{p_{B}}^{\prime}-\overrightarrow{p_{B}}\)                .............(ii)
According to Newton's third law,
\(\overrightarrow{F_{A B}}=-\overrightarrow{F_{B A}}\)
ஃ From (i) and (ii),
\(\overrightarrow{p_{A}}^{\prime}-\overrightarrow{p_{A}}=-\left(\overrightarrow{p_{B}^{\prime}}-\overrightarrow{p_{B}}\right)\)
Which shows that total final momentum of the isolated system is equal to its total initial momentum.
\(\Rightarrow \overrightarrow{p_{A}}+\overrightarrow{p_{B}}=\overrightarrow{p_{A}}+\overrightarrow{p_{B}}\)
This proves the law of conservation of linear momentum.

If the speed of the particle in circular path remains constant, the motion is uniform circular motion. We know, the arc length x covered with an angular displacement θ is x = rθ.
Differentiating,
\(\frac{d x}{d t}=r \frac{d \theta}{d t} \quad \because r \text { is constant, } \quad \therefore v=r \omega \text { . }\)

(a) \(h_{1}=\frac{u^{2}}{2 g}=\frac{20^{2}}{2 \times 10}=20 \mathrm{~m}\)
Total height = 25 + 20 = 45 m
(b) \(h=u t-\frac{1}{2} g t^{2}\)
\(-25=20 t-\frac{1}{2} \times 10 \times t^{2}\)
5t² - 20t - 25 = 0
(t - 5) (t + 1) = 0
⇒ t = 5 sec
(c) Trajectory will be a vertical line
\(y=u t+\frac{1}{2} g t^{2}\)
x = 0, y ≠ 0 

Parallelogram law of vector addition. If two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) represent two adjacent sides of a parallelogran. the sum of the vectors is represented by the diagonal of the parallelogram.

Let \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\) be two vectors at an angle θ between them. According to the law of parallelogram of vectors, the diagonal of the parallelogram indicates the sum of the other two sides/vectors \(\overrightarrow{\mathrm{A}} \text { and } \overrightarrow{\mathrm{B}}\)
\(|\overrightarrow{\mathrm{OQ}}|=|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{OT}^{2}+\mathrm{TQ}^{2}}\)
\(|\vec{R}|=\sqrt{(A+B \cos \theta)^{2}+(B \sin \theta)^{2}}\)
\(\mathrm{R}=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos \theta}\)
The resultant R is at an angle α to \(\overrightarrow{\mathrm{A}}\) given by
\(\alpha=\tan ^{-1}\left(\frac{\mathrm{B} \sin \theta}{\mathrm{A}+\mathrm{B} \cos \theta}\right)\) 

(i) [M⁰L²T^-2]
(ii) [MLT^-2]
(iii) [ML^-2T^-2]
(iv) [ML ²T⁰]
(v) [MLT²]
(vi) [ML²T^-1]
(vii) [ML²T^-2]
(viii) [ML²T^-1]
(ix) [ML²T^-2]
(x) [ML²T^-2]

Average velocity: Average velocity is the displacement divided by the time interval in which the displacement occurs.
\(\overrightarrow { v_{ av } } =\frac { \Delta \overrightarrow { x } }{ \Delta t } \)
Instantaneous velocity: Instantaneous velocity is defined as the limit of the average velocity as the time interval M becomes infinitesimally small.
\(\overrightarrow { v } =\lim _{ \Delta t\rightarrow 0 }{ \frac { \Delta \overrightarrow { x } }{ \Delta t } } =\frac { dx }{ dt } \)

No, escape velocity is independent of the direction of projection