From the figure, the distance of m₁ from the origin = a
From the figure, the distance of m₂ from the origin = a
Gravitational field \(\mathrm{E}=\frac{G M}{r^{2}} \hat{r}\)
At the origin (Point O) the change in gravitational field is
\(\vec{E}=\frac{G M}{a^{2}}\left[\left(m_{1}-m_{2}\right) \hat{i}+m_{3} \hat{j}\right]\)
It is given that
\(\mathrm{m}_{1} =\mathrm{m}_{2} \)
\(\therefore \vec{E} =\frac{G M}{a^{2}}\left[m_{3} \hat{j}\right]\)

Mass of the earth M = 6.024 5 10²⁴ kg
Radius of the earth R = 6.4\(\times\)10⁶ m
Gravitational constant G' = 6.67\(\times\)10¹¹
Gravitational constant G = 6.67\(\times\)10¹¹
Acceleration due to gravity g' =?
g' = 
g' = 9.8\(\times\)10²² m/s²
g = 9.8 m/s²
∴ g' = g\(\times\)10²² (or) 10²²
g = g' m/s²
Weight W = mg
W' = mg'
 = 10²² .W
W = 10²²

Relative velocity of B with respect to A
 \({ v }_{ BA }={ v }_{ B }-{ v }_{ A }\)
= 50 km h^-1 + (-50) km h^-1 
= 0 kmh^-1
Similarly, relative velocity of A with respect to B i.e., v_AB is also zero.
Thus each train will appear to be at rest with respect to the other.

Relative velocity of A with respect to B, \( { v } _{ AB }\)= 80 km h^-1 due east
Thus to a passenger in train B, the train A will appear to move east with a velocity of 80 km h^-1.The relative velocity of B with respect to A, V_BA = 80 km h^-1 due west.
To a passenger in train A, the train B will appear to move westwards with a velocity of 80 km h^-1

Final momentum on each object Δp = FΔt = 100\(\times\)5 = 500 kg ms^-1
Final speed on the object of mass 5 kg = 500/5 =100 m s^-1
Final speed on the object of mass 20 kg = 500/20 =25 m s^-1
Note that momentum on each object is the same after 5 seconds but speed is not the same after 5 seconds. The heavier mass acquires lesser speed than the one with lower mass.

Radius of the curve r = 36 m.
Coefficient of friction μ= 0.53
Acceleration due to gravity g = 10 ms^-2
vmax=\(\sqrt { \mu rg } =\sqrt { 0.53\times 36\times 10 } \)=13.81 ms^-1.

We know that a=\(\left[ \frac { F }{ { m }_{ 1 }+{ m }_{ 2 }+{ m }_{ 3 } } \right] =\frac { 50N }{ 10kg+7kg+2kg } =\frac { 50 }{ 19 } \)= 2.63 ms^-2

a=\(\frac { { m }_{ 1 }-{ m }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \times g=\frac { 5-4 }{ 5+4 } \times 10=\frac { 1 }{ 9 } \)=1.1 ms^-2.

As \(\vec L\)= \(\vec r\) \(\times\) \(\vec mv\) i.e., \(\vec L\) magnitude decreases but direction remains constant.

\(K=\frac { { L }^{ 2 } }{ 2I } \Rightarrow K_{ A }>K_{ A }\)

By comparing the components both sides, we can write
5 - 6 = 3T + T
 -1 = 4T
 T = \(-\frac{1}{4}\)

(a) \(\vec F=m\vec a\)
\(F_x\hat i+F_y\hat j+F_z\hat k=ma_x\hat i+ma_y\hat j+ma_z\hat k\)
By comparing the components, we get
\(F_x=ma_x,F_y=ma_y,F_z=ma_z\)
This implies that one vector equation is equivalent to three scalar equations.
(b) \(\vec p=0\)
\(p_x\hat i+p_y\hat j+p_z\hat k=0\hat i+0\hat j+0\hat k\)
By comparing the components, we get
 \(p_x=0,p_y=0,p_z=0\)

T = \(\frac { mv^{ 2 } }{ R } =\frac { 1\times (10)^{ 2 } }{ 1 } \) = 100N

Given: L = Iω= Constant
Formula: Rotational K.E. is given by, 
K = \(\frac { 1 }{ 2 } \) Iω²=\(\frac { 1 }{ 2 } \frac { { I }^{ 2 }\omega ^{ 2 } }{ I } \)= \(\frac { 1 }{ 2 } .\frac { L^{ 2 } }{ I } \)
For constant L, \(K\propto\frac{1}{1}\) 
So when the moment of inertia decreases, the rotational K.E. increases. Hence rotational K.E. is not conserved.

Here f = μR = μmg cosፀ = 0.7\(\times\)2\(\times\)9.8 cos30⁰
=0.7\(\times\)9.8\(\times\)0.866 = 11.9 N

Formula: By conservation of angular momentum
l₁ω ₁ = I₂ω₂
or mr². \(\frac { 2\pi }{ T_{ 1 } } \) = m \(\left( \frac { r }{ 2 } \right) ^{ 2 }.\frac { 2\pi }{ T_{ 2 } } \)
\({ T }_{ 2 }=\frac { 1 }{ 4 } { T }_{ 1 }\)
= \(\frac { 1 }{ 4 } \times 4\)=1s

(a) I_xx' = 4 \(\times\) (0.3)² + 1 \(\times\) (0.8)² = 1 kgm² 
(b) I_xx'= 4 \(\times\) (3)² + 2 \(\times\) (2)² + 3 \(\times\) (4)² = 92 kgm²

A vector which denotes the position of a particle at any instant of time, with respect to origin of coordinate system.
The position vector \(\vec{r}\) of the particle at a point P is given by \(\vec{r}=x\vec{i}+y\vec{j}+z\vec{k}\)

A vector \(\vec{A}\) multiplied by a scalar \(\lambda\) results in another vector, \(\lambda\vec{A}\) .
Eg: Force F = m\(\vec{a}\) . Here mass om is a scalar, and a is the acceleration. Since 'm' is always a positive scalar, the direction of force is always in the direction of acceleration.

Similarly the subtraction of two vectors is equivalent to subtracting the corresponding x, y and z components:
\(\vec{A}-\vec{B}=(A_x-B_{x})\hat{i}+(A_y-B_y)\hat{j}+(A_z-B_z)\hat{k}\)

Two vectors have same direction.

No.

The angle of friction is defined as the angle between the normal force (N) and the resultant force (R) of normal force and maximum friction force (f_s^max).

(i) The graph shows that static friction increases linearly with external applied force till it reaches the maximum.
(ii) If the object begins to move then the kinetic friction is slightly lesser than the maximum static friction.
(iii) Note that the kinetic friction is constant and it is independent of applied force.

(i) Static friction is the force which opposes the initiation of motion of an object on the surface.
(ii) When the object is at rest on the surface, only two forces act on it.
(iii) They are the downward gravitational force and upward normal force
(iv) The resultant of these two forces on the object is zero.

(i) Inertia of rest: When a stationary bus starts to move, the passengers experience a sudden backward push.
(ii) Due to inertia, the body (of a passenger) will try to continue in the state of rest, while the bus moves forward. This appears as a backward push.

The inability of objects to move on its own or change its state of motion is called inertia.

(i) Aristotle said that 'Force causes motion'.
(ii) Galileo said force is not required to maintain motion.

If the x, y and z axes are drawn in anticlockwise direction then the coordinate system is called as "right-handed Cartesian coordinate system".

A coordinate system and the position of an object is described relative to it, then such a coordinate system is called frame of reference.

When the round object moves, it always tends to roll on any surface which has a coefficient of friction any value greater than zero (μ > 0). The friction that enabling the rolling motion is called rolling friction.

Power delivered is the work done per unit time. If we differentiate the expression for work done with respect to time, we get the instantaneous power (P).
p =\(\frac { dw }{ dt } =\tau \frac { d\theta }{ dt } \)   \(\because (dw=\tau d\theta )\)
p =ፒω

(i) A quantity which is described by both its magnitude and direction is called a vector quantity.
(ii) Geometrically, a vector is a directed line segment
Examples: Force, velocity displacement, acceleration, position vector, linear momentum and angular momentum.

(i) Cartesian Coordinate system is a frame of reference in which the position of an object at any given instant is described in terms of its distances along x, y and z axes.
(ii) Conventionally right - handed Cartesian Coordinate system where the x, y and z axes are drawn in anti clockwise direction is followed in physics.

The relative velocity of B with respect to A, \(\vec { { V }_{ BA } } =\vec { { V }_{ B } } -\vec { { V }_{ A } } \)= 5 km h^-1 due west similarly, the relative velocity of A with respect to B i.e., \(\vec { { V }_{ AB } } =\vec { { V }_{ A } } -\vec { { V }_{ B } } \) = 5 km h^-1. To a passenger in the car A, the car B will appear to be moving east with a velocity 5 km h^-1. To a passenger in train B, the train A will appear to move westwards with a velocity of 5 km h^-1

\(\vec A.\vec B\) = 2 + 12 + 30 = 44
Magnitude A = \(\sqrt{4+16+25}=\sqrt{45}\) units
Magnitude B = \(\sqrt{1+9+36}=\sqrt{46}\) units
The angle between the two vectors is given by
  \(\theta=\cos^{-1}(\frac{\vec A.\vec B}{AB})=\cos^{-1}(\frac{44}{\sqrt{45}\times \sqrt{46}})=\cos^{-1}(\frac{44}{45.49})=\cos^{-1}(0.967)\)
\(\therefore \theta \cong { 15 }^{ o }\)

In physics, certain vector quantities can be defined as a scalar times another vector quantity.

The two vectors \(\vec{A}\) and \(\vec{B}\) in a Cartesian coordinate system can be expressed as
\(\overrightarrow { A } ={ A }_{ x }\hat { i } +{ A }_{ y }\hat { j } +{ A }_{ z }\hat { k } \)
\(\overrightarrow { B } ={ B }_{ x }\hat { i } +{ B }_{ y }\hat { j } +{ B }_{ z }\hat { k } \)
Then the addition of two vectors is equivalent to adding their corresponding x, y and z components.
\(\vec{A}+\vec{B}=(A_x+B_x)\hat{i}+(A_y+B_y)\hat{j}+(A_z+B_z)\hat{k}\)

Every object continues to be in the state of rest or of uniform motion (constant velocity) unless there is external force acting on it.

360^o = 27 days
1 day = \(\frac{360^o}{27}=13^o.3'\)