Consider an inclined plane on which an object is placed as shown in the figure, Let the angle which this plane makes with the horizontal be ፀ. For small angle of ፀ, the object may not slide down. As ፀ is increased, for a particular value of ፀ, the object begins to slide down. This value is called angle of repose. Hence, the angle of repose is the angle of the inclined plane with the horizontal such that an object placed on it begins to slide.

(i) Consider the case of a whirling motion of a stone tied to a string. Assume that the stone has angular velocity ω in the inertial frame (at rest).
(ii) If the motion of the stone is observed from a frame which is also rotating along with the stone with same angular velocity ω then, the stone appears to be at rest.
(iii) This implies that in addition to the inward centripetal force - mω²r there must be an equal and opposite force that acts on the stone outward with value + mω²r.
(iv) So the total force acting on the stone in a rotating frame is equal to zero (-mω²r + mω²r=0).
(v) This outward force + mω²r is called the centrifugal force.

i. Every object continues to be in the state of rest or of uniform motion unless there is external force acting on it.
ii. The force acting on an object is equal to the rate of change of its momentum.
iii. For every action there is an equal and opposite reaction.
Discussion:
i) Newton's laws are vector laws. The equation \(\vec{F}=\mathrm{m} \vec{a}\) can be written in cartesian coordinates as
\(\mathrm{F}_{x} \hat{\mathrm{i}}+\mathrm{f}_{\mathrm{y}} \hat{\mathrm{j}}+\mathrm{F}_{z} \hat{\mathrm{k}}=\operatorname{ma}_{x} \hat{\mathrm{i}}+\operatorname{ma}_{\mathrm{y}} \hat{\mathrm{j}}+\operatorname{ma}_{\mathrm{z}} \hat{\mathrm{k}}\)
On comparing both sides,
\(\mathrm{F}_{x}=m \mathrm{~m}_{x}  \)
\(\mathrm{F}_{\mathrm{y}}=m \mathrm{ma}_{\mathrm{y}}  \)
\(\mathrm{F}_{\mathrm{z}}=m \mathrm{ma}_{\mathrm{z}}\)
From the above equations, we can infer that the force acting along y direction cannot alter the acceleration along n direction. In the same way, F₂ can not affect a_y and a_n
ii) The acceleration experienced by the body at a time t depends on the force which acts on the body at that instant of time. Thus, \(\vec{F}(\mathrm{t})=\mathrm{m} \vec{a}(\mathrm{t})\)
when a bowler throws the ball to a batsman the acceleration of the ball is determined by the gravitational and air frictional forces and not by the speed which it is thrown.
iii) The direction of force may be different from the direction of force. The following motions are possible.
a) Force and motion in the same direction.
Example: Falling of an apple from the tree.
b) Force and motion are not in the same direction.
Example: The Moon experiences a force towards the Earth but Moon moves in the elliptical path.
c) Force and motion are in the opposite direction.
Example: The motion of a body thrown vertically upward.
d) Zero net force, but there is motion.
Example: The falling of rain drops.
4) If multiple forces \(\vec{F}_{1}, \vec{F}_{2}, \vec{F}_{3} \ldots\) act on the same body then total force is equal to the vectorial sum of the individual forces \(\vec{F}_{\text {net }}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}+\ldots\)
5) Newton's second law can be written in the second derivative of position vector as  \(\vec{F}=\mathrm{m} \frac{\mathrm{d}^{2}\overrightarrow{\mathrm{r}}}{\mathrm{dt}^{2}}\). It means that whenever the second derivative of position vector is not zero, there must be a force acting on the body.
6) If no force acts on the body then \(m \frac{d \vec{v}}{\mathrm{dt}}=0\). It implies that \(\vec{v}\) is constant. Thus second law is consistent with the first law even though they are independent to each other.
7) Newton's second law is cause and effect relation force is the cause and acceleration is effect.

Horizontal motion:
(i) In this case, mass m₂ is kept on a horizontal table and mass m₁ is hanging through a small pulley. Assume that there is no friction on the surface.
(ii) As both the blocks are connected to them unstretchable string, if m, moves with an acceleration a downward then m₂ also moves with the same acceleration a horizontally.
The forces acting on mass m₂ are
(i) Downward gravitational force (m₂g)
(ii) Upward normal force (N) exerted by the surface
(iii) Horizontal tension (T) exerted by the string.
The forces acting on mass m₁ are
(i) Downward gravitational force (m₁g)
(ii) Tension (T) acting upwards
The free body diagrams for both the masses.

\(T\hat{j}-m_1g\hat{j}=m_1a\hat{j}\)    ( along y direction)
(iii) By comparing the components on both sides of the above equation,
Applying Newton's second law for m2
\(T\hat{i}-m_2a\hat{i}\) (along x direction)
By comparing the components on both sides of above equation,
\(N\hat{j}-m_2g\hat{i}=0\)
By comparing the components on both sides
of the above equation
N - m₂g = 0
N = m₂g ........(3)
By substituting equation (2) in equation (1), we can find the tension T.
(iv) Comparing motion in both cases, it is clear that the tension in the string for horizontal motion is half of the tension for vertical motion for same set of masses and strings.

Conservation of linear momentum is equivalent to Newton's third law of motion. If no external force acts on a system (called isolated) of constant mass, the total momentum of the system remains constant with time.
According to this law for a system of particles \(\vec{F}=\frac{d \vec{p}}{d t}\)
In the absence of external force F = 0 and \(\vec{p}\) = constant
i.e.,  \(p_{1}+\vec{p}_{2}+\vec{p}_{3}+\dots=\) constant
or \(m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2}+m_{3} \vec{v}_{3}+\dots=\) constant
This equation shows that in absence of external force for a closed system the linear momentum of individual particles may change but their sum remains unchanged with time. For a system of two particles in absence of external force by law of conservation of linear momentum.
\(\vec{p}_{1}+\vec{p}_{2} =\text { constant }  \)
\(m_{1} \vec{v}_{1}+m_{2} \vec{v}_{2} =\text { constant }  \)
\(m_{1} \frac{d \vec{v}_{1}}{d t}+m_{2} \frac{d \vec{v}_{2}}{d t} =0 \Rightarrow m_{1} \vec{a}_{1}+m_{2} \vec{a}_{2}=0 \Rightarrow \vec{F}_{1}+\vec{F}_{2}=0  \)
\(\therefore \vec{F}_{2} =-\vec{F}_{1}\)
For every action there is equal and opposite reaction which is Newton's third law of motion.
Recoiling of a gun: For bullet and gun system, the force exerted by trigger will be internal so the momentum of the system remains unaffected.
Let m_G= mass of gun, m_B= mass of bullet, v_G velocity of gun, v_B= velocity of bullet Initial momentum of system =0
Final momentum of system \(=m_{G} \vec{v}_{G}+m_{B} \vec{v}_{B}\)
By the law of conservation linear momentum \(m_{G} \vec{v}_{G}+m_{B} \vec{v}_{B}=0\)
So recoil velocity \(v_{G}=-\frac{m_{B}}{m_{G}} \vec{v}_{B}\)
Here negative sign indicates that the velocity of recoil \(\vec{v}_{G}\) is opposite to the velocity of the bullet.
\(v_{G} \propto \frac{1}{m_{G}}\) i.e. higher the mass of gun, lesser the velocity of recoil of gun.
While firing the gun must be held tightly to the shoulder, this would save hurting the shoulder because in this condition the body of the shooter and the gun behave as one body. Total mass become large and recoil velocity becomes too small.
\(\vec{v}_{G} \propto \frac{1}{m_{G}+m_{\operatorname{man}}}\)