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.
\(\vec{v}_{G} \propto \frac{1}{m_{G}+m_{\operatorname{man}}}\)

Horizontal component of the normal reaction of the road provides the necessary centripetal force for the vehicle to move it along the curved path.

The coefficient of static friction is equal to tangent of the angle of friction.

It is not a vector equation. The relation f_s = μ_s.N is not a vector relation. This is because the normal force N and f_s are not in the same direction even though f_s is equal to  μ_s times the normal force. This is also true in the case of kinetic friction.

Rolling friction < Sliding friction

No

If the object is pressed hard on the surface then the normal force acting on the object will increase. As a consequence it is more difficult to move the object.

True

If an object is thrown vertically upward, the direction of motion is upward, but gravitational force is downward.

Yes, μ > 1, friction is stronger than normal force

False

The common acceleration of the masses is,
 \(a=\frac { \left( { m }_{ 1 }-{ m }_{ 2 } \right) }{ \left( { m }_{ 1 }+{ m }_{ 2 } \right) } \times g\)     [\(\therefore\) g \(\longrightarrow \) acceleration due to gravity]
= \(\frac { \left( 0.5-0.5 \right) }{ \left( 0.5+0.5 \right) } \times 9.8\)
= \(\frac { 0 }{ 1 } \times 9.8\)
The acceleration of the masses = 0

(i) Choose a suitable inertial coordinate system to analyse the problem. For most of the cases we can take Earth as an inertial coordinate system.
(ii) Identify the system to which Newton's laws need to be applied. The system can be a single object or more than one object.
(iii) Draw the free body diagram
(iv) Once the forces acting on the system are identified, and the free body diagram is drawn, apply Newton's second law. In the left hand side of the equation, write the forces acting on the system in vector notation and equate it to the right hand side of equation which is the product of mass and acceleration. Here, acceleration should also be in vector notation.
(v) If acceleration is given, the force can be calculated. If the force is given, acceleration can be calculated.