Newton's third law of motion states that for any action, there is equal and opposite reaction.
So, if a body applies a force F₁₂ on body 2 (action), then body 2 also applies a force F₂₁ on body 1 but in opposite direction, then
F₂₁ = - F₁₂
In terms of magnitude
|F₂₁| = |F₁₂|
It is very important to note that F₁₂ and F₂₁ though are equal in magnitude and opposite in direction yet act on different points or else no motion will be possible.
For example, hands pull up a chest expander (spring) and spring in turn exerts force on the arms. A football pressed reacts on the foot with the same force and so on.
The most important consequence of the third law of motion is the law of conservation of linear momentum and its application in collision problems.
Since F₁₂ - F₂₁ and F = m\(\triangle v\over \triangle t\)
\(\therefore m_1{\triangle v_1\over \triangle t}=-m_2{\triangle v_2\over \triangle t}\)
Here \(\triangle t\) is the time for which the bodies come in contact during impact. This is same for the two bodies of masses m₁and m₂ and having velocity changes \(\triangle v_1\) and\(\triangle v_2\) respectively.
Therefore,
m₁\(\triangle v_1\)= m₂ \(\triangle v_2\)
or m₁ \(\triangle v_1\)+ m₂ \(\triangle v_2\) = 0
Let u₁, u₂ and v₁,v₂ be the initial and final velocities of the two masses before and after collision, then,
m₁ (v₁ -u₁) = - m₂ (v₂ - u₂)
or m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
Momentum before impact = Momentum after impact.
(This is known as the law of conservation of momentum).

Newton's second law of motion states that the rate of change of momentum of a rigid body is directly proportional to the force applied on it.
The law implies that when a bigger force is applied on a body of given mass, its linear momentum changes faster and vice-versa. The momentum will change in the direction of the applied force.
Let, m = mass of a body,
\(\overrightarrow{v}\)= velocity of the body
\(\therefore\)The linear momentum of the body
\(\overrightarrow{p}=m\overrightarrow{v}\) .............(i)
Now, suppose \(\overrightarrow{F}\) = external force applied on the body in the direction of motion of the body.
\(\triangle \overrightarrow{p}\)= a small change in linear momentum of the body in a small time \(\triangle t\).
Rate of change of linear momentum of the body = \({\triangle \overrightarrow{p}\over \triangle t}\)
According to Newton's second law,
\({\triangle \overrightarrow{p}\over \triangle t} \propto \overrightarrow{F} \ or \ \overrightarrow{F} \propto {\triangle \overrightarrow{p}\over \triangle t} \)
or  \(\overrightarrow {F}=k{\triangle \overrightarrow{p}\over \triangle t}\)   ..................(ii)
where k is a constant of proportionality.
Taking the\(\triangle t \rightarrow 0,\)  the term \(={\triangle \overrightarrow{p}\over \triangle t}\) becomes the derivative or differential coefficient of \(\overrightarrow{p}\) w.r.t. time t. It is denoted by \({d\overrightarrow{p}\over dt}\) .
\(\therefore \overrightarrow{F}=k{d\overrightarrow p \over dt}\)
Using eqn (i), \(\overrightarrow{F}=k{d \over dt}(m\overrightarrow{v})=km{d\overrightarrow{v}\over dt}\)
\(\overrightarrow {F}=km\overrightarrow{a}\)
where \(\overrightarrow{a}={d\overrightarrow{v}\over dt}\) represents acceleration of the body
The value of constant of proportionality k depends on the units adopted for measuring the force.
Now, putting k=1
\(\overrightarrow{F}=m\overrightarrow{a},\) This gives mean of measuring force.
Units of Force: Force in SI units is measured in 'newton' or N. From the relation \(\overrightarrow{F}=m\overrightarrow{a},\) we can see that a newton force is that force which produces 1 ms^-2 acceleration in a body of mass 1 kg.
1 newton = 1 kilogram x 1 metre/ second²
\(\Rightarrow\) 1 N = 1 kg x 1 ms^-2 = 1 kg x ms^-2=1kg ms^-2
In CGS system, force is measured in 'dyne'.
1 dyne = 1 gram x 1 cm s^-2 = 1 g cm s^-2
Since 1 N = 1 kg ms^-2 = 1000 g x 100 cm s^-2
= 10⁵ g cm s^-2= 10⁵ dyne.
\(\Rightarrow\)IN = 10⁵ dyne.
or 1 dyne = 10^-5 N.

If the coin is to revolve with the record, then the force of friction must be enough to provide the necessary centripetal force.
\(\therefore mr \ \omega^2 \le \mu_s mg \ or \ r \le {\mu_s mg\over m \omega^2}or \ r \le {\mu_s g\over \omega^2}\)
frequency = \(33{1\over2}rpm={100\over 3}rpm ={100\over 3\times 60}rps\)
The problems in which centripetal force is obtained from force of friction, start with the following equation:
m r\(\omega^2 \le \mu_s \ mg\)
\(\omega =2\pi \times {100\over 3\times 60}rad s^{-1}={10\over 9}\pi rad s^{-1}\)
\({\mu_s g\over \omega ^2}={0.15\times 10\over ({10\over 9}\pi)^2}m=0.12m=12cm\)
The condition (r \(\le\)12cm) is satisfied by the coin placed at 4cm from the centre of the record. So, the coin at 4 cm will revolve with the record.

u = 0, a = 2 ms^-2, t = 10s
Using equation, v = u + at, we get
v = 0 + 2 x 10 = 20 ms^-1
(a) Let us first consider horizontal motion. The only force acting on the stone is force of gravity which acts vertically downwards.
Its horizontal component is zero. Moreover, air resistance is to be neglected. So, horizontal motion is uniform motion.
\(\therefore\)v _x = v = 20 ms^-1
Let us now consider vertical motion which is controlled by force of gravity.
u=0, a = g = 10 ms^-2, t = (11 - 10) s = 1s
Using v =u + at, v_y = 0 + 10 x 1 = 10 ms^-1
Resultant velocity,
\(v=\sqrt{V^2_x+V^2_y}\)
\(v=\sqrt{20^2+10^2}ms^{-1}\)
\(=\sqrt{500}ms^{-1}\)
= 22.36 ms^-1

\(tan \beta ={v_y\over v_x}={10\over20}={1\over2}=0.5\)
or \(\beta =\)tan^-1 (0.5) = 26.56°
or \(\beta =\) 26° 34'. This angle is with the horizontal.
(b) The moment the stone is dropped from the car, horizontal force on the stone is zero. The only acceleration of the stone is that due to gravity. This gives a vertically downward acceleration of 10 ms^-2. This is also the net acceleration of the stone.

Here m = 0.2 kg
r = 80 cm = 0.8 m
\(\therefore\)V_min at the lowest point so that the stone is just able to complete the circle.
\(=\sqrt{5gr}\)
\(=\sqrt{5\times 10\times 0.8}\)
\(=\sqrt{40}\)
= 6.32 m/s^-1
Now tension in the string at the lower point of the circular path = \({m V^2_{min}\over r}+mg\)
\(=5mg+mg [\because {mV^2_{min}\over r}=5mg]\)
= 6mg
= 6 x 0.2 x 10
= 12 N.

Force F = ma, therefore force acting on a particle in unaccelerated (a = 0) motion is zero.
As kite is held stationary in the sky, therefore acceleration of the kite is zero. Therefore, net force acting on the car F = ma = 0.

Force F = ma, therefore force acting on a particle in unaccelerated (a = 0) motion is zero.
In floating condition, the weight of the body is balanced by the upthrust. Therefore, net force acting on a cork floating on water = 0

Here, m = 1kg, u = 20 m/s
s = 10 cm = 0.1 m
v = 0, a= ? 
(a) As,
\(v^2=u^2+2 a s\)
\( \therefore 0=(20)^2+2 . a \cdot(0.1) \)
\( \Rightarrow a=-\frac{400}{2 \times 0.1}=-2000 \mathrm{~m} / \mathrm{s}^2 \)
b) v=u+a t
\( \therefore 0=20+(-2000) t \)
\( \Rightarrow t=\frac{20}{2000} \)
\(=\frac{1}{100}=0.01 \mathrm{sec}\)
c) Impulse = F. t
 =m(v-u)
 =1(0-20) 
 =-20 N.s.