Application of Newton's law of gravitation
(i) Dimensions of the heavenly objects can be measured using gravitation law. Mass of the earth, radius of the earth, acceleration due to gravity etc. can be calculated with a higher accuracy.
(ii) Helps in discovering new stars and planets. Mass of the double stars can be calculated.
(iii) One of the irregularities in the motion of stars is called "Wobble" which leads to the disturbance in the motion of planet nearby. In this condition mass of the star can be calculated using law of gravitation.
(iv) Helps to explain germination of roots due to the property of geotropism, which is the property of root responding to the gravity.
(v) Helps to predict the path of the astronomical bodies.

Statement:
Universal law of gravitation states that, 'every particle of matter in this universe attracts every other particle with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between centers of these masses. The direction of the force acts along the line joining the masses'.
Deviation: Force between the masses is always attractive and it does not depend on the medium where they are placed.

Let m₁ and m₂ be the masses of two bodies A and B placed at r meter apart in space 
Force \(\mathrm{F} \propto \mathrm{m}_{1} \times \mathrm{m}_{2}\)
\(\mathrm{F} \propto 1 / r^{2}\)
On combining the above two expressions,
\(\mathrm{F} \propto \frac{\mathrm{m}_{1} \times \mathrm{m}_{2}}{\mathrm{r}^{2}} \)
\(F=\frac{G m_{1} m_{2}}{r^{2}}\)
Where G is the universal gravitational constant.
Its value in SI unit is \(6.674 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\).

(i) The Newton second law tells us that applying a force on an object produces an acceleration proportional to the object's mass.
(ii) When you're wearing your seat belt, it supplies the force to decelerate you in the event of a crash so that you don't hit the wind shield.
(iii) According to Newton's first law an object in motion continues in motion with the same speed and in same direction, unless acted upon by a force.
(iv) If the motor cycle were to abruptly stop, then the rider in motion would continue in motion.
(v) The rider would likely be propelled from the motor cycle, the rider becomes a projectile.
(vi) If the person is not wearing the helmet, the injury would be severe.
(vii) Thus "wearing helmet and fastening the seat belt is highly recommended for safe journey".

Given: K₁ = K₂ = K, m₁ = 4m₂
The kinetic energy of the truck \(=\frac{1}{2} m_{1} v_{1}^{2} \Rightarrow v_1= \sqrt\frac{2 k}{m_{1}}\)
The kinetic energy of the bike \(=\frac{1}{2} \mathrm{~m}_{2} \mathrm{v}_{2}^{2}\Rightarrow v_2= \sqrt\frac{2 k}{m_{2}}\)
∴ Momentum p = mv
∴ Momentum of the two bodies are given by,
\(P_1=\sqrt{2m_1K,}\)
\(P_2=\sqrt{2m_2K,}\)
\(\therefore \frac{P_1}{P_2} =\sqrt{\frac{2 m_1K}{2m_{2}K}}=\sqrt{\frac{m_1}{m_{2}}}=\sqrt{\frac{4m_2}{m_{2}}}=\frac{\sqrt { 4}}{\sqrt 1}= \frac{2}{1}\)
Ratio of momenta = 2: 1

Given: Let m₁ = 8 kg, m₂ =2 kg. F = 15 N
Consider both the masses as a unit system as they will move with common acceleration,
\({\mathrm{F}}_{1}=\mathrm{M_1}a\)
 \(F_2=m_2a\)
\(F=F_1+F_2\)
\(=\left(\mathrm{m}_{1} +\mathrm{~m}_{2}\right){a}\)
\(15=(8+2) {a}  \)
\(15=10 a \)
\(a=15 / 10=1.5 \mathrm{~ms}^{-2}\)
a = 1.5 ms^-2
Let \(\mathrm{F}_{2}\) be the force exerted on 2 kg mass, then
\({\mathrm{F}_{2}}=\mathrm{m_2} {a}  \)
\({\mathrm{F}_{2}}=2 \times 1.5=3 \mathrm{~N}\)
So, the force exerted on 2 kg mass is 3 N.

Mass of body 1, m₁ = 50 kg
Mass of body 2, m₂ = 10kg
Distance, R = 10m
Universal gravitation
constant, G = 6.67 x 10^-11 Nm² kg^-2
To find :Force of gravitation, F =\(\frac { G{ m }_{ 1 }{ m }_{ 2 } }{ { R }^{ 2 } } \)
F=\(\frac { 6.67\times 10^{ -11 }\times 50\times 10 }{ 10^{ 2 } } \)
Force, F = 33.35 x 10^-11 N

Given:
Weight = 50 kg
To find: Acceleration, a = ? (downward)
Reaction, R = 400 N
R = m (g - a)
400 = 50 (10 - a)
400 = 500 - 50 a
500 = 500 - 400
50 a = 100
a=\(\frac{100}{50}\)
Downward acceleration, a = 20 ms^-1

Force, F = 50 N
Distance, d = 5 cm
To find: Momentum of force= F x d
Momentum of force, 50 x 5 x 10^-2
= 250 x 10^-2
= 2.5 Nm

Given:
Mass, m 50 g = 50 x 10^-3 kg
Final speed, v = 0
Initial speed, u = 250 ms^-1
Time, t = 1ms = 10^-3 s
To find: J = F x t = ? Average force, F = ma = ?
F=\(\frac { m(v-u) }{ t } \)
=\(\frac { 50\times { 10 }^{ -3 }[0-250] }{ 1\times { 10 }^{ -3 } } \)
F = 12500 = 1.25 x 10⁴ N
J = F x t
= 1.2 x 10⁴ x 10^-3
J = 12.50 Ns

Time, t = 10 s
Mass of body, m = 10 kg
Initial velocity of the body, u₁ = 0
Distance, d = 500 cm
= 500 x 10^-2 m
To find : Force, F = ?
F=m₁\(\frac { ({ u }_{ 1 }-{ v }_{ 1 }) }{ t } \)

v₁=\(\frac { distance }{ time\quad taken } \)
F=v₁=\(\frac { 500\times 10^{ -2 } }{ 10 } \)
Force, F=0.5 N