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 \(6.674 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\).

Statement:
i) There is no change in the linear momentum of a system of bodies as long as no net external force acts on them.
ii) Let us prove the law of conservation of linear momentum with the following illustration:

Proof:
(i) Let two bodies A and B having masses m₁ and m₂ move with initial velocity u₁ and u₂ in a straight line.
(ii) Let the velocity of the first body be higher than that of the second body. i.e., u₁ > u₂
(iii) During an interval of time t second, they tend to have a collision.
(iv) After the impact, both of them move along the same straight line with a velocity v₁ and v₂ respectively.
Force on body B due to A, F_A = m₂ (v₂ - u₂) / t
Force on body A due to B, F_B = m₁ (v₁ - u₁) / t
By Newton's III law of motion,
Action force = Reaction force
F_A = - F_B
m₁ (v₁ - u₁) / t = - m₂ (v₂ - u₂) / t
m₁v₁ + m₂v₂ = m₁u₁ + m₂u₂
i) The above equation confirms in the absence of an external force, the algebraic sum of the momentum after collision is numerically equal to the algebraic sum of the momentum before collision.
ii) Hence the law of conservation of linear momentum is proved.

(i) Newton's First law: Newton's first law states that, everybody continues to be in its state of rest or the state of uniform motion along a straight line unless it is acted upon by some external force.
(ii) Newton's second law of motion: Newton's second law states that, the force acting on an object is directly proportional to the rate of change of linear momentum of the object and the change in momentum takes place in the direction of force.
(iii) Newton's third law of motion: Newton's third law states that, for every action there is an equal and opposite reaction. They always act on two different bodies.

(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: 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.