Example of the Law of Universal Gravitation

The force of attraction between two bodies separated by a distance is proportional to the product of these masses and inversely proportional to the square of the distance.

m₁= kg
m = kg
r = m
G = 6.67 x 10^-11 Nm² / kg²
F = N

F = G (m₁m₂/ r₂)

EXAMPLE OF THE PROBLEM OF THE LAW OF UNIVERSAL GRAVITATION:

Find the distance at which two masses of one kilogram each must be placed so that they attract each other with a force of 1 N.
F = 1N
G = 6.67 x10^-11 Nm²/kg²
m₁= 1kg
m₂= 1kg
r =?

Impulse: It is the force exerted for the duration in which said force acts.

l = N / s
t = s
F = N

EXAMPLE:

Calculate the momentum required by a particle with a mass of one kg carrying a constant acceleration of 5 m / s² in 20 s.

l = ft = mat = (1kg) (5m / s²) (20s) = 100 N / s 

Amount of linear motion: It is the product of the mass and the speed of the mobile.
P = kgm / s P = mv
m = kg
v = m / s

EXAMPLE:

On a body of mass of 2 kg a constant force is applied during 10 s, producing a change of velocity from 10 m / s to 30 m / s. What force was applied to produce this effect?

To solve this problem, we must take as a basis the formula for momentum and momentum in relation to 2nd. Newton law.
l = Ft = P = mv
Since t, m and the change in velocity are known, we have:

Ft = m (vf-vi)
F (10s) = 2kg (30m / s-10m / s)
F = 2kg (20m / s) / 10s = 4N

Principle of conservation of momentum: It is determined by the relation that when two mobiles collide, the momentum of the sum of the two will not vary, that is, their total momentum is constant.
m₁or₁ + m₂ or₂ = m₁ v₁ + m₂v₂
mu = amount of movement before impact.
mv = amount of movement after impact.