How to deduce the centripetal acceleration formula a = R ω ^ 2?

How to deduce the centripetal acceleration formula a = R ω ^ 2?

The velocity at time t is EC, that is, the velocity at time V, t + DT is GF. If parallel line ED is made, the velocity difference is vector CD,
That is, DV, angle a = angle B, and CD = angle a * ce, that is, DV = angle B * V, divided by DT on both sides, there is a = DV / dt = ω * v,
And V = R ω, so a = R ω ^ 2

V1 '= (m1-m2) V1 / (M1 + m2) and how to deduce a momentum formula of V2'?

a. Complete elastic collision: the total kinetic energy of the system before and after the collision remains unchanged. For a system composed of two objects, the following conditions are satisfied:
m1v1＋m2v2＝m1v1′＋m2v2′
1 / 2m1v12 + 1 / 2m2v22 = 1 / 2m1v1 ′ 2 + 1 / 2m2v2 ′ 2 (kinetic energy conservation)
The combination of the two forms can lead to:
V1 ′ = [(m1-m2) V1 + 2m2v2] / (M1 + m2) when V2 = 0, V1 ′ = (m1-m2) V1 / (M1 + m2)
V2 ′ = [(m2-m1) V2 + 2m1v1] / (M1 + m2) when V2 = 0, V2 ′ = 2m1v1 / (M1 + m2)
·If M1 > > m2, the mass of the first object is much larger than that of the second object
Then m1-m2 ≈ M1, M1 + M2 ≈ M1. Then V1 '= V1 V2' = 2v1
·If M1

V1 '= (m1-m2) V1 / (M1 + m2) and how to deduce a momentum formula of V2'?

This is the formula for perfectly elastic collision
Specifically, the energy conservation before and after collision is 1 / 2 (m1v1 ^ 2 + m2v2 ^ 2) = 1 / 2 (m1v1 '^ 2 + m2v2' ^ 2)
Sum momentum conservation m1v1 + m2v2 = m1v1 '+ m2v2'
We can solve V1 'and V2' by using the two equations