i cant figure this out for the life of me and i know it is easy

**Answer 1**

If the masses are m[a], m[b], the kinetic energies are K[a], K[b], and the speeds are v[a], v[b] then:

K[a] = m[a] v[a]^2 / 2 …(1)

K[b] = m[b] v[b]^2 / 2 …(2)

Dividing (2) by (1):

(v[a] / v[b])^2 (m[b] / m[a]) = K[b] / K[a]

(v[a] / v[b])^2 * 2 = 1 / 8

v[a] / v[b] = sqrt(1 / 16)

= 1 / 4.

**Answer 2**

A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N.

**Answer 3**

K=1/2mv^2

For particle A:

8K=1/2(1/2m)v^2 –> 8K=1/4mv^2

v^2=8K/(1/4m)

v=sqrt(8K/(1/4m))

For particle B:

K=1/2mv^2

v^2=K/(1/2m)

v=sqrt(K/(1/2m))

Thus vA/vB:

sqrt(8K/(1/4m))

——————-

sqrt(K/(1/2m))

Using some algebra to simplify (eliminating K and m),

vA/vB= sqrt32/sqrt2

vA/vB=4

Good luck