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