A `0.098-kg` block slides down a frictionless track as shown in (Fig. 5.208).
.
The time taken by the block to move from `A` to `C` is.

Options

A. `sqrt((3)/(g))`
B. `sqrt((2)/(g))`
C. `sqrt((1)/(g))`
D. `(1 + sqrt(3))/(sqrt(g))`

Answer

D

Explanation

(d) `(1)/(2) mv^2 = mg(3 - 1) = 2 mg` [from conservation of energy]
or `v = sqrt(4 g) = 2 sqrt(g)`
Vertical component at A is `2 sqrt(g) sin 30^@ = sqrt(g)`
Time of flight,
`T = (2 v sin theta)/(g) = (2 sqrt(g))/(g) = (2)/(sqrt(g))`
Using `S = ut + (1)/(2) at^2`, we get
`-1 = sqrt(g) t - (1)/(2) "gt"^2 or (1)/(2) "gt"^2 - sqrt(g) t - 1 = 0`
`t = (sqrt(g) +- sqrt(g + 4 xx (1)/(2) g))/(2 xx (g)/(2)) t = (1 +- sqrt(3))/(sqrt(g))`
Neglecting negative time, `t = (1 +- sqrt(3))/(sqrt(g))`
`x = 2 sqrt(g) cos 30^@ [(1 +- sqrt(3))/(sqrt(g))] or x = (sqrt(3) + 3) m`.