In the abjoining figure, wire `PQ` is smooth, ring `A` has a mass `1kg` and block `B,2kg`. If system is released from rest with `theta =60^(@)`,find

(a)constraint relation between their velocities as a function of `theta`.
(b) constrain relation between their accelerations just after the release at `theta=60^(@)`.
(c) tension in the string and the value of these accelerations at this instant.

Explanation

`M` and `Q` are two fixed fixed points. Therefore.
`MQ=contant=c`
`l= "length of strin"g = "contant"`
(a) In traingle `MQA, (l-y)^(2) =x^(2)+c^(2)`
Differentiating w.r.t time, we get
`2(l-y) (-(dy)/(dt)) =2x (+(dx)/(dt))+0`
or `(l-y) (+(dy)/(dt)) =x (-(dx)/(dt))` ...(i)
`:. (dy)/(dt) =((x)/(l-y)) (-(dx)/(dt))` ...(ii)
y is increasing with time,
`:. +(dy)/(dt) =u_(2)`
x is increasing with time,
`:. -(dx)/(dt) =u_(1)` and `(x)/(l-y)= cos theta`
Substituting these value in Eq (ii) we have
`u_(2) =u_(1) cos theta`
(b) Further differentiating Eq. (i) we have,
`(d^(2)y)/dt^(2) (l-y)-((dy)/(dt))^(2) =-[x (d^(2)x)/dt^(2)+((dx)/(dt))^(2)]` ...(iii)
Just after the release, `u_(1),u_(2), (dx)/(dt)` and `(dy)/(dt)` all are zero.Substituting in Eq. (iii), we have ,
`(d^(2)y)/dt^(2)=((x)/(l-y)) ((d^(2)x)/dt^(2))` ...(iv)
Here `(d^(2)y)/dt^(2) = a_(2) and -(d^(2)x)/dt^(2) = a_(1)`
`(x)/(l-y) = cos 60^(@) =(1)/(2)`
Substituting in Eq. (iv), we have `a_(2) =(a_(1))/(2)` ...(v)
(c) For `A` Equation is

`T = cos 60^(@) =m_(A) a_(1)`
or `(T)/(2) =(1) a_(1) = a_(1)` ...(vi)
For `B` `20-T =m_(B)a_(2)`
`20-T =2a_(2)` ...(vii)
Solving Eqs.`(v),(vi)` and `(vii)`, we get
`T =(40)/(3)N rArr a_(1) =(20)/(3)m//s^(2)`
`a_(2) =(10)/(3)m//s^(2)`