constraint..........

An ATTEMPT:

At the particular given instant ... the system can be considered to be momentarily at rest ... The tension in the string can be broken in sine and cosine components ... Tcosθ moves the m₁ block [ friction not present ]. Tsinθ balances the weight m₁g

Hence Tsinθ = m₁g ..... (1)

Now on m₂ block .... T = m₂g .... (2)

Hence .. by solving sinθ = m₁m₂

Now considering acceleration of block m₁ ...

Since T = m₂g ... from (2) ...

Tcosθ/m₁ provides the acceleration of the block m₁

m₂gcosθm₁ = reqd. acceleration ...

cosθ = √(1 - sin²Î¸) = √(m₂² - m₁²)m₂

Hence the reqd. soln... is g√(m₂² - m₁²)m₁

yeah....
if u take vel. of m1 as u nd that of m2 as v u will get a relation
v= u cos (theta).....
but d problem is that we cannot take the same relation for their acceleration.....since it has a rotational component as well....

i think normal reaction also has to be included...

\vec{a_{1}}=a_{1}cos\theta (\hat{t)}+a_{1}sin\theta (\hat{n})
\vec{a_{2}}=-a_{2}(\hat{j})

where \hat{t}=unit vector along the string

and \hat{n}=unit vector perpendicular to string

now since string is inextensible and it doesnt slack , acceleration of both ends of the strings along the string shud be equal .

hence a₁cosθ = a₂

m₂g-T=m₂a₂=m₂a₁cosθ

Tcosθ=m₁a₁

thus m₂gcosθ=m₁a₁+m₂a₁cos²Î¸

hence