Two blocks A and B of equal masses are attached to a string passing over a smooth pulley fixed to a wedge as shown in figure. Find the magnitude of acceleration of centre of mass of the two blocks when they are released from rest. Neglect friction.

Options

A. `(sqrt(3) - 1)/(4 sqrt(2)) g`
B. `(sqrt(3) -1)g`
C. `(g)/(2)`
D. `((sqrt(3) -1)/(sqrt(2))) g`

Answer

A

Explanation

(a) `mg sin 60^@ - T = ma, T - mg sin 30^@ = ma`
Solving them, we get `a = (g sqrt(3) - 1)/(4)`
`vec a_1 = -a cos 60^0 hat i- a sin 60^0 hat j = - (a)/(2) hat i - (a sqrt(3))/(2) hat j`
`vec a_2 = -a cos 30^0 hat i + a cos 30^0 hat j = -(a sqrt(3))/(2) hat i + (a)/(2) hat j`
Now
`vec a_(cm) = (m_1 vec a_1 + m_2 vec a_2)/(m_1 + m_2) = (-(ma)/(2)(a + sqrt(3)) hat i + (ma)/(2) (1 - sqrt(3)) hat j)/(2 m)`
=`(a)/(4)[(1+ sqrt(3))hat i + (1 - sqrt(3)) hat j]`
`|vec a_(cm)| = (a)/(4) [sqrt((1 + sqrt(3))^2 + (1 - sqrt(3))^2)] = (a)/(sqrt(2)) = g ((sqrt(3 - 1))/(4 sqrt(2)))`.
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