Consider the setup presented in the figure. All surfaces are...

The given wedge forms a triangle with sides $3\text{ m}$, $4\text{ m}$, and $5\text{ m}$. Since $3^2 + 4^2 = 5^2$, it is a right-angled triangle with the right angle at the top vertex.

Let $\theta_1$ be the angle of inclination of the left surface (length $3\text{ m}$)$and \theta_2$ be the angle of inclination of the right surface (length $4\text{ m}$).

Using the geometric properties of the triangle, the altitude $h$ to the base is given by equating the area:

$\dfrac{1}{2} \times 3 \times 4 = \dfrac{1}{2} \times 5 \times h \Rightarrow h = \dfrac{12}{5}\text{ m}$

The sine of the angles of inclination are:

$\sin \theta_1 = \dfrac{h}{3} = \dfrac{12/5}{3} = \dfrac{4}{5}$

$\sin \theta_2 = \dfrac{h}{4} = \dfrac{12/5}{4} = \dfrac{3}{5}$

Let $a$ be the common acceleration of the two blocks and $T$ be the tension in the string. The equations of motion for the left block ($m_1 = 1.0\text{ kg}$) and the right block ($m_2 = 1.0\text{ kg}$) are:

$m_1 g \sin \theta_1 - T = m_1 a$

$T - m_2 g \sin \theta_2 = m_2 a$

Adding the two equations, we get:

$a = \dfrac{m_1 g \sin \theta_1 - m_2 g \sin \theta_2}{m_1 + m_2}$

Substituting the given values:

$a = \dfrac{1.0 \times g \times \left(\dfrac{4}{5}\right) - 1.0 \times g \times \left(\dfrac{3}{5}\right)}{1.0 + 1.0}$

$a = \dfrac{\dfrac{4g}{5} - \dfrac{3g}{5}}{2} = \dfrac{\dfrac{g}{5}}{2} = \dfrac{g}{10}$

Answer: $g/10$