1. **Problem statement:** A particle of mass $m_1$ lies on a smooth horizontal table connected by a light inextensible string over a smooth fixed pulley to a freely hanging particle of mass $m_2$. Initially, the system is at rest with $m_1$ at a distance $d$ from the edge of the table. We need to show that the acceleration of the system is $$a=\frac{m_2 g}{m_1 + m_2}$$ and the time taken for $m_1$ to reach the edge of the table is $$t=\sqrt{\frac{2 d (m_1 + m_2)}{m_2 g}}.$$ 2. **Set up the forces and equations:** - Let the acceleration of $m_1$ towards the pulley be $a$. - Since the string is inextensible, $m_2$ moves downward with acceleration $a$. - Tension in the string is $T$. 3. **Apply Newton's second law to $m_1$ (horizontal motion):** $$T = m_1 a$$ 4. **Apply Newton's second law to $m_2$ (vertical motion):** $$m_2 g - T = m_2 a$$ 5. **Solve the system of equations:** From step 3, $T = m_1 a$. Substitute into step 4: $$m_2 g - m_1 a = m_2 a$$ Rearranged: $$m_2 g = m_1 a + m_2 a = (m_1 + m_2) a$$ Therefore, $$a = \frac{m_2 g}{m_1 + m_2}$$ 6. **Find the time $t$ for $m_1$ to reach the edge:** - Initial velocity $u = 0$ (system starts from rest). - Distance to cover $s = d$. - Acceleration $a$ as above. Using the equation of motion: $$s = ut + \frac{1}{2} a t^2$$ Substitute values: $$d = 0 + \frac{1}{2} a t^2 = \frac{1}{2} \cdot \frac{m_2 g}{m_1 + m_2} t^2$$ Solve for $t^2$: $$t^2 = \frac{2 d (m_1 + m_2)}{m_2 g}$$ Hence, $$t = \sqrt{\frac{2 d (m_1 + m_2)}{m_2 g}}$$ **Final answers:** - Acceleration: $$a = \frac{m_2 g}{m_1 + m_2}$$ - Time to reach edge: $$t = \sqrt{\frac{2 d (m_1 + m_2)}{m_2 g}}$$