Question

14 mark for review a block of mass $m_0$ is on a horizontal surface and attached by a string to a hanging block, also of mass $m_0$, as shown in the figure. friction between the block and the horizontal surface is negligible. the masses of the string and pulley are negligible, and the pulley can rotate with negligible friction around its axle. when the hanging block is released, it has an acceleration of magnitude $a_1$. a second trial is performed after replacing the hanging block with a different block of mass $3m_0$. the hanging block of mass $3m_0$ has an acceleration of magnitude $a_2$ after being released. what is the ratio $\frac{a_2}{a_1}$? a $\frac{1}{2}$ b 1 c $\frac{3}{2}$ d 3

Explanation:

Step1: Apply Newton's second - law for the first case

For the two - block system with both blocks of mass $M_0$. Let the tension in the string be $T_1$. For the hanging block, $M_0g - T_1 = M_0a_1$ (using $F = ma$, where the net force on the hanging block is its weight minus the tension). For the block on the horizontal surface, $T_1 = M_0a_1$. Substituting $T_1 = M_0a_1$ into $M_0g - T_1 = M_0a_1$, we get $M_0g-M_0a_1 = M_0a_1$. Simplifying gives $M_0g=2M_0a_1$, so $a_1=\frac{g}{2}$.

Step2: Apply Newton's second - law for the second case

For the two - block system with the hanging block of mass $3M_0$ and the block on the horizontal surface of mass $M_0$. Let the tension in the string be $T_2$. For the hanging block, $3M_0g - T_2 = 3M_0a_2$. For the block on the horizontal surface, $T_2 = M_0a_2$. Substitute $T_2 = M_0a_2$ into $3M_0g - T_2 = 3M_0a_2$, we get $3M_0g-M_0a_2 = 3M_0a_2$. Combining like terms: $3M_0g = 4M_0a_2$, so $a_2=\frac{3g}{4}$.

Step3: Calculate the ratio $\frac{a_2}{a_1}$

Substitute $a_1=\frac{g}{2}$ and $a_2=\frac{3g}{4}$ into $\frac{a_2}{a_1}$. $\frac{a_2}{a_1}=\frac{\frac{3g}{4}}{\frac{g}{2}}=\frac{3}{2}$.

Answer:

C. $\frac{3}{2}$