Question

In the given figure mass of block A is m and that of B is 2m. They are attached at the ends of a spring and kept on a smooth horizontal surface. Now, the spring is compressed by $x_0$ and released. Find displacement of block B by the time compression of the spring is reduced to $\frac{x_0}{2}$ :-

• A. $\frac{x_0}{3}$
• B. $\frac{x_0}{2}$
• C. $\frac{x_0}{6}$
• D. $\frac{x_0}{4}$

Answer 1

Answer: (C)

$\frac{x_0}{6}$

Answer 2

The system consists of two blocks of masses m and 2m attached to a spring, on a smooth horizontal surface. Since there are no external horizontal forces, the total horizontal momentum of the system is conserved. Initially, the system is at rest, so the total momentum is zero. Therefore, the total momentum remains zero throughout the motion.

Let $\Delta x_A$ and $\Delta x_B$ be the displacements of block A and block B from their initial positions, respectively. Since the initial velocities are zero and the total momentum is conserved and equal to zero, we have: $m v_A + 2m v_B = 0$. Integrating this equation with respect to time from the initial state to the final state, we get: $\int m v_A dt + \int 2m v_B dt = 0$ $m \int_{x_{A,i}}^{x_{A,f}} dx_A + 2m \int_{x_{B,i}}^{x_{B,f}} dx_B = 0$ $m (x_{A,f} - x_{A,i}) + 2m (x_{B,f} - x_{B,i}) = 0$ $m \Delta x_A + 2m \Delta x_B = 0$ $\Delta x_A = -2 \Delta x_B$.

Let the natural length of the spring be $L$. Initially, the spring is compressed by $x_0$, so the initial length of the spring is $L_i = L - x_0$. Let the initial positions of A and B be $x_{A,i}$ and $x_{B,i}$. Then $x_{B,i} - x_{A,i} = L - x_0$.

Finally, the compression of the spring is reduced to $\frac{x_0}{2}$, so the final length of the spring is $L_f = L - \frac{x_0}{2}$. Let the final positions of A and B be $x_{A,f}$ and $x_{B,f}$. Then $x_{B,f} - x_{A,f} = L - \frac{x_0}{2}$.

The displacements are $\Delta x_A = x_{A,f} - x_{A,i}$ and $\Delta x_B = x_{B,f} - x_{B,i}$. Subtracting the initial length equation from the final length equation: $(x_{B,f} - x_{A,f}) - (x_{B,i} - x_{A,i}) = (L - \frac{x_0}{2}) - (L - x_0)$ $(x_{B,f} - x_{B,i}) - (x_{A,f} - x_{A,i}) = L - \frac{x_0}{2} - L + x_0$ $\Delta x_B - \Delta x_A = \frac{x_0}{2}$.

Now we have a system of two equations with two unknowns $\Delta x_A$ and $\Delta x_B$:

1. $\Delta x_A = -2 \Delta x_B$
2. $\Delta x_B - \Delta x_A = \frac{x_0}{2}$

Substitute equation (1) into equation (2): $\Delta x_B - (-2 \Delta x_B) = \frac{x_0}{2}$ $\Delta x_B + 2 \Delta x_B = \frac{x_0}{2}$ $3 \Delta x_B = \frac{x_0}{2}$ $\Delta x_B = \frac{x_0}{6}$.

The displacement of block B is $\Delta x_B = \frac{x_0}{6}$.