Question

The system is released from rest. 2 second after the release the heavier block is held for an instant and again released. Find the time elapsed after the momentary stopage of heavier block that the string again become tight.

• A. $\frac{4}{3}$ sec
• B. $\frac{2}{3}$ sec
• C. $\frac{1}{3}$ sec
• D. $\frac{5}{3}$ sec

Answer 1

Answer: (B)

2/3 sec

Answer 2

The problem involves a system of blocks and pulleys. We need to analyze the motion in two phases:

1. Initial motion until 2 seconds.
2. Motion after the heavier block is momentarily stopped and released.

Phase 1: Initial motion (0 to 2 seconds)

Let $a_m$ be the acceleration of mass 'm' and $a_{4m}$ be the acceleration of mass '4m'. Let 'T' be the tension in the string.

From the pulley configuration, the acceleration constraint is: If mass 'm' moves up by a distance $x$, the movable pulley (and thus mass '4m') moves down by $x/2$. So, if $a_m$ is the upward acceleration of 'm', and $a_{4m}$ is the downward acceleration of '4m', then: $a_m = 2a_{4m}$

Now, apply Newton's second law: For mass 'm' (upward positive): $T - mg = ma_m$ (Equation 1)

For mass '4m' (downward positive): The movable pulley is supported by two segments of the string, each having tension 'T'. So, the upward force on '4m' is $2T$. $4mg - 2T = 4ma_{4m}$ (Equation 2)

Substitute $a_m = 2a_{4m}$ into Equation 1: $T - mg = m(2a_{4m}) \implies T = mg + 2ma_{4m}$

Substitute this expression for T into Equation 2: $4mg - 2(mg + 2ma_{4m}) = 4ma_{4m}$ $4mg - 2mg - 4ma_{4m} = 4ma_{4m}$ $2mg = 8ma_{4m}$ $a_{4m} = \frac{2mg}{8m} = \frac{g}{4}$ (downwards)

Now find $a_m$: $a_m = 2a_{4m} = 2\left(\frac{g}{4}\right) = \frac{g}{2}$ (upwards)

After 2 seconds, the velocities of the blocks are: Initial velocity $u=0$ for both. $v_m = u + a_m t = 0 + \left(\frac{g}{2}\right)(2) = g$ (upwards) $v_{4m} = u + a_{4m} t = 0 + \left(\frac{g}{4}\right)(2) = \frac{g}{2}$ (downwards)

Phase 2: Motion after momentary stoppage

At $t=2$ seconds, the heavier block (4m) is held for an instant and released. This means its velocity instantaneously becomes zero. The string becomes slack because mass 'm' still has an upward velocity.

Let $t'=0$ be the instant the heavier block is released (which corresponds to $t=2$ seconds in the overall timeline). At $t'=0$: $v_m(0) = g$ (upwards) $v_{4m}(0) = 0$

Since the string is slack, both blocks move independently under gravity. For mass 'm': It moves upwards with initial velocity $g$ and is acted upon by gravity (downwards). So, its acceleration is $a'_m = -g$ (if upward is positive). The displacement of 'm' from its position at $t'=0$ is $\Delta y_m$. Let upward displacement be positive. $\Delta y_m(t') = v_m(0)t' + \frac{1}{2}a'_m t'^2 = gt' - \frac{1}{2}gt'^2$

For mass '4m': It starts from rest and falls under gravity. So, its acceleration is $a'_{4m} = g$ (downwards). The displacement of '4m' from its position at $t'=0$ is $\Delta y_{4m}$. Let downward displacement be positive. $\Delta y_{4m}(t') = v_{4m}(0)t' + \frac{1}{2}a'_{4m} t'^2 = 0 + \frac{1}{2}gt'^2 = \frac{1}{2}gt'^2$

The string will become tight again when the relative positions of the blocks satisfy the length constraint of the string. Let's define the positions from a fixed reference point (e.g., the ceiling). Let $y_m$ be the position of mass 'm' (downwards positive). Let $y_P$ be the position of the movable pulley (downwards positive). The total length of the string $L$ is constant: $L = (y_P - y_{fixed\_support}) + (y_P - y_{fixed\_pulley}) + (y_m - y_{fixed\_pulley})$ $L = 2y_P + y_m - (y_{fixed\_support} + 2y_{fixed\_pulley})$ Since $L$, $y_{fixed\_support}$, and $y_{fixed\_pulley}$ are constants, we have: $2y_P + y_m = \text{constant}$ Differentiating twice with respect to time: $2a_P + a_m = 0$ (where $a_P$ is acceleration of pulley, $a_m$ is acceleration of mass 'm', both positive downwards). Since $a_P = a_{4m}$, $2a_{4m} + a_m = 0$. This is the constraint when the string is tight.

Let's use the displacements from their positions at $t'=0$. Let $\Delta y_m$ be the displacement of 'm' (downward positive). Let $\Delta y_{4m}$ be the displacement of '4m' (downward positive). The constraint implies that the change in length of the string must be zero for the string to be taut. The total length of the string is $L = 2y_P + y_m - C$. So, $2\Delta y_P + \Delta y_m = 0$ (where $\Delta y_P$ is the displacement of the movable pulley, same as $\Delta y_{4m}$). Thus, $2\Delta y_{4m} + \Delta y_m = 0$ is the condition for the string to become tight again.

From our earlier calculations for motion during slack: $\Delta y_m(t') = -(gt' - \frac{1}{2}gt'^2) = -gt' + \frac{1}{2}gt'^2$ (converted to downward positive displacement) $\Delta y_{4m}(t') = \frac{1}{2}gt'^2$ (downward positive displacement)

Substitute these into the condition $2\Delta y_{4m} + \Delta y_m = 0$: $2\left(\frac{1}{2}gt'^2\right) + \left(-gt' + \frac{1}{2}gt'^2\right) = 0$ $gt'^2 - gt' + \frac{1}{2}gt'^2 = 0$ $\frac{3}{2}gt'^2 - gt' = 0$ $gt'\left(\frac{3}{2}t' - 1\right) = 0$

This equation gives two solutions:

1. $t' = 0$: This corresponds to the instant the heavier block was stopped.
2. $\frac{3}{2}t' - 1 = 0 \implies \frac{3}{2}t' = 1 \implies t' = \frac{2}{3}$ sec.

This is the time elapsed after the momentary stoppage of the heavier block that the string again becomes tight.