Question

In the figure shown, pulleys and string are ideal. The system is at rest. Suddenly block $p_2$ is imparted an impulse of 5 N-S downwards. What is the resulting motion of the system? Take mass of p = 2kg.

• A. $p_2$ moves downwards with a constant speed of 5 m/s.
• B. Both $p_1$ and $p_2$ move downwards with a constant speed of 5 m/s.
• C. $p_2$ moves downwards with a constant speed of 3.75 m/s.
• D. $p_2$ moves upwards with a constant speed of 2.5m/s.

Answer 1

D

Answer 2

The problem describes a pulley system where block $p_2$ is given an impulse. We need to determine the resulting motion.

1. Analyze the Pulley System and Constraints:

Let $m_1$ be the mass of block $p_1$ and $m_2$ be the mass of block $p_2$. The problem states "mass of p = 2kg", which implies $m_1 = m_2 = 2 \text{ kg}$. Pulleys and string are ideal.

Let's trace the string and determine the relationship between the velocities of $p_1$ and $p_2$. The diagram shows a fixed pulley (A) and a movable pulley (C).

• A string starts from a fixed point (ceiling), goes around the movable pulley (C), then goes over the fixed pulley (A), and finally connects to block $p_1$.
• Block $p_2$ is attached to the movable pulley (C).

Let $y_1$ be the downward displacement of $p_1$ and $y_2$ be the downward displacement of $p_2$ (and pulley C). If pulley C moves down by a distance $dy_2$, then the length of the string segment from the fixed point to C decreases by $dy_2$, and the length of the string segment from C to pulley A also decreases by $dy_2$. So, a total of $2dy_2$ length of string is effectively pulled from the side of $p_1$. This means $p_1$ must move upwards by $2dy_2$.

Therefore, if $v_2$ is the downward speed of $p_2$, then $v_1 = 2v_2$ is the upward speed of $p_1$. Similarly, their accelerations are related by $a_1 = 2a_2$ (where $a_1$ is upward acceleration of $p_1$ and $a_2$ is downward acceleration of $p_2$).

2. Check Initial Equilibrium Condition:

The problem states "The system is at rest". This means the net force on each block is zero. Let $T$ be the tension in the string.

For block $p_1$: Since $p_1$ would tend to move upwards in response to $p_2$ moving downwards, the upward tension $T$ must balance its weight $m_1 g$.

$T - m_1 g = 0 \implies T = m_1 g$.

For block $p_2$ and movable pulley C: The downward force is $m_2 g$. The upward force is due to two segments of the string pulling up the movable pulley, so $2T$.

$2T - m_2 g = 0 \implies 2T = m_2 g$.

Substitute $T = m_1 g$ into the second equation:

$2(m_1 g) = m_2 g$

$2m_1 = m_2$.

Given that "mass of p = 2kg", if we assume $m_1 = m_2 = 2 \text{ kg}$, then the condition $2m_1 = m_2$ becomes $2(2) = 2$, which is $4 = 2$. This is false.

This implies that if $m_1 = m_2 = 2 \text{ kg}$, the system cannot be at rest. This is a common situation in competitive exams where the "at rest" condition implies the system is in equilibrium, meaning the derived mass ratio must hold. If the given masses contradict this, it means the system is not truly at rest in equilibrium, or the problem implies a special case.

However, if the system is stated to be "at rest", it must be in equilibrium. For this to be true with $m_1=2kg$, $m_2$ must be $4kg$. But the problem states "mass of p = 2kg", which is ambiguous. The most reasonable interpretation for "system at rest" when an impulse is applied is that the system is in equilibrium, meaning the net force is zero. If the net force is zero, then after an impulse, the system will move with a constant velocity.

Let's assume the "at rest" implies that the system is in equilibrium, and therefore, the acceleration after the impulse is zero. This would mean $m_2 = 2m_1$. Since $m_1 = 2kg$, this would imply $m_2 = 4kg$. If this is the case, then the net force on the system is zero.

3. Calculate Initial Velocity after Impulse:

An impulse $J$ imparted to a block changes its momentum: $J = \Delta p = m \Delta v$. The impulse is given to $p_2$ downwards, $J = 5 \text{ N-s}$. Initial velocity of $p_2$ is $v_{2,initial} = 0$ (since the system is at rest).

$J = m_2 (v_{2,final} - v_{2,initial})$

$5 = 2 \text{ kg} \times (v_{2,final} - 0)$

$v_{2,final} = \frac{5}{2} = 2.5 \text{ m/s}$ downwards.

4. Determine Resulting Motion:

Since the system is stated to be "at rest" initially, it implies that it is in equilibrium. In an ideal pulley system, if a system is in equilibrium, the net force on it is zero. If an impulse is applied to such a system, it will acquire an initial velocity and continue to move with that constant velocity because there are no net forces to cause acceleration or deceleration.

So, $p_2$ will move downwards with a constant speed of $2.5 \text{ m/s}$. And $p_1$ will move upwards with a constant speed of $2 \times 2.5 = 5 \text{ m/s}$.

The calculated speed of $p_2$ immediately after the impulse is $2.5 \text{ m/s}$ downwards. If the system is in equilibrium, it will move with this constant speed. Option D has the correct magnitude of speed, but the direction is stated as upwards. Given the impulse is downwards, it is likely a typo in the option.