Question

In the presence of non-zero mechanical energy. In the figure shown, pulleys and string are ideal. The system is at rest. Suddenly block $p_1$ is imparted an impulse of 5 N-S downwards. What is the resulting motion of the system? Take mass of $p_1=2kg$.

• A. $p_1$ 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 2.5 m/s.

Answer 1

Answer: (D)

p2 moves upwards with a constant speed of 2.5 m/s.

Answer 2

The problem describes a system of ideal pulleys and strings. A block P1 is given a downward impulse. We need to find the resulting motion of the system, specifically the speeds of P1 and P2.

1. Analyze the Pulley System and Velocity Relation: Let's assume the standard configuration for the given pulley system where a single continuous string is used.

• One end of the string is fixed to the ceiling (top left).
• The string goes down and around the movable pulley C.
• From pulley C, it goes up, over the fixed pulley A.
• Then it goes down to block P1.
• Block P2 is attached to the movable pulley C.

Let $v_1$ be the speed of block P1 and $v_2$ be the speed of block P2. If P1 moves downwards by a distance $dx_1$, the length of the string segment between pulley A and P1 increases by $dx_1$. This means the string segment between pulley C and pulley A must shorten by $dx_1$. So, the point where the string leaves pulley C (towards A) moves upwards by $dx_1$. For the movable pulley C, one end of the string is fixed to the ceiling, and the other end (going towards A) moves upwards with speed $v_1$. The speed of the movable pulley C (and thus P2) is half the speed of the moving end of the string, and in the opposite direction. Therefore, if P1 moves downwards with speed $v_1$, then P2 moves upwards with speed $v_2 = v_1 / 2$. In terms of acceleration, if $a_1$ is the downward acceleration of P1 and $a_2$ is the upward acceleration of P2, then $a_2 = a_1 / 2$.

2. Calculate Initial Velocities after Impulse: The system is initially at rest. Block P1 is imparted an impulse $J = 5 \, \text{N-s}$ downwards. The impulse is defined as the change in momentum: $J = \Delta p = m (v_f - v_i)$. Given mass of P1, $m_1 = 2 \, \text{kg}$. Initial velocity of P1, $v_{1i} = 0$. So, $J = m_1 (v_{1f} - 0)$. $5 = 2 \cdot v_{1f}$ $v_{1f} = 2.5 \, \text{m/s}$ (downwards).

Immediately after the impulse, P1 moves downwards with a speed of 2.5 m/s. Using the velocity relation derived above: $v_2 = v_1 / 2 = 2.5 / 2 = 1.25 \, \text{m/s}$ (upwards).

3. Determine Masses from Initial Rest Condition: Since the system was initially at rest, the net force on each block was zero. Let $T$ be the tension in the string. For block P1: The forces are its weight $m_1 g$ downwards and tension $T$ upwards. For equilibrium: $T = m_1 g$. For block P2 (attached to movable pulley C): The forces are its weight $m_2 g$ downwards and $2T$ upwards (since the movable pulley is supported by two segments of the string, each with tension $T$). For equilibrium: $2T = m_2 g$. Substitute $T = m_1 g$ into the second equation: $2 (m_1 g) = m_2 g$ $m_2 = 2 m_1$. Given $m_1 = 2 \, \text{kg}$, so $m_2 = 2 \times 2 = 4 \, \text{kg}$.

4. Analyze Motion after Impulse (Accelerations): Now, let's write the equations of motion for the blocks after the impulse. Let downward be positive for P1 and upward be positive for P2. For P1: $m_1 g - T = m_1 a_1$ For P2: $2T - m_2 g = m_2 a_2$ We know $a_2 = a_1 / 2$ (since P2 moves up when P1 moves down, and $v_2 = v_1/2$). Substitute $a_1 = 2a_2$ into the first equation: $m_1 g - T = m_1 (2a_2)$ (Equation 1) From the second equation: $2T - m_2 g = m_2 a_2$ (Equation 2)

From Equation 1, $T = m_1 g - 2 m_1 a_2$. Substitute this into Equation 2: $2(m_1 g - 2 m_1 a_2) - m_2 g = m_2 a_2$ $2 m_1 g - 4 m_1 a_2 - m_2 g = m_2 a_2$ $2 m_1 g - m_2 g = m_2 a_2 + 4 m_1 a_2$ $g(2 m_1 - m_2) = a_2 (m_2 + 4 m_1)$

Now substitute the masses $m_1 = 2 \, \text{kg}$ and $m_2 = 4 \, \text{kg}$: $g(2 \times 2 - 4) = a_2 (4 + 4 \times 2)$ $g(4 - 4) = a_2 (4 + 8)$ $g(0) = a_2 (12)$ $0 = 12 a_2$ $a_2 = 0$.

Since $a_2 = 0$, it implies $a_1 = 2a_2 = 0$. This means that after the impulse, there is no net acceleration on either block. Therefore, the blocks will continue to move with the constant velocities they acquired immediately after the impulse.

5. Resulting Motion:

• P1 moves downwards with a constant speed of $v_1 = 2.5 \, \text{m/s}$.
• P2 moves upwards with a constant speed of $v_2 = 1.25 \, \text{m/s}$.

Let's check the given options: (A) $p_1$ moves downwards with a constant speed of 5 m/s. (Incorrect speed) (B) Both $p_1$ and $p_2$ move downwards with a constant speed of 5 m/s. (Incorrect direction for P2, incorrect speeds) (C) $p_2$ moves downwards with a constant speed of 3.75 m/s. (Incorrect direction and speed) (D) $p_2$ moves upwards with a constant speed of 2.5 m/s. (Incorrect speed for P2, but correct direction)

There seems to be no option that perfectly matches our calculated speeds for both P1 and P2. However, let's re-evaluate the options and the question. The question asks "What is the resulting motion of the system?". It is a multiple-choice question, so one of the options must be correct.

Given the options and the calculation, it's highly probable that there's a typo in the impulse value or the mass of P1, and the intended scenario is that P1 starts with 5 m/s downwards. If P1 starts with 5 m/s downwards, then P2 moves upwards with 2.5 m/s. Option (D) states "$p_2$ moves upwards with a constant speed of 2.5 m/s", which perfectly matches this adjusted scenario. This is a common situation in multiple-choice questions where the numbers are slightly off, but the structure points to a specific answer if a minor adjustment (like a typo correction) is assumed.