Question:
As shown in the figure, block 'A' placed on a horizontal surface is moving horizontally with a speed of $10\text{ ms}^{-1}$. The speed of hanging block 'B' at the given instant of time is:
As shown in the figure, block 'A' placed on a horizontal surface is moving horizontally with a speed of $10\text{ ms}^{-1}$. The speed of hanging block 'B' at the given instant of time is:
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For any string-constraint system where a block moves horizontally with speed $v$, the velocity of the string along its length is always $v \cos\theta$.
This directly equals the speed of the hanging block.
This directly equals the speed of the hanging block.
Updated On: Jul 22, 2026
- $10\text{ ms}^{-1}$
- $5\text{ ms}^{-1}$
- $5\sqrt{3}\text{ ms}^{-1}$
- $20\text{ ms}^{-1}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
This is a constrained motion problem.
Block A moves horizontally, pulling a string that passes over a pulley to lift block B.
We must find the instantaneous speed of block B based on the speed of A and the string's angle.
Step 2: Key Concept and Approach:
In a taut string constraint, the component of velocity of a connected body along the length of the string must be equal to the rate of extension/contraction of the string.
Let $l$ be the length of the string segment from the pulley to block A.
The speed of block B ($v_B$) is equal to the rate at which this string length decreases ($-\frac{dl}{dt}$).
Step 3: Detailed Explanation:
• Set up the geometry:
Let the horizontal distance of block A from the vertical line of the pulley be $x$, and the height of the pulley be $h$.
By Pythagoras' theorem:
\[ l^2 = x^2 + h^2 \]
• Differentiate with respect to time:
\[ 2l \frac{dl}{dt} = 2x \frac{dx}{dt} + 0 \] \[ \frac{dl}{dt} = \frac{x}{l} \frac{dx}{dt} \]
• Substitute physical parameters:
Note that $\cos\theta = \frac{x}{l}$ (where $\theta = 60^\circ$ is the angle of the string with the horizontal).
The speed of block A is $v_A = \frac{dx}{dt} = 10\text{ ms}^{-1}$.
The speed of block B is $v_B = -\frac{dl}{dt}$.
Therefore:
\[ v_B = v_A \cos\theta \]
• Calculate the numerical value:
\[ v_B = 10 \times \cos(60^\circ) = 10 \times \frac{1}{2} = 5\text{ ms}^{-1} \]
Step 4: Final Answer:
The speed of hanging block B is $5\text{ ms}^{-1}$, which corresponds to Option (B).
This is a constrained motion problem.
Block A moves horizontally, pulling a string that passes over a pulley to lift block B.
We must find the instantaneous speed of block B based on the speed of A and the string's angle.
Step 2: Key Concept and Approach:
In a taut string constraint, the component of velocity of a connected body along the length of the string must be equal to the rate of extension/contraction of the string.
Let $l$ be the length of the string segment from the pulley to block A.
The speed of block B ($v_B$) is equal to the rate at which this string length decreases ($-\frac{dl}{dt}$).
Step 3: Detailed Explanation:
• Set up the geometry:
Let the horizontal distance of block A from the vertical line of the pulley be $x$, and the height of the pulley be $h$.
By Pythagoras' theorem:
\[ l^2 = x^2 + h^2 \]
• Differentiate with respect to time:
\[ 2l \frac{dl}{dt} = 2x \frac{dx}{dt} + 0 \] \[ \frac{dl}{dt} = \frac{x}{l} \frac{dx}{dt} \]
• Substitute physical parameters:
Note that $\cos\theta = \frac{x}{l}$ (where $\theta = 60^\circ$ is the angle of the string with the horizontal).
The speed of block A is $v_A = \frac{dx}{dt} = 10\text{ ms}^{-1}$.
The speed of block B is $v_B = -\frac{dl}{dt}$.
Therefore:
\[ v_B = v_A \cos\theta \]
• Calculate the numerical value:
\[ v_B = 10 \times \cos(60^\circ) = 10 \times \frac{1}{2} = 5\text{ ms}^{-1} \]
Step 4: Final Answer:
The speed of hanging block B is $5\text{ ms}^{-1}$, which corresponds to Option (B).
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