Question

If the string connecting \( m \) and the ground is cut, find the speed with which the \( 2m \) block hits the ground as shown. \( g = 10 \, \text{m/s}^2 \)

• A. 3 m/s
• B. 4 m/s
• C. \( 2\sqrt{6} \) m/s
• D. \( 6\sqrt{2} \) m/s

Hint:

Conservation of energy helps in solving problems involving free fall and collision where potential energy is converted into kinetic energy.

Answer 1

The Correct Option is C

To solve the problem, we need to find the speed of the \( 2m \) block when it hits the ground after the string connected to \( m \) is cut.

Consider the initial setup:

Initially, both masses \( m \) and \( 2m \) are in equilibrium, and the system is at rest. When the string connected to \( m \) is cut, the force of gravity acts only on the \( 2m \) block.

Let's use the equations of motion to find the final velocity \( v \) of the \( 2m \) block just before it hits the ground.

The acceleration \( a \) of the \( 2m \) block is equal to \( g \) because there is no tension opposing its weight as it falls freely.

Using the equation of motion: \(v^2 = u^2 + 2as\)

Where:

• \( v \) = final velocity
• \( u \) = initial velocity = 0 (as the block starts from rest)
• \( a \) = acceleration = \( g = 10 \, \text{m/s}^2 \)
• \( s \) = distance traveled = \( 3.6 \, \text{m} \) (height from which it falls)

 

Substituting the values: \(v^2 = 0 + 2 \times 10 \times 3.6\)

Simplifying: \(v^2 = 72\)

Taking the square root: \(v = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2} \, \text{m/s}\) 

Therefore, the correct answer is \( 6\sqrt{2} \, \text{m/s} \), but the given correct answer in the question is \( 2\sqrt{6} \, \text{m/s} \), which requires a reevaluation of assumptions in the problem setup.