A mass of `3kg` descending vertically downwards supports a mass of `2kg` by means of a light string passing over a pulley At the end of `5s` the string breaks How much high from now the `2kg` mass will go `(g = 9.8m//^(2))`.

Options

A. 4.9 m
B. 9.8m
C. 10.6 m
D. 2.45 m

Explanation

To solve the problem step by step, we will analyze the motion of the two masses connected by a string over a pulley.

### Step 1: Understand the system
We have two masses:
- Mass \( m_1 = 3 \, \text{kg} \) (descending)
- Mass \( m_2 = 2 \, \text{kg} \) (ascending)

The mass \( m_1 \) is descending, which means it is accelerating downwards, while \( m_2 \) is moving upwards with the same acceleration due to the tension in the string.

### Step 2: Write the equations of motion
For mass \( m_1 \) (3 kg):
- The net force acting on it is \( 3g - T \) (where \( T \) is the tension in the string).
- According to Newton's second law, this net force equals the mass times acceleration:
\[ 3g - T = 3a \quad \text{(1)} \]

For mass \( m_2 \) (2 kg):
- The net force acting on it is \( T - 2g \).
- Again, applying Newton's second law:
\[ T - 2g = 2a \quad \text{(2)} \]

### Step 3: Solve for acceleration \( a \)
We can add equations (1) and (2) to eliminate \( T \):
\[ (3g - T) + (T - 2g) = 3a + 2a \]
This simplifies to:
\[ 3g - 2g = 5a \]
Thus:
\[ g = 5a \quad \Rightarrow \quad a = \frac{g}{5} \]

### Step 4: Calculate the upward velocity of \( m_2 \) after 5 seconds
Now, we need to find the velocity of mass \( m_2 \) after 5 seconds. Using the formula:
\[ v = u + at \]
where \( u = 0 \) (initial velocity), \( a = \frac{g}{5} \), and \( t = 5 \, \text{s} \):
\[ v = 0 + \left(\frac{g}{5}\right) \cdot 5 = g \]
Thus, the velocity of mass \( m_2 \) after 5 seconds is \( g \).

### Step 5: Analyze the motion after the string breaks
When the string breaks, mass \( m_2 \) will continue to move upwards with the velocity \( g \) but will be acted upon only by gravity, which will decelerate it.

Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \) (final velocity at the highest point),
- \( u = g \) (initial velocity just after the string breaks),
- \( a = -g \) (acceleration due to gravity, acting downwards).

Substituting these values:
\[ 0 = g^2 + 2(-g)s \]
This simplifies to:
\[ 0 = g^2 - 2gs \quad \Rightarrow \quad 2gs = g^2 \quad \Rightarrow \quad s = \frac{g}{2} \]

### Step 6: Calculate the height \( s \)
Substituting \( g = 9.8 \, \text{m/s}^2 \):
\[ s = \frac{9.8}{2} = 4.9 \, \text{m} \]

### Final Answer
The height from now that the \( 2 \, \text{kg} \) mass will go after the string breaks is **4.9 meters**.

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