A spring is compressed between two toy carts of masses `m_(1)` and `m_(2)`. When the toy carts are released then spring exerts equal and opposite average force on each for the same time t. Assume that there is no friction between the toy carts and the ground, the velocities of toy carts are

Options

A. In opposite directions but in inverse ratio of their masses
B. In opposite directions but in direct ratio of their masses
C. Equal and in opposite direction
D. Equal and in same direction.

Answer

To solve the problem of the velocities of the toy carts after the spring is released, we can follow these steps:

### Step 1: Understand the System
Initially, both toy carts of masses \( m_1 \) and \( m_2 \) are at rest with a compressed spring between them. When the spring is released, it exerts equal and opposite forces on both carts.

### Step 2: Apply Newton's Third Law
According to Newton's Third Law, the force exerted by the spring on cart 1 is equal in magnitude and opposite in direction to the force exerted on cart 2. Therefore, if the force exerted by the spring is \( F \), we have:
\[ F = kx \]
where \( k \) is the spring constant and \( x \) is the compression of the spring.

### Step 3: Calculate the Acceleration of Each Cart
Using Newton's second law, we can find the acceleration of each cart:
- For cart 1:
\[ a_1 = \frac{F}{m_1} = \frac{kx}{m_1} \]
- For cart 2:
\[ a_2 = \frac{F}{m_2} = \frac{kx}{m_2} \]

### Step 4: Use the Kinematic Equation
Since both carts start from rest, we can use the kinematic equation to find their velocities after time \( t \):
- For cart 1:
\[ v_1 = u + a_1 t = 0 + \left(\frac{kx}{m_1}\right)t = \frac{kxt}{m_1} \]
- For cart 2:
\[ v_2 = u + a_2 t = 0 + \left(\frac{kx}{m_2}\right)t = \frac{kxt}{m_2} \]

### Step 5: Determine the Direction of Velocities
Since the spring pushes the carts in opposite directions, we can assign directions:
- Let \( v_1 \) be in the positive direction.
- Let \( v_2 \) be in the negative direction.

Thus, we can write:
\[ v_1 = \frac{kxt}{m_1} \]
\[ v_2 = -\frac{kxt}{m_2} \]

### Step 6: Find the Ratio of Velocities
To find the ratio of the velocities:
\[ \frac{v_1}{v_2} = \frac{\frac{kxt}{m_1}}{-\frac{kxt}{m_2}} = -\frac{m_2}{m_1} \]
This shows that the velocities are inversely proportional to their masses.

### Conclusion
The velocities of the toy carts after the spring is released are:
- \( v_1 = \frac{kxt}{m_1} \) (in one direction)
- \( v_2 = -\frac{kxt}{m_2} \) (in the opposite direction)