Two blocks of masses `5 kg` and `2 kg` are placed on a frictionless surface and connected by a spring. An external kick gives a velocity of `14 m//s` to the heavier block in the direction of lighter one. The magnitudes of velocities of two blocks in the centre of mass frame after the kick are, respectively,

Options

A. `4m//s,4m//s`
B. `10m//s,4m//s`
C. `4m//s,10m//s`
D. `10m//s,10m//s`

Explanation

To solve the problem, we need to find the magnitudes of the velocities of two blocks in the center of mass (CM) frame after the external kick is applied to the heavier block. Here’s a step-by-step solution:

### Step 1: Identify the masses and initial conditions
- Mass of the heavier block (m1) = 5 kg
- Mass of the lighter block (m2) = 2 kg
- Initial velocity of the heavier block (V1) = 14 m/s (to the right)
- Initial velocity of the lighter block (V2) = 0 m/s (at rest)

### Step 2: Calculate the velocity of the center of mass (V_cm)
The velocity of the center of mass for the system can be calculated using the formula:
\[ V_{cm} = \frac{m_1 V_1 + m_2 V_2}{m_1 + m_2} \]
Substituting the values:
\[ V_{cm} = \frac{(5 \, \text{kg} \times 14 \, \text{m/s}) + (2 \, \text{kg} \times 0 \, \text{m/s})}{5 \, \text{kg} + 2 \, \text{kg}} = \frac{70 \, \text{kg m/s}}{7 \, \text{kg}} = 10 \, \text{m/s} \]

### Step 3: Determine the velocities of the blocks in the center of mass frame
In the center of mass frame, the center of mass is at rest. Therefore, we need to subtract the velocity of the center of mass from the velocities of each block.

- For the heavier block (5 kg):
\[ V_{1, \text{CM}} = V_1 - V_{cm} = 14 \, \text{m/s} - 10 \, \text{m/s} = 4 \, \text{m/s} \]

- For the lighter block (2 kg):
\[ V_{2, \text{CM}} = V_2 - V_{cm} = 0 \, \text{m/s} - 10 \, \text{m/s} = -10 \, \text{m/s} \]
(The negative sign indicates that the lighter block is moving in the opposite direction to the heavier block.)

### Step 4: Find the magnitudes of the velocities
The magnitudes of the velocities are:
- Magnitude of velocity of the heavier block = |4 m/s| = 4 m/s
- Magnitude of velocity of the lighter block = |-10 m/s| = 10 m/s

### Final Answer
The magnitudes of the velocities of the two blocks in the center of mass frame after the kick are:
- 4 m/s (for the 5 kg block)
- 10 m/s (for the 2 kg block)

### Options
The correct option is: 4 m/s and 10 m/s.

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