A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of `0.70 m s^(-1)` with respect to the man. The speed of the man with respect to the surface is:

Options

A. `0.14 m s^(-1)`
B. `0.47 m s^(-1)`
C. `0.28 ms^(-1)`
D. `0.20 ms^(-1)`

Explanation

To solve the problem, we will use the principle of conservation of momentum. Let's break down the steps:

### Step 1: Understand the System
- We have a man with mass \( m_1 = 50 \, \text{kg} \) and his son with mass \( m_2 = 20 \, \text{kg} \).
- They are initially at rest on a frictionless surface, so their initial velocities are both 0.

### Step 2: Define the Velocities
- Let \( v \) be the velocity of the man after he pushes his son.
- The son moves away from the man with a speed of \( 0.70 \, \text{m/s} \) relative to the man. Therefore, if the man moves with speed \( v \), the speed of the son \( v_s \) with respect to the surface will be:
\[ v_s = v + 0.70 \, \text{m/s} \]

### Step 3: Apply Conservation of Momentum
- The initial momentum of the system is 0 (since both are at rest).
- The final momentum of the system must also equal 0. Thus, we can write:
\[ m_1 \cdot (-v) + m_2 \cdot (v + 0.70) = 0 \]
Here, the man's momentum is negative because he moves in the opposite direction to his son.

### Step 4: Substitute the Masses
- Substituting the values of \( m_1 \) and \( m_2 \):
\[ 50 \cdot (-v) + 20 \cdot (v + 0.70) = 0 \]

### Step 5: Expand and Rearrange the Equation
- Expanding the equation gives:
\[ -50v + 20v + 14 = 0 \]
Combining like terms:
\[ -30v + 14 = 0 \]

### Step 6: Solve for \( v \)
- Rearranging the equation:
\[ 30v = 14 \]
\[ v = \frac{14}{30} = \frac{7}{15} \approx 0.4667 \, \text{m/s} \]

### Step 7: Find the Speed of the Man
- The speed of the man with respect to the surface is:
\[ v \approx 0.4667 \, \text{m/s} \]

### Final Answer
The speed of the man with respect to the surface is approximately \( 0.4667 \, \text{m/s} \).

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