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

Options

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

Answer

To solve the problem, we will apply the principle of conservation of momentum. Here's a step-by-step breakdown of the solution:

### Step 1: Understand the system
Initially, both the man and his son are at rest on a frictionless surface. Therefore, the initial momentum of the system is zero.

### Step 2: Define the variables
Let:
- Mass of the man, \( m_m = 50 \, \text{kg} \)
- Mass of the son, \( m_s = 20 \, \text{kg} \)
- Velocity of the man with respect to the ground, \( v_m \)
- Velocity of the son with respect to the ground, \( v_s \)

### Step 3: Relate the velocities
The son moves with a velocity of \( 0.70 \, \text{m/s} \) with respect to the man. Therefore, we can express the velocity of the son in terms of the velocity of the man:
\[ v_s = v_m + 0.70 \]

### Step 4: Apply conservation of momentum
According to the conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]
Since the initial momentum is zero, we have:
\[ 0 = m_m \cdot v_m + m_s \cdot v_s \]
Substituting the values of \( m_m \) and \( m_s \):
\[ 0 = 50 \cdot v_m + 20 \cdot (v_m + 0.70) \]

### Step 5: Simplify the equation
Expanding the equation gives:
\[ 0 = 50 v_m + 20 v_m + 14 \]
Combining like terms:
\[ 0 = 70 v_m + 14 \]

### Step 6: Solve for \( v_m \)
Rearranging the equation to solve for \( v_m \):
\[ 70 v_m = -14 \]
\[ v_m = -\frac{14}{70} = -0.2 \, \text{m/s} \]

### Step 7: Interpret the result
The negative sign indicates that the direction of the man's velocity is opposite to the assumed positive direction (which we took as rightward). Thus, the man is moving at a speed of \( 0.2 \, \text{m/s} \) to the left.

### Final Answer
The speed of the man with respect to the surface is \( 0.2 \, \text{m/s} \) towards the left.

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