A running man has the same kinetic energy as that of a boy of half his mass. The man speed up by `2 ms^(-1)` and the boy changes his speed by `x ms^(-1)` so that the kinetic energies of the boy and the man are again equal. Then `x` in `ms^(-1)` is

Options

A. `-2sqrt2`
B. `+2sqrt2`
C. `sqrt2`
D. 2

Answer

To solve the problem step by step, let's denote the following:

- Let the mass of the man be \( M \).
- Let the mass of the boy be \( m = \frac{M}{2} \).
- Let the initial speed of the man be \( V_m \).
- Let the initial speed of the boy be \( V_b \).

### Step 1: Set up the initial kinetic energy equality

The kinetic energy of the man and the boy are equal initially:

\[ KE_m = KE_b \]

This can be expressed as:

\[ \frac{1}{2} M V_m^2 = \frac{1}{2} m V_b^2 \]

Substituting \( m = \frac{M}{2} \):

\[ \frac{1}{2} M V_m^2 = \frac{1}{2} \left(\frac{M}{2}\right) V_b^2 \]

### Step 2: Simplify the equation

Canceling \( \frac{1}{2} \) and \( M \) from both sides gives:

\[ V_m^2 = \frac{1}{2} V_b^2 \]

### Step 3: Solve for \( V_b \)

Rearranging gives:

\[ V_b^2 = 2 V_m^2 \]

Taking the square root:

\[ V_b = \sqrt{2} V_m \]

This is our **Equation 1**.

### Step 4: Set up the new kinetic energy equality after speed changes

After the man speeds up by \( 2 \, \text{m/s} \) and the boy speeds up by \( x \, \text{m/s} \), their new kinetic energies are equal:

\[ KE'_m = KE'_b \]

This can be expressed as:

\[ \frac{1}{2} M (V_m + 2)^2 = \frac{1}{2} m (V_b + x)^2 \]

Substituting \( m = \frac{M}{2} \):

\[ \frac{1}{2} M (V_m + 2)^2 = \frac{1}{2} \left(\frac{M}{2}\right) (V_b + x)^2 \]

### Step 5: Simplify the new equation

Canceling \( \frac{1}{2} \) and \( M \) from both sides gives:

\[ (V_m + 2)^2 = \frac{1}{2} (V_b + x)^2 \]

### Step 6: Substitute \( V_b \) from Equation 1

Substituting \( V_b = \sqrt{2} V_m \) into the equation:

\[ (V_m + 2)^2 = \frac{1}{2} (\sqrt{2} V_m + x)^2 \]

### Step 7: Expand both sides

Expanding both sides:

Left side:

\[ (V_m + 2)^2 = V_m^2 + 4V_m + 4 \]

Right side:

\[ \frac{1}{2} (\sqrt{2} V_m + x)^2 = \frac{1}{2} (2 V_m^2 + 2\sqrt{2} V_m x + x^2) = V_m^2 + \sqrt{2} V_m x + \frac{1}{2} x^2 \]

### Step 8: Set the equations equal

Setting the left side equal to the right side:

\[ V_m^2 + 4V_m + 4 = V_m^2 + \sqrt{2} V_m x + \frac{1}{2} x^2 \]

### Step 9: Cancel \( V_m^2 \) and rearrange

Cancelling \( V_m^2 \) from both sides gives:

\[ 4V_m + 4 = \sqrt{2} V_m x + \frac{1}{2} x^2 \]

### Step 10: Rearranging gives a quadratic equation in \( x \)

Rearranging yields:

\[ \frac{1}{2} x^2 + \sqrt{2} V_m x - 4V_m - 4 = 0 \]

### Step 11: Solve for \( x \)

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = \frac{1}{2} \), \( b = \sqrt{2} V_m \), and \( c = -4V_m - 4 \).

Calculating the discriminant and solving will yield the value of \( x \).

After solving, we find that:

\[ x = 2\sqrt{2} \, \text{m/s} \]

### Final Answer

Thus, the value of \( x \) is:

\[ \boxed{2\sqrt{2} \, \text{m/s}} \]