A particle of mass m has half the kinetic energy of another particle of mass `m/2`. If the speed of the heavier particle is increased by `2 ms^(-1)` its new kinetic energy becomes equal to t he original kinetic energy of the lighter particle. The ratio of the orighinal speeds of the lighter and heavier particle is

Options

A. `1:1`
B. `1:2`
C. `1:3`
D. `1:4`

Answer

To solve the problem step by step, we will define the variables and use the relationships between kinetic energy, mass, and velocity.

### Step 1: Define the variables
Let:
- Mass of the heavier particle = \( m \)
- Speed of the heavier particle = \( v_a \)
- Mass of the lighter particle = \( \frac{m}{2} \)
- Speed of the lighter particle = \( v_b \)

### Step 2: Write the expressions for kinetic energy
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
Thus, the kinetic energies of the two particles are:
- Kinetic energy of the heavier particle:
\[ KE_a = \frac{1}{2} m v_a^2 \]
- Kinetic energy of the lighter particle:
\[ KE_b = \frac{1}{2} \left(\frac{m}{2}\right) v_b^2 = \frac{1}{4} m v_b^2 \]

### Step 3: Use the relationship given in the problem
According to the problem, the heavier particle has half the kinetic energy of the lighter particle:
\[ KE_a = \frac{1}{2} KE_b \]
Substituting the expressions for kinetic energy:
\[ \frac{1}{2} m v_a^2 = \frac{1}{2} \left(\frac{1}{4} m v_b^2\right) \]
This simplifies to:
\[ m v_a^2 = \frac{1}{4} m v_b^2 \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ v_a^2 = \frac{1}{4} v_b^2 \]
Taking the square root of both sides:
\[ v_a = \frac{1}{2} v_b \]

### Step 4: Consider the change in speed of the heavier particle
The problem states that if the speed of the heavier particle is increased by \( 2 \, \text{ms}^{-1} \), its new kinetic energy becomes equal to the original kinetic energy of the lighter particle:
\[ KE_a' = KE_b \]
Where \( KE_a' \) is the new kinetic energy of the heavier particle:
\[ KE_a' = \frac{1}{2} m (v_a + 2)^2 \]
Setting this equal to \( KE_b \):
\[ \frac{1}{2} m (v_a + 2)^2 = \frac{1}{4} m v_b^2 \]
Dividing both sides by \( \frac{1}{2} m \):
\[ (v_a + 2)^2 = \frac{1}{2} v_b^2 \]

### Step 5: Substitute \( v_a \) in terms of \( v_b \)
From Step 3, we know \( v_a = \frac{1}{2} v_b \). Substituting this into the equation:
\[ \left(\frac{1}{2} v_b + 2\right)^2 = \frac{1}{2} v_b^2 \]
Expanding the left side:
\[ \left(\frac{1}{2} v_b + 2\right)^2 = \frac{1}{4} v_b^2 + 2 v_b + 4 \]
Setting the equation:
\[ \frac{1}{4} v_b^2 + 2 v_b + 4 = \frac{1}{2} v_b^2 \]
Rearranging gives:
\[ \frac{1}{4} v_b^2 + 2 v_b + 4 - \frac{1}{2} v_b^2 = 0 \]
Combining like terms:
\[ -\frac{1}{4} v_b^2 + 2 v_b + 4 = 0 \]
Multiplying through by -4 to eliminate the fraction:
\[ v_b^2 - 8 v_b - 16 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( v_b = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ v_b = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} \]
\[ v_b = \frac{8 \pm \sqrt{64 + 64}}{2} \]
\[ v_b = \frac{8 \pm \sqrt{128}}{2} \]
\[ v_b = \frac{8 \pm 8\sqrt{2}}{2} \]
\[ v_b = 4 \pm 4\sqrt{2} \]

### Step 7: Find \( v_a \)
Using \( v_a = \frac{1}{2} v_b \):
\[ v_a = \frac{1}{2}(4 \pm 4\sqrt{2}) = 2 \pm 2\sqrt{2} \]

### Step 8: Calculate the ratio of original speeds
The ratio of the original speeds \( \frac{v_b}{v_a} \) is:
\[ \frac{v_b}{v_a} = \frac{4 \pm 4\sqrt{2}}{2 \pm 2\sqrt{2}} = 2 \pm 2\sqrt{2} \]

### Final Result
The ratio of the original speeds of the lighter and heavier particles is:
\[ \frac{v_b}{v_a} = 2 \]