Two bodies A and B have masses 20 kg and 5 kg respectively. Each one is acted upon by a force of 4 kg wt. If they acquire the same kinetic energy in times `t_A` and `t_B`, then the ratio `t_A/t_B` is

Options

A. `1/2`
B. 2
C. `2/5`
D. `5/6`

Explanation

To solve the problem, we need to find the ratio of the times \( t_A \) and \( t_B \) taken by two bodies A and B to acquire the same kinetic energy when acted upon by a force. Here are the steps to derive the solution:

### Step 1: Identify the given data
- Mass of body A, \( m_A = 20 \, \text{kg} \)
- Mass of body B, \( m_B = 5 \, \text{kg} \)
- Force acting on each body, \( F = 4 \, \text{kg wt} = 4 \times 9.8 \, \text{N} = 39.2 \, \text{N} \) (using \( g \approx 9.8 \, \text{m/s}^2 \))

### Step 2: Calculate the acceleration of each body
Using Newton's second law, \( F = m \cdot a \), we can find the acceleration for each body.

- For body A:
\[ a_A = \frac{F}{m_A} = \frac{39.2}{20} = 1.96 \, \text{m/s}^2 \]

- For body B:
\[ a_B = \frac{F}{m_B} = \frac{39.2}{5} = 7.84 \, \text{m/s}^2 \]

### Step 3: Write the expression for kinetic energy
The kinetic energy \( KE \) acquired by each body can be expressed as:
\[ KE = \frac{1}{2} m a t^2 \]

For body A:
\[ KE_A = \frac{1}{2} m_A a_A t_A^2 = \frac{1}{2} \cdot 20 \cdot 1.96 \cdot t_A^2 \]

For body B:
\[ KE_B = \frac{1}{2} m_B a_B t_B^2 = \frac{1}{2} \cdot 5 \cdot 7.84 \cdot t_B^2 \]

### Step 4: Set the kinetic energies equal
Since both bodies acquire the same kinetic energy:
\[ \frac{1}{2} \cdot 20 \cdot 1.96 \cdot t_A^2 = \frac{1}{2} \cdot 5 \cdot 7.84 \cdot t_B^2 \]

### Step 5: Simplify the equation
Canceling \( \frac{1}{2} \) from both sides:
\[ 20 \cdot 1.96 \cdot t_A^2 = 5 \cdot 7.84 \cdot t_B^2 \]

### Step 6: Rearrange to find the ratio \( \frac{t_A}{t_B} \)
Rearranging gives:
\[ \frac{t_A^2}{t_B^2} = \frac{5 \cdot 7.84}{20 \cdot 1.96} \]

### Step 7: Calculate the numerical values
Calculating the right-hand side:
\[ \frac{t_A^2}{t_B^2} = \frac{39.2}{39.2} = 1 \]
Thus,
\[ \frac{t_A}{t_B} = \sqrt{1} = 1 \]

### Final Step: Conclusion
The ratio \( \frac{t_A}{t_B} \) is \( 1:1 \).