In case of a freely falling body, the ratio of kinetic energy at the end of the third second to increase in kinetic energy in th next three seconds is

Options

A. `1:1`
B. `1:2`
C. `1:3`
D. `1:9`

Answer

To solve the problem of finding the ratio of kinetic energy at the end of the third second to the increase in kinetic energy in the next three seconds for a freely falling body, we can follow these steps:

### Step 1: Determine the velocity at the end of the third second
Since the body is freely falling, we can use the first equation of motion:
\[ v = u + at \]
Here,
- \( u = 0 \) (initial velocity),
- \( a = g \) (acceleration due to gravity),
- \( t = 3 \) seconds.

Substituting the values:
\[ v_3 = 0 + g \cdot 3 = 3g \]

### Step 2: Calculate the kinetic energy at the end of the third second
The kinetic energy (KE) at the end of the third second can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
Substituting \( v = 3g \):
\[ KE_3 = \frac{1}{2} m (3g)^2 = \frac{1}{2} m \cdot 9g^2 = \frac{9}{2} mg^2 \]

### Step 3: Determine the velocity at the end of the sixth second
Using the same equation of motion for \( t = 6 \) seconds:
\[ v_6 = u + at \]
Substituting the values:
\[ v_6 = 0 + g \cdot 6 = 6g \]

### Step 4: Calculate the kinetic energy at the end of the sixth second
Using the kinetic energy formula again:
\[ KE_6 = \frac{1}{2} m (6g)^2 = \frac{1}{2} m \cdot 36g^2 = 18mg^2 \]

### Step 5: Calculate the increase in kinetic energy from the third to the sixth second
The increase in kinetic energy is given by:
\[ \Delta KE = KE_6 - KE_3 \]
Substituting the values:
\[ \Delta KE = 18mg^2 - \frac{9}{2} mg^2 \]
To perform the subtraction, convert \( 18mg^2 \) into a fraction:
\[ 18mg^2 = \frac{36}{2} mg^2 \]
Now, substituting:
\[ \Delta KE = \frac{36}{2} mg^2 - \frac{9}{2} mg^2 = \frac{27}{2} mg^2 \]

### Step 6: Find the ratio of kinetic energy at the end of the third second to the increase in kinetic energy
We need to find the ratio:
\[ \text{Ratio} = \frac{KE_3}{\Delta KE} \]
Substituting the values:
\[ \text{Ratio} = \frac{\frac{9}{2} mg^2}{\frac{27}{2} mg^2} \]
The \( mg^2 \) and \( \frac{1}{2} \) cancel out:
\[ \text{Ratio} = \frac{9}{27} = \frac{1}{3} \]

### Final Answer
The ratio of kinetic energy at the end of the third second to the increase in kinetic energy in the next three seconds is:
\[ \text{Ratio} = 1 : 3 \]

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