If a body looses half of its velocity on penetrating 3 cm in a wooden block, then how much will it penetrate more before coming to rest?

Options

A. 1 cm
B. 2 cm
C. 3 cm
D. 4 cm

Explanation

To solve the problem step by step, we can break it down as follows:

### Step 1: Understand the Initial Conditions
The body loses half of its velocity after penetrating 3 cm into the wooden block. Let:
- Initial velocity \( V \)
- After penetrating 3 cm, the velocity becomes \( \frac{V}{2} \)

### Step 2: Use the Equation of Motion
We can use the second equation of motion, which is:
\[ V^2 = U^2 + 2aS \]
Where:
- \( V \) = final velocity
- \( U \) = initial velocity
- \( a \) = acceleration (negative in this case since it's retardation)
- \( S \) = distance traveled

#### For the first case (penetration of 3 cm):
- Initial velocity \( U = V \)
- Final velocity \( V = \frac{V}{2} \)
- Distance \( S_1 = 3 \, \text{cm} \)

Substituting these values into the equation:
\[ \left(\frac{V}{2}\right)^2 = V^2 + 2(-a)(3) \]
This simplifies to:
\[ \frac{V^2}{4} = V^2 - 6a \]
Rearranging gives:
\[ 6a = V^2 - \frac{V^2}{4} = \frac{3V^2}{4} \]
Thus:
\[ a = \frac{3V^2}{24} = \frac{V^2}{8} \]

### Step 3: Calculate the Second Penetration
Now we need to find how much further the body will penetrate before coming to rest.

#### For the second case (from \( \frac{V}{2} \) to rest):
- Initial velocity \( U = \frac{V}{2} \)
- Final velocity \( V = 0 \)
- Distance \( S_2 \) (unknown)

Using the same equation of motion:
\[ 0 = \left(\frac{V}{2}\right)^2 + 2(-a)S_2 \]
Substituting for \( a \):
\[ 0 = \frac{V^2}{4} - 2\left(\frac{V^2}{8}\right)S_2 \]
This simplifies to:
\[ 0 = \frac{V^2}{4} - \frac{V^2}{4}S_2 \]
Rearranging gives:
\[ \frac{V^2}{4}S_2 = \frac{V^2}{4} \]
Thus:
\[ S_2 = 1 \, \text{cm} \]

### Final Answer
The body will penetrate an additional **1 cm** before coming to rest.

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