Given information:
- A body penetrates 3 cm in a wooden block.
- The body loses half of its velocity while penetrating the block.

To find:
- How much more will the body penetrate before coming to rest?

Assumptions:
- The body is moving in a straight line without any external forces acting on it except for the resistance provided by the wooden block.
- The resistance force is constant and opposite to the direction of motion.

Analysis:
Let's consider the initial velocity of the body as v. After penetrating 3 cm, the body loses half of its velocity, which means its final velocity becomes v/2.

Now, we need to find the distance the body travels with a final velocity of v/2 until it comes to rest.

Using the first equation of motion:
v^2 = u^2 - 2as

Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance

Let's assume the distance the body travels after penetrating the block is S.

For the first part of the motion (penetrating 3 cm):
Initial velocity (u1) = v
Final velocity (v1) = v/2
Distance (s1) = 3 cm = 0.03 m

Using the equation:
(v/2)^2 = v^2 - 2a1(0.03)

Simplifying the equation:
v^2/4 = v^2 - 0.06a1

Calculating acceleration during the first part of the motion:
0.06a1 = 3v^2/4

Simplifying the equation:
a1 = 50v^2/2

Now, for the second part of the motion (from penetrating 3 cm until coming to rest):
Initial velocity (u2) = v/2
Final velocity (v2) = 0 (since it comes to rest)
Distance (s2) = S (to be determined)

Using the equation:
0^2 = (v/2)^2 - 2a2S

Simplifying the equation:
0 = v^2/4 - 2a2S

Calculating acceleration during the second part of the motion:
2a2S = v^2/4

Simplifying the equation:
a2 = v^2/(8S)

Equating the accelerations:
Since the body is the same and only the velocity changes, the accelerations during both parts should be the same.

a1 = a2
50v^2/2 = v^2/(8S)

Simplifying the equation:
400v^2 = v^2S

Calculating the distance S:
S = 400

Therefore, the body will penetrate an additional 400 cm (or 4 meters) before coming to rest.

Conclusion:
The body will penetrate an additional 400 cm (or 4 meters) before coming to rest after losing half of its velocity while penetrating 3 cm in a wooden block.