M1=10g V1=400m/s M2=900g V2=? M1v1=m2v2( according to the law of conservation of momentum) 10x400=900 x v2 4000/900=v2 Therefore, V2 = 4.4444 m/s = 4.4 m/s

**Solution:**

To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

**Step 1: Calculate the momentum of the bullet before the collision**

Momentum (p) is given by the product of mass (m) and velocity (v).

The mass of the bullet is 10g, which is equal to 0.01kg. The velocity of the bullet is 400m/s.

So, the momentum of the bullet before the collision is:

p1 = m1 * v1 = 0.01kg * 400m/s = 4 kg·m/s

**Step 2: Calculate the momentum of the bullet after the collision**

Since the bullet gets embedded in the wooden block, the total mass after the collision is the sum of the masses of the bullet and the block.

The mass of the block is 900g, which is equal to 0.9kg.

The velocity of the block after the collision is denoted by v2.

So, the momentum of the bullet and the block after the collision is:

p2 = (m1 + m2) * v2 = (0.01kg + 0.9kg) * v2 = 0.91kg * v2

**Step 3: Apply the principle of conservation of momentum**

According to the principle of conservation of momentum, the total momentum before the collision (p1) is equal to the total momentum after the collision (p2).

Therefore, we can equate the two expressions for momentum:

p1 = p2

4 kg·m/s = 0.91kg * v2

**Step 4: Solve for the velocity of the block**

To find the velocity of the block (v2), we rearrange the equation:

v2 = 4 kg·m/s / 0.91kg

v2 ≈ 4.3956 m/s

Therefore, the velocity acquired by the wooden block after the collision is approximately 4.3956 m/s.