**Answer:**

(a) Final speed of block = 3.2896 m/s

(b) 6.7350 m/s is the speed of the bullet-block center of mass?

**Explanation:**

Given that:

Mass of bullet (m₁) = 6.20 g

Initial Speed of bullet (u₁) = 929 m/s

Final speed of bullet (v₁) = 478 m/s

Mass of wooden block (m₂) = 850g

Initial speed of block initial (u₂) = 0 m/s

Final speed of block (v₂) = ?

<u>**By the law of conservation of momentum as:**</u>

<u>**m₁×u₁ + m₂×u₂ = m₁×v₁ + m₂×v₂**</u>

6.20×929** **+ 850×0 = 6.20×478** **+ 850×v₂

Solving for v₂, we get:

<u>**v₂ = 3.2896 m/s**</u>

Let the V be the speed of the bullet-block center of mass. So,

V = [m₁* u₁]/[m₁ + m₂] (p before collision = p after collision)

= [6.2 *929]/[5.2+850]

<u>**V = 6.7350 m/s
**</u>