A 6.20 g bullet moving at 929 m/s strikes a 850 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 478 m/s.

(a) What is the resulting speed of the block?

(b) What is the speed of the bullet-block center of mass?

(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) = ?

By the law of conservation of momentum as:

m₁*u₁ + m₂*u₂ = m₁*v₁ + m₂*v₂

6.20*929 + 850*0 = 6.20*478 + 850*v₂

Solving for v₂, we get:

v₂ = 3.2896 m/s

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]

V = 6.7350 m/s