Question

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?

Answer 1

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

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

Step-by-step 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