A6.20 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

answer: the correct answer for the fill in the blank is rods.

rods are the specialized photoreceptor cells that are present in the retina of human eye. they are responsible for vision in the night time ( also called scotopic vision). they have very low visual acuity and color distinguishing ability.

as they are responsible for vision in night time, therefore they are found predominantly in the nocturnal animals.