Question

I just need number 2

Answer 1

We will apply the conservation of linear momentum to answer this question.

Whenever there is an interaction between any number of objects, the total momentum before is the same as the total momentum after. For simplicity's sake we mostly use this equation to keep track of the momenta of two objects before and after a collision:

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Note that v₁ and v₁' is the velocity of m₁ before and after the collision.

Let's choose m₁ and v₁ to represent the bullet's mass and velocity.

m₂ and v₂ represents the wood block's mass and velocity.

The bullet and wood will stick together after the collision, so their final velocities will be the same. v₁' = v₂'. We can simplify the equation by replacing these terms with a single term v'

m₁v₁ + m₂v₂ = m₁v' + m₂v'

m₁v₁ + m₂v₂ = (m₁+m₂)v'

Let's assume the wood block is initially at rest, so v₂ is 0. We can use this to further simplify the equation.

m₁v₁ = (m₁+m₂)v'

Here are the given values:

m₁ = 0.005kg

v₁ = 500m/s

m₂ = 5kg

Plug in the values and solve for v'

0.005×500 = (0.005+5)v'

v' = 0.4995m/s

v' ≅ 0.5m/s