A 12.0-g bullet is fired horizontally into a 109-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 152 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 78.0 cm, what was the speed of the bullet at impact with the block?

v₀ = 280.6 m / s

Explanation:

we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,

We write the mechanical energy when the shock has passed the bodies

   Em₀ = K = ½ (m + M) v²

We write the mechanical energy when the spring is in maximum compression

½ (m + M) v² = ½ k x²

Let's calculate the system speed

   v = √ [k x² / (m + M)]

   v = √[152 ×0.78² / (0.012 +0.109) ]

   v = 27.65 m / s

This is the speed of the bullet + Block system

Now let's use the moment to solve the shock

Before the crash

   p₀ = m v₀

After the crash

The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved

  m v₀ = (m + M) v

  v₀ = v (m + M) / m

let's calculate

v₀ = 27.83 (0.012 +0.109) /0.012

  v₀ = 280.6 m / s

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.

f is prop to m1xm2 double m1 double force