Question

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?

Answer 1

Answer:

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

$Em_{f} = K_{e} \\= \frac{1}{2} kx^2\\ Em_0 = Em_{f}$

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

$p_{f} = (m + M) v$

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

 $p_0 = p_{f}$

  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