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vagabundo [1.1K]
3 years ago
13

A 12.0-g bullet is fired horizontally into a 109-g wooden block that is initially at rest on a frictionless horizontal surface a

nd 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?
Physics
1 answer:
kykrilka [37]3 years ago
4 0

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

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