Use the observations made in the virtual lab to complete this table. (pg. 5)
1 answer:
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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
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