Yea, gonna need some help. Thanks
1 answer:
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As per FBD while its accelerating upwards
we can say that
here normal force is given as
now mass is given as
now we will have
Now while accelerating downwards we can say by FBD
again plug in all values
Answer:
v₀ = 292.3 m / s
Explanation:
Let's analyze the situation, on the one hand 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, as the data they give us are Let's start with this second part.
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 k x²
Em₀ =
½ (m + M) v² = ½ k x²
Let's calculate the system speed
v = √ [k x² / (m + M)]
v = √[154 0.83² / (0.012 +0.104) ]
v = 30.24 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
= (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
p₀ =
m v₀ = (m + M) v
v₀ = v (m + M) / m
let's calculate
v₀ = 30.24 (0.012 +0.104) /0.012
v₀ = 292.3 m / s