Answer:
block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s
Explanation:
We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.
Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)
Before the crash
p₀ = m v₀ + 0
After the crash
= (m + M) v
p₀ = 
m v₀ = (m + M) v (1)
Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring
Initial
Em₀ = K = ½ m v2
Final
E
= Ke = ½ k x2
Emo = E
½ m v² = ½ k x²
v² = k/m x²
Let's look for the spring constant (k), with Hook's law
F = -k x
k = -F / x
k = - 0.75 / -0.25
k = 3 N / m
Let's calculate the speed
v = √(k/m) x
v = √ (3/8.00) 0.15
v = 0.09186 = 9.18 10⁻² m/s
This is the spped of the block plus bullet rsystem right after the crash
We substitute calculate in equation (1)
m v₀ = (m + M) v
v₀ = v (m + M) / m
v₀ = 0.09186 (0.008 + 0.992) /0.008
v₀ = 11.5 m / s
Fission and fusion are two processes that release very little amounts of energy.
C, at this moment it is directly 90 degrees to the right and is only travelling downwards
Answer:
|Ax| =1.03 m (directed towards negative x-axis)
|Ay|= 5.30 (directed towards negative y-axis)
Explanation:
Let A is a vector = 5.4 m
θ = 259°
to Find Ax, Ay
Sol:
according the condition it lies in 3rd quadrant
we know that Horizontal Component Ax = A cos θ
Ax = 5.4 Cos 259°
Ax = - 1.03 m
|Ax| =1.03 m (directed towards negative x-axis)
Now Ay = A sin θ
Ay = 5.4 Sin 259°
Ay = -5.30
|Ay|= 5.30 (directed towards negative y-axis)