Answer:
vo = 277.7m / s
Step-by-step explanation:
We will solve this problem in two parts, the first one as an inelastic shock where we will find the speed of the most bullet block system and the second with the conservation of the system's energy
We create a system formed by the bullet and block assembly, in this system the forces of the shock are internal and the moment is preserved, call us to the mass of the bullet (m) and the mass of the block (M)
before the crash
p₀ = m v₀
just after the crash
pf = (m + M) vf
Like the inelastic shock
p₀ = pf
m v₀= (m + M) vf
vf = m / (m + M) v₀ (1)
Now we use energy conservation for this system, let's write mechanical energy at two points; an initial point at the lowest part and an end point at the height part 0.700m
Initial
Em₀ = K = ½ (m + M) vf²
Final
We do not know if this is the highest point of the trajectory
Em₂ = U + K = mgy + ½ (m + M) v²
Em₀ = Em₂
½ (m + M) vf² = (m + M) g + ½ (m + M) v² (2)
The movement of the system is circulated, so with Newton's second law we can find the speed at this point
sin θ = Tₓ / T
cos θ = Ty / T
Tₓ = T sin θ
Ty = T cos θ
Y Axis
Ty -W = 0
X axis
Tₓ = m a
Where the acceleration is centripetal
a = v² / r
T sin θ = m v² / r
v² = T r sin θ
Let's calculate
sin θ = y / L
sin θ = 0.700 / 1.68
sin θ = 0.416
θ = 24.6º
v² = 5 1.68 0.416
v² = 3.4944 m2 / s2
v = 1.869 m/s
We substitute in equation 2
½ (m + M) vf² = (m + M) g y + ½ (m + M) 1.869²
vf² = 2 g y + 3,494
vf² = 2 9.8 0.700 + 3.494
vf² = 17.214 m/s
vf = 4.15 m/s
This is the speed with which the system leaves at the lowest point of the path
Now we substitute in equation 1 and calculate the initial velocity (vo)
vf = m/(m + M) vo
vo = vf (m + M)/m
vo = 4.15 (0.0110 + 0.725) / 0.0110
vo = 277.7m / s