Determine the minimum radius of a horizontal circular curve for a route having a 70 mph design speed, super-elevation (e) 3%, an
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Answer:
1.176
Explanation:
When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.
v2 * (M + mb) = v1 * mb
Where
v1: muzzle velocity of the bullet
M: mass of the bob
mb: mass of the bullet
v2: mass of the bob with the bullet after being hit
v2 = v1 * mb / (M + mb)
Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.
Ek = 1/2 (M + mb) * v2^2
Ep = (M + mb) * g * h
Ek = Ep
1/2 (M + mb) * v2^2 = (M + mb) * g * h
1/2 * (v1 * mb / (M + mb))^2 = g * h
1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h
v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)
The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.
h = L * (1 - cos(a))
For the 9 mm:
For the 0.44 caliber:
The ratio is 460 / 391 = 1.176