ME222 (HW 1)
Fundamentals
1. Conservation of momentum
$mv_0 = (M + m)v_1$
2. Energy Conservation
$\frac{1}{2}(M + m)v_1^2 = (M + m)g(L - Lcos(\theta))$
Given
m = 0.125 lb
M = 10 lb
g = 32.2 ft/s$^2$
L = 2 ft
Questions)
1. Find the maximum angle $\theta$ when the initial
velocity ($v_0$) of the bullet varies from 0 to
1200 ft/s. (set the step size as 1 ft/s)
2. Plot and Label axis, $v_0$ in ft vs. $\theta$ in degree
Solution: $v_1 = \frac{mv_0}{(M+m)}$
$\theta = acos(1 - v_1^2/(2g)/L)$
