A ballistic pendulum consists of an arm of mass \(M\) and length \(L=0.48 \mathrm{~m} .\) One end of the arm is pivoted so that the arm rotates freely in a vertical plane. Initially, the arm is motionless and hangs vertically from the pivot point. A projectile of the same mass \(M\) hits the lower end of the arm with a horizontal velocity of \(V=3.6 \mathrm{~m} / \mathrm{s}\). The projectile remains stuck to the free end of the arm during their subsequent motion. Find the maximum angle to which the arm and attached mass will swing in each case: a) The arm is treated as an ideal pendulum, with all of its mass concentrated as a point mass at the free end. b) The arm is treated as a thin rigid rod, with its mass evenly distributed along its length.

In this case, we can assume that all the mass of the arm is concentrated at the free end. The initial energy of the system is entirely kinetic energy from the projectile: \(E_{i}=\frac{1}{2}Mv^2\) When the pendulum reaches its maximum height, all kinetic energy has been converted to potential energy: \(E_{f}=Mgh\) Where \(h\) is the increase in height of the center of mass. We can find the height by using geometry. When the pendulum is at its maximum angle, we have: \(h=L(1-\cos(\theta_{max}))\) Now, we use the conservation of energy principle: \(E_i = E_f\) \(\frac{1}{2}Mv^2 = MgL(1-\cos(\theta_{max}))\) Since every term in the equation has \(M\), we can cancel it out: \(\frac{1}{2}v^2 = gL(1-\cos(\theta_{max}))\) Now, by rearranging the terms and plugging in the given values, we can solve for \(\theta_{max}\): \(\theta_{max}=\cos^{-1}\left(1-\frac{v^2}{2gL}\right)\) \(\theta_{max}=\cos^{-1}\left(1-\frac{(3.6)^2}{2\times9.81\times0.48}\right) \approx 18.38^\circ\)

In this case, the arm is treated as a thin, rigid rod whose mass is distributed evenly along its length. We will still use the conservation of energy principle. The initial energy is still the kinetic energy of the projectile: \(E_{i}=\frac{1}{2}Mv^2\) Now, when the arm reaches its maximum height, the center of mass will be at a different position compared to when it was an ideal pendulum. Since the mass is evenly distributed, the center of mass will be located in the middle of the arm. The height of the center of mass can be expressed as: \(h=L\frac{1}{2}(1-\cos(\theta_{max}))\) The final energy, which will be the potential energy of the center of mass, is given by: \(E_{f}=\frac{3}{2}Mgh\) Using the conservation of energy principle, we have: \(E_i = E_f\) \(\frac{1}{2}Mv^2 = \frac{3}{2}MgL\frac{1}{2}(1-\cos(\theta_{max}))\) After canceling out \(M\), we can rearrange the terms and solve for \(\theta_{max}\): \(\theta_{max}=\cos^{-1}\left(1-\frac{v^2}{3gL}\right)\) \(\theta_{max}=\cos^{-1}\left(1-\frac{(3.6)^2}{3\times9.81\times0.48}\right) \approx 11.11^\circ\) So, the maximum angles for the two cases are as follows: a) \(\theta_{max} \approx 18.38\,^\circ\) (Ideal Pendulum) b) \(\theta_{max} \approx 11.11\,^\circ\) (Thin, Rigid Rod)