Question 16

Step 1: Determine the velocities of particles 1 and 2 just before the collision at $t_0 = 0$. The velocities are given by differentiating the position equations: $v_1(t) = \frac{dx_1}{dt} = a\omega \cos \omega t$ $v_2(t) = \frac{dx_2}{dt} = -a\omega \cos \omega t$ At $t_0 = 0$: $v_1(0) = a\omega$ $v_2(0) = -a\omega$

Step 2: Analyze the collision. Particle 3 (mass $m$) hits particle 2 (mass $m$) elastically with speed $u_0 = a\omega/2$. In an elastic collision between two identical masses, the velocities are exchanged. Therefore, the velocity of particle 2 after collision ($v'_2$) becomes the velocity of particle 3 before collision: $v'_2 = u_0 = \frac{a\omega}{2}$ The velocity of particle 1 remains unchanged: $v_1 = a\omega$.

Step 3: Calculate the center of mass speed $v_{cm}$ for the system of particles 1 and 2. $v_{cm} = \frac{m v_1 + m v'_2}{m + m} = \frac{a\omega + a\omega/2}{2} = \frac{3a\omega/4}{1} = 0.75 a\omega$ Thus, $\frac{v_{cm}}{a\omega} = 0.75$.