GSO002 Problem 16

No external forces act, so the center of mass moves at constant velocity. Total mass $M = 16 \text{ kg}$ .

\vec{V}_{CM} = \frac{m_A\vec{v}_A(0) + m_B\vec{v}_B(0)}{M} = \frac{12(5,0)+4(-20,10)}{16} = \left(-\frac{5}{4}, \frac{5}{2}\right) \text{ m/s}

The relative position $\vec{r}(t) = \vec{r}_B(t) - \vec{r}_A(t)$ satisfies the equation of motion with reduced mass $\mu$ :

\mu = \frac{m_Am_B}{m_A+m_B} = \frac{12\times4}{16} = 3 \text{ kg}

3\ddot{\vec{r}} = -75\vec{r} \implies \ddot{\vec{r}} = -25\vec{r}

This is 2D harmonic oscillation with angular frequency $\omega = 5 \text{ rad/s}$ .

Initial conditions:

\vec{r}(0) = (0, 6), \quad \dot{\vec{r}}(0) = (-25, 10)

General solution: $\vec{r}(t) = \vec{C}_1\cos(5t) + \vec{C}_2\sin(5t)$

From initial conditions: $\vec{C}_1 = (0, 6)$ , $\vec{C}_2 = (-5, 2)$ , giving:

\vec{r}(t) = (-5\sin(5t),\ 6\cos(5t) + 2\sin(5t))

r^2(t) = 25\sin^2(5t) + (6\cos(5t)+2\sin(5t))^2 = \frac{65}{2} + \frac{25}{2}\sin(10t + \alpha)

where $\cos\alpha = \frac{24}{25}$ , $\sin\alpha = \frac{7}{25}$ .

Maximum when $\sin(10t_1 + \alpha) = 1$ , giving $\cos(5t_1) = \frac{4}{5}$ , $\sin(5t_1) = \frac{3}{5}$ .

Relative velocity at $t_1$ :

\dot{\vec{r}}(t_1) = (-25\cos(5t_1),\ -30\sin(5t_1)+10\cos(5t_1)) = (-20, -10)

Velocity of puck B in the stationary frame:

\vec{v}_B(t_1) = \vec{V}_{CM} + \frac{m_A}{m_A+m_B}\dot{\vec{r}}(t_1) = \left(-\frac{5}{4},\frac{5}{2}\right) + \frac{3}{4}(-20,-10) = \left(-\frac{65}{4}, -5\right) \text{ m/s}

Kinetic energy:

E_K = \frac{1}{2}m_B|\vec{v}_B(t_1)|^2 = 2\left(\frac{4225}{16}+25\right) = 2\times\frac{4625}{16} = \frac{4625}{8} \text{ J}

$\gcd(4625, 8) = 1$ (4625 is odd), so $A = 4625$ , $B = 8$ .

A + B = 4625 + 8 = 4633

Answer: 4633

Center-of-Mass Frame 2D Harmonic Oscillation: Trajectory of a Puck Under Special Attractive Force