Solution
Step 1: Work-Energy Theorem
We apply the Work-Energy Theorem to the system of the two discs. The only external force doing work on the system is the applied force $F$. The tension in the cord is an internal force and does no net work on the system. $$ W_{ext} = \Delta K $$
We apply the Work-Energy Theorem to the system of the two discs. The only external force doing work on the system is the applied force $F$. The tension in the cord is an internal force and does no net work on the system. $$ W_{ext} = \Delta K $$
Step 2: Calculate Work Done by F
The force $F$ is applied at the midpoint of the cord.
The force $F$ is applied at the midpoint of the cord.
- Initially, the cord is straight with length $l$. The midpoint is on the line joining the centers.
- Finally, when the discs are about to collide, the cord is folded such that the two halves are parallel to the direction of the force. The midpoint has been pulled forward.
Step 3: Calculate Change in Kinetic Energy
Let $v$ be the speed of each disc just before collision. Since the discs are identical and the motion is symmetric, they both have the same speed $v$. $$ K_{final} = \frac{1}{2}m v^2 + \frac{1}{2}m v^2 = m v^2 $$ $$ K_{initial} = 0 $$
Let $v$ be the speed of each disc just before collision. Since the discs are identical and the motion is symmetric, they both have the same speed $v$. $$ K_{final} = \frac{1}{2}m v^2 + \frac{1}{2}m v^2 = m v^2 $$ $$ K_{initial} = 0 $$
Step 4: Solve for Speed
Equating Work and Change in Kinetic Energy: $$ \frac{Fl}{2} = m v^2 $$ $$ v = \sqrt{\frac{Fl}{2m}} $$ Substituting the given values: $F = 2.0$ N, $l = 1.0$ m, $m = 1.0$ kg. $$ v = \sqrt{\frac{2.0 \times 1.0}{2 \times 1.0}} = \sqrt{1} = 1.0 \text{ m/s} $$
Equating Work and Change in Kinetic Energy: $$ \frac{Fl}{2} = m v^2 $$ $$ v = \sqrt{\frac{Fl}{2m}} $$ Substituting the given values: $F = 2.0$ N, $l = 1.0$ m, $m = 1.0$ kg. $$ v = \sqrt{\frac{2.0 \times 1.0}{2 \times 1.0}} = \sqrt{1} = 1.0 \text{ m/s} $$
Conclusion:
The speed of each disc when they are about to collide is 1.0 m/s. Since they are approaching symmetrically, this is the speed at which they close in on the collision point.
The speed of each disc when they are about to collide is 1.0 m/s. Since they are approaching symmetrically, this is the speed at which they close in on the collision point.