Solution
Step 1: Analyze Energy Loss
The kinetic energy of the disc is consumed by the work done against friction. Since the surface texture (and thus friction) is a function of position $s$ only, the work done by friction over a specific distance is constant regardless of the initial speed.
The total work done by friction in stopping the first particle (from $v=5$ to $v=0$) is equal to its initial kinetic energy: $$ W_{friction}(0 \to 10) = \frac{1}{2}m(5)^2 = 25 \left(\frac{m}{2}\right) $$
The kinetic energy of the disc is consumed by the work done against friction. Since the surface texture (and thus friction) is a function of position $s$ only, the work done by friction over a specific distance is constant regardless of the initial speed.
The total work done by friction in stopping the first particle (from $v=5$ to $v=0$) is equal to its initial kinetic energy: $$ W_{friction}(0 \to 10) = \frac{1}{2}m(5)^2 = 25 \left(\frac{m}{2}\right) $$
Step 2: Second Particle Scenario
The second particle starts with speed $u = 4$ m/s. It will stop when the work done by friction equals its initial kinetic energy: $$ K_{initial, 2} = \frac{1}{2}m(4)^2 = 16 \left(\frac{m}{2}\right) $$ We need to find the distance $s$ such that the accumulated work of friction is $16 (m/2)$.
The second particle starts with speed $u = 4$ m/s. It will stop when the work done by friction equals its initial kinetic energy: $$ K_{initial, 2} = \frac{1}{2}m(4)^2 = 16 \left(\frac{m}{2}\right) $$ We need to find the distance $s$ such that the accumulated work of friction is $16 (m/2)$.
Step 3: Relate to the Graph
Consider the first particle again. At the point $s$ where the second particle stops, the first particle would have done the exact same amount of work against friction ($16 (m/2)$).
Therefore, at this specific position $s$, the remaining kinetic energy of the first particle would be: $$ K_{remaining} = K_{initial, 1} – \text{Work Done} $$ $$ K_{remaining} = 25\left(\frac{m}{2}\right) – 16\left(\frac{m}{2}\right) = 9\left(\frac{m}{2}\right) $$ This corresponds to a velocity $v$ for the first particle such that: $$ \frac{1}{2}mv^2 = 9\left(\frac{m}{2}\right) \implies v^2 = 9 \implies v = 3 \text{ m/s} $$
Consider the first particle again. At the point $s$ where the second particle stops, the first particle would have done the exact same amount of work against friction ($16 (m/2)$).
Therefore, at this specific position $s$, the remaining kinetic energy of the first particle would be: $$ K_{remaining} = K_{initial, 1} – \text{Work Done} $$ $$ K_{remaining} = 25\left(\frac{m}{2}\right) – 16\left(\frac{m}{2}\right) = 9\left(\frac{m}{2}\right) $$ This corresponds to a velocity $v$ for the first particle such that: $$ \frac{1}{2}mv^2 = 9\left(\frac{m}{2}\right) \implies v^2 = 9 \implies v = 3 \text{ m/s} $$
Step 4: Read the Graph
We look at the provided graph for the first particle to find the position $s$ where its velocity is $3$ m/s.
Observing the grid carefully: The curve passes exactly through the intersection of $s=5$ m and $v=3$ m/s.
We look at the provided graph for the first particle to find the position $s$ where its velocity is $3$ m/s.
Observing the grid carefully: The curve passes exactly through the intersection of $s=5$ m and $v=3$ m/s.
Conclusion:
The second disc will cover a total distance of 5 m.
The second disc will cover a total distance of 5 m.