COM CYU 6

Physics Solution – Question 6

Solution to Question 6

Problem Analysis

We are dealing with a system of two identical bars (mass $m$) on a frictionless floor. The collision with the wall is elastic, but there is friction between the bars surfaces. We define the unit vector $\hat{i}$ pointing to the right (towards the wall).

Lower (m) Upper (m) v v f_k f_k Intermediate State: After 1st Collision

Step-by-Step Solution

1. First Collision (Upper Bar with Wall)

Initially, both bars move with velocity $\vec{u} = u\hat{i}$. The upper bar is offset by $l_0$ towards the wall, so it strikes the wall first. Since the collision is perfectly elastic, the velocity of the upper bar reverses instantly.

  • Velocity of Upper bar: $\vec{v}_1 = -u\hat{i}$
  • Velocity of Lower bar: $\vec{v}_2 = +u\hat{i}$

2. Motion Between Collisions

Now the bars are slipping past each other. The relative velocity is $\vec{v}_{rel} = \vec{v}_1 – \vec{v}_2 = -2u\hat{i}$. Kinetic friction $f_k = \mu N = \mu mg$ acts to oppose this relative motion.

  • Force on Upper bar: Acts to the right ($+\hat{i}$). Acceleration $\vec{a}_1 = +\mu g \hat{i}$.
  • Force on Lower bar: Acts to the left ($-\hat{i}$). Acceleration $\vec{a}_2 = -\mu g \hat{i}$.
Symmetry Note: Both bars have equal mass and start with the same speed $u$ (though in opposite directions). They experience the same magnitude of deceleration $\mu g$. Therefore, their speeds decrease at the exact same rate. At any instant $t$, $|\vec{v}_1| = |\vec{v}_2|$.

3. Lower Bar Reaches the Wall

The lower bar must travel a distance $l_0$ to hit the wall. We apply the Work-Energy Theorem to the lower bar to find its velocity just before impact.

$$ K_{final} – K_{initial} = W_{friction} $$ $$ \frac{1}{2}mv^2 – \frac{1}{2}mu^2 = (-\mu mg) \cdot l_0 $$ $$ v^2 = u^2 – 2\mu g l_0 $$

4. Final Analysis

We must consider two cases based on the value of $u^2$:

Case I: $u^2 \le 2\mu g l_0$
The friction is sufficient to stop the lower bar before it traverses the distance $l_0$. By symmetry, the upper bar also comes to a halt at the same time.
Final Velocity = 0

Case II: $u^2 > 2\mu g l_0$
The lower bar hits the wall with velocity magnitude $v’ = \sqrt{u^2 – 2\mu g l_0}$.

  • Just before 2nd collision: $\vec{v}_{lower} = v’\hat{i}$ and $\vec{v}_{upper} = -v’\hat{i}$.
  • Collision: The lower bar hits the wall elastically. Its velocity reverses to $-v’\hat{i}$.
  • Just after 2nd collision: $\vec{v}_{lower} = -v’\hat{i}$ and $\vec{v}_{upper} = -v’\hat{i}$.

Since both bars now have the same velocity (magnitude and direction), there is no longer any relative motion. Friction ceases, and they continue moving left together.

Answer

The final velocities of both bars are:

$$ \vec{v}_{final} = \begin{cases} 0 & \text{if } u^2 \le 2\mu g l_0 \\ \sqrt{u^2 – 2\mu g l_0} \text{ (away from wall)} & \text{if } u^2 \ge 2\mu g l_0 \end{cases} $$

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