Solution
1. Property of Elastic Cords
Since the cord is elastic, obeys Hooke’s law, and is assumed to be uniform, the strain is distributed uniformly along its length. This implies that any point $C$ on the cord divides the total length $AB$ in a constant ratio throughout the motion.
Initial State ($t=0$):
$A = (0, 2)$, $B = (0, -8)$, $C = (0, 0)$.
Length $AC = \sqrt{(0-0)^2 + (2-0)^2} = 2$.
Length $CB = \sqrt{(0-0)^2 + (0 – (-8))^2} = 8$.
Ratio $AC : CB = 2 : 8 = 1 : 4$.
Therefore, at any time $t$, the coordinates of $C$ are given by the section formula dividing the segment $AB$ in the ratio $1:4$:
$\vec{r}_C = \frac{4\vec{r}_A + 1\vec{r}_B}{4+1} = \frac{4\vec{r}_A + \vec{r}_B}{5}$.
2. Analyzing Motion
Motion of A:
Point A moves in the positive $x$-direction with constant velocity $v_A = 1.0$ m/s.
$\vec{r}_A(t) = (v_A t) \hat{i} + 2 \hat{j} = t \hat{i} + 2 \hat{j}$.
Motion of B:
Point B moves in the negative $y$-direction with constant acceleration $a_B$ (starting from rest, as implied by “made to move”).
$\vec{r}_B(t) = 0 \hat{i} + (-8 – \frac{1}{2}a_B t^2) \hat{j}$.
Point C:
We are given that at a specific instant, $C$ passes through $(4, -2)$.
$\vec{r}_C = 4 \hat{i} – 2 \hat{j}$.
3. Calculation
Substitute the coordinates into the section formula:
X-coordinate:
$x_C = \frac{4 x_A + x_B}{5}$
$4 = \frac{4(t) + 0}{5}$
$20 = 4t \implies t = 5$ s.
Y-coordinate:
$y_C = \frac{4 y_A + y_B}{5}$
$-2 = \frac{4(2) + y_B}{5}$
$-10 = 8 + y_B \implies y_B = -18$ m.
Now, use the kinematic equation for B to find acceleration $a_B$.
The coordinate $y_B$ changed from $-8$ to $-18$.
Displacement $S_y = -18 – (-8) = -10$ m.
Using $S = ut + \frac{1}{2}at^2$ (with $u=0$):
$-10 = 0 + \frac{1}{2} (-a_{B,mag}) (5)^2$
$-10 = – \frac{25}{2} a_{B,mag}$
$a_{B,mag} = \frac{20}{25} = 0.8$ m/s$^2$.