COM BYU 3

Step 1: Calculate the distance fallen

The bead starts at nail A and falls vertically. The string length is $l$, and the nails are separated by $d = 0.5l$. The string becomes taut when the sum of distances from the bead to the nails equals $l$. Let the bead fall a vertical distance $h$.

At position P directly below A:

$$AP = h$$ $$BP = \sqrt{h^2 + (0.5l)^2}$$

Condition for tautness:

$$AP + BP = l \implies h + \sqrt{h^2 + 0.25l^2} = l$$ $$\sqrt{h^2 + 0.25l^2} = l – h$$

Squaring both sides:

$$h^2 + 0.25l^2 = l^2 + h^2 – 2lh$$ $$2lh = 0.75l^2 \implies h = \frac{3}{8}l$$

Step 2: Calculate velocity just before the jerk

Using free fall kinematics ($v^2 = 2gh$):

$$v = \sqrt{2g \left(\frac{3l}{8}\right)} = \sqrt{\frac{3gl}{4}} = \frac{\sqrt{3gl}}{2}$$

This velocity is directed vertically downwards.

Step 3: Analyze the Impulse and Final Velocity

When the string becomes taut, the bead is constrained to move along an ellipse with foci at A and B. The tension in the string provides an impulsive force directed along the normal to this ellipse. The normal bisects the angle $\angle APB$.

The tangential velocity component is conserved. We need the angle $\theta$ between the vertical (direction of $v$) and the tangent to the path.

Let $\phi$ be the angle $\angle APB$. The vector $\vec{PA}$ is vertical. $\vec{PB}$ makes an angle with the vertical. $$\tan(\text{angle of PB}) = \frac{0.5l}{h} = \frac{0.5l}{0.375l} = \frac{4}{3}$$ This corresponds to an angle of $53^\circ$. Thus, the angle between the two string segments is $53^\circ$.

The normal (Impulse direction) bisects this angle, so it is at $\alpha = \frac{53^\circ}{2} = 26.5^\circ$ to the vertical.

The tangent is perpendicular to the normal. Therefore, the angle between the vertical velocity vector and the tangent is $90^\circ – 26.5^\circ$. However, simpler logic applies: The component of velocity along the normal is destroyed; the component perpendicular to the normal (tangential) remains.

$$v_{\text{final}} = v_{\text{initial}} \sin(\alpha)$$

(Here $\sin(\alpha)$ projects the vertical velocity onto the tangent, because the angle between Vertical and Normal is $\alpha$, so angle between Vertical and Tangent is $90-\alpha$, and $\cos(90-\alpha) = \sin \alpha$).

Step 4: Final Calculation

We need $\sin(26.5^\circ)$. We know $\cos(53^\circ) = 3/5$. Using half-angle identity $\cos \theta = 1 – 2\sin^2(\theta/2)$:

$$\frac{3}{5} = 1 – 2\sin^2(26.5^\circ)$$ $$2\sin^2(26.5^\circ) = 1 – 0.6 = 0.4$$ $$\sin^2(26.5^\circ) = 0.2 = \frac{1}{5} \implies \sin(26.5^\circ) = \frac{1}{\sqrt{5}}$$

So,

$$v_{\text{final}} = v \times \frac{1}{\sqrt{5}} = \frac{\sqrt{3gl}}{2} \times \frac{1}{\sqrt{5}} = \sqrt{\frac{3gl}{20}}$$