Solution 20: Wax Bar on Moving Wedge
Given:
- Wedge acceleration $a_w = 0.5$ mm/s$^2$ (Leftwards/Towards wall).
- Rate of melting $v_{melt} = 1.0$ mm/s (Length reduction).
- Wedge angle $\theta = 37^\circ$ ($\tan 37^\circ = 3/4$).
Mathematical Analysis
Let $y$ be the vertical coordinate of the bar (measured from the wedge tip level downwards) and $x$ be the position of the wedge.
The length of the bar $L$ fills the gap between the wall and the wedge face.
Gap width $L = y \cot \theta + x_{tip}$.
Also, the bar length decreases due to melting: $L(t) = L_0 – v_{melt} t$.
The wedge moves towards the wall: $x_{gap}(t) = d_{initial} – \frac{1}{2} a_w t^2$.
Equating the geometry: $$ L(t) = y(t) \cot 37^\circ + (d – \frac{1}{2} a_w t^2) $$ $$ L_0 – v_{melt} t = y(t) (\frac{4}{3}) + d – \frac{1}{2} a_w t^2 $$ Differentiating with respect to time ($L_0, d$ are constants): $$ -v_{melt} = \frac{4}{3} v_y – a_w t $$ $$ \frac{4}{3} v_y = a_w t – v_{melt} $$ $$ v_y = \frac{3}{4} (0.5 t – 1) $$
Analyzing the Motion
Velocity Analysis: $v_y = \frac{3}{4} (0.5 t – 1)$
- At $t=0$, $v_y = -0.75$ mm/s (Negative indicates downward motion).
- At $t=2$ s, $v_y = 0$ (Bar stops momentarily).
- At $t > 2$ s, $v_y > 0$ (Bar moves upwards).
This confirms Option (a) and Option (b).
Distance Calculation (First 4s):
Displacement at $t$: $y(t) = \int v_y dt = \frac{3}{4} (\frac{0.5 t^2}{2} – t) = \frac{3}{4} (0.25 t^2 – t)$.
- Position at $t=0$: $y(0) = 0$.
- Position at $t=2$ (turning point): $y(2) = \frac{3}{4} (0.25(4) – 2) = \frac{3}{4} (1-2) = -0.75$ mm.
- Position at $t=4$: $y(4) = \frac{3}{4} (0.25(16) – 4) = 0$ mm.
Modulus of Displacement: $|y(4) – y(0)| = 0$. (Option c says 1.5, so c is false).
Distance Travelled:
The bar goes down 0.75 mm and comes back up 0.75 mm.
Total Distance = $|-0.75| + |0.75| = 1.5$ mm. Option (d) is True.
Correct Options: (a), (b), and (d)