Solution for Question 43
1. Calculate Initial Relative Velocity
The disc moves perpendicular to the belt. Let the belt move in the $x$-direction and the disc enter in the $y$-direction.
- Velocity of belt, $v_b = 3.0$ m/s.
- Velocity of disc, $v_0 = 4.0$ m/s.
The magnitude of the initial relative velocity $v_{rel}$ is:
$$ v_{rel} = \sqrt{v_b^2 + v_0^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5.0 \text{ m/s} $$2. Stopping Distance on the Belt
The deceleration due to friction is $a = \mu g$, where $\mu = 0.5$.
$$ a = 0.5 \times 10 = 5 \text{ m/s}^2 $$The total distance $s$ traveled relative to the belt before stopping is given by $v^2 = u^2 + 2as$ (using magnitudes):
$$ 0 = v_{rel}^2 – 2as \implies s = \frac{v_{rel}^2}{2a} = \frac{(5)^2}{2(5)} = \frac{25}{10} = 2.5 \text{ m} $$3. Determine Minimum Width
The path relative to the belt is a straight line at an angle $\theta$ relative to the belt’s length (x-axis). From the velocity vector triangle:
$$ \sin \theta = \frac{v_0}{v_{rel}} = \frac{4}{5} $$The transverse displacement (width required) $w$ is the component of the stopping distance perpendicular to the belt:
$$ w = s \sin \theta = 2.5 \times \frac{4}{5} = 0.5 \times 4 = 2.0 \text{ m} $$Answer: (c) 2.0 m