Solution for Question 42
Analysis of Frames of Reference
To determine the path of the disc, we must analyze its motion in two different reference frames: the frame of the conveyor belt and the ground frame.
1. Motion Relative to the Belt:
Let the velocity of the belt be $\vec{v}_b$ and the initial velocity of the disc be $\vec{v}_0$.
The initial velocity of the disc relative to the belt is:
$$ \vec{v}_{rel} = \vec{v}_{disc} – \vec{v}_{belt} = \vec{v}_0 – \vec{v}_b $$
Since the belt is moving in the $+x$ direction and the disc enters in the $+y$ direction (perpendicular), the relative velocity vector points diagonally backwards. The kinetic friction force $\vec{f}_k$ acts opposite to the direction of this relative velocity ($ \vec{f}_k \propto -\vec{v}_{rel} $).
In the belt’s frame, the only force acting on the disc is friction, which opposes the motion. Since the direction of motion relative to the belt does not change (the force is directly opposite to velocity), the disc decelerates along a straight line until it stops relative to the belt.
2. Motion Relative to the Ground:
In the ground frame, the frictional force is constant in magnitude ($f_k = \mu mg$) and acts in a constant direction (opposite to the initial relative velocity). This implies the disc experiences a constant acceleration vector $\vec{a}$.
The equation of motion is: $$ \vec{r}(t) = \vec{r}_0 + \vec{u}t + \frac{1}{2}\vec{a}t^2 $$ Since the initial velocity relative to the ground ($\vec{v}_0$) is not collinear with the constant acceleration vector ($\vec{a}$), the trajectory is a parabola.
Conclusion
- Relative to the belt: Straight line (Option b).
- Relative to the ground: Parabola (Option c).
Correct Answer: (b) and (c)