Solution to Question 10
Concept: The motion of a body in a viscous fluid is governed by the interplay between gravity, buoyancy, and the drag force. The drag force is proportional to velocity ($F_d = kv$).
1. Equation of Motion
Let’s define the forces acting on the stone moving downwards in the water:
- Weight (downwards): $mg$
- Buoyant Force (upwards): $F_B$
- Drag Force (upwards, resisting motion): $F_d = kv$
The net acceleration $a$ is given by Newton’s Second Law:
$$ma = (mg – F_B) – kv$$Terminal velocity ($v_0$) is reached when acceleration is zero ($a=0$):
$$kv_0 = mg – F_B \implies v_0 = \frac{mg – F_B}{k}$$2. Analyzing the Cases
The stone is dropped from air into water. Its velocity just before hitting the water depends on the height of the drop.
Case 1: Dropped from a large height
The stone gains significant speed in the air. If it enters the water with $v > v_0$, the drag force will be larger than the net weight ($kv > mg – F_B$). The net force is upwards (retarding). The stone decelerates until it reaches $v_0$.
Result: The graph starts high and decays down to an asymptote. (Matches the 1st graph).
Case 2: Dropped from a small height
The stone enters the water with low speed $v < v_0$. The drag force is small ($kv < mg - F_B$). The net force is downwards (accelerating). The stone accelerates until it reaches $v_0$.
Result: The graph starts low (or at 0) and rises to an asymptote. (Matches the 2nd graph).
Case 3: Dropped from specific height
If the height is chosen such that the entry velocity is exactly $v = v_0$, then net force is zero immediately. Velocity remains constant.
Result: A horizontal line. (Matches the 3rd graph).
- Option (b) claims the 1st graph is for a large height (deceleration) and the 2nd for a small height (acceleration). This is physically correct.
- Option (c) claims the 3rd graph occurs if the drop height provides exactly $v_0$ at entry. This is also physically correct.