To solve this problem, we need to consider how the motion of the stone appears to both Student 1 (on the cart) and Student 2 (on the ground). The key concepts here involve relative velocity and relative acceleration, particularly when the frame of reference is moving with respect to another.

Key Points:

• Student 1's frame of reference (cart's frame): Since Student 1 is moving with the cart, the stone is initially at rest with respect to the cart. When Student 1 releases the stone, the stone begins accelerating downward due to gravity.

• Student 2's frame of reference (ground's frame): Student 2 is stationary with respect to the ground, so they observe both the motion of the cart and the stone relative to the ground.

1. Speeds:

• In the cart’s frame (Student 1's view):

  • The stone starts from rest with respect to the cart and accelerates downward. At the moment just before the stone hits the ground, Student 1 measures the speed of the stone to be $2u_1$.

• In the ground’s frame (Student 2's view):

  • From Student 2's perspective, the stone has an initial horizontal velocity equal to the velocity of the cart, $v_1$, when released. This is because, initially, the stone is moving along with the cart horizontally.
  • As the stone falls, it accelerates downward (due to gravity), just as observed by Student 1. However, in Student 2's frame, the stone also retains the horizontal component of its velocity $v_1$.
  • Therefore, the total speed $v_2$ measured by Student 2 will be the vector sum of the horizontal velocity $v_1$ and the vertical component measured by Student 1, which is $2u_1$. Using the Pythagorean theorem: $v_2 = \sqrt{v_1^2 + (2u_1)^2}$

2. Accelerations:

• In the cart’s frame (Student 1's view):

  • Student 1 measures the acceleration of the stone as $a_1$, which would likely be equal to $g$, the acceleration due to gravity, if we assume no other forces (air resistance, etc.).

• In the ground’s frame (Student 2's view):

  • The only force acting on the stone is gravity, and gravity acts vertically. Both Student 1 and Student 2 observe the same gravitational force, so the vertical component of the acceleration (due to gravity) is the same in both frames.
  • Since there is no horizontal acceleration (the cart moves at constant velocity), Student 2 also measures the same vertical acceleration $a_2 = a_1 = g$.

Summary of Relationships:

1. The speed measured by Student 2, $v_2$, is the vector sum of the cart’s speed $v_1$ and the speed of the stone in the vertical direction as measured by Student 1, $2u_1$: $v_2 = \sqrt{v_1^2 + (2u_1)^2}$

2. The acceleration measured by both students is the same: $a_2 = a_1 = g$

This means that:

• The speed $v_2$ is related to $v_1$ and $u_1$ by the Pythagorean relationship.
• The acceleration $a_2$ is equal to $a_1$.

Would you like more details or have any questions on this?

Relative Questions:

1. What would happen to the measured speed if the cart's velocity increased?
2. How does relative velocity apply to objects moving in different directions?
3. Can this problem be extended to include air resistance?
4. How does the reference frame affect the perception of acceleration?
5. What role does gravity play in both students' measurements?

Tip: Always break down the components of velocity and acceleration when shifting between different frames of reference.