Problem 28 - Thomas Calculus

Question

Colliding marbles The accompanying figure shows an experiment with two marbles. Marble \(A\) was launched toward marble \(B\) with launch angle \(\alpha\) and initial speed \(v_{0 .}\) At the same instant, marble \(B\) was released to fall from rest at \(R \tan \alpha\) units directly above a spot \(R\) units downrange from \(A .\) The marbles were found to collide regardless of the value of \(v_{0}\) . Was this mere coincidence, or must this happen? Give reasons for your answer.

Step-by-Step Solution

Answer

The marbles always collide; it's not a coincidence. Their relative motion ensures they meet.

Step 1: Understand the problem

We have two marbles, A and B. Marble A is launched at an angle \( \alpha \) and with an initial speed \( v_0 \). Marble B is dropped from a height \( R \tan \alpha \), positioned \( R \) units horizontally from A's starting point. We need to determine if they will always collide, regardless of \( v_0 \).

Step 2: Analyze the Position of Marble A (Projectile Motion)

The horizontal range \( R_x \) and vertical distance \( R_y \) covered by marble A at time \( t \) can be expressed as: \[ R_x(t) = v_0 \cos(\alpha) t \] \[ R_y(t) = v_0 \sin(\alpha) t - \frac{1}{2}gt^2 \] Where \( g \) is the acceleration due to gravity.

Step 3: Analyze the Position of Marble B (Free Fall)

Marble B falls directly downward from the height \( R \tan \alpha \). The vertical position at time \( t \) is: \[ y_B(t) = R \tan(\alpha) - \frac{1}{2}gt^2 \] The horizontal position stays constant at \( R \).

Step 4: Determine Collision Condition

For a collision, both marbles must be at the same horizontal and vertical positions at the same time \( t \): - Horizontal equilibrium: \( R_x(t) = R \) implies \( v_0 \cos(\alpha) t = R \). - Vertical equilibrium needs \( R_y(t) = y_B(t) \): \[ v_0 \sin(\alpha) t - \frac{1}{2}gt^2 = R \tan(\alpha) - \frac{1}{2}gt^2 \] Removing \( \frac{1}{2}gt^2 \) yields: \[ v_0 \sin(\alpha) t = R \tan(\alpha) \]

Step 5: Solve for Time of Collision

From horizontal equilibrium: \[ t = \frac{R}{v_0 \cos(\alpha)} \] Substitute into the vertical equilibrium equation: \[ v_0 \sin(\alpha) \left( \frac{R}{v_0 \cos(\alpha)} \right) = R \tan(\alpha) \] This simplifies to: \[ R \tan(\alpha) = R \tan(\alpha) \] It's an identity, true regardless of \( v_0 \), confirming the collision.

Key Concepts

Collision Condition

When analyzing the interaction between two objects in motion, such as marbles in projectile motion, understanding their collision condition is crucial. For two objects to collide, they must occupy the same space at the same time.
In this experiment, marble A is launched with a certain speed and angle, while marble B is dropping from rest. To predict their collision, two conditions must be met:

• **Horizontal Condition**: Both marbles must be at the same horizontal position at the same time. For marble A, this is determined by its initial speed and launch angle. So, we find its horizontal position using its velocity in the x direction.


• **Vertical Condition**: Both must also align vertically. This means the vertical position of marble A, influenced by its initial speed and the effect of gravity, must match marble B's falling position.

If these conditions are satisfied, as the mathematical proof below shows, collision is inevitable, no matter the launch speed of marble A.

Free Fall

Free fall is an essential concept in physics, describing the motion of an object under the influence of gravity alone. In this scenario, marble B is in free fall from its starting point, as it is dropped without any initial vertical velocity.

• As it falls, marble B's vertical position changes solely due to gravity, which acts as a constant downwards force.


• The equation that describes this motion is \[ y_B(t) = R \tan(\alpha) - \frac{1}{2}gt^2 \]This formula shows how the height decreases over time as gravity pulls the marble downward.
• Throughout its motion, the horizontal position of marble B remains static at the value\[ R \],not affected by gravity since no horizontal force is at play.

Understanding free fall helps predict and synchronize the positions needed for collision at a specific time.

Mathematical Proof

Mathematical proof provides a logical framework that verifies if conditions like collisions in projectile motion can consistently happen. With mathematics, we dissect and ensure each component of the problem behaves as expected. In the case of our marbles, we aim to prove the inevitable collision, regardless of the initial speed of marble A.

• **Horizontal Proof**: By setting the horizontal motion of marble A equal to marble B, \[ R_x(t) = R \].Through this equality, we solve for time \[ t = \frac{R}{v_0 \cos(\alpha)} \].


• **Vertical Proof**: Now equate their vertical motions. From this, \[ v_0 \sin(\alpha) t = R \tan(\alpha) \],the condition remains true,\[ R \tan(\alpha) = R \tan(\alpha)\] This redundancy confirms a collision happens without reliance on the initial speed \( v_0 \).

Each of these alignments confirms the necessary condition for collision, showcasing how mathematics can predict physical phenomena with certainty.