A 90-g Apple Is Falling From A Tree. What Is The Impulse That Earth Exerts On It During The First 0.50 Seconds?
Introduction
Understanding the concept of impulse in physics provides insight into the forces acting on objects during motion, especially during events like free fall. When a 90-gram apple detaches from a tree and begins to fall under Earth's gravity, it experiences a force exerted by the planet. The question here is: during the initial 0.50 seconds of its fall, what is the magnitude and nature of the impulse that Earth exerts on the apple?
This article explores the physics behind the apple's fall, focusing on the concept of impulse, the forces involved, and the calculations necessary to determine the impulse during the first half-second of free fall. We will analyze the problem step by step, considering relevant formulas, assumptions, and real-world considerations.
Understanding the Key Concepts
What Is Impulse?
Impulse is a fundamental concept in physics that relates to the change in momentum of an object when a force is applied over a period of time. It is defined as:
- Impulse (J) = Force (F) × Time interval (Δt)
- Or equivalently, the change in momentum: J = Δp
Where:
- Impulse is measured in Newton-seconds (N·s).
- Force is measured in Newtons (N).
- Time interval is measured in seconds (s).
- Momentum (p) is mass × velocity.
Understanding impulse helps us analyze how forces change an object's momentum over specified time intervals.
Forces Acting on the Falling Apple
During free fall, the primary force acting on the apple is gravity, which exerts a downward force equal to:
\[
F_g = m \times g
\]
Where:
- \(m\) is the mass of the apple.
- \(g\) is the acceleration due to gravity (~9.8 m/s²).
In an idealized scenario, neglecting air resistance, the force exerted by Earth on the apple is simply the weight of the apple, which remains constant during the fall.
Analyzing the Problem Step by Step
Step 1: Convert the Mass to Kilograms
Since SI units are standard in physics calculations:
\[
m = 90\, \text{g} = 0.090\, \text{kg}
\]
Step 2: Calculate the Force Earth Exerts on the Apple
The force due to gravity (Earth's force on the apple):
\[
F_g = m \times g = 0.090\, \text{kg} \times 9.8\, \text{m/s}^2 = 0.882\, \text{N}
\]
This force acts downward, and in the absence of air resistance, it remains constant during the initial phase of fall.
Step 3: Determine the Acceleration and Velocity After 0.50 Seconds
Since the apple is in free fall:
- Acceleration \(a = g = 9.8\, \text{m/s}^2\)
- Initial velocity \(u = 0\, \text{m/s}\) (assuming it starts from rest)
Using kinematic equations:
\[
v = u + a \times t
\]
\[
v = 0 + 9.8\, \text{m/s}^2 \times 0.50\, \text{s} = 4.9\, \text{m/s}
\]
So, after 0.50 seconds, the apple's velocity is approximately 4.9 m/s downward.
Step 4: Calculate the Change in Momentum (Δp)
Momentum is given by:
\[
p = m \times v
\]
Initial momentum:
\[
p_{initial} = 0\, \text{kg·m/s}
\]
Final momentum after 0.50 seconds:
\[
p_{final} = 0.090\, \text{kg} \times 4.9\, \text{m/s} = 0.441\, \text{kg·m/s}
\]
Change in momentum:
\[
Δp = p{final} - p{initial} = 0.441\, \text{kg·m/s} - 0 = 0.441\, \text{kg·m/s}
\]
This change in momentum is directly related to the impulse exerted by Earth:
\[
J = Δp = 0.441\, \text{kg·m/s}
\]
Note: The impulse exerted by Earth during this interval equals the change in the apple's momentum, assuming negligible air resistance and no other forces.
Implication of the Force and Impulse in Real-World Context
Force During the Fall
The force exerted by Earth is approximately 0.882 N downward, which remains essentially constant during free fall. Because this force is sustained over the interval, the impulse can also be calculated as:
\[
J = F \times Δt = 0.882\, \text{N} \times 0.50\, \text{s} = 0.441\, \text{N·s}
\]
This aligns perfectly with the change in momentum calculated earlier, confirming the physical consistency.
Role of Air Resistance
In real-world scenarios, air resistance opposes the motion of the falling apple, slightly reducing the net acceleration from \(g\). For simplicity, this analysis neglects air resistance, but in practice:
- The actual impulse exerted by Earth may be slightly less.
- The velocity after 0.50 seconds would be slightly less than 4.9 m/s.
- The force exerted by Earth remains close to the weight, but the net force experienced by the apple would be reduced due to air drag.
Summary of Key Calculations
- Mass of apple: 0.090 kg
- Force due to gravity: 0.882 N
- Velocity after 0.50 s: 4.9 m/s
- Change in momentum: 0.441 kg·m/s
- Impulse exerted by Earth: 0.441 N·s
Conclusion
The impulse exerted by Earth on the apple during the first 0.50 seconds of its fall is approximately 0.441 Newton-seconds. This impulse corresponds to the change in the apple's momentum due to the gravitational force acting over that period. The understanding of impulse in this context emphasizes how continuous forces produce measurable changes in an object's motion and highlights the interconnectedness of force, time, and momentum in classical mechanics.
While the idealized calculations assume no air resistance, real-world factors could slightly alter the precise values. Nevertheless, the fundamental physics principles remain the same, illustrating the power of impulse analysis in understanding motion under gravity.
