Problem 48 - University Physics with Modern Physics

Question

A \(0.100-\mu g\) speck of dust is accelerated from rest to a speed of 0.900\(c\) by a constant \(1.00 \times 10^{6} \mathrm{N}\) force. (a) If the nonrelativistic form of Newton's second law \((\Sigma F=m a)\) is used, how far does-the object travel to reach its final speed? (b) Using the correct relativistic treatment of Section 37.8 , how far does the object travel to reach its final speed? (c) Which distance is greater? Why?

Step-by-Step Solution

Answer

(a) 1.215 x 10^7 m; (b) 4.96 x 10^6 m; (c) Nonrelativistic is greater due to no relativistic inertia increase.

Step 1: Calculate acceleration using nonrelativistic Newton's second law

First, convert the mass of the dust speck from micrograms to kilograms: \(0.100 \mu g = 0.100 \times 10^{-6}g = 0.100 \times 10^{-9} kg\). Using \(F = ma\), solve for the acceleration \(a\):\(a = \frac{F}{m} = \frac{1.00 \times 10^6 \text{ N}}{0.100 \times 10^{-9} \text{ kg}} = 1.00 \times 10^{16} \text{ m/s}^2\).

Step 2: Calculate nonrelativistic distance to reach final speed

Use the formula for distance with constant acceleration, \(d = \frac{v^2}{2a}\), where \(v = 0.900c\) and \(c = 3.00 \times 10^8 \text{ m/s}\). Substitute \(v\) and \(a\) to find \(d\): \(d = \frac{(0.900 \times 3.00 \times 10^8)^2}{2 \times 1.00 \times 10^{16}} = 1.215 \times 10^7 \text{ m}\).

Step 3: Calculate the relativistic acceleration

Use the relativistic force relation \(F = \frac{dp}{dt}\), where \(p = \gamma mv\) is the relativistic momentum, \(\gamma = \frac{1}{\sqrt{1-(v/c)^2}}\). The differentiation of \(p\) with respect to time gives \(a_{rel} = \frac{F}{\gamma^3 m}\). Substitute the mass, \(F\), and find \(\gamma\) at \(v = 0.900c\), \(\gamma = 2.294\).

Step 4: Calculate relativistic distance to reach final speed

Use the relativistic result for distance \(d = \frac{mc^2}{F} \left(\gamma - 1\right)\). Substitute values: \(m = 0.100 \times 10^{-9} \text{ kg}\), \(\gamma = 2.294\), \(F = 1.00 \times 10^{6} \text{ N}\), and \(c = 3.00 \times 10^8 \text{ m/s}\) to find \(d\): \(d = \frac{0.100 \times 10^{-9} \times (3.00 \times 10^8)^2}{1.00 \times 10^6} (2.294 - 1) = 4.96 \times 10^6 \text{ m}\).

Step 5: Compare the distances

Compare the results from the nonrelativistic and relativistic calculations. The nonrelativistic distance \(d = 1.215 \times 10^7 \text{ m}\) is greater than the relativistic distance \(d = 4.96 \times 10^6 \text{ m}\). In the relativistic case, the increasing inertia slows down the acceleration process, leading to a smaller distance.

Key Concepts

Newton's Second Law

Newton's second law is a cornerstone of classical physics. It states that the force \( F \) acting on an object is equal to the mass \( m \) of the object multiplied by its acceleration \( a \) (i.e., \( F = ma \)). This fundamental principle explains how an object's motion changes when subjected to a force.
In the context of the exercise, using this law in a nonrelativistic framework, we calculated the acceleration experienced by a tiny speck of dust. By applying the force of \( 1.00 \times 10^6 \) N to the speck with a mass of \( 0.100 \times 10^{-9} \) kg, we found that it experiences an enormous acceleration, \( a = 1.00 \times 10^{16} \) m/s².
However, this approach assumes that the acceleration can increase indefinitely, which isn't the case at velocities approaching the speed of light.

Relativistic Momentum

As objects approach the speed of light, they no longer obey the classic momentum equation \( p = mv \). Instead, relativistic momentum must be considered, incorporating the Lorentz factor \( \gamma \), where \( p = \gamma mv \) and \( \gamma = \frac{1}{\sqrt{1-(v/c)^2}} \). This is crucial for high-speed scenarios.
In our exercise, when the speck's speed approaches \( 0.900c \), the Lorentz factor becomes significant, affecting its momentum. At this speed, the Lorentz factor \( \gamma \) was calculated to be approximately 2.294, showing how the momentum increases non-linearly with velocity.
Relativistic momentum ensures that even as more force is applied, the object cannot accelerate indefinitely, reflecting the increase in inertial resistance.

Relativistic Acceleration

With relativistic speeds, acceleration also deviates from classical predictions. Instead of simply \( a = \frac{F}{m} \), it becomes \( a_{rel} = \frac{F}{\gamma^3 m} \), due to the dependency on the object's velocity through the Lorentz factor. This insight changes how we perceive movement at high speeds.
In this exercise, the relativistic treatment required us to determine \( \gamma \) to correctly compute acceleration. The increased inertia, evident from the factor \( \gamma^3\), dramatically reduces the acceleration as the particle's speed approaches \( c \). This means that the more force applied, the more significantly \( \gamma^3\) impacts the acceleration, thus slowing down the particle much more than expected classically.

Constant Force Dynamics

Applying a constant force to an object results in different outcomes, depending on whether classical or relativistic physics are considered. As observed, under classical physics, the speck should travel farther due to constant acceleration. However, relativistically speaking, the same force applied sees diminishing returns on acceleration at higher speeds.
For this problem, interpreting the dynamics of constant force led to differing travel distances—roughly \( 1.215 \times 10^7 \) m for the classical approach and \( 4.96 \times 10^6 \) m when incorporating relativity.
This contrast arises because, in relativistic terms, as objects gain speed, their effective inertia increases, diminishing the effect of constant force on additional acceleration and thus reducing total travel distance.