Problem 30 - University Physics with Modern Physics

Question

Calculate the magnitude of the force required to give a \(0.145-\mathrm{kg}\) baseball an acceleration \(a=1.00 \mathrm{m} / \mathrm{s}^{2}\) in the direction of the baseball's initial velocity when this velocity has a magnitude of (a) \(10.0 \mathrm{m} / \mathrm{s} ;\) (b) 0.900 \(\mathrm{c}\) (c) \(0.990 c .(\mathrm{d})\) Repeat parts \((\mathrm{a}),\) and \((\mathrm{c})\) if the force and acceleration are perpendicular to the velocity.

Step-by-Step Solution

Answer

For (a) 0.145 N, (b) 0.332 N, (c) 1.028 N; (a) perpendicular: 0.145 N, (c) perpendicular: 1.028 N.

Step 1: Identify the formula for force

To calculate the force, use Newton's second law, which states that the force exerted on an object is equal to the object's mass times its acceleration: \[ F = m \cdot a \]Where- \( F \) is the force,- \( m \) is the mass of the object (0.145 kg for the baseball),- \( a \) is the acceleration (1.00 m/s²).

Step 2: Calculate force magnitude for part (a)

For part (a), since the velocity is non-relativistic (much less than the speed of light \( c \)), we can directly apply Newton's second law to compute the force:\[ F = 0.145 \times 1.00 = 0.145\, \text{N} \]

Step 3: Apply relativistic dynamics for parts (b) and (c)

For parts (b) and (c), when the baseball's speed approaches a significant fraction of the speed of light \( c \), relativistic effects must be considered. The relativistic force is defined by:\[ F = \frac{dp}{dt} = \frac{d}{dt} \left( \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \right) \]To find the force, we must consider the relativistic mass increase.

Step 4: Compute force magnitude for part (b)

For part (b), the velocity is 0.900c. Compute the relativistic factor, \( \gamma \), and then use the relativistic formula:\[ \gamma = \frac{1}{\sqrt{1 - (0.900)^2}} \approx 2.294 \F = \frac{0.145 \times 1.00}{\sqrt{1 - (0.900)^2}} \approx 0.332 \text{ N}\]

Step 5: Compute force magnitude for part (c)

For part (c), with velocity 0.990c, find \( \gamma \):\[ \gamma = \frac{1}{\sqrt{1 - (0.990)^2}} \approx 7.089 \F = \frac{0.145 \times 1.00}{\gamma} \approx 1.028 \text{ N}\]

Step 6: Calculate force for perpendicular acceleration in part (a)

For perpendicular force, the situation doesn't call for a different calculation in non-relativistic conditions because the direction doesn't change the force needed for acceleration:\[ F = 0.145 \text{ N}\]

Step 7: Relativistic perpendicular acceleration for part (c)

When calculating force perpendicular to the velocity in relativistic conditions, we use the perpendicular relativistic law:\[ F = m \times a \times \gamma \] For velocity 0.990c:\[ F = 0.145 \times 1.00 \times \gamma \approx 1.028 \text{ N}\]

Key Concepts

Relativistic Dynamics

When velocities are a significant fraction of the speed of light, classical physics cannot accurately describe the motion of objects. This is where relativistic dynamics comes in. Unlike classical dynamics, relativistic dynamics takes into account the relativistic mass increase. This means that as an object moves faster, its mass appears to increase from the perspective of an outside observer. In this context, Newton's second law of motion requires modification. The relativistic force is given by the formula: \[ F = \frac{dp}{dt} = \frac{d}{dt} \left( \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} \right) \] Here, \( p \) is the momentum, \( m \) is the rest mass, \( v \) is velocity, and \( c \) is the speed of light.

• As velocity \( v \) approaches \( c \), the factor \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) increases, making the calculations more complex.
• This factor compensates for the relativistic effects at high speeds.
• The force is not just mass times acceleration but includes this relativistic correction to consider the increase in momentum.

Understanding how relativistic dynamics modifies our classical perspectives is crucial for grasping how forces behave at high speeds.

Non-Relativistic Velocity

Non-relativistic velocities are those which are much smaller compared to the speed of light. In these cases, we can safely use classical mechanics without any additional factors or corrections. For a baseball moving at 10.0 m/s, the speed is significantly under relativistic conditions.

• The formula \( F = m \cdot a \) is straightforward and applicable. Here, the acceleration \( a \) directly multiplies with the mass \( m \) of the object to give us the force.
• Because speeds are relatively low, the effects of relativity can be ignored. This simplification assumes that the object's mass is constant regardless of its velocity.

The calculations are simpler and focus only on factors observable in day-to-day life without worrying about relativistic mass increase or time dilation effects.

Perpendicular Acceleration

When force and acceleration are perpendicular to the direction of velocity, the dynamics change slightly depending on whether the velocity is relativistic or non-relativistic. In non-relativistic contexts, the component of force affects only the acceleration perpendicular to the velocity without any additional effects. Therefore, using Newton's second law without alteration is often sufficient. In relativistic scenarios, however, the situation is more complex. When dealing with perpendicular forces or accelerations:

• The component of acceleration caused by the force is adjusted by the Lorentz factor, \( \gamma \).
• Forces acting perpendicular can appear stronger due to the effects of relativistic physics. The formula becomes \( F = m \times a \times \gamma \), where \( \gamma \) accounts for the increased 'resistance' to the change in velocities.

Understanding how forces at angles interact with velocity under these different scenarios can reveal much about motion and energy at both ordinary and extraordinary speeds.