91Ó°ÊÓ

Momentum change is a crucial concept in physics. It helps us understand how an object's motion alters when subjected to external forces.
In this specific problem, we have bullets that initially travel towards Superman then bounce back.
Momentum (\(p\)) is the product of mass (\(m\)) and velocity (\(v\)).
Initially, each bullet has a momentum calculated as \(p_i = m \, v\), where \(m = 0.003 \, \text{kg}\) and \(v = 500 \, \text{m/s}\). Thus, \(p_i = 1.5 \, \text{kg m/s}\).
The bullet rebounds with opposite momentum, \(p_f = -1.5 \, \text{kg m/s}\).
The change in momentum (\(\Delta p\)) is the final momentum minus the initial momentum: \(\Delta p = p_f - p_i = -3 \, \text{kg m/s}\). This value occurs per bullet hit.

When a bullet rebounds, it changes direction but maintains speed. This is indicated by a change in the sign of its momentum.
In the exercise, the bullets hit Superman's chest with a certain momentum and then bounce back with the same speed, hence the same magnitude of momentum but in the opposite direction.
This means the bullet's velocity remains \(500 \, \text{m/s}\) but reverses direction, resulting in a negative sign for the final momentum.

• Initial Momentum: \(1.5 \, \text{kg m/s}\)
• Rebound Momentum: \(-1.5 \, \text{kg m/s}\)

This complete switch in momentum is responsible for a significant magnitude of force when considered over many bullets.

Force calculation begins by understanding that force is derived from the momentum change over time.
For this mechanics problem, the average force on Superman’s chest is decided by how much momentum change happens each second.
Using the formula for force, \(F = \frac{\Delta p}{\Delta t}\), where \(\Delta p\) is the total momentum change and \(\Delta t\) is the time duration, the force exerted by the bullets becomes clearer.
Here, \(\Delta t = 60 \, \text{s}\) since the bullet rate is 100 per minute.
The total momentum change over that minute is \(-300 \, \text{kg m/s}\)
for all bullets combined.
Plugging values into the formula gives \(F = \frac{-300 \, \text{kg m/s}}{60 \, \text{s}} = -5 \, \text{N}\).
However, since force's direction doesn't affect its magnitude, the magnitude is \(5 \, \text{N}\).

Mechanics problems involve studying forces and motion, very much like bullets rebounding off a surface.
They are often used to illustrate the concept of momentum and force relationships.
These problems are not only theoretical but have real-world applications, such as in safety equipment testing or analyzing vehicle collisions.
Understanding how mechanical principles apply to bullet impacts can give insights into more extensive systems.

• Focus on Momentum: Changes reflect force applications.
• Rebounding: Illustrates conservation concepts.
• Force Calculation: Shows direct application of Newton's Laws.

By breaking down and solving these problems, mechanics become much easier to comprehend and apply to various fields.