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Chapter 14: Problem 43

ssm In \(10.0 \mathrm{s}, 200\) bullets strike and embed themselves in a wall. The bullets strike the wall perpendicularly. Each bullet has a mass of \(5.0 \times 10^{-3} \mathrm{kg}\) and a speed of 1200 \(\mathrm{m} / \mathrm{s}\) . (a) What is the average change in momentum per second for the bullets? (b) Determine the average force exerted on the wall. (c) Assuming the bullets are spread out over an area of \(3.0 \times 10^{-4} \mathrm{m}^{2}\) , obtain the average pressure they exert on this region of the wall.

Short Answer

Expert verified
(a) -120 kg m/s虏, (b) -120 N, (c) 4.0 脳 10鈦� Pa.

Step by step solution

01

Calculate Total Momentum Change Per Bullet

The momentum change for each bullet can be determined using the formula \( \Delta p = m \times (v_f - v_i) \). Given that each bullet comes to a stop upon impact, \( v_f = 0 \). Therefore, \( \Delta p = 5.0 \times 10^{-3} \text{ kg} \times (0 - 1200 \text{ m/s}) = -6.0 \text{ kg m/s} \). The negative sign indicates a reduction in momentum.
02

Calculate Total Momentum Change for All Bullets

The total momentum change for all 200 bullets is \( \Delta P = 200 \times (-6.0 \text{ kg m/s}) = -1200 \text{ kg m/s} \).
03

Determine Average Change in Momentum per Second

The average change in momentum per second (also known as the average rate of momentum change) is given by the formula: \( \frac{\Delta P}{\Delta t} \). Thus, it is \( \frac{-1200 \text{ kg m/s}}{10.0 \text{ s}} = -120 \text{ kg m/s}^2 \).
04

Calculate the Average Force on the Wall

The average force \( F \) exerted on the wall is equal to the average change in momentum per second. Hence, \( F = -120 \text{ N} \).
05

Determine the Average Pressure on the Wall

Pressure is defined as force per unit area. Using the formula \( P = \frac{F}{A} \), where \( A = 3.0 \times 10^{-4} \text{ m}^2 \), the pressure is \( P = \frac{120 \text{ N}}{3.0 \times 10^{-4} \text{ m}^2} = 4.0 \times 10^{5} \text{ Pa} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

average force
Understanding the concept of average force can simplify various physics problems. It's essentially the total force applied distributed over the time during which it acts. In the context of a bullet collision, each bullet impacts the wall and comes to a stop, exerting a force. This force isn't constant because it's strongest at the moment of impact and decreases until the bullet is at rest. The average force, however, gives us a simplified number to work with by averaging out these variations. It's calculated by dividing the total change in momentum over time. Key aspects include:
  • Average force represents the mean force applied during the time of impact.
  • It's measured in Newtons (N).
  • The formula used is derived from Newton's second law, where force equals the rate of change of momentum: \( F = \frac{\Delta P}{\Delta t} \).
pressure
Pressure plays a crucial role when examining the effects of force on surfaces. It's defined as the force applied per unit area. In our scenario involving bullet collisions, understanding pressure helps to quantify the impact over the wall's surface.Pressure is calculated using the formula \( P = \frac{F}{A} \, \ \text{where F is the force and A is the area it acts on} \). By determining the average force exerted on the wall, one can further calculate how distributed this force is across the area impacted by the bullets.Some important points to consider:
  • Pressure is measured in Pascals (Pa).
  • It gives insights into structural stress and potential damage on materials.
  • Higher pressure implies higher potential for material deformation or damage.
bullet collision
Every time a bullet strikes a wall, it undergoes a collision process. Collisions, particularly in physics, are classified based on the conservation of momentum and energy. In this problem, bullets embed into the wall, indicating inelastic collisions.During the inelastic collision, the bullets lose all their velocity and kinetic energy as they become part of the wall. Essential points about bullet collisions involve:
  • Change in momentum, a key factor as discussed in \( \Delta p = m \times (v_f - v_i) \, \ \text{where v_f = 0 m/s} \).
  • The concept of momentum conservation, although kinetic energy is not conserved here.
  • The impact generates forces and pressures which we evaluate in the problem.
physics problem
Physics problems often involve breaking down real-world phenomena into simpler calculations and principles. For instance, to solve scenarios like the bullet collision, a clear understanding of mechanics is crucial. This exercise touches on several core principles: momentum, force, and pressure, all fundamental to analyzing motion and impacts. A systematic approach often involves:
  • Starting with defining known values and equations pertinent to momentum and force.
  • Breaking down the problem step-by-step, as seen in the solution process.
  • Utilizing the right formulas to bridge concepts, such as linking momentum to force and force to pressure.
By handling each component separately, we simplify the complexity of physical interactions involved, making it easier to understand the entire scenario.

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Most popular questions from this chapter

Compressed air can be pumped underground into huge caverns as a form of energy storage. The volume of a cavern is \(5.6 \times 10^{5} \mathrm{m}^{3}\) , and the pressure of the air in it is \(7.7 \times 10^{6}\) Pa. Assume that air is a diatomic ideal gas whose internal energy \(U\) is given by \(U=\frac{5}{2} n R T\) . If one home uses 30.0 \(\mathrm{kW}\) \cdoth of energy per day, how many homes could this internal energy serve for one day?

Two moles of an ideal gas are placed in a container whose volume is \(8.5 \times 10^{-3} \mathrm{m}^{3} .\) The absolute pressure of the gas is \(4.5 \times 10^{5} \mathrm{Pa}\) . What is the average translational kinetic energy of a molecule of the gas?

When you push down on the handle of a bicycle pump, a piston in the pump cylinder compresses the air inside the cylinder. When the pressure in the cylinder is greater than the pressure inside the inner tube to which the pump is attached, air begins to flow from the pump to the inner tube. As a biker slowly begins to push down the handle of a bicycle pump, the pressure inside the cylinder is \(1.0 \times 10^{5} \mathrm{Pa},\) and the piston in the pump is 0.55 \(\mathrm{m}\) above the bottom of the cylinder. The pressure inside the inner tube is \(2.4 \times 10^{5}\) Pa. How far down must the biker push the handle before air begins to flow from the pump to the inner tube? Ignore the air in the hose connecting the pump to the inner tube, and assume that the temperature of the air in the pump cylinder does not change.

ssm Oxygen for hospital patients is kept in special tanks, where the oxygen has a pressure of 65.0 atmospheres and a temperature of 288 \(\mathrm{K}\) . The tanks are stored in a separate room, and the oxygen is pumped to the patient's room, where it is administered at a pressure of 1.00 atmosphere and a temperature of 297 \(\mathrm{K}\) . What volume does 1.00 \(\mathrm{m}^{3}\) of oxygen in the tanks occupy at the conditions in the patient's room?

ssm Hemoglobin has a molecular mass of 64500 u. Find the mass (in \(\mathrm{kg}\) ) of one molecule of hemoglobin.
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