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A bullet is fired into a large stationary absorber and comes to rest. Temperature measurements of the absorber show that the bullet lost \(1960 \mathrm{~J}\) of kinetic energy, and high-speed photos of the bullet show that it was moving at \(965 \mathrm{~m} / \mathrm{s}\) just as it struck the absorber. What is the mass of the bullet?

Short Answer

Expert verified
The mass of the bullet is approximately 0.0042 kg.

Step by step solution

01

Understand the Problem

We need to find the mass of the bullet given that it lost 1960 J of kinetic energy upon coming to rest. We also know the initial velocity of the bullet is 965 m/s.
02

Use Kinetic Energy Formula

The kinetic energy of the bullet initially is given by the formula \(KE = \frac{1}{2}mv^2\). Here, \(KE = 1960\) J, \(v = 965\) m/s, and \(m\) is the mass of the bullet, which we need to find.
03

Rearrange Formula to Solve for Mass

We rearrange the formula to solve for mass \(m\). The equation becomes \(m = \frac{2 \times KE}{v^2}\).
04

Plug Values into the Formula

Substitute the known values into the formula: \(m = \frac{2 \times 1960}{965^2}\). Simplify the numbers to find the mass.
05

Calculate the Mass

Calculate \(m = \frac{3920}{931225}\). After division, the mass of the bullet is approximately \(0.0042\) kg.

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Key Concepts

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

Bullet Mass Calculation
Calculating the mass of a bullet using kinetic energy and velocity is a straightforward task in physics. When a bullet is fired, it possesses kinetic energy due to its motion. This energy can be calculated using the formula:\[ KE = \frac{1}{2} m v^2 \]Where:- \(KE\) is the kinetic energy- \(m\) is the mass of the bullet- \(v\) is the velocity of the bullet
To find the mass of the bullet, we can rearrange the formula to:\[ m = \frac{2 \times KE}{v^2} \]By substituting the given kinetic energy and velocity into this formula, you can solve for the mass of the bullet. For our specific example, by calculating with the values provided, the mass turns out to be approximately 0.0042 kg. This process illustrates how energy principles can help measure physical properties.
Initial Velocity
Initial velocity is a crucial parameter in problems involving motion and energy. It sets the starting conditions for any calculations. In the context of a bullet shot from a gun, the initial velocity is the speed at which the bullet was moving just before any external interference, such as hitting an absorber.
In physics problems, initial velocity is usually denoted by \(v_i\). For this problem:- The initial velocity \(v_i\) is given as 965 m/s.This information is essential as it allows us to calculate other quantities, such as kinetic energy and further down the line, the mass of the bullet. Proper understanding of initial velocity helps in predicting outcomes based on the laws of motion.
Energy Loss
Energy loss in physical systems occurs when kinetic energy is transformed into other forms, such as heat or sound. In the case of the bullet being absorbed, the bullet's initial kinetic energy is lost to the absorber, increasing the absorber's temperature. This scenario is a real-world example of energy loss as per the conservation of energy principle.
The given information tells us that the bullet loses \(1960 \mathrm{~J}\) of kinetic energy upon impact and halts. This lost energy is absorbed and often measured as a temperature rise in the absorber. Understanding how energy dissipates helps us comprehend various physical interactions, such as how airbags in cars work by reducing speed and absorbing energy.
Physics Problem Solving
Physics problem solving involves systematically breaking down a question into understandable parts and applying relevant principles to find a solution. The process often includes:
  • Identifying given data and what needs to be found.
  • Choosing appropriate formulas and rearranging them to solve for unknowns.
  • Substituting known values and calculating results step by step.
  • Reviewing and interpreting the results in the context of the problem.
Solving this specific problem about the bullet requires applying kinetic energy formulas and understanding motion principles. By practicing these problem-solving steps, one becomes adept at tackling various physics challenges efficiently.

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Most popular questions from this chapter

The power of the human heart. The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about \(7500 \mathrm{~L}\) of blood. Assume that the work done by the heart is equal to the work required to lift that amount of blood a height equal to that of the average American female, approximately \(1.63 \mathrm{~m} .\) The density of blood is \(1050 \mathrm{~kg} / \mathrm{m}^{3} .\) (a) How much work does the heart do in a day? (b) What is the heart's power output in watts? (c) In fact, the heart puts out more power than you found in part (b). Why? What other forms of energy does it give the blood?

A wooden rod of negligible mass and length \(80.0 \mathrm{~cm}\) is pivoted about a horizontal axis through its center. A white rat with mass \(0.500 \mathrm{~kg}\) clings to one end of the stick, and a mouse with mass \(0.200 \mathrm{~kg}\) clings to the other end. The system is released from rest with the rod horizontal. If the animals can manage to hold on, what are their speeds as the rod swings through a vertical position?

A \(20.0 \mathrm{~kg}\) box is pulled along a rough horizontal surface with a rope at a constant speed of \(8 \mathrm{~m} / \mathrm{s}\). The rope is parallel to the floor, and the coefficient of kinetic friction between the box and the floor is 0.200 . What power must the rope supply to the box? Where does this energy go?

You are rearranging the furniture in your living room. In doing so, you push a \(30 \mathrm{~kg}\) sofa to the left a total distance of \(2 \mathrm{~m}\) and then you push it back to the right a distance of \(1 \mathrm{~m}\). The coefficient of kinetic friction between the sofa and the hardwood floor is 0.2 . (a) What is the work done by friction in moving the sofa? (b) Suppose that you had instead simply moved the sofa to its final position by moving it directly \(1 \mathrm{~m}\) to the left. What is the work done by friction in this case? (c) From the results you obtained in parts (a) and (b), what can you conclude about the nature of the friction force? Is it a conservative force? Explain.

A good workout. You overindulged in a delicious dessert, so you plan to work off the extra calories at the gym. To accomplish this, you decide to do a series of arm raises while holding a \(5.0 \mathrm{~kg}\) weight in one hand. The distance from your elbow to the weight is \(35 \mathrm{~cm}\), and in each arm raise you start with your arm horizontal and pivot it until it is vertical. Assume that the weight of your arm is small enough compared with the weight you are lifting that you can ignore it. As is typical, your muscles are \(20 \%\) efficient in converting the food energy into mechanical energy, with the rest going into heat. If your dessert contained 350 food calories, how many arm raises must you do to work off these calories? Is it realistic to do them all in one session?

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