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\(\cdot\) 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 bullet's mass is approximately 0.00421 kg.

Step by step solution

01

Understand the Relationship

When the bullet comes to rest in the absorber, all of its kinetic energy is converted to thermal energy within the absorber. Hence, we can use the equation for kinetic energy loss to find the mass of the bullet.
02

Recall the Kinetic Energy Formula

The kinetic energy (KE) of any object in motion is given by the formula: \[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass of the object and \( v \) is its velocity.
03

Set Up the Equation

We know that the bullet loses 1960 J of kinetic energy when it comes to a stop. Therefore, we set the kinetic energy equation equal to 1960 J:\[ \frac{1}{2} m (965\, \mathrm{m/s})^2 = 1960 \]Our task is to solve for the mass \( m \).
04

Solve for Mass

Rearrange the equation to solve for mass \( m \):\[ m = \frac{2 \times 1960}{965^2} \]Calculate the value by substituting in the numbers:\[ m = \frac{3920}{931225} \approx 0.00421 \text{ kg} \]Hence, the mass of the bullet is approximately 0.00421 kg.

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Key Concepts

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

Thermal Energy
Thermal energy refers to the internal energy within a system due to the random motions of its molecules. When the bullet mentioned in the problem comes to rest in the absorber, its kinetic energy is converted into thermal energy. This conversion process is crucial in understanding energy transformations.
Thermal energy is a byproduct when kinetic energy dissipates, especially in scenarios where objects come to a stop. The increased molecular activity in the absorber results in a temperature rise, so measuring temperature changes can give clues about the energy initially present in the system.
In this example, the bullet's kinetic energy of 1960 Joules becomes thermal energy within the absorber, raising its temperature.
Mass Calculation
Determining the mass of an object involves understanding the relationship between energy and motion. In our example, the bullet's loss of kinetic energy helped determine its mass using known variables like velocity and energy loss.
Kinetic energy has a straightforward formula:
  • \[ KE = \frac{1}{2} mv^2 \]
Where
  • \( KE \) is kinetic energy,
  • \( m \) is mass, and
  • \( v \) is velocity.
Since kinetic energy and velocity are known, rearranging the formula allows solving for the mass \( m \). Plug the numbers into this equation:
  • \[ m = \frac{2 \times KE}{v^2} \]
  • \[ m = \frac{2 \times 1960}{965^2} \approx 0.00421 \text{ kg} \]
This calculation shows that the bullet's mass is about 0.00421 kg.
Motion Equations
Motion equations describe how objects move in space over time, and they're pivotal in problems involving kinetic energy. The main equation here is the kinetic energy equation:
  • \[ KE = \frac{1}{2} mv^2 \]
This equation tells us how an object's mass and velocity contribute to its kinetic energy. In this exercise, it's crucial to converting kinetic energy into another form, highlighting energy conservation principles.
To solve motion problems, understand these components:
  • Mass \( m \) and how it correlates to inertia, determining how an object responds to forces.
  • Velocity \( v \), which reflects the object's speed in a specific direction.
  • Using these, calculate kinetic energy to evaluate energy transformations and related mass, as seen with our bullet's mass determination.
These principles form the core of many physics problems, granting insights into the dynamics of objects in motion.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) Ski jump ramp. You are designing a ski jump ramp for the next Winter Olympics. You need to calculate the vertical height \(h\) from the starting gate to the bottom of the ramp. The skiers push off hard with their ski poles at the start, just above the starting gate, so they typically have a speed of 2.0 \(\mathrm{m} / \mathrm{s}\) as they reach the gate. For safety, the skiers should have a speed of no more than 30.0 \(\mathrm{m} / \mathrm{s}\) when they reach the bottom of the ramp. You determine that for a 85.0 -kg skier with good form, friction and air resistance will do total work of magnitude 4000 J on him during his run down the slope. What is the max- imum height \(h\) for which the maximum safe speed will not be exceeded?

\(\cdot\) \(\cdot\) An 8.00 kg package in a mail-sorting room slides 2.00 \(\mathrm{m}\) down a chute that is inclined at \(53.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.40 . Calculate the work done on the package by (a) friction (b) gravity, and (c) the normal force. (d) What is the net work done on the package?

\(\bullet\) \(\bullet\) \(\bullet\) Pendulum. A small 0.12 kg metal ball is tied to a very light (essentially massless) string 0.80 \(\mathrm{m}\) long to form a pendu- lum that is then set swinging by releasing the ball from rest when the string makes a \(45^{\circ}\) angle with the vertical. Air drag and other forms of friction are negligible. What is the speed of the ball when the string passes through its vertical position and what is the tension in the string at that instant?

\(\bullet\) \(\bullet\) A 250 g object on a frictionless, horizontal lab table is pushed against a spring of force constant 35 \(\mathrm{N} / \mathrm{cm}\) and then released. Just before the object is released, the spring is com- pressed 12.0 \(\mathrm{cm} .\) How fast is the object moving when it has gained half of the spring's original stored energy?

\(\bullet\) \(\bullet\) The aircraft carrier \(U S S\) George Washington has mass \(1.0 \times 10^{8} \mathrm{kg} .\) When its engines are developing their full power of \(260,000\) hp, the George Washington travels at its top speed of 35 knots \((65 \mathrm{km} / \mathrm{h}) .\) If 70\(\%\) of the power output of the engines is applied to pushing the ship through the water, what is the magnitude of the force of water resistance that opposes the carrier's motion at this speed?

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