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Chapter 7: Problem 21

IP A 9.50-g bullet has a speed of \(1.30 \mathrm{km} / \mathrm{s}\). (a) What is its \(k i\) netic energy in joules? (b) What is the bullet's kinetic energy if its speed is halved? \((\mathrm{c})\) If its speed is doubled?

Short Answer

Expert verified
(a) 8035.75 J, (b) 2008.94 J, (c) 32143 J

Step by step solution

01

Convert Mass and Velocity to Appropriate Units

The mass of the bullet is given as 9.50 g. We need to convert this to kilograms for use in kinetic energy calculations. Thus, the mass is \( m = \frac{9.50}{1000} \) kg = 0.0095 kg. The speed of the bullet is given as 1.30 km/s. We need it in m/s, so \( v = 1.30 \times 1000 \) m/s = 1300 m/s.
02

Calculate Initial Kinetic Energy

The formula for kinetic energy is \( KE = \frac{1}{2} mv^2 \). Using the converted values from Step 1, \( KE = \frac{1}{2} \times 0.0095 \times (1300)^2 \). Perform the calculation to find the kinetic energy: \( KE = \frac{1}{2} \times 0.0095 \times 1690000 = 8035.75 \) Joules.
03

Calculate Kinetic Energy If Speed Is Halved

If the speed is halved, the new velocity \( v' = \frac{1300}{2} = 650 \) m/s. The kinetic energy with this new speed is \( KE' = \frac{1}{2} \times 0.0095 \times (650)^2 \). Calculate this to get \( KE' = \frac{1}{2} \times 0.0095 \times 422500 = 2008.94 \) Joules.
04

Calculate Kinetic Energy If Speed Is Doubled

If the speed is doubled, the new velocity \( v'' = 2 \times 1300 = 2600 \) m/s. The kinetic energy with this new speed is \( KE'' = \frac{1}{2} \times 0.0095 \times (2600)^2 \). Perform the calculation to find the kinetic energy: \( KE'' = \frac{1}{2} \times 0.0095 \times 6760000 = 32143.00 \) Joules.

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Key Concepts

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

Mass Conversion
When working on kinetic energy problems, it's important to ensure the mass of any objects involved is in the correct units鈥攌ilograms (kg). This is because standard formulas use kilograms, not grams, and other incorrect units could lead to significant errors in your calculations.

To convert mass from grams to kilograms, you simply divide the mass in grams by 1000. For example, a bullet weighing 9.50 grams is converted as follows:

\[ m = \frac{9.50 \, \text{g}}{1000} = 0.0095 \, \text{kg} \]

By using the correct units from the start, you ensure that any further calculations with this mass are accurate and aligned with physics principles.
Velocity Conversion
Similarly to mass, the velocity must also be in standard SI units, which are meters per second (m/s) for kinetic energy calculations. Often, velocities are given in kilometers per second (km/s) or even miles per hour, so they must be converted appropriately.

To convert from kilometers per second to meters per second, multiply by 1000, because there are 1000 meters in a kilometer. Using this method, a velocity of 1.30 km/s is converted to:

\[ v = 1.30 \, \text{km/s} \times 1000 = 1300 \, \text{m/s} \]
This ensures the values may be plugged into any physics equation without any unit issues.
Kinetic Energy Formula
The kinetic energy (KE) of an object is a measure of its energy due to its motion and is calculated using the formula:

\[ KE = \frac{1}{2} mv^2 \]

Here's a breakdown of the formula:
  • The term \( m \) denotes mass in kilograms.
  • \( v \) represents velocity in meters per second.
  • \( v^2 \) highlights the point that kinetic energy is proportional to the square of velocity. This means if velocity doubles, kinetic energy increases by a factor of four.
  • The coefficient \( 1/2 \) is a constant that arises from the integration of velocity in the derivation of kinetic energy.
A practical use of this formula can compute kinetic energy as seen in the exercise: a bullet with a mass of 0.0095 kg moving at 1300 m/s has a kinetic energy of:
\[ KE = \frac{1}{2} \times 0.0095 \times (1300)^2 = 8035.75 \, \text{Joules} \]

Keep these ideas in mind to solve variable scenarios, like halving or doubling the velocity, which reflect the quadratic relationship of kinetic energy with velocity.

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Most popular questions from this chapter

You throw a glove straight upward to celebrate a victory. Its initial kinetic energy is \(K\) and it reaches a maximum height \(h\) What is the kinetic energy of the glove when it is at the height \(h / 2 ?\)

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IP A certain car can accelerate from rest to the speed \(v\) in \(T\) seconds. If the power output of the car remains constant, (a) how long does it take for the car to accelerate from \(v\) to \(2 v ?\) (b) How fast is the car moving at \(2 T\) seconds after starting?

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