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Chapter 5: Problem 15

The damage done by a projectile on impact is correlated with its kinetic energy. Calculate and compare the kinetic energies of these three projectiles: a) a \(10.0 \mathrm{~kg}\) stone at \(30.0 \mathrm{~m} / \mathrm{s}\) b) a \(100.0 \mathrm{~g}\) baseball at \(60.0 \mathrm{~m} / \mathrm{s}\) c) a \(20.0 \mathrm{~g}\) bullet at \(300 . \mathrm{m} / \mathrm{s}\)

Short Answer

Expert verified
Question: Compare the kinetic energies of the three projectiles: a 10.0 kg stone at 30.0 m/s, a 100.0 g baseball at 60.0 m/s, and a 20.0 g bullet at 300.0 m/s. Answer: The stone has the highest kinetic energy (4500 J), followed by the bullet (900 J), and lastly the baseball (180 J).

Step by step solution

01

Understand the exercise and given information

We are supposed to calculate the kinetic energies of three projectiles and compare them. We have given masses and velocities for all three projectiles: a) 10.0 kg stone at 30.0 m/s b) 100.0 g baseball at 60.0 m/s c) 20.0 g bullet at 300.0 m/s
02

Convert masses to appropriate units

Since the base unit of mass in the SI system is kilograms, we should convert the grams (baseball and bullet) to kg before calculating the kinetic energy: b) 100.0 g to kg: \(\frac{100.0}{1000}=0.1\,kg\) c) 20.0 g to kg: \(\frac{20.0}{1000}=0.02\,kg\)
03

Calculate kinetic energies

Now we can calculate the kinetic energies (\(KE = \frac{1}{2}mv^2\)) for each projectile: a) Stone: \(KE_{stone} = \frac{1}{2}(10.0\,kg)(30.0\,m/s)^2 = 4500 \,J\) b) Baseball: \(KE_{baseball} = \frac{1}{2}(0.1\,kg)(60.0\,m/s)^2 = 180\,J\) c) Bullet: \(KE_{bullet} = \frac{1}{2}(0.02\,kg)(300.0\,m/s)^2 = 900\,J\)
04

Compare the kinetic energies

We have found the kinetic energies for all projectiles, now let's compare their values: - Stone: 4500 J - Baseball: 180 J - Bullet: 900 J In comparison, the stone has the highest kinetic energy, followed by the bullet, and lastly the baseball.

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Most popular questions from this chapter

A force given by \(F(x)=5 x^{3} \hat{x}\left(\right.\) in \(\left.\mathrm{N} / \mathrm{m}^{3}\right)\) acts on a \(1.00-\mathrm{kg}\) mass moving on a frictionless surface. The mass moves from \(x=2.00 \mathrm{~m}\) to \(x=6.00 \mathrm{~m}\) a) How much work is done by the force? b) If the mass has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\) at \(x=2.00 \mathrm{~m},\) what is its speed at \(x=6.00 \mathrm{~m} ?\)

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A car of mass \(m\) accelerates from rest along a level straight track, not at constant acceleration but with constant engine power, \(P\). Assume that air resistance is negligible. a) Find the car's velocity as a function of time. b) A second car starts from rest alongside the first car on the same track, but maintains a constant acceleration. Which car takes the initial lead? Does the other car overtake it? If yes, write a formula for the distance from the starting point at which this happens. c) You are in a drag race, on a straight level track, with an opponent whose car maintains a constant acceleration of \(12.0 \mathrm{~m} / \mathrm{s}^{2} .\) Both cars have identical masses of \(1000 . \mathrm{kg} .\) The cars start together from rest. Air resistance is assumed to be negligible. Calculate the minimum power your engine needs for you to win the race, assuming the power output is constant and the distance to the finish line is \(0.250 \mathrm{mi}\)
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