A car of mass \(1200 \mathrm{~kg}\), moving with a speed of \(72 \mathrm{mph}\) on a highway, passes a small SUV with a mass \(1 \frac{1}{2}\) times bigger, moving at \(2 / 3\) of the speed of the car. a) What is the ratio of the momentum of the SUV to that of the car? b) What is the ratio of the kinetic energy of the SUV to that of the car?

Short Answer

Expert verified
Answer: The ratio of the momentum of the SUV to that of the car is approximately 1.00, and the ratio of the kinetic energy of the SUV to that of the car is approximately 0.89.

Step by step solution

01

Convert velocity of the car to meters per second

To convert the car's velocity from miles per hour (mph) to meters per second (m/s), we can use the following conversion factors: 1 mile = 1609.34 meters 1 hour = 3600 seconds So, the velocity in m/s is: $$ velocity_{car} = 72 \frac{miles}{hour} \times \frac{1609.34 \space meters}{1 \space mile} \times \frac{1 \space hour}{3600 \space seconds} \approx 32.2\frac{meters}{second} $$
02

Calculate the mass and velocity of the SUV

According to the given information, the SUV has a mass of \(1.5\) times the mass of the car, and the velocity is \(2 / 3\) the velocity of the car. Therefore, we can write: $$ mass_{SUV} = 1.5 \times mass_{car} = 1.5 \times 1200 kg = 1800 kg $$ and $$ velocity_{SUV} = \frac{2}{3} \times velocity_{car} = \frac{2}{3} \times 32.2 \frac{m}{s} \approx 21.5\frac{m}{s} $$
03

Calculate the momentum of the car and the SUV

Now we can calculate the momentum for both the car and the SUV using the formula: $$ momentum = mass \times velocity $$ For the car: $$ momentum_{car} = 1200 kg \times 32.2 \frac{m}{s} = 38640 \frac{kg \cdot m}{s} $$ For the SUV: $$ momentum_{SUV} = 1800 kg \times 21.5 \frac{m}{s} = 38700 \frac{kg \cdot m}{s} $$
04

Calculate the ratio of the momentum of the SUV to that of the car

We can now find the ratio of the momentum of the SUV to that of the car by dividing the momentum of the SUV by the momentum of the car: $$ \frac{momentum_{SUV}}{momentum_{car}} = \frac{38700}{38640} \approx 1.00 $$ So, the ratio of the momentum is approximately \(1.00\).
05

Calculate the kinetic energy of the car and the SUV

Next, we can calculate the kinetic energy for both the car and the SUV using the formula: $$ kinetic~energy = \frac{1}{2} \times mass \times velocity^2 $$ For the car: $$ KE_{car} = \frac{1}{2} \times 1200 kg \times (32.2\frac{m}{s})^2 \approx 626304\ J $$ For the SUV: $$ KE_{SUV} = \frac{1}{2} \times 1800 kg \times (21.5\frac{m}{s})^2 \approx 556650\ J $$
06

Calculate the ratio of the kinetic energy of the SUV to that of the car

Finally, we can find the ratio of the kinetic energy of the SUV to that of the car by dividing the kinetic energy of the SUV by the kinetic energy of the car: $$ \frac{KE_{SUV}}{KE_{car}} = \frac{556650}{626304} \approx 0.89 $$ So, the ratio of the kinetic energy is approximately \(0.89\).

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Key Concepts

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

Momentum Calculation
Momentum is a measure of the quantity of motion that an object has, directly dependent on its mass and velocity. In physics, it's represented by the formula:
\[ momentum = mass \times velocity \]
The greater the mass or velocity of an object, the greater its momentum. For our educational exercise, we calculated the momentum of a moving car and an SUV. It's crucial to note that even if two objects have the same momentum, they might not necessarily have the same mass and velocity; it’s the product of these two variables that matters. In this case, the car's momentum came out to be 38,640 kg·m/s, and the SUV's was 38,700 kg·m/s, making the ratio roughly 1:1. This simple yet profound calculation is foundational to understanding motion and collision in physics.
Kinetic Energy Calculation
Kinetic energy quantifies how much work an object can do by virtue of its motion and is given by the equation:
\[ kinetic\ energy = \frac{1}{2} \times mass \times velocity^2 \]
This tells us that kinetic energy increases with the square of the velocity; hence, even a small increase in speed significantly raises an object’s kinetic energy. When performing the kinetic energy calculation in our exercise, we found that the car's kinetic energy was approximately 626,304 Joules, while the SUV's was 556,650 Joules. Thus, despite the SUV having slightly higher momentum than the car, its kinetic energy was about 89% of the car's. This illustrates how mass and velocity affect an object's energy differently and is vital for students to understand energy principles in mechanics.
Velocity Conversion
  • Understanding how to convert velocity units from miles per hour to meters per second is crucial for solving physics problems
  • In our example, the car's speed was given in miles per hour and needed to be converted into meters per second for standard momentum and kinetic energy calculations

The formula used for conversion is:
\[ 1 \, mph = \frac{1609.34 \, meters}{3600 \, seconds} \]
The car's velocity in meters per second, after conversion, is approximately 32.2 m/s. This step is not just a formality but a fundamental practice that aligns various measurements to universal scientific standards, ensuring accuracy in calculations and making it easier for students worldwide to understand and reproduce results.
Mass and Velocity Relationship
In physics, mass and velocity have direct and intricate relationships with both momentum and kinetic energy. As we've seen in the example above, their product defines momentum. For kinetic energy, while mass is directly proportional, velocity impacts energy quadratically. This distinction is crucial:
  • A change in mass will alter momentum and kinetic energy linearly
  • A change in velocity will affect momentum linearly but influence kinetic energy quadratically

The interplay of mass and velocity showcases the dynamic nature of physical quantities. Recognizing these relationships can help students make predictions about the effects of changing an object's mass or velocity, which is especially important in subjects such as engineering, aerospace, and crash safety.

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Most popular questions from this chapter

The electron-volt, \(\mathrm{eV},\) is a unit of energy \((1 \mathrm{eV}=\) \(\left.1.602 \cdot 10^{-19} \mathrm{~J}, 1 \mathrm{MeV}=1.602 \cdot 10^{-13} \mathrm{~J}\right) .\) Since the unit of \(\mathrm{mo}\) mentum is an energy unit divided by a velocity unit, nuclear physicists usually specify momenta of nuclei in units of \(\mathrm{MeV} / c,\) where \(c\) is the speed of light \(\left(c=2.998 \cdot 10^{9} \mathrm{~m} / \mathrm{s}\right) .\) In the same units, the mass of a proton \(\left(1.673 \cdot 10^{-27} \mathrm{~kg}\right)\) is given as \(938.3 \mathrm{MeV} / \mathrm{c}^{2} .\) If a proton moves with a speed of \(17,400 \mathrm{~km} / \mathrm{s}\) what is its momentum in units of \(\mathrm{MeV} / \mathrm{c}\) ?

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In the movie Superman, Lois Lane falls from a building and is caught by the diving superhero. Assuming that Lois, with a mass of \(50.0 \mathrm{~kg}\), is falling at a terminal velocity of \(60.0 \mathrm{~m} / \mathrm{s}\), how much force does Superman exert on her if it takes \(0.100 \mathrm{~s}\) to slow her to a stop? If Lois can withstand a maximum acceleration of \(7 g^{\prime}\) s, what minimum time should it take Superman to stop her after he begins to slow her down?

Here is a popular lecture demonstration that you can perform at home. Place a golf ball on top of a basketball, and drop the pair from rest so they fall to the ground. (For reasons that should become clear once you solve this problem, do not attempt to do this experiment inside, but outdoors instead!) With a little practice, you can achieve the situation pictured here: The golf ball stays on top of the basketball until the basketball hits the floor. The mass of the golf ball is \(0.0459 \mathrm{~kg}\), and the mass of the basketball is \(0.619 \mathrm{~kg}\). you can achieve the situation pictured here: The golf ball stays on top of the basketball until the basketball hits the floor. The mass of the golf ball is \(0.0459 \mathrm{~kg}\), and the mass of the basketball is \(0.619 \mathrm{~kg}\). a) If the balls are released from a height where the bottom of the basketball is at \(0.701 \mathrm{~m}\) above the ground, what is the absolute value of the basketball's momentum just before it hits the ground? b) What is the absolute value of the momentum of the golf ball at this instant? c) Treat the collision of the basketball with the floor and the collision of the golf ball with the basketball as totally elastic collisions in one dimension. What is the absolute magnitude of the momentum of the golf ball after these collisions? d) Now comes the interesting question: How high, measured from the ground, will the golf ball bounce up after its collision with the basketball?

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