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Chapter 7: Problem 2

How fast must an \(816 \mathrm{~kg}\) VW Beetle travel to have the same translational momentum as a \(2650 \mathrm{~kg}\) Cadillac going \(16 \mathrm{~km} / \mathrm{h} ?\)

Short Answer

Expert verified
The VW Beetle must travel approximately 51.91 km/h to have the same momentum as the Cadillac.

Step by step solution

01

- Understand the concept of momentum

Momentum (\textbf{p}) of an object is given by the product of its mass (m) and velocity (v). The equation for momentum is: \[ \textbf{p} = m \times v \]We need to ensure that the momentum of the VW Beetle equals the momentum of the Cadillac.
02

- Calculate the momentum of the Cadillac

First, convert the Cadillac's speed from km/h to m/s. 16 km/h = 16 \times \frac{1000}{3600} m/s = \frac{16 \times 1000}{3600} m/s = \frac{16000}{3600} m/s \approx 4.44 m/s.Now, calculate the Cadillac's momentum using its mass (2650 kg) and velocity (4.44 m/s): \[ \textbf{p}_{Cadillac} = 2650 \times 4.44 \approx 11766 \text{ kg} \text{ m/s} \]
03

- Set up the equation for the VW Beetle's momentum

Let the velocity of the VW Beetle be \( v_{VW} \). We know its mass is 816 kg. According to the problem, the momentum of the VW Beetle must match the momentum of the Cadillac:\[ 816 \times v_{VW} = 11766 \]
04

- Solve for the VW Beetle's velocity

Divide both sides of the equation by 816 to isolate \( v_{VW} \): \[ v_{VW} = \frac{11766}{816} \approx 14.42 \text{ m/s} \]
05

- Convert the velocity from m/s back to km/h

To convert \( v_{VW} \) from m/s to km/h, multiply by \( \frac{3600}{1000} \): \[ v_{VW} \approx 14.42 \times \frac{3600}{1000} \approx 51.91 \text{ km/h} \]

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Key Concepts

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

momentum equation
Momentum is a vital concept in physics. It's a measure of the motion of an object and is calculated as the product of the object's mass and its velocity. The momentum equation is fairly simple: \( p = m \times v \). This compact equation communicates a lot of information in a tiny expression. Here, \( p \) represents momentum, \( m \) stands for mass, and \( v \) is the velocity of the object. We can use the momentum equation to solve various problems in mechanics, including collisions and other interactions.
unit conversion
Unit conversion is an essential skill in physics because measurements are often given in different units. For this problem, the first step was to convert the Cadillac's speed from kilometers per hour (km/h) to meters per second (m/s). Conversion helps ensure consistency in the units used for calculations. The conversion factor between these two speeds is 1 km/h = \( \frac{1000}{3600} \) m/s. Therefore, to convert 16 km/h to m/s, we multiply: 16 km/h \( = 16 \times \frac{1000}{3600} \approx 4.44 \) m/s. This consistency is crucial for accurate calculations, especially when tackling problems involving momentum or energy.
velocity calculation
The final step in solving the problem involves calculating the required velocity for the VW Beetle to match the Cadillac's momentum. Given the mass of the VW Beetle and using the momentum calculated for the Cadillac, we can set the two momenta equal to each other: \( 816 \times v_{VW} = 11766 \). Solving for \( v_{VW} \), we divide both sides by 816: \( v_{VW} \approx 14.42 \) m/s. To make the answer more practical, we convert this velocity back to kilometers per hour: \( v_{VW} \approx 14.42 \times \frac{3600}{1000} \approx 51.91\) km/h. This step-by-step approach ensures clarity and accuracy in the final solution.

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Most popular questions from this chapter

A \(150 \mathrm{~g}\) baseball pitched at a speed of \(40 \mathrm{~m} / \mathrm{s}\) is hit straight back to the pitcher at a speed of \(60 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the average force on the ball from the bat if the bat is in contact with the ball for \(5.0 \mathrm{~ms}\) ?

In Edmund Rostand's famous play, Cyrano de Bergerac, Cyrano, in an attempt to distract a suitor from visiting Roxanne, claims to have descended to Earth from the Moon and proclaims to have invented six novel and fantastical methods for traveling to the Moon. One is as follows. Sitting on an iron platform- thence To throw a magnet in the air. This is A method well conceived - the magnet flown, Infallibly the iron will pursue: Then quick! relaunch your magnet, and you thus Can mount and mount unmeasured distances! \(^{*}\) In an old cartoon, there is another version of this method. A character in the old West is on a hand-pumped, two-person rail car. After getting tired of pumping the handle up and down to make the car move along the rails, he takes out a magnet, hangs it from a fishing pole, and holds it in front of the cart. The magnet pulls the cart toward it, which pushes the magnet forward, and so on, so the cart moves forward continually. What do you think of these methods? Can some version of them work? Discuss in terms of the physics you have learned.

A spacecraft is separated into two parts by detonating the explosive bolts that hold them together. The masses of the parts are \(1200 \mathrm{~kg}\) and \(1800 \mathrm{~kg}\); the magnitude of the impulse on each part from the bolts is \(300 \mathrm{~N} \cdot \mathrm{s}\). With what relative speed do the two parts separate because of the detonation?

Vessel at Rest Explodes A vessel at rest explodes, breaking into three pieces. Two pieces, having equal mass, fly off perpendicular to one another with the same speed of \(30 \mathrm{~m} / \mathrm{s}\). The third piece has three times the mass of each other piece. What are the magnitude and direction of its velocity immediately after the explosion?

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