/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 (a) What is the magnitude of the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) What is the magnitude of the momentum of a \(10,000-\mathrm{kg}\) truck whose speed is 12.0 \(\mathrm{m} / \mathrm{s} ?\) (b) What speed would a \(2000-\mathrm{kg}\) SUV have to attain in order to have (i) the same momentum? (ii) the same kinetic energy?

Short Answer

Expert verified
(a) 120,000 kg·m/s; (b) (i) 60 m/s; (ii) 26.83 m/s.

Step by step solution

01

Understanding Momentum

Momentum (\(p\)) is calculated as the product of mass (\(m\)) and velocity (\(v\)). The formula is \(p = mv\).
02

Calculate Momentum of the Truck

For the truck, we know that \(m = 10,000 \, \text{kg}\) and \(v = 12.0 \, \text{m/s}\). Applying the formula: \(p = 10,000 \, \text{kg} \times 12.0 \, \text{m/s} = 120,000 \, \text{kg} \cdot \text{m/s}\).
03

Determine SUV's speed for same momentum

To have the same momentum, the SUV's speed is given by \(v = \frac{p}{m}\). With \(p = 120,000 \, \text{kg} \cdot \text{m/s}\) and \(m = 2000 \, \text{kg}\):\[v = \frac{120,000}{2000} = 60 \, \text{m/s}.\]
04

Understand Kinetic Energy

Kinetic energy (\(KE\)) is calculated using the formula \(KE = \frac{1}{2}mv^2\).
05

Calculate Kinetic Energy of the Truck

The truck's kinetic energy is \(KE = \frac{1}{2} \times 10,000 \, \text{kg} \times (12.0 \, \text{m/s})^2 = 720,000 \, \text{J}\).
06

Determine SUV's speed for same kinetic energy

For the SUV to have the same kinetic energy, set its kinetic energy equal to the truck's and use \(\frac{1}{2} \times 2000 \, \text{kg} \times v^2 = 720,000 \, \text{J}\). Solving for \(v\):\[2000v^2 = 1,440,000 \v^2 = 720 \v = \sqrt{720} \approx 26.83 \, \text{m/s}.\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It's a key concept in physics, especially when solving problems related to moving objects. The formula to calculate kinetic energy (\(KE\)) is:
  • \(KE = \frac{1}{2}mv^2\)
where \(m\) is the mass of the object, and \(v\) is its velocity.

For instance, in our original exercise, we calculated the kinetic energy of a truck by plugging in its mass and velocity. We further used this understanding to solve for the velocity of a different vehicle when it has the same kinetic energy. This demonstrates how kinetic energy relates to both mass and velocity.

Understanding how to manipulate the kinetic energy formula can help you solve many real-world physics problems, as well as develop a deeper intuition for how energy changes with speed and mass.
Velocity
Velocity is a vector quantity that describes the speed of an object and its direction of motion. It is crucial for understanding momentum and kinetic energy because it determines how fast an object is moving and in which direction.

In our truck and SUV example from the exercise, velocity is used to determine both the momentum and kinetic energy of the vehicles. When calculating momentum (\(p = mv\)) or kinetic energy (\(KE = \frac{1}{2}mv^2\)), velocity is an essential piece of the puzzle.

When we wanted the SUV to match the truck's momentum, we needed to find the required velocity given its mass. Similarly, to match the kinetic energy, velocity played a pivotal role. Recognizing velocity's role in these formulas can greatly improve problem-solving skills in physics.
Mass
Mass is a fundamental property of an object that quantifies the amount of matter it contains. It is one of the components in calculating both momentum and kinetic energy.

In the given exercise, mass plays a critical role as it multiplies with velocity to determine momentum (\(p = mv\)) and is a factor in the kinetic energy formula (\(KE = \frac{1}{2}mv^2\)). The truck's mass is used to calculate its momentum and kinetic energy, while the SUV's mass is needed to find the required velocity to match these quantities.

Understanding mass's role helps solve various physics problems by giving insight into how differently sized objects move and interact. It's important to remember that mass and weight are not the same; mass is a scalar quantity measured in kilograms, whereas weight measures the gravitational force on an object.
Physics Problem Solving
Physics problem-solving often involves applying fundamental principles to reach a solution. It requires breaking down a problem into manageable steps and systematically applying known formulas.

In the original exercise, we started by understanding the concept of momentum and how to calculate it. This foundational step allows us to solve part (a) and (b). We used the known formulas for momentum and kinetic energy to understand the relationships between mass, velocity, and energy.

Here's a helpful approach for solving physics problems:
  • Identify what is given and what needs to be found.
  • Write down relevant formulas.
  • Substitute the known values into these formulas.
  • Solve for the unknowns.
  • Check the units and ensure physical accuracy.
By practicing these steps regularly, you can become proficient at solving complex physics problems with confidence and accuracy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\mathrm{A}^{232} \mathrm{Th}\) (thorium) nucleus at rest decays to a \(^{228} \mathrm{Ra}\) (radium) nucleus with the emission of an alpha particle. The total kinetic energy of the decay fragments is \(6.54 \times 10^{-13} \mathrm{J}\) . An alpha particle has 1.76\(\%\) of the mass of a \(^{228} \mathrm{Ra}\) nucleus. Calculate the kinetic energy of (a) the recoiling 228 nucleus and (b) the alpha particle.

Block \(A\) in Fig. E8.24 has mass \(1.00 \mathrm{kg},\) and block \(B\) has mass 3.00 \(\mathrm{kg}\) . The blocks are forced together, compressing a spring \(S\) between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block \(B\) acquires a speed of 1.20 \(\mathrm{m} / \mathrm{s}\) . (a) What is the final speed of block \(A\) ? (b) How much potential energy was stored in the compressed spring?

On a frictionless, horizontal air table, puck \(A\) (with mass 0.250 \(\mathrm{kg}\) is moving toward puck \(B\) (with mass \(0.350 \mathrm{kg} ),\) which is initially at rest. After the collision, puck \(A\) has a velocity of 0.120 \(\mathrm{m} / \mathrm{s}\) to the left, and puck \(B\) has a velocity of 0.650 \(\mathrm{m} / \mathrm{s}\) to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

A 10.0 -g marble slides to the left with a velocity of magnitude 0.400 \(\mathrm{m} / \mathrm{s}\) on the frictionless, horizontal surface of an icy New York side- walk and has a head- on, elastic collision with a larger 30.0 -g marble sliding to the right with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}(\mathrm{Fig} . \mathrm{E} 8.48) .\) (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line.) (b) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision for each marble. Compare the values you get for each marble. (c) Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare the values you get for each marble.

A 7.0 -kg shell at rest explodes into two fragments, one with a mass of 2.0 \(\mathrm{kg}\) and the other with a mass of 5.0 \(\mathrm{kg}\) . If the heavier fragment gains 100 \(\mathrm{J}\) of kinetic energy from the explosion, how much kinetic energy does the lighter one gain?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.