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Chapter 6: Problem 1

An object of mass 2 \(\mathrm{kg}\) has a linear momentum of magnitude 6 \(\mathrm{kg}\) .\(\mathrm{m} / \mathrm{s} .\) What is this object's kinetic energy? (A) 3 \(\mathrm{J}\) (B) 6 \(\mathrm{J}\) (C) 9 \(\mathrm{J}\) (D) 12 \(\mathrm{J}\)

Short Answer

Expert verified
(C) 9 J

Step by step solution

01

Identify the Given Values

Firstly, identify the values given in the problem. The mass (m) of the object is 2 kg and its linear momentum (p) is 6 kg.m/s.
02

Substitute into the Formula

Next, substitute these given values into the formula for kinetic energy. So, K.E = \( \frac{p^2}{2m} \) = \( \frac{(6)^2}{2*2} \) = \( \frac{36}{4} \) = 9 Joules.
03

Match the Answer

Finally, match the calculated kinetic energy with the options given in the problem to find the correct answer. The kinetic energy is 9 Joules, so the correct answer is (C) 9 J.

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Key Concepts

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

Kinetic Energy Calculation
When dealing with objects in motion, understanding kinetic energy calculation is fundamental. Kinetic energy, a form of mechanical energy, quantifies how much work an object in motion can do as a result of its velocity. The formula used to calculate kinetic energy ( K.E. ) from momentum ( p ) and mass ( m ) is:\[ K.E. = \frac{p^2}{2m} \]This equation arises because momentum, defined as the product of mass and velocity ( p = mv ), influences how energy is dissipated as motion.Let's break down this formula:
  • **Numerator: \( p^2 \) ** - Represents the square of momentum, capturing both the magnitude and direction of motion.
  • **Denominator: \( 2m \) ** - Spreads this squared momentum over twice the mass, equating to kinetic energy given mass is in kilograms and velocity in meters per second .
Using our example, when a mass of 2 kg and momentum of 6 \( \text{kg.m/s} \) is substituted in, the kinetic energy calculates as 9 Joules .
Linear Momentum
Linear momentum is an essential concept in physics, closely related to kinetic energy. It is defined as the product of an object's mass and its velocity, reflected in the equation:\[ p = mv \]Here, p is momentum, m is mass, and v is velocity. Linear momentum portrays the energy possessed by a moving object due to its mass and velocity.Let's explore:
  • **Mass Contribution:** Increasing the mass increases momentum when velocity stays constant.
  • **Velocity Contribution:** Higher velocities significantly boost momentum due to the direct proportionality.
In the given problem, a 2 kg object's momentum is 6 \( \text{kg.m/s} \) , informing us of the object's velocity and potential kinetic energy. Knowing how momentum behaves helps in various applications, from sports to vehicle safety.
Mass and Energy Relationship
Understanding the mass and energy relationship helps unveil how objects store and utilize energy. This relationship becomes apparent in both kinetic energy and linear momentum formulations:
  • **Kinetic Energy:** Higher mass generally increases kinetic energy if velocity (and thus momentum) remains unchanged, due to the dependence of kinetic energy on both mass and velocity squared in \( K.E. = \frac{1}{2}mv^2 \).
  • **Energy Representation:** Mass more significantly affects the energy content when objects move at high speeds, amplifying how collisions and movements affect energy distribution.
To further clarify, in our scenario with a 2 kg object, the mass allows us to determine how energy is partitioned in motion. Recognizing this influence aids in designing efficient systems, from mechanical tools to space mission calculations.
Physics Problem Solving Steps
Solving physics problems effectively involves a methodical approach. Whether calculating kinetic energy or determining momentum, following structured steps ensures precision.Here鈥檚 a practical breakdown:
  • **Identify Given Values:** Isolate known quantities from the problem statement鈥攎ass, velocity, momentum.
  • **Understand Formulas Involved:** Familiarize with relevant equations, such as kinetic energy \( K.E = \frac{p^2}{2m} \) or momentum \( p = mv \).
  • **Perform Substitutions:** Insert identified values into chosen formulas, performing arithmetic or algebraic manipulations as needed.
  • **Verify Against Options:** Compare calculated results with provided solutions or expected theories to confirm accuracy.
In our example, laying out each step鈥攆rom parsing given values to validating against multiple-choice answers (like confirming 9 Joules )鈥攈elps in attaining confidence when tackling diverse physics scenarios. Embracing structured problem-solving enhances critical reasoning, essential for academic success and real-world applications.

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Most popular questions from this chapter

Object 1 moves with an initial speed of \(v_{0}\) toward Object \(2,\) which has a mass half that of Object 1. If the final speed of both objects after colliding is o, what must have been the initial speed of Object 2 (assuming no external forces)? (A) \(v_{\mathrm{o}}\) (B) 2\(v_{\mathrm{o}}\) (C) \(\frac{1}{2} v_{0}\) (D) 4\(v_{0}\)

A box with a mass of 2 \(\mathrm{kg}\) accelerates in a straight line from 4 \(\mathrm{m} / \mathrm{s}\) to 8 \(\mathrm{m} / \mathrm{s}\) due to the application of a force whose duration is 0.5 \(\mathrm{s}\) . Find the average strength of this force. (A) 4 \(\mathrm{N}\) (B) 8 \(\mathrm{N}\) (C) 12 \(\mathrm{N}\) (D) 16 \(\mathrm{N}\)

Object 1 moves toward Object \(2,\) whose mass is twice that of Object 1 and which is initially at rest. After their impact, the objects lock together and move with what fraction of Object 1 s initial kinetic energy? (A) 1\(/ 18\) (B) 1\(/ 9\) (C) 1\(/ 6\) (D) 1\(/ 3\)

A wooden block of mass \(M\) is moving at speed \(V\) in a straight line. How fast would the bullet of mass \(m\) need to travel to stop the block (assuming that the bullet became embeded inside)? (A) \(m V /(m+M)\) (B) \(m V / M\) (C) \(M V / m\) (D) \((m+M) V / m\)

Two objects move toward each other, collide, and separate. If there was no net external force acting on the objects, but some kinetic energy was lost, then (A) the collision was elastic and total linear momentum was conserved (B) the collision was elastic and total linear momentum was not conserved (C) the collision was not elastic and total linear momentum was conserved (D) the collision was not elastic and total linear momentum was not conserved
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