/*! 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 4 Father Racing Son A father racin... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Father Racing Son A father racing his son has half the kinetic energy of the son, who has half the mass of the father. The father speeds up by \(1.0 \mathrm{~m} / \mathrm{s}\) and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?

Short Answer

Expert verified
The original speed of the father is 1 m/s and the son's speed is \(\sqrt{2}\) m/s.

Step by step solution

01

Express Kinetic Energy

Use the formula for kinetic energy: \[ K = \frac{1}{2} m v^2 \]
02

Set up Relationship Between Father and Son's Kinetic Energy and Mass

Let the mass of the father be \(m_f\) and his speed be \(v_f\). Let the mass of the son be \(m_s\) and his speed be \(v_s\). Given that the son’s mass is half the father's, so \(m_s = \frac{1}{2}m_f\).
03

Write the Father's Kinetic Energy in terms of Son's Kinetic Energy

Since the father’s kinetic energy is half that of the son's, \[ \frac{1}{2} m_f v_f^2 = \frac{1}{2} \left( \frac{1}{2} m_f \right) v_s^2 \]which simplifies to \[ m_f v_f^2 = \frac{1}{2} m_f v_s^2 \]
04

Simplify the Equation for Speeds

Cancel out \(m_f\) from both sides: \[ v_f^2 = \frac{1}{2} v_s^2 \] Taking the square root of both sides: \[ v_f = \frac{v_s}{\sqrt{2}} \]
05

Express Father's Speed After Speeding Up

When the father speeds up by \(1.0 \text{ m/s}\), his new speed is \(v_f + 1\). His kinetic energy now equals the son's kinetic energy:\[ \frac{1}{2} m_f (v_f + 1)^2 = \frac{1}{2} \left( \frac{1}{2}m_f \right) v_s^2 \]
06

Set Up the Equation with Speeds

Since the kinetic energies are equal: \[ m_f (v_f + 1)^2 = \frac{1}{2}m_f v_s^2 \] Cancel out \(m_f\): \[ (v_f + 1)^2 = \frac{1}{2} v_s^2 \]
07

Substitute Father’s Original Speed

Substitute \(v_f = \frac{v_s}{\sqrt{2}}\) into the equation: \[ \left( \frac{v_s}{\sqrt{2}} + 1 \right)^2 = \frac{1}{2} v_s^2 \]
08

Solve for Son's Speed

Expand and simplify: \[ \left( \frac{v_s}{\sqrt{2}} + 1 \right)^2 = \frac{v_s^2}{2} + \sqrt{2} v_s + 1 \] equals \(\frac{1}{2} v_s^2\). Equate and solve for \(v_s\): \[ \sqrt{2} v_s + 1 = 0 \] giving \(v_s = \sqrt{2}\).
09

Solve for Father's Speed

Recall \(v_f = \frac{v_s}{\sqrt{2}}\) with \(v_s = \sqrt{2}\): \[ v_f = \frac{\sqrt{2}}{\sqrt{2}} = 1 \text{ m/s} \].

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 that an object possesses due to its motion. In physics, it is defined by the formula \[ K = \frac{1}{2} m v^2 \]. Here, **K** represents kinetic energy, **m** is the mass of the object, and **v** is the speed of the object. This equation tells us that the kinetic energy of an object rises with the square of its speed, making speed a crucial factor in determining kinetic energy.

Understanding kinetic energy can help solve problems involving movement and forces. Often, kinetic energy changes when an object's speed changes. In our exercise, the father's kinetic energy is half that of the son's, making it important to accurately apply the formula.
mass-speed relationship
In physics, the relationship between an object's mass and its speed directly influences its kinetic energy. This relationship becomes particularly interesting when comparing the kinetic energies of two objects with different masses.

In our problem, we know that:
  • The father's mass (**\(m_f\)**) is twice that of the son's.
  • Thus, the son's mass (**\(m_s\)**) is: \[ m_s = \frac{1}{2} m_f \]
Speed also plays a key role here. Initially, the father has half the kinetic energy of the son even though he has greater mass. This implies that the father's speed must be lower for his kinetic energy to be half the son's.

Mathematically, this is shown through the equation:
  • Father's kinetic energy: \[ K_f = \frac{1}{2} m_f v_f^2 \]
  • Son's kinetic energy: \[ K_s = \frac{1}{2} m_s v_s^2 \]
Simplification of these equations confirms that initially, the father's speed, **\(v_f\)**, is: \[ v_f = \frac{v_s}{\sqrt{2}} \].

Therefore speed and mass are inversely proportionate in affecting kinetic energy in this scenario.
physics equations
Solving physics problems often involves using equations deriving from fundamental principles. For our kinetic energy puzzle:
  • We start with the kinetic energy formula: \[ K = \frac{1}{2} m v^2 \]
  • Given relationships like: \[ m_s = \frac{1}{2} m_f \] and \[ K_f = \frac{1}{2} K_s \], we substitute and simplify.
Let's step through a key rearrangement:

We know: \[ \frac{1}{2} m_f v_f^2 = \frac{1}{2} \left( \frac{1}{2} m_f \right) v_s^2 \]
  • Removing constants and masses, we're left with: \[ v_f^2 = \frac{1}{2} v_s^2 \]
Taking the square root results in: \[ v_f = \frac{v_s}{\sqrt{2}} \].

Using this speed **\(v_f\)**, solving the new kinetic equation when the father speeds up helps show the full depth in understanding physics equations:
\[ (v_f + 1)^2 = \frac{1}{2} v_s^2 \]. This results in solving for the son's and father's exact speeds effectively through logical steps.

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

Two Pieces from One An object, with mass \(m\) and speed \(v\) relative to an observer, explodes into two pieces, one three times as massive as the other; the explosion takes place in deep space. The less massive piece stops relative to the observer. How much kinetic energy is added to the system in the explosion, as measured in the observer's reference frame? Hint: Translational momentum is conserved.

Swimmer A swimmer moves through the water at a constant speed of \(0.22 \mathrm{~m} / \mathrm{s}\). The average drag force opposing this motion is \(110 \mathrm{~N}\). What average power is required of the swimmer?

Luge Rider A luge and rider, with a total mass of \(85 \mathrm{~kg}\), emerge from a downhill track onto a horizontal straight track with an initial speed of \(37 \mathrm{~m} / \mathrm{s}\). If they slow at a constant rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) (a) what magnitude \(F\) is required for the slowing force, (b) what distance \(d\) do they travel while slowing, and (c) what work \(W\) is done on them by the slowing force? What are (d) \(F,(\mathrm{e}) d\), and (f) \(W\) if the luge and the rider slow at a rate of \(4.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

Large Meteorite vs. TNT On August 10,1972, a large meteorite skipped across the atmosphere above western United States and Canada, much like a stone skipped across water. The accompanying fireball was so bright that it could be seen in the daytime sky (see Fig. \(9-22\) for a similar event). The meteorite's mass was about \(4 \times 10^{6} \mathrm{~kg}\) : its speed was about \(15 \mathrm{~km} / \mathrm{s}\). Had it entered the atmosphere vertically, it would have hit Earth's surface with about the same speed. (a) Calculate the meteorite's loss of kinetic energy (in joules) that would have been associated with the vertical impact. (b) Express the energy as a multiple of the explosive energy of 1 megaton of \(\mathrm{TNT}\), which is \(4.2 \times 10^{15} \mathrm{~J}\). (c) The energy associated with the atomic bomb explosion over Hiroshima was equivalent to 13 kilotons of TNT. To how many Hiroshima bombs would the meteorite impact have been equivalent?

A Locomotive A locomotive with a power capability of \(1.5 \mathrm{MW}\) can accelerate a train from a speed of \(10 \mathrm{~m} / \mathrm{s}\) to \(25 \mathrm{~m} / \mathrm{s}\) in \(6.0 \mathrm{~min}\). (a) Calculate the mass of the train. Find (b) the speed of the train and (c) the force accelerating the train as functions of time (in seconds) during the \(6.0\) min interval. (d) Find the distance moved by the train during the interval.

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.