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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
Father's original speed is 2 m/s, son's is 2.83 m/s.

Step by step solution

01

Understand the Given Information

The father's kinetic energy initially is half that of the son's kinetic energy, and the father's mass is double the son's mass. When the father speeds up by 1.0 m/s, his kinetic energy equals the son's kinetic energy.
02

Set Up Initial Equations

Let the mass of the son be \(m\) and the velocity of the son be \(v_s\). Then the mass of the father is \(2m\), and his initial velocity is \(v_f\). The kinetic energy of the son is \(\frac{1}{2}m v_s^2\), and the initial kinetic energy of the father is \(\frac{1}{2} \times \frac{1}{2}m v_s^2 = \frac{1}{4}m v_s^2\).
03

Develop the Father's Kinetic Energy Equation

Since the father has half the kinetic energy of the son, we can establish that:\[ \frac{1}{2}(2m)v_f^2 = \frac{1}{4}mv_s^2 \]This simplifies to:\[ 2m v_f^2 = mv_s^2 \]Therefore:\[ v_f^2 = \frac{1}{2}v_s^2 \]
04

Calculate Change in Velocity for the Father

The father speeds up by 1.0 m/s, so his new velocity is \((v_f + 1)\). The new kinetic energy is equal to the son's kinetic energy:\[ \frac{1}{2} (2m)(v_f + 1)^2 = \frac{1}{2}m v_s^2 \]Simplifying, we have:\[ (v_f + 1)^2 = v_s^2 \]
05

Solve for Velocities

From equation 3, \(v_f^2 = \frac{1}{2}v_s^2\). Since \((v_f + 1)^2 = v_s^2\), substituting \(v_f = \sqrt{\frac{1}{2}}v_s\), we solve:\[ (\sqrt{\frac{1}{2}}v_s + 1)^2 = v_s^2 \]This simplifies to:\[ \frac{1}{2}v_s^2 + 2\sqrt{\frac{1}{2}}v_s + 1 = v_s^2 \]Subtracting \(\frac{1}{2}v_s^2\) from both sides:\[ 2\sqrt{\frac{1}{2}}v_s + 1 = \frac{1}{2}v_s^2 \]
06

Solve Quadratic Equation

This simplies to a quadratic in terms of \(v_s\):\[ v_s^2 - 4\sqrt{\frac{1}{2}}v_s - 2 = 0 \]Solve quadratic using the quadratic formula \(v_s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\sqrt{\frac{1}{2}}\), and \(c = -2\).
07

Explain the Quadratic Formula Solution

Calculate \(b^2 - 4ac = 4\times(\frac{1}{2})\times4 \times 1 + 8 = 16 \), thus solutions will be real. Therefore:\[ v_s = \frac{4\sqrt{\frac{1}{2}} \pm \sqrt{16}}{2} \]Simplifying gives \(v_s\) as either 0 or positive \(4\sqrt{\frac{1}{2}} \approx 2.828\). However, as velocity cannot be zero, the son moves with a speed \(v_s = -b\approx 2.83 \text{ m/s}\).
08

Final Calculation and Verification

With son’s speed \(v_s=2.83 \text{ m/s}\), calculate father's speed using equation:\( v_f = \sqrt{\frac{1}{2}}v_s = 2 \). Increase father’s velocity by 1 m/s, verify father’s speed inward check: father = new son’s kinetic energy equals by plugging same kinetic energy as current grandson.

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Key Concepts

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

Mass and Velocity
In physics, two crucial factors that influence an object's kinetic energy are its mass and velocity. Mass refers to the amount of matter in an object, typically measured in kilograms (kg). Velocity is a bit more dynamic; it represents the speed of an object in a particular direction and is measured in meters per second (m/s). Both mass and velocity are fundamental in determining the kinetic energy of an object.

In the context of our exercise, the father's mass is twice that of his son. This means that if the son's mass is denoted as \(m\), the father's mass would be \(2m\). Despite having a larger mass, the father's kinetic energy is initially less than the son's due to differences in their velocity.

Velocity changes can have a significant impact on the kinetic energy, as seen when the father increases his velocity by 1.0 m/s to match his son's kinetic energy. This illustrates how velocity adjustments play a key role in energy dynamics.
Kinetic Energy Equations
Kinetic energy is a measure of the energy possessed by an object due to its motion. It is given by the equation \( KE = \frac{1}{2}mv^2 \), where \(m\) is the mass and \(v\) is the velocity. This formula indicates that kinetic energy is directly proportional to the mass of an object and the square of its velocity, highlighting the quadratic nature of velocity's influence on energy.

In our problem, the son's kinetic energy can be written as \( \frac{1}{2}mv_s^2 \). The father's initial kinetic energy is given by \( \frac{1}{4}mv_s^2 \), because it’s half that of his son. As the father increases his speed by 1.0 m/s, his new kinetic energy matches that of his son, showing how sensitive kinetic energy is to changes in velocity.

Understanding these equations helps us predict how kinetic energy will change with alterations in mass or velocity, which is fundamental in solving physics problems.
Quadratic Equation
When solving problems involving kinetic energy, especially those with changing velocities, quadratic equations often come into play. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \). It arises in physics when handling relationships where velocity is squared, as in the kinetic energy equation.

In this exercise, to find the velocity of the son, we end up with the quadratic equation \( v_s^2 - 4\sqrt{\frac{1}{2}}v_s - 2 = 0 \). To solve it, we use the quadratic formula: \[ v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a\), \(b\), and \(c\) are constants from the equation. This approach allows us to find the velocity values that satisfy both the initial and changed conditions of kinetic energy.

Solving quadratic equations not only helps in understanding physics problems but also strengthens mathematical skills, which are valuable in various scientific contexts.

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Most popular questions from this chapter

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During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of \(100 \mathrm{~N} / \mathrm{m} .\) If the hose is stretched by \(5.00 \mathrm{~m}\) and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?

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