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Chapter 7: Problem 25

Kevin has a mass of \(87 \mathrm{~kg}\) and is skating with in-line skates. He sees his \(22-\mathrm{kg}\) younger brother up ahead standing on the sidewalk, with his back turned. Coming up from behind, he grabs his brother and rolls off at a speed of \(2.4 \mathrm{~m} / \mathrm{s}\). Ignoring friction, find Kevin's speed just before he grabbed his brother.

Short Answer

Expert verified
Kevin's initial speed was approximately \(3.01\text{ m/s}\).

Step by step solution

01

Understand the concept of Conservation of Momentum

The principle of Conservation of Momentum states that the total momentum of a closed system is constant if no external forces act on it. Mathematically, this can be written as \(m_1v_1 + m_2v_2 = (m_1+m_2)v_f\), where \(m_1\) and \(v_1\) are the mass and velocity of the first object, \(m_2\) and \(v_2\) are the mass and velocity of the second object, and \(v_f\) is the final velocity of both objects combined.
02

Identify known values

From the problem, we know Kevin's mass \(m_1 = 87 \text{ kg}\), Kevin's brother's mass \(m_2 = 22 \text{ kg}\), and the final velocity after Kevin has grabbed his brother \(v_f = 2.4 \text{ m/s}\). Since his brother is initially standing still, his initial velocity \(v_2 = 0 \text{ m/s}\).
03

Set up the momentum equation

Using the conservation of momentum \(m_1v_1 + m_2v_2 = (m_1 + m_2)v_f\), substitute the known values. This gives us the equation \(87v_1 + 22 \times 0 = (87 + 22) \times 2.4\).
04

Solve for Kevin's initial velocity \(v_1\)

Simplify the equation: \(87v_1 = 109 \times 2.4\). Solve for \(v_1\) by dividing both sides by 87, \(v_1 = \frac{261.6}{87}\). Calculating this gives \(v_1 = 3.0069\). Kevin's speed just before grabbing his brother was approximately \(3.01\text{ m/s}\).

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Key Concepts

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

Momentum Calculation
The concept of momentum is fundamental in physics and helps us understand motion in terms of mass and velocity. Momentum, simply put, is the product of an object's mass and velocity. It is calculated using the formula:
\[ p = mv \]where:
  • \( p \) is the momentum,
  • \( m \) is the mass, and
  • \( v \) is the velocity of the object.

In problems involving more than one object, such as Kevin and his brother, we apply the conservation of momentum. This principle tells us that the total momentum before an event is equal to the total momentum after, provided no external forces act on the system. For Kevin's skating scenario, the equation would look like:
\[ m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \]
Here, \( m_1 \) and \( m_2 \) represent the masses of Kevin and his brother respectively, \( v_1 \) and \( v_2 \) are their initial velocities, and \( v_f \) is their combined velocity after Kevin catches up with his brother. This helps us solve for unknown velocities by rearranging the equation and substituting known values.
Initial Velocity
Initial velocity is the speed at which an object must be moving before the event or interaction takes place. In Kevin's case, we need to determine how fast he was skating before grabbing his brother.
Using conservation of momentum, we can calculate this unknown initial velocity. Initially, his brother is at rest, so his velocity is zero. By substituting the known values into the formula:
\[ 87v_1 + 22 \times 0 = (87 + 22) \times 2.4 \]
we simplify to find Kevin's initial velocity:
\[ 87v_1 = 109 \times 2.4 \]
Solve for \( v_1 \) by dividing both sides by 87:
  • \( v_1 = \frac{261.6}{87} \)

This gives an initial velocity of approximately \( 3.01 \text{ m/s} \), revealing Kevin was skating at this speed before grabbing his brother.
In-Line Skating Physics
In-line skating is not just a fun activity; it involves fascinating physics principles. Skaters experience concepts like momentum and velocity that dictate their motion on wheels.
Understanding momentum in this context helps explain why Kevin's speed changes once he grabs his younger brother. Without external forces like friction or air resistance considered, any interaction between skaters, like catching up on the sidewalk, involves the exchange and conservation of momentum. Each skater's mass and velocity contribute to the system's total momentum, conserving it as energy transfers between the skaters.
Some other key aspects in skating physics include:
  • Center of Mass: Balancing the body's weight on skates affects motion.
  • Friction: While neglected in this problem, friction generally influences speed and control.
  • Acceleration and Deceleration: Changes in speed require forces applied or balanced by the skater.
Understanding these principles helps skaters improve performance while appreciating the science behind their movements.

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Most popular questions from this chapter

A space probe is traveling in outer space with a momentum that has a magnitude of \(7.5 \times 10^{7} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). A retrorocket is fired to slow down the probe. It applies a force to the probe that has a magnitude of \(2.0 \times 10^{6} \mathrm{~N}\) and a direction opposite to the probe's motion. It fires for a period of 12 s. Determine the momentum of the probe after the retrorocket ceases to fire.

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. (a) Is it possible that the two-object system has a final total momentum of zero after the collision? (b) Roughly, what is the direction of the final total momentum of the two-object system after the collision? Problem Object A has a mass of \(m_{A}=17.0 \mathrm{~kg}\) and an initial velocity of \(\overrightarrow{\mathbf{v}}_{0 \mathrm{~A}}=8.00 \mathrm{~m} / \mathrm{s},\) due east. Object \(\mathrm{B},\) however, has a mass of \(m_{\mathrm{B}}=29.0 \mathrm{~kg}\) and \(\mathrm{an}\) initial velocity of \(\overrightarrow{\mathbf{v}}_{0 \mathrm{~B}}=5.00 \mathrm{~m} / \mathrm{s},\) due north. Find the magnitude and direction of the total momentum of the two-object system after the collision. Make sure that your answers are consistent with your answers to the Concept Questions.

A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. (a) During the downward motion of the ball, are any of the following conserved: its momentum, its kinetic energy, its total mechanical energy? (b) During the collision with the block, are any of the following conserved: the horizontal component of the total momentum of the ball/block system, the total kinetic energy of the system? Provide reasons for your choices.

The drawing shows a human figure approximately in a sitting position. For purposes of this problem, there are three parts to the figure, and the center of mass of each one is shown in the drawing. These parts are: (1) the torso, neck, and head (total mass \(=41 \mathrm{~kg}\) ) with a center of mass located on the \(y\) axis at a point \(0.39 \mathrm{~m}\) above the origin, (2) the upper legs (mass \(=17 \mathrm{~kg}\) ) with a center of mass located on the \(x\) axis at a point \(0.17 \mathrm{~m}\) to the right of the origin, and (3) the lower legs and feet (total mass \(=9.9 \mathrm{~kg}\) ) with a center of mass located \(0.43 \mathrm{~m}\) to the right of and \(0.26 \mathrm{~m}\) below the origin. Find the \(x\) and \(y\) coordinates of the center of mass of the human figure. Note that the mass of the arms and hands (approximately \(12 \%\) of the wholebody mass) has been ignored to simplify the drawing.

At presents a method for modeling this problem. The carbon monoxide molecule (CO) consists of a carbon atom and an oxygen atom separated by a distance of \(1.13 \times 10^{-10} \mathrm{~m}\). The mass \(m_{\mathrm{C}}\) of the carbon atom is 0.750 times the mass \(m_{\mathrm{O}}\) of the oxygen atom, or \(\mathrm{m}_{\mathrm{c}}=0.750 \mathrm{~m}_{0} .\) Determine the location of the center of mass of this molecule relative to the carbon atom.
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