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Stuck in the middle of a frozen pond with only your physics book, you decide to put physics in action and throw the \(5.00-\mathrm{kg}\) book. If your mass is \(62.0 \mathrm{~kg}\) and you throw the book at \(13.0 \mathrm{~m} / \mathrm{s}\), how fast do you then slide across the ice? (Assume the absence of friction.)

Short Answer

Expert verified
Answer: The final speed of the student after throwing the book is approximately 1.05 m/s in the opposite direction of the book's motion.

Step by step solution

01

Identify the initial and final momenta of the system

Initially, both the student and the book are at rest, so the initial momentum of the system is zero: \(p_{initial} = m_{student}v_{student} + m_{book}v_{book} = 0\) After throwing the book, the book gains velocity, and the student gains velocity in the opposite direction. We'll denote the final velocities of the student and the book as \(v'_{student}\) and \(v'_{book}\) respectively.
02

Apply the conservation of momentum principle

Since no external forces are acting on the system, the total momentum of the system must be conserved. In other words, the initial momentum of the system equals its final momentum: \(p_{initial} = p_{final}\) This means: \(0 = m_{student}v'_{student} + m_{book}v'_{book}\)
03

Plug in the known values and solve for the unknown velocity

We know the mass of the student (\(m_{student} = 62.0\mathrm{~kg}\)), the mass of the book (\(m_{book} = 5.00\mathrm{~kg}\)), and the final velocity of the book (\(v'_{book} = 13.0\mathrm{~m}/\mathrm{s}\)). We can now plug these values into the conservation of momentum equation and solve for \(v'_{student}\): \(0 = (62.0\mathrm{~kg})v'_{student} + (5.00\mathrm{~kg})(13.0\mathrm{~m}/\mathrm{s})\) To solve for \(v'_{student}\), first move the second term on the right side to the left side: \((62.0\mathrm{~kg})v'_{student} = -(5.00\mathrm{~kg})(13.0\mathrm{~m}/\mathrm{s})\) Now, divide by the mass of the student to find \(v'_{student}\): \(v'_{student} = \frac{(5.00\mathrm{~kg})(13.0\mathrm{~m}/\mathrm{s})}{62.0\mathrm{~kg}}\)
04

Calculate the final answer

By plugging the numbers into the equation, we get the final velocity of the student: \(v'_{student} = \frac{(5.00\mathrm{~kg})(13.0\mathrm{~m}/\mathrm{s})}{62.0\mathrm{~kg}} \approx 1.05\mathrm{~m}/\mathrm{s}\) So, after throwing the book, the student slides across the ice with a speed of approximately \(1.05\mathrm{~m}/\mathrm{s}\) in the opposite direction of the book's motion.

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics, encapsulating the quantity of motion an object has. Defined by the product of an object's mass and velocity, represented as \( p = mv \), it's a vector quantity, meaning it has both magnitude and direction.

As demonstrated in the frozen pond scenario, when a person throws a book, they apply force to the book, accelerating it and imparting momentum in the process. The book's momentum after throwing is \( p_{book} = m_{book}v'_{book} \), where \( m_{book} \) is the mass of the book and \( v'_{book} \) its velocity post-throw. Understanding the relationship between force, mass, and velocity is crucial for solving problems involving momentum.
Conservation Laws
Conservation laws are pivotal to our understanding of the physical universe, with the conservation of momentum being particularly relevant in closed systems. This principle states that within a closed system, the total momentum remains constant unless acted upon by external forces.

In the example where the book is thrown while on ice, assuming no friction, the system (book plus student) is closed. Thus, the total momentum before and after the book is thrown must be equal. These laws simplify calculations in isolated systems and illustrate the intrinsic symmetry of the physical world.
Physics Problem Solving
Effective problem-solving in physics requires a step-by-step approach: identifying knowns and unknowns, choosing the relevant physics principles, applying equations, and solving for the desired quantities.

In the textbook problem where a student uses a book to propel themselves across ice, the steps demonstrate this methodical approach. By understanding the principles at play, such as conservation of momentum, and carefully manipulating equations, one can solve for unknown variables, such as the student's final velocity after tossing the book. This structured problem-solving technique is a valuable skill in physics and beyond.
Kinematics
Kinematics is the branch of mechanics that focuses on the motion of objects without considering the forces that cause such motion. It involves the analysis of velocity, acceleration, displacement, and time.

In the context of the stranded student, kinematics can explain the subsequent motion of the student and book separately, after considering the initial action. By solving the problem, one can find the velocity of the student, which is a kinematic quantity. Understanding kinematics is essential in predicting how objects will move, which is fundamental in fields ranging from astrophysics to everyday engineering problems.

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Most popular questions from this chapter

For a totally elastic collision between two objects, which of the following statements is (are) true? a) The total mechanical energy is conserved. b) The total kinetic energy is conserved. c) The total momentum is conserved. d) The momentum of each object is conserved. e) The kinetic energy of each object is conserved.

A hockey puck with mass \(0.170 \mathrm{~kg}\) traveling along the blue line (a blue-colored straight line on the ice in a hockey rink) at \(1.50 \mathrm{~m} / \mathrm{s}\) strikes a stationary puck with the same mass. The first puck exits the collision in a direction that is \(30.0^{\circ}\) away from the blue line at a speed of \(0.750 \mathrm{~m} / \mathrm{s}\) (see the figure). What is the direction and magnitude of the velocity of the second puck after the collision? Is this an elastic collision?

Tennis champion Venus Williams is capable of serving a tennis ball at around 127 mph. a) Assuming that her racquet is in contact with the \(57.0-\mathrm{g}\) ball for \(0.250 \mathrm{~s}\), what is the average force of the racquet on the ball? b) What average force would an opponent's racquet have to exert in order to return Williams's serve at a speed of \(50.0 \mathrm{mph}\), assuming that the opponent's racquet is also in contact with the ball for \(0.250 \mathrm{~s} ?\)

A \(1439-\mathrm{kg}\) railroad car traveling at a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed (in \(\mathrm{m} / \mathrm{s}\) ) afterward?

Moessbauer spectroscopy is a technique for studying molecules by looking at a particular atom within them. For example, Moessbauer measurements of iron (Fe) inside hemoglobin, the molecule responsible for transporting oxygen in the blood, can be used to determine the hemoglobin's flexibility. The technique starts with X-rays emitted from the nuclei of \({ }^{57} \mathrm{Co}\) atoms. These X-rays are then used to study the Fe in the hemoglobin. The energy and momentum of each X-ray are \(14.0 \mathrm{keV}\) and \(14.0 \mathrm{keV} / \mathrm{c}\) (see Example 7.5 for an explanation of the units). \(A^{57}\) Co nucleus recoils as an X-ray is emitted. A single \({ }^{57}\) Co nucleus has a mass of \(9.52 \cdot 10^{-26}\) kg. What are the final momentum and kinetic energy of the \({ }^{57}\) Co nucleus? How do these compare to the values for the X-ray?

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