If one wanted to calculate the velocity of the ramp as the box slid off, one would need to do which of the following? i. Equate the momentum of the box plus the momentum of the ramp to zero ii. Equate the kinetic energy of the crate to \(m g L\) iii. Equate the kinetic energy of the crate plus ramp to \(m g L\) a. i b. ii c. iii d. i and ii e. i and iii

Short Answer

Expert verified
a) The momentum of the box plus the momentum of the ramp should be zero. b) The kinetic energy of the box should be equal to m * g * L. c) The kinetic energy of the box plus the ramp should be equal to m * g * L. e) i and iii Answer: e) i and iii

Step by step solution

01

Review the law of conservation of momentum

According to the law of conservation of momentum, the total momentum of a closed system of objects remains constant before and after any interaction between the objects. As the box and ramp are interacting, we can apply this principle to the system.
02

Review the law of conservation of energy

According to the law of conservation of energy, the total energy of a closed system remains constant. If the kinetic energy of the crate plus the ramp should be equal to \(m g L\), then this condition signifies that mechanical energy is conserved in this system.
03

Analyze option i

This option suggests equating the momentum of the box plus the momentum of the ramp to zero. This is accurate as the total momentum of the system (box and ramp) remains conserved before and after the interaction. Hence, option i is correct.
04

Analyze option ii

This option suggests equating the kinetic energy of the crate to \(m g L\). However, this does not consider the energy or momentum of the ramp in any way and does not help in calculating the velocity of the ramp. Thus, option ii is not correct.
05

Analyze option iii

This option suggests equating the kinetic energy of the crate plus the ramp to \(m g L\). In this case, the total mechanical energy of the system is being conserved. Therefore, option iii is also correct for solving the problem.
06

Determine the correct answer

Based on the analysis of the given options, both options i and iii are correct. Thus, the answer to this problem is (e) i and iii.

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Key Concepts

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

Law of Conservation of Momentum
The law of conservation of momentum is a fundamental principle in physics that states that the total momentum of an isolated system remains constant if there is no external force acting on it. Momentum itself is the product of an object's mass and its velocity, and is a vector quantity, meaning it has both magnitude and direction.

In the classroom or during SAT or AP physics preparations, this concept is vital when studying collisions or interactions between objects. Imagine a skater pushing off from a wall or two bumper cars colliding; in both scenarios, the total momentum before the event is equal to the total momentum afterwards. This means if you are calculating velocities after an interaction, like a box sliding off a ramp, you'd apply this law to ensure that the momentum with the ramp and the box combined before sliding equals their momentum after sliding, regardless of the direction in which they move.
Kinetic Energy Calculation
Kinetic energy is the energy an object possesses because of its motion. Calculating kinetic energy is crucial for problems involving motion and is given by the formula: \( KE = \frac{1}{2}mv^{2} \), where \(m\) is mass and \(v\) is velocity of the object.

Understanding how to compute kinetic energy is beneficial for students preparing for standardized tests like the SAT or AP exams, where they may encounter scenarios requiring them to connect how objects' speeds translate into energy. By mastering kinetic energy calculations, students can solve a variety of physics problems, including those related to work-energy principles and mechanical energy conservation.
Mechanical Energy Conservation
Conservation of mechanical energy is an essential principle dictating that the total mechanical energy in an isolated system—sum of potential energy and kinetic energy—remains constant if only conservative forces, like gravity, are doing work.

No mechanical energy is lost when only conservative forces are involved; it simply transforms from one type to another, such as potential becoming kinetic. In the case of a box sliding down a ramp, if we assume no frictional losses, the gravitational potential energy lost by the box (dependent on its height and weight) is converted into kinetic energy as it slides down, which can be calculated using the formula given in the kinetic energy section.
SAT Physics Preparation
SAT Physics subject test preparation involves a deep understanding of fundamental physics concepts, including the law of conservation of momentum and energy. High school students preparing for the SAT should practice problems that integrate these principles, such as calculating the velocities of objects after collisions or determining possible energy transformations in various systems.

Effective SAT prep should also incorporate strategies for eliminating incorrect answer choices and for mastering the content-specific vocabulary. To gain proficiency, students might tackle practice problems that simulate the test – time-constrained and varying in difficulty – and thoroughly review solutions to understand each step taken.
AP Physics Concepts
AP Physics courses cover a wide range of topics, such as mechanics, electricity, magnetism, and waves. Important concepts like the conservation of momentum and energy are cornerstones of the AP curriculum. Students aiming for success on the AP exams must be able to apply these principles across multiple contexts, from simple systems in mechanics to complex applications in electromagnetism and beyond.

In particular, understanding and working with energy and momentum conservation allows students to analyze isolated systems and predict outcomes as they tackle problems of varying complexity. Strengthening these foundational concepts can lead to better problem-solving skills, essential for achieving higher scores on the AP Physics exams.

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Most popular questions from this chapter

Model rocket engines burn for only a short time before using up all their propellant. Suppose a model rocket is launched from its stand at an angle \(60^{\circ}\) above the horizontal. The mass of the rocket is \(0.25 \mathrm{~kg}\) and its final speed is \(50 \mathrm{~m} / \mathrm{s}\). If the engine burns for \(1.25 \mathrm{~s}\), the impulse it gives to the rocket is most nearly (A) \(15.6 \mathrm{~N} \mathrm{~s}\) (B) \(12.5 \mathrm{~N} \mathrm{~s}\) (C) \(6.25 \mathrm{~N} \mathrm{~s}\) (D) \(\frac{\sqrt{3}}{2} \times 12.5 \mathrm{~N} \mathrm{~s}\) (E) \(12.5 \mathrm{~N}\)

A crate of mass \(m\) sits atop a frictionless ramp that has a height and length \(L\). The ramp has mass \(2 m\), which is concentrated at the lower left corner, and it sits on a frictionless ice pond. The crate is released and slides down the ramp. By the instant that the crate reaches the bottom, how far has the ramp moved and in what direction? (A) It hasn't moved. (B) It has moved to the left a distance \(L / 3\). (C) It has moved to the right a distance \(L / 3\). (D) It has moved to the left a distance \(L / 2\). (E) It has moved to the right a distance \(L / 2\).

cant copy graph Two students, Alice and Bob, decide to compute the power that the Earth's gravitational field expends on a block of mass \(m\) as the block slides down a frictionless inclined plane. Alice reasons: "The gravitational force pulling the block down the incline is \(F=m g \sin \theta\). The block's velocity at any given height \(h\) from the top of the incline is \(v=\sqrt{2 g h}\). Power is defined as force \(\times\) velocity. Therefore, the power is \(P=m g \sin \theta \sqrt{2 g h}\)." Bob reasons: "Power is \(\Delta W / \Delta t\). By the work-energy theorem, the change in work is the change in kinetic energy, but without friction \(\Delta W=m g h\). The change in time is \(\Delta t=\Delta v / a, \Delta v=\sqrt{2 g h}\) and \(a=g \sin \theta\). So \(\Delta t=\sqrt{2 g h} /(g \sin \theta)\). Therefore, \(P=m g h \times \frac{g \sin \theta}{\sqrt{2 g h}}=m g \sin \theta \sqrt{\frac{g h}{2}}\)." Alice and Bob look at each other and scratch their heads. Who is correct? \(\begin{array}{llll}\text { i. Alice } & \text { ii. Bob } & \text { iii. Neither iv. Both } & \text { v. The problem is imprecisely worded. }\end{array}\) (A) \(\mathrm{i}\) (B) ii (C) iii (D) iv (E) iv and \(v\)

A mass \(m\) is attached to a mass \(M\) by a light string that passes over a frictionless pulley of negligible mass, so that it can pull \(M\) up the incline. The incline is frictionless and tilted at an angle of \(\theta=30^{\circ}\) from the horizontal. \(M\) is initially at rest at the bottom of the incline; then \(m\) is released. When \(M\) reaches the top it hits a stop and launches a small ball of negligible mass from a height \(h=1 \mathrm{~m}\) above the starting position. The ball lands on a shelf at the same height at a range \(R=1.8 \sqrt{3} \mathrm{~m}\) from its launch point. Assume the acceleration of gravity is \(g=10 \mathrm{~m} / \mathrm{s}^2\). cant copy graph a) What is the launch speed of the ball? b) What is the acceleration of \(M\) ? c) What is the mass \(M\) in terms of \(m\) ? d) If the ball is given a larger mass, will \(R\) increase, decrease, or stay the same? Explain your reasoning. e) The shelf is removed and the experiment repeated. This time the ball falls to the floor. If \(H=1 \mathrm{~m}\), what is the time the ball spends in the air? f) What is the new range, \(R\) ?

The units of momentum can be expressed as (A) \(\mathrm{N} \cdot \mathrm{S}\) (B) \(\sqrt{\mathrm{kg} \cdot \mathrm{J}}\) (C) \(\sqrt{\mathrm{kg} \cdot \mathrm{W} \cdot \mathrm{s}}\) (D) all of the above (E) none of the above

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