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Cart \(A\) has a mass of \(5 \mathrm{~kg}\) and is moving in the \(+x\) direction at \(2 \mathrm{~m} / \mathrm{s}\). Cart \(B\) has a mass of \(2 \mathrm{~kg}\) and is moving in the \(+y\) direction at \(5 \mathrm{~m} / \mathrm{s}\). (a) Do the two carts have the same momentum? Explain. (b) Is the magnitude of the momentum of each cart the same? Explain. (c) Is the kinetic energy of each cart the same? Explain.

Short Answer

Expert verified
(a) No, the carts have different momentum vectors. (b) Yes, magnitudes are equal. (c) No, kinetic energies differ.

Step by step solution

01

Calculate Momentum of Cart A

The momentum of Cart A is calculated using the formula: \( p_A = m_A \times v_A \), where \( m_A = 5 \text{ kg} \) and \( v_A = 2 \text{ m/s} \). Therefore, \( p_A = 5 \times 2 = 10 \text{ kg m/s} \). Since Cart A is moving in the \(+x\) direction, the momentum vector is \( \mathbf{p_A} = (10, 0) \text{ kg m/s} \).
02

Calculate Momentum of Cart B

The momentum of Cart B is calculated using the formula: \( p_B = m_B \times v_B \), where \( m_B = 2 \text{ kg} \) and \( v_B = 5 \text{ m/s} \). Therefore, \( p_B = 2 \times 5 = 10 \text{ kg m/s} \). Since Cart B is moving in the \(+y\) direction, the momentum vector is \( \mathbf{p_B} = (0, 10) \text{ kg m/s} \).
03

Compare Momentum Vectors

Comparing the two momentum vectors \( \mathbf{p_A} = (10, 0) \) and \( \mathbf{p_B} = (0, 10) \), they have the same magnitude but point in different directions. Thus, the vectors are not the same, and the carts do not have the same momentum.
04

Calculate the Magnitude of Momentum for Each Cart

The magnitude of the momentum of Cart A is \( |\mathbf{p_A}| = \sqrt{10^2 + 0^2} = 10 \text{ kg m/s} \).\ The magnitude of the momentum of Cart B is \( |\mathbf{p_B}| = \sqrt{0^2 + 10^2} = 10 \text{ kg m/s} \).\ Since both magnitudes are 10 kg m/s, the magnitude of the momentum of each cart is the same.
05

Calculate Kinetic Energy of Cart A

The kinetic energy of Cart A is given by \( KE_A = \frac{1}{2} m_A v_A^2 \). Plug in the values \( m_A = 5\text{ kg} \) and \( v_A = 2\text{ m/s} \), yielding \( KE_A = \frac{1}{2} \times 5 \times 2^2 = 10\text{ J}\).
06

Calculate Kinetic Energy of Cart B

The kinetic energy of Cart B is given by \( KE_B = \frac{1}{2} m_B v_B^2 \). Plug in the values \( m_B = 2\text{ kg} \) and \( v_B = 5\text{ m/s} \), yielding \( KE_B = \frac{1}{2} \times 2 \times 5^2 = 25\text{ J}\).
07

Compare Kinetic Energies

Compare \( KE_A = 10\text{ J} \) with \( KE_B = 25\text{ J} \). Since the kinetic energies are different, each cart does not have the same kinetic energy.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a measure of the energy an object possesses due to its motion. It is calculated using the formula: \[ KE = \frac{1}{2} m v^2 \]where \(m\) is the mass of the object and \(v\) is its velocity. For Cart A with a mass of \(5\, \mathrm{kg}\) and velocity of \(2\, \mathrm{m/s},\) the kinetic energy can be calculated as \(10\, \mathrm{J}\). Similarly, for Cart B with a mass of \(2\, \mathrm{kg}\) and a velocity of \(5\, \mathrm{m/s},\) the kinetic energy is \(25\, \mathrm{J}\). While the velocities and masses differ, it's evident that the greater velocity of Cart B drastically impacts its kinetic energy, making it significantly higher than that of Cart A. This disparity highlights how the kinetic energy is not just dependent on the mass or speed alone but on how these two factors combine.
Momentum Comparison
Momentum in physics is defined as the product of an object's mass and its velocity. The formula is simple:\[ p = m \, v \]where \(p\) is momentum, \(m\) is mass, and \(v\) is velocity. Both Cart A and B have a momentum value of \(10\, \mathrm{kg \cdot m/s}\). However, momentum is a vector quantity, meaning it has both magnitude and direction. While the scalar magnitudes of momentum for both Carts are equal, the directions differ. Cart A moves along the \(+x\) direction, resulting in a vector of \((10, 0)\). Cart B, moving along the \(+y\) axis, has a momentum vector of \((0, 10)\). Comparing these vectors reveals that they point in orthogonal directions, meaning the total momentum involves not just comparing magnitudes but also considering the direction allowing us to analyze the momentum difference beyond just numbers.
Vector Analysis
Vector analysis in physics helps us understand quantities that have both magnitude and direction, such as momentum and forces. When dealing with vectors, it’s crucial to consider not just the size but also the direction in which they act.
  • Each cart's momentum is represented as a vector in 2D space: \(\mathbf{p_A} = (10, 0)\) and \(\mathbf{p_B} = (0, 10)\).
  • These vector components reveal that although both carts possess momentum, their vectors are different due to their orthogonal directions in space.
Consequently, vector analysis aids in comparing objects moving along different axes. Even if they have equal momentum magnitudes, their real-world effects can be vastly different based on direction. Understanding this concept furthers the comprehension of how forces and movements interact in a multidimensional environment.
Physics Problem-Solving
Solving physics problems often involves understanding and simultaneously applying various concepts such as force, motion, energy, and vectors. In this problem:
  • We start with identifying what needs solving - in this case, momentum and energy.
  • Breaking down the problem into smaller steps helps: calculate momentum, find vectors, compare them, and then calculate kinetic energies.
Using a structured approach not only clarifies each component of the problem but also aids in detecting errors in the process. A clear understanding of ideas like vectors, kinetic energy, and their real-world implications is fundamental. Learning to handle these concepts and problems effectively makes physics both fascinating and approachable, equipping students with strategies for tackling complex scenarios.

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Most popular questions from this chapter

A ball with a mass of \(0.600 \mathrm{~kg}\) is initially at rest. It is struck by a second ball having a mass of \(0.400 \mathrm{~kg}\), initially moving with a velocity of \(0.250 \mathrm{~m} / \mathrm{s}\) toward the right along the \(x\) axis. After the collision, the \(0.400 \mathrm{~kg}\) ball has a velocity of \(0.200 \mathrm{~m} / \mathrm{s}\) at an angle of \(36.9^{\circ}\) above the \(x\) axis in the first quadrant. Both balls move on a frictionless, horizontal surface. (a) What are the magnitude and direction of the velocity of the \(0.600 \mathrm{~kg}\) ball after the collision? (b) What is the change in the total kinetic energy of the two balls as a result of the collision?

To protect their young in the nest, peregrine falcons will fly into birds of prey (such as ravens) at high speed. In one such episode, a \(600 \mathrm{~g}\) falcon flying at \(20.0 \mathrm{~m} / \mathrm{s}\) flew into a \(1.5 \mathrm{~kg}\) raven flying at \(9.0 \mathrm{~m} / \mathrm{s}\). The falcon hit the raven at right angles to its original path and bounced back with a speed of \(5.0 \mathrm{~m} / \mathrm{s} .\) By what angle did the falcon change the raven's direction of motion?

Tennis players sometimes leap into the air to return a volley. (a) If a \(57 \mathrm{~g}\) tennis ball is traveling horizontally at \(72 \mathrm{~m} / \mathrm{s}\) (which does occur), and a \(61 \mathrm{~kg}\) tennis player leaps vertically upward and hits the ball, causing it to travel at \(45 \mathrm{~m} / \mathrm{s}\) in the reverse direction, how fast will her center of mass be moving horizontally just after hitting the ball? (b) If, as is reasonable, her racket is in contact with the ball for \(30.0 \mathrm{~ms}\), what force does her racket exert on the ball? What force does the ball exert on the racket?

You (mass \(55 \mathrm{~kg}\) ) are riding your frictionless skateboard (mass \(5.0 \mathrm{~kg}\) ) in a straight line at a speed of \(4.5 \mathrm{~m} / \mathrm{s}\) when a friend standing on a balcony above you drops a \(2.5 \mathrm{~kg}\) sack of flour straight down into your arms. (a) What is your new speed, while holding the flour sack? (b) since the sack was dropped vertically, how can it affect your horizontal motion? Explain. (c) Suppose you now try to rid yourself of the extra weight by throwing the flour sack straight up. What will be your speed while the sack is in the air? Explain.

Biomechanics. The mass of a regulation tennis ball is \(57.0 \mathrm{~g}\) (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for \(30 \mathrm{~ms}\). (This number can also vary, depending on the racket and swing.) We assume a \(30.0 \mathrm{~ms}\) contact time in this problem. In the 2011 Davis Cup competition, Ivo Karlovic made one of the fastest recorded serves in history, which was clocked at \(156 \mathrm{mph}(70 \mathrm{~m} / \mathrm{s}) .\) (a) What impulse and what average force did Karlovic exert on the tennis ball in his record serve? (b) If his opponent returned this serve with a speed of \(55.0 \mathrm{~m} / \mathrm{s},\) what impulse and what average force did his opponent exert on the ball, assuming purely horizontal motion?

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