/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A \(2140 \mathrm{~kg}\) railroad... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 34

A \(2140 \mathrm{~kg}\) railroad flatcar, which can move with negligible friction, is motionless next to a platform. A \(242 \mathrm{~kg}\) sumo wrestler runs at \(5.3 \mathrm{~m} / \mathrm{s}\) along the platform (parallel to the track) and then jumps onto the flatcar. What is the speed of the flatcar if he then (a) stands on it, (b) runs at \(5.3 \mathrm{~m} / \mathrm{s}\) relative to the flatcar in his original direction, and (c) turns and runs at \(5.3 \mathrm{~m} / \mathrm{s}\) relative to the flatcar opposite his original direction?

Short Answer

Expert verified
(a) 0.538 m/s, (b) 1.137 m/s, (c) 0.061 m/s

Step by step solution

01

Understand Conservation of Momentum

The total momentum of the system (sumo wrestler + flatcar) must be conserved before and after the sumo wrestler jumps onto the flatcar. Use the principle of conservation of momentum, which states that the total momentum before an event is equal to the total momentum after the event.
02

Define Variables

Let: \( m_w = 242 \text{ kg} \) (mass of the sumo wrestler), \( m_c = 2140 \text{ kg} \) (mass of the flatcar), \( v_w = 5.3 \text{ m/s} \) (velocity of the sumo wrestler relative to the ground), \( v_c \) (velocity of the flatcar after the sumo wrestler jumps on).
03

Set Up Initial Momentum Equation

Initially, the flatcar is motionless. Thus, the initial momentum of the system is only due to the sumo wrestler: \[ p_{\text{initial}} = m_w \times v_w \]
04

Calculate Initial Momentum

Substitute the values: \[ p_{\text{initial}} = 242 \text{ kg} \times 5.3 \text{ m/s} = 1282.6 \text{ kg} \text{ m/s} \]
05

Set Up Final Momentum Equations for Each Scenario

Consider three cases to find \( v_c \): (a) sumo wrestler stands, (b) sumo wrestler runs in the same direction as initially, (c) sumo wrestler runs in the opposite direction.
06

Case (a) - Sumo Wrestler Stands

When the sumo wrestler stands on the flatcar, they move together with a common velocity \( v_c \). Using conservation of momentum: \[ m_w \times v_w = (m_w + m_c) \times v_c \], solve for \( v_c \): \[ v_c = \frac{m_w \times v_w}{m_w + m_c} \]. Substituting the values: \[ v_c = \frac{242 \text{ kg} \times 5.3 \text{ m/s}}{242 \text{ kg} + 2140 \text{ kg}} = \frac{1282.6 \text{ kg} \text{ m/s}}{2382 \text{ kg}} = 0.538 \text{ m/s} \].
07

Case (b) - Sumo Wrestler Runs at 5.3 m/s Relative to Flatcar in Original Direction

Here, the final velocities are such that, relative to the ground, the sumo wrestler has a velocity of \( v_w + v_c \). Using conservation of momentum: \[ 1282.6 \text{ kg} \text{ m/s} = m_w \times (v_w + v_c) + m_c \times v_c \], solving for \( v_c \): \[ v_c = \frac{m_w \times v_w}{m_w + m_c} + \frac{m_w \times v_w}{m_c} = \frac{1282.6}{2382} + \frac{1282.6}{2140} = 0.538 + 0.599 = 1.137 \text{ m/s} \].
08

Case (c) - Sumo Wrestler Runs at 5.3 m/s in Opposite Direction

Now, the final velocities relative to the ground: sumo wrestler's final velocity = \( v_c - v_w \). Use conservation of momentum: \[ 1282.6 \text{ kg} \text{ m/s} = m_w \times (v_c - v_w) + m_c \times v_c \], solving for \( v_c \): \[ v_c = \frac{m_w \times v_w}{m_c} - \frac{m_w \times v_w}{m_w + m_c} = \frac{1282.6}{2140} - 0.538 = 0.599 - 0.538 = 0.061 \text{ m/s} \].

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Key Concepts

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

momentum
Momentum is the product of an object's mass and its velocity. It is a measure of how much motion an object has and is represented by the formula: \( p = m \times v \). Here, \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. Momentum is a vector quantity, meaning it has both magnitude and direction. In the given exercise, the momentum of the system involving the sumo wrestler and the flatcar is calculated at various scenarios using their masses and velocities.
physics
Physics is the branch of science concerned with the nature and properties of matter and energy. It uses fundamental principles to explain natural phenomena. In this exercise, the principle of conservation of momentum from physics is applied. The total momentum before and after the sumo wrestler jumps onto the flatcar remains the same, illustrating a common physical principle.
inelastic collisions
An inelastic collision is a type of collision where part of the kinetic energy is transformed into other forms of energy, meaning objects may stick together or deform. In the exercise, when the sumo wrestler stands on the flatcar, it is an example of an inelastic collision. The wrestler and flatcar move together with a common velocity after the collision. This is different from an elastic collision where objects bounce off each other without losing kinetic energy.
relative velocity
Relative velocity describes the velocity of an object as observed from another moving object's frame of reference. It is crucial when analyzing the motion in our exercise. For instance, when the sumo wrestler runs on the flatcar, his velocity relative to the flatcar affects the overall final speed. If he runs at 5.3 m/s relative to the flatcar in his original direction, it means his total speed relative to the ground is the sum of his running speed and flatcar鈥檚 speed.
motion equations
Motion equations are mathematical formulations used to study and predict the positions, velocities, and accelerations of moving objects. In this exercise, we apply equations derived from the conservation of momentum formula: \( p_{initial} = p_{final} \). Each part of the problem requires setting up and solving different motion equations based on different scenarios, such as the sumo wrestler standing still or running in the same or opposite direction. Each equation's solution gives us the specific flatcar velocity.

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Most popular questions from this chapter

A big olive \((m=0.50 \mathrm{~kg})\) lies at the origin and a big Brazil nut \((M=1.5 \mathrm{~kg})\) lies at the point \((1.0,2.0) \mathrm{m}\) in an \(x y\) plane. At \(t_{1}=0\), a force \(\vec{F}_{o}=(2.0 \mathrm{~N}) \hat{\mathrm{i}}+(3.0 \mathrm{~N}) \hat{\mathrm{j}}\) begins to act on the olive, and a force \(\vec{F}_{n}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}+(-2.0 \mathrm{~N}) \hat{\mathrm{j}}\) begins to act on the nut. In unit-vector notation, what is the displacement of the center of mass of the olive-nut system at \(t_{2}=4.0 \mathrm{~s}\), with respect to its position at \(t_{1}=0\) ?

Two identical containers of sugar are connected by a massless cord that passes over a massless, frictionless pulley with a diameter of \(50 \mathrm{~mm}\) (Fig. 8-27). The two containers are at the same level. Each originally has a mass of 500 g. (a) What is the horizontal position of their center of mass? (b) Now \(20 \mathrm{~g}\) of sugar is transferred from one container to the other, but the containers are prevented from moving. What is the new horizontal position of their center of mass, relative to the central axis through the lighter container? (c) The two containers are now released. In what direction does the center of mass move? (d) What is its acceleration?

Ricardo, of mass \(80 \mathrm{~kg}\), and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a \(30 \mathrm{~kg}\) canoe. When the canoe is at rest in the placid water, they exchange seats, which are \(3.0 \mathrm{~m}\) apart and symmetrically located with respect to the canoe's center. Ricardo notices that the canoe moves \(40 \mathrm{~cm}\) relative to a submerged log during the exchange and calculates Carmelita's mass, which she has not told him. What is it?

In the ammonia \(\left(\mathrm{NH}_{3}\right)\) molecule (see Fig. 8-19), the three hydrogen (H) atoms form an equilateral triangle; the center of the triangle is \(9.40 \times\) \(10^{-11} \mathrm{~m}\) from each hydrogen atom. The nitrogen (N) atom is at the apex of a pyramid, with the three hydrogen atoms forming the base. The nitrogen-to-hydrogen atomic mass ratio is \(13.9\). and the nitrogen-to-hydrogen distance is \(10.14 \times 10^{-11} \mathrm{~m}\). Locate the center of mass of the molecule relative to the nitrogen atom.

A suspicious package is sliding on a frictionless surface when it explodes into three pieces of equal masses and with the velocities (1) \(7.0 \mathrm{~m} / \mathrm{s}\), north, (2) \(4.0 \mathrm{~m} / \mathrm{s}, 30^{\circ}\) south of west, and (3) \(4.0 \mathrm{~m} / \mathrm{s}, 30^{\circ}\) south of east. (a) What is the velocity (magnitude and direction) of the package before it explodes? (b) What is the displacement of the center of mass of the three- piece system (with respect to the point where the explosion occurs) \(3.0 \mathrm{~s}\) after the explosion?
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