/*! 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 26 ecp A cannon is mounted on a rai... [FREE SOLUTION] | 91Ó°ÊÓ

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ecp A cannon is mounted on a railroad flatcar, the muzzle elevated to \(30.0^{\circ}\) and pointed in the direction of the track. The cannon fires a \(1.00-\) metric-ton projectile at\(1.00 \mathrm{~km} / \mathrm{s} .\) (a) If the flatcar and cannon together have a mass of \(36.0\) metric tons (not including the projectile), what is the initial recoil speed of the flatcar? (b) In this problem, it appears that momentum in the \(y\) -direction is not conserved. Explain what happens to it.

Short Answer

Expert verified
The initial recoil speed of the flatcar is \(v_f = m_p \cdot v_o \cdot \cos(30^{\circ}) / m_f\). The momentum in the y-direction is actually conserved since the Earth experiences a small downward momentum.

Step by step solution

01

Calculate Horizontal Momentum

Establish the momentum before the cannon fires. This is 0 because the system is at rest. After the cannon fires, the total momentum must remain at zero. Calculate the horizontal momentum of the projectile (towards +x direction) using its mass and velocity: \(p_x = m_p \cdot v_o \cdot \cos(30^{\circ})\) where \(m_p\) is the mass of the projectile and \(v_o\) is the velocity of it.
02

Find Recoil Speed of the Flatcar

Since the total momentum remains constant (0), the momentum of the recoiling flatcar must equal the momentum of the projectile (but in opposite direction). Thus, the flatcar moves in -x direction with a speed \(v_f\) that satisfies: \(m_f \cdot v_f = m_p \cdot v_o \cdot \cos(30^{\circ})\). You can compute \(v_f\) by simply rearranging the equation. Here, \(m_f\) denotes the mass of the flatcar.
03

Analyze Momentum in the Y-Direction

The cannonball is fired at an angle, so it has an initial vertical (upward) momentum \(p_y = m_p \cdot v_o \cdot \sin(30^{\circ})\). However, the flatcar remains on the track, so it doesn't have any vertical momentum. It may seem that the y-direction momentum isn't conserved, but the Earth experiences a tiny downward momentum. The Earth is so massive that its recoil speed is unnoticeable.

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Key Concepts

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

Conservation of Momentum
One of the fundamental principles in physics is the conservation of momentum. It states that if no external forces are acting on a system, its total momentum remains constant over time. This principle is crucial when analyzing systems where objects interact, such as collisions or explosions. In the given exercise, when the cannon fires the projectile, the system of the flatcar, cannon, and projectile is considered isolated, assuming no external forces like friction. Initially, the system is at rest, so the total momentum is zero. After the projectile is fired, to maintain the total momentum at zero, the flatcar must recoil with a momentum equal in magnitude and opposite in direction to the momentum of the projectile.

Understanding momentum as a product of an object's mass and velocity helps us explore interactions within a system. For instance, the heavier flatcar will recoil with a smaller speed compared to the high velocity of the lighter projectile. This inverse relationship is vital and is mathematically represented as \(m_f\cdot v_f = m_p\cdot v_o\cdot\cos(30^\circ)\), a formula derived from the law of conservation of momentum.

This concept has profound implications in everyday phenomena, from playing pool to understanding car crashes, and is a cornerstone in predicting the outcome of dynamic systems.
Projectile Motion
Projectile motion describes the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The movement of a projectile is a two-dimensional motion that can be broken into horizontal (\(x\)) and vertical (\(y\)) components. To analyze projectile motion, one typically decomposes the motion into these components since, according to Newton's first law of motion, the horizontal and vertical motions are independent of each other.

In the context of our exercise, the projectile is launched at an angle, so its initial velocity has both horizontal and vertical components. The horizontal component, calculated using \(v_o\cdot\cos(30^\circ)\), affects the recoil speed of the flatcar and is governed by the conservation of horizontal momentum. Meanwhile, the vertical component \(v_o\cdot\sin(30^\circ)\) is responsible for the projectile's ascent and descent, influenced by gravity, but does not affect the flatcar's recoil since the flatcar cannot move vertically.

By understanding these components of projectile motion, we can predict the trajectory of the projectile and the recoil of the firing mechanism, whether it's a cannon or a rifle, highlighting the importance of angular measurements and initial velocities in determining the behavior of projectiles.
Momentum Conservation in Two Dimensions
When dealing with two-dimensional motions, such as in the case of a cannon firing a projectile, we must consider the conservation of momentum along both horizontal and vertical axes separately. In two dimensions, the conservation of momentum implies that the momentum components in each direction are conserved independently, assuming no external forces act upon the system in any of those directions.

The exercise provided exemplifies this concept. Horizontally, the momentum is transferred from the projectile to the flatcar in the opposite direction, maintaining the overall horizontal momentum at zero. Vertically, the initial vertical momentum of the projectile might seem to violate conservation of momentum since the flatcar does not move up or down. However, the Earth itself receives the counteracting momentum. While this change in Earth’s momentum is imperceptible due to its massive size, it ensures that total vertical momentum is also conserved.

It's this principle that allows us to tackle complex movements, such as those seen in various sports, vehicular dynamics, and even astronomical events. By applying momentum conservation in two dimensions, we get a comprehensive grasp of how objects interact and influence each other's motion across a plane, providing a solid foundation for predicting the final outcome of two-dimensional events.

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Most popular questions from this chapter

The front \(1.20 \mathrm{~m}\) of a \(1400-\mathrm{kg}\) car is designed as a "crumple zone" that collapses to absorb the shock of a collision. If a car traveling \(25.0 \mathrm{~m} / \mathrm{s}\) stops uniformly in \(1.20 \mathrm{~m}\), (a) how long does the collision last, (b) what is the magnitude of the average force on the car, and (c) what is the acceleration of the car? Express the acceleration as a multiple of the acceleration of gravity.

GP A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass \(m_{1}=48.0 \mathrm{~kg}\) travels in the positive \(x\) -direction at \(12.0 \mathrm{~m} / \mathrm{s}\), and a second piece of mass \(m_{2}=62.0 \mathrm{~kg}\) travels in the \(x y\) -plane at an angle of \(105^{\circ}\) at \(15.0 \mathrm{~m} / \mathrm{s}\). The third piece has mass \(m_{\mathrm{g}}=112 \mathrm{~kg}\). (a) Sketch a diagram of the situation, labeling the different masses and their velocities. (b) Write the general expression for conservation of momentum in the \(x\) - and \(y\) -directions in terms of \(m_{1}, m_{2}, m_{3}, v_{1}, v_{2}\), and \(v_{3}\) and the sines and cosines of the angles, taking \(\theta\) to be the unknown angle. (c) Calculate the final \(x\) -components of the momenta of \(m_{1}\) and \(m_{2}\). (d) Calculate the final \(y\) -components of the momenta of \(m_{1}\) and \(m_{2} .\) (e) Substitute the known momentum components into the general equations of momentum for the \(x\) - and \(y\) -directions, along with the known mass \(m_{3} .\) (f) Solve the two momentum equations for \(v_{3} \cos \theta\) and \(v_{3} \sin \theta\), respectively, and use the identity \(\cos ^{2} \theta+\sin ^{2} \theta=1\) to obtain \(v_{3}\). (g) Divide the equation for \(v_{3} \sin \theta\) by that for \(v_{3} \cos \theta\) to obtain \(\tan \theta\), then obtain the angle by taking the inverse tangent of both sides. (h) In general, would three such pieces necessarily have to move in the same plane? Why?

A billiard ball rolling across a table at \(1.50 \mathrm{~m} / \mathrm{s}\) makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision (a) when the second ball is initially at rest, (b) when the second ball is moving toward the first at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\), and (c) when the second ball is moving away from the first at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\)

ecp A railroad car of mass \(M\) moving at a speed \(v_{1}\) collides and couples with two coupled railroad cars, each of the same mass \(M\) and moving in the same direction at a speed \(v_{2}\). (a) What is the speed \(v_{f}\) of the three coupled cars after the collision in terms of \(v_{1}\) and \(v_{2} ?\) (b) How much kinetic energy is lost in the collision? Answer in terms of \(M, v_{1}\), and \(v_{2}\).

A \(20.0\) -kg toboggan with \(70.0\) -kg driver is sliding down a frictionless chute directed \(30.0^{\circ}\) below the horizontal at \(8.00 \mathrm{~m} / \mathrm{s}\) when a \(55.0-\mathrm{kg}\) woman drops from a tree limb straight down behind the driver. If she drops through a vertical displacement of \(2.00 \mathrm{~m}\), what is the subsequent velocity of the toboggan immediately after impact?

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