/*! 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 An empty 15000 -kg coal car is c... [FREE SOLUTION] | 91Ó°ÊÓ

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An empty 15000 -kg coal car is coasting on a level track at \(5.00\) \(\mathrm{m} / \mathrm{s}\). Suddenly \(5000 \mathrm{~kg}\) of coal is dumped into it from directly above it. The coal initially has zero horizontal velocity with respect to the ground. Find the final speed of the car.

Short Answer

Expert verified
The final speed of the coal car is 3.75 m/s.

Step by step solution

01

Understand the Problem

We need to find the final speed of a coal car after 5000 kg of coal drops into it. The car initially moves at 5.00 m/s, and the coal has an initial horizontal velocity of 0 m/s. Conservation of momentum will help us solve this problem.
02

Identify the Given Variables

The mass of the empty coal car is 15000 kg, initial speed is 5.00 m/s, and mass of the coal is 5000 kg. The coal has zero initial horizontal velocity.
03

Apply Conservation of Momentum

According to the principle of conservation of momentum, the momentum before the event (dumping coal) should equal the momentum after the event. Therefore, 15000 kg * 5.00 m/s = (15000 kg + 5000 kg) * final velocity.
04

Solve for Final Velocity

Substitute the known values into the momentum equation: \[ 15000 ext{ kg} * 5.00 ext{ m/s} = 20000 ext{ kg} * v_f \] Simplify and solve for \(v_f\): \[ v_f = \frac{15000 ext{ kg} * 5.00 ext{ m/s}}{20000 ext{ kg}} = 3.75 ext{ m/s} \]
05

Conclusion

The final speed of the coal car, after the coal is added and momentum is conserved, is 3.75 m/s.

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Key Concepts

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

Momentum Equation
The momentum equation is a vital tool in physics used for solving problems involving collisions or interactions, like the one where coal is dumped into the moving coal car. Momentum is defined as the product of an object's mass and its velocity, expressed mathematically as \( p = m \times v \). In this exercise, two masses are involved: the moving coal car and the stationary coal.

The principle of conservation of momentum states that in a closed system, the total momentum before an event must equal the total momentum after the event. This equation helps us maintain a balance of forces in isolated systems and provides us with a way to calculate unknown variables, such as the final velocity of the coal car after it collects additional mass.
  • The initial momentum of the system is entirely due to the moving coal car since the coal is at rest.
  • By plugging variables into the equation \( p_{\text{initial}} = p_{\text{final}} \), we can equate the initial moving momentum of the coal car with the combined momentum of car and coal after the coal is added.
Velocity
Velocity is a key concept when considering the motion of objects such as a coal car. In physics, velocity is described as the speed of something in a given direction and is a vector quantity. Here, the coal car initially coasts at 5.00 m/s.

After the coal is dumped, the additional mass affects the velocity of this system. Although the coal has no initial horizontal velocity, when it becomes part of the coal car system, the conservation of momentum dictates that the final velocity would necessarily change to accommodate the new mass distribution.
  • To find the new velocity \( v_f \), we use the momentum equation to ensure momentum is conserved.
  • The solution gives us the speed, but keep in mind direction remains unchanged since all initial motion was horizontal.
Mass
Mass is an important factor in the momentum equation and significantly influences the final speed of the coal car in this scenario. Mass, being a scalar quantity, measures the amount of matter in an object and directly affects momentum. In this exercise:
  • The coal car weighs 15000 kg initially, which influences its initial momentum.
  • When 5000 kg of coal is added, the total mass becomes 20000 kg.
By acknowledging this change, we better understand how the momentum remains conserved despite altered dynamics. The increased mass results in a decreased velocity to keep the total momentum balanced, showcasing mass's crucial role in momentum conservation.
Physics Problem Solving
Problem-solving in physics often requires a clear understanding of fundamental principles like the conservation of momentum. By systematically breaking down the steps:
  • First, interpret the problem to understand what is being asked.
  • Identify known variables and equations relevant to the scenario.
  • Apply these equations, like the momentum equation, diligently to relate these variables.
This systematic approach ensures accuracy and understanding across problems. Solving physics problems extends beyond math; it involves critical thinking and logic. This helps ensure that the calculation of the final velocity of 3.75 m/s accurately reflects real-world behavior and interactions.

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Most popular questions from this chapter

Two girls (masses \(m_{1}\) and \(m_{2}\) ) are on roller skates and stand at rest, close to each other and face to face. Girl-1 pushes squarely against girl-2 and sends her moving backward. Assuming the girls move freely on their skates, write an expression for the speed with which girl-1 moves. We take the two girls to comprise the system under consideration. The problem states that girl-2 moves "backward," so let that be the negative direction; therefore, the "forward" direction is positive. There is no resultant external force on the system (the push of one girl on the other is an internal force), and so momentum is conserved: 1 from which Girl-1 recoils with this speed. Notice that if \(m_{2} / m_{1}\) is very large, \(v_{1}\) is much larger than \(v_{2}\). The velocity of girl-1, \(\overrightarrow{\mathbf{v}}_{1}\), points in the positive forward direction. The velocity of girl-2, \(\overrightarrow{\mathbf{v}}_{2}\), points in the negative backward direction. If we put numbers into the equation, \(\mathrm{v}_{2}\) would have to be negative and \(u_{1}\) would come out positive.

A rocket standing on its launch platform points straight upward. Its engines are activated and eject gas at a rate of \(1500 \mathrm{~kg} / \mathrm{s}\). The molecules are expelled with an average speed of \(50 \mathrm{~km} / \mathrm{s}\). How much mass can the rocket initially have if it is slow to rise because of the thrust of the engines? The problem provides mass flow and speed, the product of which is equivalent to the time rate-of-change of momentum. That should bring to mind the impulse- momentum relationship, which, of course, is Newton's Second Law. Since the initial motion of the rocket itself is negligible in comparison to the speed of the expelled gas, we can assume the gas is accelerated from rest to a speed of \(50 \mathrm{~km} / \mathrm{s}\). The impulse required to provide this acceleration to a mass \(m\) of gas is from which $$ \begin{array}{c} F \Delta t=m v_{f}-m v_{i}=m(50000 \mathrm{~m} / \mathrm{s})-0 \\ F=(50000 \mathrm{~m} / \mathrm{s}) \frac{\mathrm{m}}{\mathrm{L}} \end{array} $$ But we are told that the mass ejected per second \((m / \Delta t)\) is 1500 \(\mathrm{kg} / \mathrm{s}\), and so the force exerted on the expelled gas is $$ F=(50000 \mathrm{~m} / \mathrm{s})(1500 \mathrm{~kg} / \mathrm{s})=75 \mathrm{MN} $$ But we are told that the mass ejected per second \((m / \Delta t)\) is 1500 \(\mathrm{kg} / \mathrm{s}\), and so the force exerted on the expelled gas is $$ F=(50000 \mathrm{~m} / \mathrm{s})(1500 \mathrm{~kg} / \mathrm{s})=75 \mathrm{MN} $$ An equal but opposite reaction force acts on the rocket, and this is the upward thrust on the rocket. The engines can therefore support a weight of \(75 \mathrm{MN}\), so the maximum mass the rocket could have is $$ M_{\text {rocket }}=\frac{\text { weight }}{g}=\frac{75 \times 10^{6} \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=7.7 \times 10^{6} \mathrm{~kg} $$

Two balls of equal mass, moving with speeds of \(3 \mathrm{~m} / \mathrm{s}\), collide head-on. Find the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, \((c)\) the coefficient of restitution is \(1 / 3\).

A 90 -g ball moving at \(100 \mathrm{~cm} / \mathrm{s}\) collides head-on with a stationary 10-g ball. Determine the speed of each after impact if \((a)\) they stick together, \((b)\) the collision is perfectly elastic, \((c)\) the coefficient of restitution is \(0.90\).

Two identical balls collide head-on. The initial velocity of one is \(0.75 \mathrm{~m} / \mathrm{s}\) - EAST, while that of the other is \(0.43 \mathrm{~m} / \mathrm{s}\) -WEST. If the collision is perfectly elastic, what is the final velocity of each ball? Because the collision is perfectly elastic, both momentum and KE are conserved. Since the collision is head-on, all motion takes place along a straight line. Take east as positive and call the mass of each ball \(m\). Momentum is conserved in a collision, so we can write Momentum before \(=\) Momentum after $$ m(0.75 \mathrm{~m} / \mathrm{s})+m(-0.43 \mathrm{~m} / \mathrm{s})=m v_{1}+m v_{2} $$ where \(v_{1}\) and \(v_{2}\) are the final values. This equation simplifies to $$ 0.32 \mathrm{~m} / \mathrm{s}=v_{1}+v_{2} $$ Because the collision is assumed to be perfectly elastic, KE is also conserved. Thus, $$ \begin{array}{c} \text { KE before }=\text { KE after } \\ \frac{1}{2} m(0.75 \mathrm{~m} / \mathrm{s})^{2}+\frac{1}{2} m(-0.43 \mathrm{~m} / \mathrm{s})^{2}=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2} \end{array} $$ This equation can be simplified to $$ 0.747=v_{1}^{2}+v_{2}^{2} $$ We can solve for \(u_{2}\) in Eq. (1) to get \(v_{2}=0.32-v_{1}\) and substitute that into Eq. (2). This yields $$ 0.747=\left(0.32-v_{1}\right)^{2}+v_{1}^{2} $$ \(2 v_{1}^{2}-0.64 v_{1}-0.645=0\) Using the quadratic formula, $$ v_{1}=\frac{0.64 \pm \sqrt{(0.64)^{2}+5.16}}{4}=0.16 \pm 0.59 \mathrm{~m} / \mathrm{s} $$ from which \(v_{1}=0.75 \mathrm{~m} / \mathrm{s}\) or \(-0.43 \mathrm{~m} / \mathrm{s}\). Substitution back into \(\mathrm{Eq} .\) (1) gives \(v_{2}=-0.43 \mathrm{~m} / \mathrm{s}\) or \(0.75 \mathrm{~m} / \mathrm{s}\). Two choices for answers are available: \(\left(v_{1}=0.75 \mathrm{~m} / \mathrm{s}, v_{2}=-0.43 \mathrm{~m} / \mathrm{s}\right) \quad\) and \(\quad\left(v_{1}=-0.43 \mathrm{~m} / \mathrm{s}, v_{2}=0.75 \mathrm{~m} / \mathrm{s} \mathrm{s}\right.\) We must discard the first choice because it implies that the balls continue on unchanged; that is to say, no collision occurred. The correct answer is therefore \(v_{1}=-0.43 \mathrm{~m} / \mathrm{s}\) and \(v_{2}=0.75 \mathrm{~m} / \mathrm{s}\), which tells us that in a perfectly elastic, head-on collision between equal masses, the two bodies simply exchange velocities. Hence, \(\overrightarrow{\mathbf{v}}_{1}=0.43 \mathrm{~m} / \mathrm{s}-\) WEST and \(\overrightarrow{\mathbf{v}}_{2}=0.75 \mathrm{~m} / \mathrm{s}\) - EAST. Alternative Method If we recall that \(e=1\) for a perfectly elastic head-on collision, then $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad 1=\frac{v_{2}-v_{1}}{(0.75 \mathrm{~m} / \mathrm{s})-(20.43 \mathrm{~m} / \mathrm{s})} $$ which gives $$ v_{2}-v_{1}=1.18 \mathrm{~m} / \mathrm{s} $$ Equations (1) and (3) determine \(v_{1}\) and \(v_{2}\) uniquely.

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