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Chapter 7: Problem 57

7.57 A 1439 -kg railroad car traveling at a speed of \(12 \mathrm{~m} / \mathrm{s}\) strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed (in \(\mathrm{m} / \mathrm{s})\) afterward?

Short Answer

Expert verified
Answer: The common speed of the two railroad cars after they collide and lock together is 6 m/s.

Step by step solution

01

Analyze the given information

We are given the mass and initial speed of the first railroad car, and that the second car is at rest. Both cars have the same mass, 1439 kg. The initial speed of the first car is 12 m/s, and the second car is at rest, so its initial speed is 0 m/s.
02

Apply the conservation of momentum principle

Since the momentum is conserved during this inelastic collision, the total momentum before the collision must equal the total momentum after the collision. Mathematically, we can express this as follows: \(m_1v_1 + m_2v_2 = (m_1 + m_2) v_f\) Where \(m_1\) and \(m_2\) are the masses of the first and second railroad cars, \(v_1\) and \(v_2\) are their initial speeds, and \(v_f\) is the final common speed of both cars after they lock.
03

Plug in the given values

Since both cars have the same mass (\(m_1 = m_2 = 1439 \mathrm{~kg}\)), the initial speed of the first car is \(v_1 = 12 \mathrm{~m/s}\), and the second car is at rest (\(v_2 = 0\)), we can plug these values into the formula from Step 2: \(1439 \mathrm{~kg} \cdot 12 \mathrm{~m/s} + 1439 \mathrm{~kg} \cdot 0 \mathrm{~m/s} = (1439 \mathrm{~kg} + 1439 \mathrm{~kg}) \cdot v_f\)
04

Simplify and solve for the final common speed \(v_f\)

This equation can now be simplified: \(1439 \mathrm{~kg} \cdot 12 \mathrm{~m/s} = (2878 \mathrm{~kg}) \cdot v_f\) We can now solve for \(v_f\) by dividing both sides by the total mass of the two cars after the collision: \(v_f = \frac{1439 \mathrm{~kg} \cdot 12 \mathrm{~m/s}}{2878 \mathrm{~kg}} = 6 \mathrm{~m/s}\) Therefore, the common speed of the two railroad cars after they collide and lock together is \(6 \mathrm{~m/s}\).

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Key Concepts

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

Inelastic Collision
When we talk about inelastic collisions in physics, we're referring to interactions where objects collide and then move together as a single system afterwards. Unlike elastic collisions where objects bounce off each other and kinetic energy is conserved, inelastic collisions do not conserve kinetic energy but the total momentum of the system is conserved.

Imagine two clay balls hurtling towards each other; upon impact, they stick together, moving at a combined speed. This is the essence of an inelastic collision. The example from the exercise with railroad cars demonstrates this: after colliding, the cars lock together and move with a common speed. This provides a perfect hands-on scenario to apply the principles of momentum conservation even when kinetic energy isn't conserved.
Physics Problem Solving
Effective physics problem solving follows a structured approach to break down complex problems into manageable pieces. First, we start by analyzing the given data and identifying the underlying physics principles applicable to the problem. In our case, it was recognizing the conservation of momentum in an inelastic collision.

Next, we apply theoretical concepts to formulate mathematical equations that represent the physical situation. By plugging in known values and applying algebraic manipulations, we isolate our unknown variables. Always make sure each step follows logically from the last and that unit consistency is maintained throughout the solving process to arrive at a reasonable and accurate solution.
Momentum Calculation
Momentum, calculated as the product of mass and velocity, is a key concept in many physics problems. For momentum calculations, the formula we use is: \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. In the scenario with the railroad cars, we applied this fundamental equation to both cars before the collision and set this equal to the total momentum after the collision.

By using the provided masses and velocities, we executed a momentum calculation to solve for the final velocity after the cars locked together. It's crucial to consider the direction of the velocity, too, because momentum is a vector quantity. Here, as both cars eventually move in the same direction, we simply add their momenta to find the collective motion post-collision.

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Most popular questions from this chapter

Astronauts are playing baseball on the International Space Station. One astronaut with a mass of \(50.0 \mathrm{~kg}\), initially at rest, hits a baseball with a bat. The baseball was initially moving toward the astronaut at \(35.0 \mathrm{~m} / \mathrm{s},\) and after being hit, travels back in the same direction with a speed of \(45.0 \mathrm{~m} / \mathrm{s}\). The mass of a baseball is \(0.14 \mathrm{~kg}\). What is the recoil velocity of the astronaut?

Consider two carts, of masses \(m\) and \(2 m\), at rest on a frictionless air track. If you push the lower-mass cart for \(3 \mathrm{~s}\) and then the other cart for the same length of time and with the same force, which cart undergoes the larger change in momentum? a) The cart with mass \(m\) has the larger change b) The cart with mass \(2 m\) has the larger change. c) The change in momentum is the same for both carts. d) It is impossible to tell from the information given.

A bored boy shoots a soft pellet from an air gun at a piece of cheese with mass \(0.25 \mathrm{~kg}\) that sits, keeping cool for dinner guests, on a block of ice. On one particular shot, his 1.2-g pellet gets stuck in the cheese, causing it to slide \(25 \mathrm{~cm}\) before coming to a stop. According to the package the gun came in, the muzzle velocity is \(65 \mathrm{~m} / \mathrm{s}\). What is the coefficient of friction between the cheese and the ice?

A sled initially at rest has a mass of \(52.0 \mathrm{~kg}\), including all of its contents. A block with a mass of \(13.5 \mathrm{~kg}\) is ejected to the left at a speed of \(13.6 \mathrm{~m} / \mathrm{s} .\) What is the speed of the sled and the remaining contents?

A soccer ball rolls out of a gym through the center of a doorway into the next room. The adjacent room is \(6.00 \mathrm{~m}\) by \(6.00 \mathrm{~m}\) with the \(2.00-\mathrm{m}\) wide doorway located at the center of the wall. The ball hits the center of a side wall at \(45.0^{\circ} .\) If the coefficient of restitution for the soccer ball is \(0.700,\) does the ball bounce back out of the room? (Note that the ball rolls without slipping, so no energy is lost to the floor.)
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