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Chapter 9: Problem 61

A cart with mass \(340 \mathrm{~g}\) moving on a frictionless linear air track at an initial speed of \(1.2 \mathrm{~m} / \mathrm{s}\) undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at \(0.66 \mathrm{~m} / \mathrm{s}\). (a) What is the mass of the second cart? (b) What is its speed after impact? (c) What is the speed of the twocart center of mass?

Short Answer

Expert verified
The mass of the second cart is 98.7 g, its speed after the collision is 1.86 m/s, and the center of mass speed is 0.93 m/s.

Step by step solution

01

Understand the Problem

Begin by identifying the given information and what needs to be found. We have a cart with mass \(m_1 = 340 \text{ g} = 0.34 \text{ kg}\) moving at an initial speed \(v_{1i} = 1.2 \text{ m/s}\). After an elastic collision, it moves at \(v_{1f} = 0.66 \text{ m/s}\). The second cart is initially stationary with \(v_{2i} = 0\text{ m/s}\), and we need to find its mass \(m_2\) and its speed \(v_{2f}\) after the collision. We are also asked for the center of mass speed of the system.
02

Apply Conservation of Momentum

Use the conservation of momentum which states that the total momentum before the collision is equal to the total momentum after the collision. So we write:\[m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}\]Substitute the known values:\[0.34 \times 1.2 + m_2 \times 0 = 0.34 \times 0.66 + m_2 v_{2f}\]
03

Solve for Second Cart's Mass

Since the collision is elastic, we can use the relative speed of separation equals relative speed of approach:\[v_{1i} - 0 = v_{2f} - v_{1f}\]Substitute the known values:\[1.2 = v_{2f} - 0.66 \]This simplifies to:\[v_{2f} = 1.2 + 0.66 = 1.86 \text{ m/s}\]Return to the momentum equation to solve for \(m_2\):\[0.408 = 0.2244 + m_2 \times 1.86\]Thus,\[m_2 = \frac{0.1836}{1.86} \approx 0.0987 \text{ kg} = 98.7 \text{ g}\]
04

Calculate Speed of Center of Mass

The speed of the center of mass \(v_{cm}\) for two carts is given by:\[v_{cm} = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2}\]Using the initial speeds:\[v_{cm} = \frac{0.34 \times 1.2 + 0.0987 \times 0}{0.34+0.0987} = \frac{0.408}{0.4387} \approx 0.93 \text{ m/s}\]
05

Complete Solution Overview

We've determined the mass of the second cart is 98.7 grams, its speed after impact is 1.86 m/s, and the speed of the center of mass of the two carts is approximately 0.93 m/s.

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Key Concepts

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

Conservation of Momentum
The principle of conservation of momentum is central in analyzing collisions, such as the one described in this exercise. This principle states that the total momentum of a closed system remains constant, provided no external forces act on it.

In formula terms, this can be written as:
  • Before collision: Total momentum = \(m_1 \cdot v_{1i} + m_2 \cdot v_{2i}\)
  • After collision: Total momentum = \(m_1 \cdot v_{1f} + m_2 \cdot v_{2f}\)
In the exercise, the first cart's momentum changes due to the elastic collision, but the total system momentum is preserved. By substituting known values into the momentum equation, we can solve for unknown variables, such as the mass and speed of the second cart.
Center of Mass
The concept of the center of mass is crucial for understanding motion in systems involving multiple objects. It is the point where the total mass of a system is considered to be concentrated. In this scenario, knowing the speed of the center of mass (COM) helps us determine the joint motion of both carts in the system.

For the two-cart system, the speed of the COM \(v_{cm}\) can be expressed as:
  • \(v_{cm} = \frac{m_1 \cdot v_{1i} + m_2 \cdot v_{2i}}{m_1 + m_2}\)
With plug-in of known values and solving, we can find that the speed of the COM is \(0.93 \text{ m/s}\), indicating how the two-cart system moves jointly as a whole.
Physics Problem Solving
Physics problem solving often involves a structured approach that breaks down the problem into manageable parts. Here, this method involves identifying relevant physical principles and equations.

Steps to follow might include:
  • Understanding what is given and what needs to be found.
  • Selecting appropriate physical laws, such as conservation of momentum and energy.
  • Substituting known values into the equations.
  • Solving the equations to find unknown quantities.
In this exercise, utilizing these steps allows us to solve for unknowns like the second cart's mass and speed, as well as the system's COM speed.
Conservation of Energy
In elastic collisions, such as the one in this problem, conservation of energy plays an important role alongside momentum conservation. This is because no kinetic energy is lost in these collisions. The total kinetic energy before and after the collision remains constant.

For this exercise, this implies:
  • Initial kinetic energy: \(\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2\)
  • Final kinetic energy: \(\frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2\)
The elastic nature of the collision ensures energy is conserved, which is key to solving for the unknowns effectively, enhancing our understanding of collision dynamics.

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Most popular questions from this chapter

In the Olympiad of 708 B.C., some athletes competing in the standing long jump used handheld weights called halteres to lengthen their jumps (Fig. \(9-56\) ). The weights were swung up in front just before liftoff and then swung down and thrown backward during the flight. Suppose a modern \(78 \mathrm{~kg}\) long jumper similarly uses two \(5.50 \mathrm{~kg}\) halteres, throwing them horizontally to the rear at his maximum height such that their horizontal velocity is zero relative to the ground. Let his liftoff velocity be \(\vec{v}=(9.5 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\) with or without the halteres, and assume that he lands at the liftoff level. What distance would the use of the halteres add to his range?

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with the cue ball's original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don't consider the rotation) conserved?

A \(2.00 \mathrm{~kg}\) particle has the \(x y\) coordinates \((-1.20 \mathrm{~m}, 0.500 \mathrm{~m})\), and a \(4.00 \mathrm{~kg}\) particle has the \(x y\) coordinates \((0.600 \mathrm{~m},-0.750 \mathrm{~m})\). Both lie on a horizontal plane. At what (a) \(x\) and (b) \(y\) coordinates must you place a \(3.00 \mathrm{~kg}\) particle such that the center of mass of the three-particle system has the coordinates \((-0.500 \mathrm{~m},-0.700 \mathrm{~m}) ?\)

A 75 kg man rides on a \(39 \mathrm{~kg}\) cart moving at a velocity of \(2.3 \mathrm{~m} / \mathrm{s}\). He jumps off with zero horizontal velocity relative to the ground. What is the resulting change in the cart's velocity, including sign?

A force in the negative direction of an \(x\) axis is applied for \(27 \mathrm{~ms}\) to a \(0.40 \mathrm{~kg}\) ball initially moving at \(14 \mathrm{~m} / \mathrm{s}\) in the positive direction of the axis. The force varies in magnitude, and the impulse has magnitude \(32.4 \mathrm{~N} \cdot \mathrm{s}\). What are the ball's (a) speed and (b) direction of travel just after the force is applied? What are (c) the average magnitude of the force and (d) the direction of the impulse on the ball?
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