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Chapter 7: Problem 53

By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor. As the plate falls, its momentum has only a vertical component, and no component parallel to the floor. After the collision, the component of the total momentum parallel to the floor must remain zero, since the net external force acting on the plate has no component parallel to the floor. Using the data shown in the drawing, find the masses of pieces 1 and 2 .

Short Answer

Expert verified
Solve the momentum equations in x and y directions using the given data to find the masses.

Step by step solution

01

Understanding the Law of Conservation of Momentum

The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. Since the plate's momentum originally has no horizontal component, the total horizontal momentum of the system must remain zero after breaking.
02

Analyzing the Momentum Components

We consider the momentum components of all three pieces along the x and y directions parallel to the floor. Let's denote the masses of the three pieces as \(m_1, m_2,\) and \(m_3\) respectively, and their velocities as \(v_1, v_2,\) and \(v_3\). The momentum equations in the x and y directions are as follows: \[m_1 v_{1x} + m_2 v_{2x} + m_3 v_{3x} = 0\] and \[m_1 v_{1y} + m_2 v_{2y} + m_3 v_{3y} = 0\].
03

Gathering Provided Data

Use the drawing's data to fill in known details such as the direction and magnitude of velocities for each mass. Assume \(v_{3x}\) and \(v_{3y}\) are given, along with the direction angles for pieces 1 and 2.
04

Set Up Equations

Based on the drawing, express velocities \(v_{1x}, v_{1y}, v_{2x},\) and \(v_{2y}\) in terms of their angles. Use trigonometric functions such as sine and cosine to split the given velocity magnitudes into directional components.
05

Solve for Unknowns

With two equations from Step 2, solve the resulting system of equations for the unknown variables \(m_1\) and \(m_2\). Substitute any known values to find numerical solutions.

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Key Concepts

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

Momentum Components
When analyzing motion, it's important to break down momentum into components. In the case of the accident where a plate breaks into three pieces, we focus on the horizontal (parallel to the floor) momentum components. Since the plate initially has only vertical momentum, we must ensure that the horizontal momentum after the collision sums to zero, due to conservation laws.

Momentum is calculated for each piece using the formula: momentum = mass 脳 velocity. Therefore, each piece's momentum is split into x (horizontal) and y (vertical) components.
  • The x-component depends on the mass and the velocity component in the x-direction.
  • The y-component depends on the mass and the velocity component in the y-direction.
Properly understanding and calculating these components is essential to solving momentum conservation problems, especially when dealing with multiple pieces flying in different directions.
Momentum Equations
Conservation of momentum lies at the heart of many physics problems. The law posits that the total momentum of a system stays the same unless an external force acts on it. When the plate shatters, conservation of momentum dictates that the sum of the horizontal momentum components for all pieces must offset to zero because no external horizontal forces acted during the breaking.

For our plates scenario:
  • In the x-direction: \[m_1 v_{1x} + m_2 v_{2x} + m_3 v_{3x} = 0\]
  • In the y-direction: \[m_1 v_{1y} + m_2 v_{2y} + m_3 v_{3y} = 0\]
These equations are essential to find unknown masses such as \(m_1\) and \(m_2\), by filling in known variables like \(v_{1x}\), \(v_{2x}\), etc., extracted from given conditions or data representations.
Velocity Components
Velocity components define how the velocity of an object is distributed in different directions. For the broken plate pieces, each has its own velocity vector with horizontal and vertical components.

To analyze these:
  • Velocity components in the x direction, \(v_{ix}\), describe the forward or backward movement parallel to the floor.
  • Velocity components in the y direction, \(v_{iy}\), describe the sideways movement parallel to the floor.
Understanding these components is crucial for applying momentum conservation laws. The velocity components ultimately determine the trajectory and speed directional breakdown needed to maintain the total horizontal momentum as zero post-collision.
Trigonometric Functions
Trigonometric functions simplify breaking down vectors like velocity into components. When dealing with angles, these functions allow us to project the total velocity into x and y directions using sine and cosine.

Typical steps include:
  • For the x-component of velocity, use cosine: \[v_{ix} = v_i \cos(\theta_i)\]
  • For the y-component of velocity, use sine: \[v_{iy} = v_i \sin(\theta_i)\]
In physics problems involving multiple directions, like broken plate pieces, trigonometry helps accurately piece together velocity components according to their specific angles. This understanding is vital in system equilibrium analyses where no additional momentum is introduced in certain directions.

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Most popular questions from this chapter

Part \(a\) of the drawing shows a bullet approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part \(b\). Note that both blocks are moving after the collision with the bullet. (a) Can the conservation of linear momentum be applied to this three-object system, even though the second collision occurs a bit later than the first one? Justify your answer. Neglect any mass removed from the first block by the bullet. (b) Is the total kinetic energy of this three-body system conserved? If not, would the total kinetic energy after the collisions be greater than or smaller than that before the collisions? Justify your answer.

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions One object is at rest, and another is moving. Furthermore, one object is more massive than the other. The two collide in a one- dimensional completely inelastic collision. In other words, they stick together after the collision and move off with a common velocity. Momentum is conserved. (a) Consider the final momentum of the two-object system after the collision. Is it greater when the large-mass object is moving initially or when the small-mass object is moving initially? (b) Is the final speed after the collision greater when the large-mass or the smallmass object is moving initially? Problem The speed of the object that is moving initially is \(25 \mathrm{~m} / \mathrm{s}\). The masses of the two objects are 3.0 and \(8.0 \mathrm{~kg}\). Determine the final speed of the two-object system after the collision for the case when the large-mass object is moving initially and the case when the small-mass object is moving initially. Be sure that your answers are consistent with your answers to the Concept Questions.

A 50.0 -kg skater is traveling due east at a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). A \(70.0-\mathrm{kg}\) skater is moving due south at a speed of \(7.00 \mathrm{~m} / \mathrm{s}\). They collide and hold on to each other after the collision, managing to move off at an angle \(\theta\) south of east, with a speed of \(v_{\mathrm{f}}\). Find (a) the angle \(\theta\) and (b) the speed \(v_{\mathrm{f}}\), assuming that friction can be ignored.

A student \((m=63 \mathrm{~kg})\) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of \(0.010 \mathrm{~s}\). The average force exerted on him by the ground is \(+18000 \mathrm{~N}\), where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.

Starting with an initial speed of \(5.00 \mathrm{~m} / \mathrm{s}\) at a height of \(0.300 \mathrm{~m},\) a \(1.50-\mathrm{kg}\) ball swings downward and strikes a \(4.60-\mathrm{kg}\) ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the \(1.50-\mathrm{kg}\) ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?
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