Problem 40 - College Physics

Question

An object initially at rest explodes and splits into three fragments. The first fragment flies off to the west, and the second fragment flies off to the south. The third fragment will fly off toward a general direction of (1) southwest, (2) north of east, (3) either due north or due east. Why? (b) If the object has a mass of \(3.0 \mathrm{~kg},\) the first fragment has a mass of \(0.50 \mathrm{~kg}\) and a speed of \(2.8 \mathrm{~m} / \mathrm{s},\) and the second fragment has a mass of \(1.3 \mathrm{~kg}\) and a speed of \(1.5 \mathrm{~m} / \mathrm{s},\) what are the speed and direction of the third fragment?

Step-by-Step Solution

Answer

(a) The third fragment moves north of east. (b) Speed: 1.74 m/s, Direction: 54.5° north of east.

Step 1: Understand the scenario

An object initially at rest explodes into three fragments. Using conservation of momentum, the total initial momentum is zero. Each fragment then carries momentum in different directions, and the vectors of these momenta should sum to zero. We need to determine the possible direction and speed of the third fragment.

Step 2: Analyze possible directions

The first fragment moves west, and the second fragment moves south. This means the combined momentum of these two fragments has a resultant direction southwest. To conserve momentum, the third fragment must have momentum opposing the southwest direction. Hence, the third fragment must move in a direction that is somewhere between northeast and due north to counteract the southwest direction of the combined first two fragments.

Step 3: Momentum equations

For conservation of momentum, the sum of momenta in each direction (north-south and east-west) should be zero. Let \( m_3 \) be the mass and \( v_3 \) be the speed of the third fragment, with its velocity projecting into east (positive x-axis) and north (positive y-axis):- West momentum (first fragment): \( -m_1 v_1 \)- South momentum (second fragment): \( -m_2 v_2 \)- East momentum (third fragment): \( m_3 v_{3x} \)- North momentum (third fragment): \( m_3 v_{3y} \)We set up these two equations:1. East-West: \( -m_1 v_1 + m_3 v_{3x} = 0 \)2. North-South: \( -m_2 v_2 + m_3 v_{3y} = 0 \)

Step 4: Solve for the third fragment's velocity components

Using the equations:1. West momentum: \( m_1 = 0.5 \, \text{kg} \) and \( v_1 = 2.8 \, \text{m/s} \), hence momentum is \( -0.5 \times 2.8 = -1.4 \, \text{kg m/s} \).2. South momentum: \( m_2 = 1.3 \, \text{kg} \) and \( v_2 = 1.5 \, \text{m/s} \), hence momentum is \( -1.3 \times 1.5 = -1.95 \, \text{kg m/s} \).Third fragment's east momentum equation: \( 1.4 = m_3 v_{3x} \).Third fragment's north momentum equation: \( 1.95 = m_3 v_{3y} \).Given \( m_3 = 3.0 - 0.5 - 1.3 = 1.2 \, \text{kg} \).\( v_{3x} = \frac{1.4}{1.2} = 1.167 \, \text{m/s} \).\( v_{3y} = \frac{1.95}{1.2} = 1.625 \, \text{m/s} \).

Step 5: Find the speed and direction of the third fragment

The speed \( v_3 \) is the resultant vector magnitude given by:\[ v_3 = \sqrt{v_{3x}^2 + v_{3y}^2} = \sqrt{1.167^2 + 1.625^2} = \sqrt{3.04} \approx 1.74 \, \text{m/s} \]To find the direction, calculate the angle \( \theta \) north of east using:\[ \theta = \tan^{-1}\left(\frac{v_{3y}}{v_{3x}}\right) = \tan^{-1}\left(\frac{1.625}{1.167}\right) \approx 54.5^\circ \]Hence, the third fragment moves at approximately 54.5 degrees north of east.

Key Concepts

Vector Components

When analyzing an object's movement in physics, especially in two-dimensional space, it's crucial to understand the concept of vector components. A vector component is a part of a vector that points in a particular direction, commonly along the x-axis (east-west) and y-axis (north-south). Each vector can be broken down into these components using trigonometry.

Consider an event like an explosion, where fragments move in various directions. If you know the speed and direction of these fragments, you can calculate their vector components. Use simple trigonometric functions:

• The east or west component (let's say the x-component) can be calculated using cosine: \[v_x = v \cdot \cos(\theta)\]Here, \( v \) is the speed and \( \theta \) is the angle in the appropriate direction.
• The north or south component (y-component) is found with sine: \[v_y = v \cdot \sin(\theta)\]

In the context of the exploding object, each fragment's movement can be described using these vector components, enabling us to apply conservation laws more effectively.

Explosions in Physics

Explosions are fascinating physical phenomena that can be studied in terms of momentum and energy. Typically, an explosion involves a rapid expansion of energy that causes an object at rest to break apart and its fragments to scatter in various directions.

In physics problems, explosions are often idealized. This means that external forces like air resistance are ignored, and it's assumed that the explosion itself is instantaneous. Despite these simplifications, these scenarios can effectively illustrate key principles.

When an object explodes, the initial momentum of the system is zero if it was initially at rest. This is because momentum, defined as mass times velocity, was initially cumulative to zero as nothing moved. After the explosion, however, the system's energy is divided among the fragments, each gaining momentum in different directions. By calculating these various momenta, we can analyze the outcome of explosions without delving into the process of explosion itself.

Momentum Conservation Equations

The principle of momentum conservation is fundamental in physics. It states that the total momentum of a closed system must remain constant over time, provided no external forces are acting on it. This principle is particularly useful in scenarios involving collisions or explosions.

For an exploding object at rest, each fragment's momentum vector must sum to zero to satisfy conservation laws. Given directions for fragments, you'll need to consider momentum equations both east-west (x-axis) and north-south (y-axis):

• In the x-axis, you'll set up the equation: \[-m_1v_1 + m_3v_{3x} = 0\]
• On the y-axis, the equation will look like: \[-m_2v_2 + m_3v_{3y} = 0\]

From these equations, unknown variables such as the speed of a fragment can be solved. In the exercise, these principles help determine the third fragment's direction and speed after parsing the known quantities of the others. By solving these equations, we find that the third fragment must counteract the momentum of the first and second fragments, finding its approximate direction and resultant speed.