Problem 47 - College Physics

Question

An explosion of a \(10.0-\mathrm{kg}\) bomb releases only two separate pieces. The bomb was initially at rest and a 4.00-kg piece travels westward at \(100 \mathrm{~m} / \mathrm{s}\) immediately after the explosion. (a) What are the speed and direction of the other piece immediately after the explosion? (b) How much kinetic energy was released in this explosion?

Step-by-Step Solution

Answer

(a) The other piece travels eastward at 66.7 m/s. (b) 33,340 J of kinetic energy was released.

Step 1: Identify the Conservation of Momentum

Since the bomb was initially at rest, its initial momentum is zero. After the explosion, the total momentum of the system remains zero due to the conservation of momentum. This means the momentum of the 4 kg piece must be equal and opposite to the momentum of the second piece.

Step 2: Calculate Momentum of the 4 kg Piece

Calculate the momentum of the 4 kg piece using the formula \( p = mv \). Here, \( m = 4.00 \) kg and \( v = 100 \) m/s.\[ p_1 = 4.00 \, \text{kg} \times 100 \, \text{m/s} = 400 \, \text{kg} \, \text{m/s} \]

Step 3: Find the Momentum of the Second Piece

Since the total momentum must be zero, the momentum of the second piece, \( p_2 \), must satisfy \( p_2 = -p_1 \). Thus, \[ p_2 = -400 \, \text{kg} \, \text{m/s} \]

Step 4: Determine the Mass of the Second Piece

The second piece has a mass of \( 10.0 - 4.00 = 6.00 \, \text{kg} \).

Step 5: Calculate the Speed of the Second Piece

Use the equation for momentum to find the speed \( v_2 \) for the second piece. \[ p_2 = m_2 v_2 \Rightarrow -400 \, \text{kg} \, \text{m/s} = 6.00 \, \text{kg} \cdot v_2 \]\[ v_2 = \frac{-400}{6.00} = -66.7 \, \text{m/s} \] The negative sign indicates the second piece is traveling eastward.

Step 6: Calculate the Initial Kinetic Energy

The initial kinetic energy is zero because the bomb is initially at rest.

Step 7: Calculate Kinetic Energy of the First Piece

Calculate the kinetic energy of the 4 kg piece using \( KE = \frac{1}{2}mv^2 \). \[ KE_1 = \frac{1}{2} \times 4.00 \, \text{kg} \times (100 \, \text{m/s})^2 = 20000 \, \text{J} \]

Step 8: Calculate the Kinetic Energy of the Second Piece

Do the same for the 6 kg piece. \[ KE_2 = \frac{1}{2} \times 6.00 \, \text{kg} \times (66.7 \, \text{m/s})^2 \approx 13340 \, \text{J} \]

Step 9: Sum the Kinetic Energies to Find Total Released

Add the kinetic energies of both pieces to find the total energy released.\[ KE_{\text{total}} = KE_1 + KE_2 = 20000 \, \text{J} + 13340 \, \text{J} = 33340 \, \text{J} \]

Key Concepts

Explosion Physics

In the world of physics, when we talk about explosions, we are not just referring to fireworks or bombs but also to any event that causes an object to break apart into separate pieces. In an explosion, the various laws of physics play a crucial role in determining the behavior of the objects involved. One critical law is the conservation of momentum, which states that within a closed system, the total momentum remains constant—meaning momentum before the explosion must equal momentum after.
When an object like a bomb explodes, it releases energy causing the fragments to move in different directions. If the bomb was initially at rest, its initial momentum is zero, setting the stage for the fragments to fly off in such a way that their combined momentum is still zero after the explosion.

• Consider how a force acts on an object to initiate an explosion, distributing force and energy to all parts involved.
• Understand that the pieces fly apart in different directions, often with different speeds based on their masses.

The study of explosions not only involves understanding the distribution of forces but also understanding how these forces translate into movement and energy for the resulting fragments.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. When we measure kinetic energy, we're looking at how much energy an object has because of its velocity and mass. This concept is vital in explosion physics because an explosion involves converting stored potential energy into kinetic energy, propelling fragments across various speeds.
Initially, the kinetic energy of the bomb is zero if it is at rest. Upon explosion, kinetic energy is generated, and each fragment of the bomb moves with a certain speed and thus has its own kinetic energy.

• Calculating kinetic energy involves using the formula: \[ KE = \frac{1}{2}mv^2 \] where \( m \) is mass and \( v \) is velocity.
• Kinetic energy increases with greater mass and higher velocity.
• Each piece after the explosion has a specific kinetic energy based on how the total energy released is divided among them.

Understanding kinetic energy is essential to grasp how energy is split among fragments and how much force each can exert—invaluable especially in assessing explosions.

Momentum Calculation

Momentum is a crucial concept in physics that denotes the quantity of motion an object has. It is calculated as the product of an object's mass and velocity and is expressed as \( p = mv \). In the context of an explosion, momentum plays a pivotal role due to the principle of conservation of momentum.
This principle assures that in the absence of external forces, the total momentum of a system remains constant before and after an event, such as an explosion. For a bomb initially at rest, its momentum is zero. Consequently, the pieces emerging post-explosion must collectively maintain this zero momentum to satisfy the conservation law.

• The momentum of individual fragments can be calculated and summed to verify the total system momentum is zero after the split.
• If one fragment moves in a positive direction, its momentum must be balanced by another fragment's momentum in the opposite direction.
• Solving problems involving momentum requires careful mass and velocity calculations for each piece to ensure conservation is maintained.

By making these calculations, we understand how explosions result in particular pieces moving in specific ways, allowing for predictive modeling and comprehensive analysis of such physical events.