Question

A body of mass \(8.0 \mathrm{~kg}\) is traveling at \(2.0 \mathrm{~m} / \mathrm{s}\) under the influence of no external force. At a certain instant an internal explosion occurs, splitting the body into two chunks of \(4.0 \mathrm{~kg}\) mass each; \(16 \mathrm{~J}\) of translational kinetic energy are imparted to the two-chunk system by the explosion. Neither chunk leaves the line of the original motion. Determine the speed and direction of motion of each of the chunks after the explosion.

Short answer

After the explosion, the two chunks will move in opposite directions with speeds of \(2.83 m/s\) and \(-2.83 m/s\), respectively.

Step-by-step solution

Conservation of Momentum

From the principle of conservation of linear momentum, we have Total momentum before explosion = Total momentum after explosion. \nTherefore, \(mass_{total} \times velocity = 2 \times mass_{chunk} \times velocity_{chunk}\), which simplifies to \(16 kg \cdot m/s = 2 \times 4.0 kg \times v\). Solve the equation to find that \(v = 16 kg \cdot m/s / 8.0 kg = 2 m/s\). This tells us that the total velocity of the chunks after the explosion should be 2 m/s.

Conservation of Energy

The additional energy given by the explosion is 16 J, and this energy becomes kinetic energy. Total kinetic energy after explosion = initial kinetic energy + energy of the explosion. From the formula of kinetic energy \(KE = 1/2 m v^2\), we have the initial kinetic energy of the body as \(1/2 \times 8 kg \times (2m/s)^2 = 16 J\). Therefore, after the explosion, the total energy becomes \(16 J + 16 J = 32 J\).

Determine the Speed of each Chunk

We know the total kinetic energy after the explosion is 32 J. This kinetic energy must be the sum of the kinetic energies of the chunks. Therefore by setting \(32 J = 1/2 \times 4 kg \times v1^2 + 1/2 \times 4kg \times v2^2\). Because the two chunks have the same mass and the total velocity remains the same, the speeds of the two chunks must be equal in magnitude but opposite in direction, so we can assume \(v1 = -v2\). Thus, we have \(v1 = sqrt(32 J / 4 kg) = sqrt(8 m^2/s^2) = 2.83 m/s\) in one direction, and similarly, \(v2 = -2.83 m/s\) in the other direction.

Key concepts

Linear Momentum

Linear momentum, a fundamental concept in physics, is a product of an object's mass and velocity, represented as \( p = m \times v \). It's a vector quantity, meaning it has both magnitude and direction. In the context of our exercise, an 8.0 kg body traveling at 2.0 m/s has a linear momentum calculated by multiplying the mass by the velocity.

When an explosion occurs, causing the body to split into two equal chunks, the linear momentum of the system remains constant if no external forces act upon it, which is an application of the conservation of linear momentum. This principle provides the baseline for solving the problem; it dictates that the combined momenta of the chunks post-explosion must equal the original momentum of the 8 kg body.

Conservation of Energy

The conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another within a closed system. In an explosion, chemical or internal energy is converted into kinetic energy, which is the energy an object possesses due to its motion.

In our example, the body had an initial kinetic energy that we calculate using the formula \( KE = \frac{1}{2} m v^2 \). Following the explosion, the internal energy that was present has been added to this system in the form of kinetic energy. Consequently, to find the new speeds of the chunks, we apply the conservation of energy, which ensures total kinetic energy post-explosion is the sum of the initial kinetic energy and the energy released during the explosion. This approach allows us to quantify the effects of the explosion on the body's motion.

Kinetic Energy

Kinetic energy is directly associated with the motion of an object and is calculated by the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity. It's a scalar quantity, meaning it only has magnitude and no direction.

In our specific problem, we initially calculate the kinetic energy of an 8 kg mass moving at 2 m/s prior to the explosion. Post-explosion, we reassess the kinetic energy to account for the energy imparted by the explosion. The total kinetic energy is then redistributed between the two chunks, allowing us to solve for their individual velocities.

Internal Explosion Physics

Internal explosion physics concerns itself with the behavior of objects when subjected to sudden forces from within. In this scenario, an internal explosion leads to the body splitting into two chunks. The energy released during this event adds to the system's kinetic energy without altering the linear momentum if there are no external forces.

This principle is crucial for understanding the aftermath of the explosion in our exercise. Since both chunks gain additional kinetic energy from the explosion, their velocities change. However, the system's total momentum remains the same. This relationship allows us to solve for the velocities of the resulting chunks by utilizing the conservation laws of physics.