Problem 78 - College Physics

Question

Rocket failure! Just as it has reached an upward speed of 5.0 \(\mathrm{m} / \mathrm{s}\) during a vertical launch, a rocket explodes into two pieces. Photographs of the explosion reveal that the lower piece, with a mass one-fourth that of the upper piece, was moving downward at 3.0 \(\mathrm{m} / \mathrm{s}\) the instant after the explosion. (a) Find the speed of the upper piece just after the explosion. (b) How high does the upper piece go above the point where the explosion occurred?

Step-by-Step Solution

Answer

The speed of the upper piece is 7.0 m/s. It reaches a maximum height of 2.5 m above the explosion point.

Step 1: Understand the problem

The problem describes an explosion of a rocket into two pieces with known speeds and masses. We need to find the speed of the upper piece and how high it travels above the explosion point.

Step 2: Identify given values

The initial upward speed of the rocket is given as 5.0 m/s. The lower piece is moving downward at -3.0 m/s after the explosion. The mass of the lower piece is one-fourth that of the upper piece.

Step 3: Apply the conservation of momentum

Use the conservation of momentum for the system. Before explosion: upward momentum is shared by the whole rocket. After explosion: total momentum must still equal the initial momentum (5.0 m/s). Let the mass of the upper piece be \(m\) and the mass of the lower piece be \(m/4\).

Step 4: Set up the conservation of momentum equation

The total momentum before the explosion is \( (m + \frac{m}{4}) \cdot 5.0 \) because both parts of the mass traveled at the initial speed of 5.0 m/s. After the explosion, the momentum equation is: \( m \cdot v_{upper} + \frac{m}{4} \cdot (-3.0) = \frac{5m}{4} \cdot 5.0 \).

Step 5: Solve for the speed of the upper piece

Solve the equation \( m \cdot v_{upper} - \frac{3m}{4} = \frac{5m}{4} \cdot 5.0 \) to find \(v_{upper}\). After canceling 'm' from each term, simplify the equation to: \( v_{upper} = \frac{25}{4} + \frac{3}{4} = 7.0 \). The speed of the upper piece is 7.0 m/s.

Step 6: Calculate the maximum height reached

Use energy conservation: initial kinetic energy of the upper piece converts to gravitational potential energy at the topmost point. Set \( \frac{1}{2} mv_{upper}^2 = mgh \). Solve for \(h\): \( h = \frac{v_{upper}^2}{2g} = \frac{7.0^2}{2 \times 9.8} \).

Step 7: Simplify and calculate height

Calculate \( h = \frac{49}{19.6} \), which simplifies to approximately 2.5 m. This is the height above the explosion point where the upper piece reaches its peak.

Key Concepts

Rocket Explosion

When a rocket explodes, it breaks apart into multiple pieces that move in various directions and speeds. An explosion is essentially a sudden release of energy that can affect the motion of the object, such as a rocket. Each piece gains its individual momentum and velocity after the explosion.
In this specific rocket explosion scenario, the rocket initially moves upward at 5.0 m/s, and then splits into two pieces. Importantly, momentum is conserved, which means the total momentum before the explosion remains equivalent to the total momentum after the explosion.

• The lower piece has a mass that is one-fourth of the upper piece.
• The lower piece moves downward at 3.0 m/s after the explosion.
• The motion of both pieces will depend on their respective masses and velocities post-explosion.

By using the conservation of momentum, one can determine the speed of each piece after the explosion. This principle helps us in finding the speed of the upper piece, which is **7.0 m/s** as calculated in the step-by-step solution.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. This type of energy is highly dependent on both the mass and velocity of the object. The formula for the kinetic energy (KE) of an object is given by \( KE = \frac{1}{2}mv^2 \). Here, \(m\) is the mass and \(v\) is the velocity of the object.
In our rocket explosion case, the upper piece immediately after the explosion has a velocity of 7.0 m/s. As it moves upward, this velocity causes it to possess a certain amount of kinetic energy.

• Kinetic energy is highest right after the explosion when speed is at its peak.
• As the piece ascends, its speed decreases due to gravitational pull, thus reducing its kinetic energy.
• At the peak of its trajectory, the kinetic energy is minimized, momentarily dropping to zero before the piece begins to descend.

This conversion of kinetic energy as the upper piece ascends is a demonstration of the excellent interplay between kinetic and potential energy (covered next).

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy stored within an object as a result of its position in a gravitational field. When an object is lifted, it accumulates GPE, which is given by the formula \( GPE = mgh \), where \( m \) is mass, \( g \) is gravitational acceleration (9.8 m/s² on Earth), and \( h \) is the height from the reference point.
During the rocket explosion problem, the upper piece, as it rises up, starts converting kinetic energy into gravitational potential energy.

• Initially, right after explosion, the upper piece has maximum kinetic energy and zero potential energy.
• As it ascends, kinetic energy is gradually converted into potential energy.
• At its highest point, all kinetic energy is converted to gravitational potential energy.

The maximum height reached by this piece, approximately **2.5 meters** above the point of explosion, is where its potential energy peaks. This entire process elegantly demonstrates the conservation of energy principle, wherein the total energy remains constant, merely transitioning between kinetic and potential forms.