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Chapter 8: Problem 78

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?

Short Answer

Expert verified
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 by step solution

01

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.
02

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.
03

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 \(m/4\).
04

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: \( m \cdot v_{upper} + \frac{m}{4} \cdot (-3.0) = \frac{5m}{4} \cdot 5.0 \).
05

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.
06

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 \(h\): \( h = \frac{v_{upper}^2}{2g} = \frac{7.0^2}{2 \times 9.8} \).
07

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.

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Key Concepts

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

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/s虏 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.

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Most popular questions from this chapter

Forensic science. Forensic scientists can measure the muzzle velocity of a gun by firing a bullet horizontally into a large hanging block that absorbs the bullet and swings upward. (See Figure 8.49.) The meas ured maximum angle of swing can be used to calculate the speed of the bullet. In one such test, a rifle fired a 4.20 g bullet into a 2.50 kg block hanging by a thin wire 75.0 \(\mathrm{cm}\) long. causing the block to swing upward to a maximum angle of \(34.7^{\circ}\) from the vertical. What was the original speed of this bullet?

\(\cdot\) On a highly polished, essentially frictionless lunch counter, a 0.500 kg submarine sandwich moving 3.00 \(\mathrm{m} / \mathrm{s}\) to the left collides with a 0.250 \(\mathrm{kg}\) grilled cheese sandwich moving 1.20 \(\mathrm{m} / \mathrm{s}\) to the right. (a) If the two sandwiches stick together, what is their final velocity? (b) How much mechanical energy, dissipates in the collision? Where did this energy go?

. Biomechanics. The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 ms. (This number can also vary, depending on the racket and swing.) We shall assume a 30.0 \(\mathrm{ms}\) contact time throughout this problem. The fastest-known served tennis ball was served by "Big Bill" Tilden in \(1931,\) and its speed was measured to be 73.14 \(\mathrm{m} / \mathrm{s}\) .(a) What impulse and what force did Big Bill exert on the ten- nis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of \(55 \mathrm{m} / \mathrm{s},\) what force and what impulse did he exert on the ball, assuming only horizontal motion?

\(\bullet\) On a frictionless air track, a 0.150 \(\mathrm{kg}\) glider moving at 1.20 \(\mathrm{m} / \mathrm{s}\) to the right collides with and sticks to a stationary 0.250 \(\mathrm{kg}\) glider. (a) What is the net momentum of this two- glider system before the collision? (b) What must be the net momentum of this system after the collision? Why? (c) Use your answers in parts (a) and (b) to find the speed of the glid- ers after the collision. (d) Is kinetic energy conserved during the collision?

A machine part consists of a thin, uniform \(4.00-\mathrm{kg}\) bar that is 1.50 \(\mathrm{m}\) long, hinged perpendicular to a similar vertical bar of mass 3.00 \(\mathrm{kg}\) and length 1.80 \(\mathrm{m} .\) The longer bar has a small but dense 2.00 -kg ball at one end (Fig. 8.42\()\) By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through \(90^{\circ}\) to make the entire part horizontal?
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