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Chapter 9: Problem 42

A \(15.0 \mathrm{~kg}\) package is moving at a speed of \(10.0 \mathrm{~m} / \mathrm{s}\) vertically upward along a \(y\) axis when it explodes into three fragments: a \(2.00\) \(\mathrm{kg}\) fragment is shot upward with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\) and a \(3.00 \mathrm{~kg}\) fragment is shot in the positive direction of a horizontal \(x\) axis with an initial speed of \(5.00 \mathrm{~m} / \mathrm{s}\). Find (a) the speed of the third fragment right after the explosion and (b) the total kinetic energy provided by the explosion.

Short Answer

Expert verified
The speed of the third fragment is approximately 11.1 m/s and the explosion provides 303.55 J of kinetic energy.

Step by step solution

01

Analyze the Conservation of Momentum in the Y Direction

The total momentum before and after the explosion must be equal. The initial momentum in the y-direction is given by the equation: \[ p_y = m \cdot v_y = 15.0 \times 10.0 = 150 \, \text{kg m/s}. \] The momentum of the 2 kg fragment in the y-direction is: \[ p_{y, \, 2} = 2.00 \times 20.0 = 40.0 \, \text{kg m/s}. \] Let the speed of the third fragment in the y-direction be \( v_{y, \, 3} \) and its mass be \( m_3 = 10.0 \mathrm{~kg} \). Thus, we write for momentum conservation in y:\[ 150 = 40 + m_3 \cdot v_{y, \, 3}. \]
02

Solve for the Y-component of the Third Fragment's Velocity

Substitute the values and solve for \( v_{y, \, 3} \):\[ 150 = 40 + 10 \cdot v_{y, \, 3} \]\[ 110 = 10 \cdot v_{y, \, 3} \]\[ v_{y, \, 3} = 11 \, \text{m/s}. \]
03

Analyze the Conservation of Momentum in the X Direction

Initially, there is no momentum in the x-direction since the package is only moving vertically:\[ p_x = 0. \] After the explosion, the 3 kg fragment contributes to the x-direction momentum. If \( v_{x, \, 3} \) is the velocity of the third fragment in the x-direction:\[ 3 \cdot 5 + 10 \times v_{x, \, 3} = 0. \]
04

Solve for the X-component of the Third Fragment's Velocity

Using the momentum conservation equation, solve for \( v_{x, \, 3} \):\[ 15 + 10 \cdot v_{x, \, 3} = 0 \] \[ v_{x, \, 3} = -1.5 \, \text{m/s}. \]
05

Calculate the Speed of the Third Fragment

The speed of the third fragment is computed from its x and y components using the Pythagorean theorem:\[ v_3 = \sqrt{v_{x, \, 3}^2 + v_{y, \, 3}^2} \] \[ = \sqrt{(-1.5)^2 + (11)^2} \] \[ = \sqrt{2.25 + 121} \] \[ = \sqrt{123.25} \] \[ \approx 11.1 \, \text{m/s}. \]
06

Calculate the Initial Total Kinetic Energy

The initial kinetic energy of the package is calculated as:\[ KE_i = \frac{1}{2} m v^2 = \frac{1}{2} \times 15 \times (10)^2 \] \[ = \frac{1}{2} \times 15 \times 100 \] \[ = 750 \, \text{J}. \]
07

Calculate the Final Total Kinetic Energy

The final kinetic energy after the explosion is the sum of the kinetic energies of all fragments:\[ KE_f = \frac{1}{2} \times 2 \times (20)^2 + \frac{1}{2} \times 3 \times (5)^2 + \frac{1}{2} \times 10 \times (11.1)^2 \] \[ = 400 + 37.5 + 616.05 \] \[ = 1053.55 \, \text{J}. \]
08

Determine the Kinetic Energy Provided by the Explosion

The total kinetic energy provided by the explosion is the difference between the final and initial kinetic energies:\[ \Delta KE = KE_f - KE_i \] \[ = 1053.55 - 750 \] \[ = 303.55 \, \text{J}. \]

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Key Concepts

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

Kinetic Energy
Understanding kinetic energy is crucial to solving problems related to explosions and conservation of momentum. Kinetic energy (\( KE \)) is the energy an object possesses due to its motion. It is calculated using the formula:
  • \[ KE = \frac{1}{2} m v^2 \]
  • where \( m \) is the mass and \( v \) is the velocity of the object.
In the given exercise, the initial kinetic energy of the package is calculated as 750 J using this formula.
This energy is distributed among the fragments after the explosion. The sum of their kinetic energies gives the final kinetic energy, which exceeds the initial due to the energy added by the explosion.
The change in kinetic energy (\( \Delta KE \)) represents the energy supplied by the explosion.
Explosion Dynamics
The dynamics of an explosion involve the sudden release of stored energy, resulting in the rapid movement of fragments. During an explosion, the total momentum is conserved, but kinetic energy is often increased.
In this exercise, a package explodes, distributing momentum and increasing kinetic energy across three fragments. The law of conservation of momentum ensures that the sum of the momenta before and after the explosion remains constant:
  • Momentum conservation helps us find the velocity of individual fragments post-explosion.
  • Vector components of momentum must be considered separately in each direction (i.e., x and y directions).
The energy supplied by the explosion comes from internal forces, providing additional kinetic energy to each fragment, reflected in the calculated \( \Delta KE \).
Vector Components
When examining motion in multiple dimensions, vector components become essential.
A vector, like velocity or momentum, can have both magnitude and direction. To analyze motion, especially in two dimensions, we break vectors into x and y components.
  • The initial situation describes a package moving vertically, meaning the initial momentum exists only along the y-axis.
  • Post-explosion fragments acquire momentum in both y and x directions simultaneously.
To solve for the velocities of the fragments, we treat them as independent vector components:
  • Use trigonometry and the Pythagorean theorem for calculating the resultant speed from components: \[ v = \sqrt{v_x^2 + v_y^2} \]
By analyzing these components, we gain insight into the direction and magnitude of each fragment's movement, crucial for understanding both momentum and energy distribution.

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Most popular questions from this chapter

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass \(3.0 \mathrm{~kg}\), is moving upward at \(20 \mathrm{~m} / \mathrm{s}\) and the other ball, of mass \(2.0 \mathrm{~kg}\), is moving downward at \(10 \mathrm{~m} / \mathrm{s}\). How high do the combined two balls of putty rise above the collision point? (Neglect air drag.)

Two bodies of masses \(m=0.30 \mathrm{~kg}\) and \(2 m\) are connected by a long string of negligible mass. The string is looped over a pulley and, with the string taut, the bodies are released at time \(t=0\) so that the heavier one descends and the lighter one ascends. At time \(t=4.0 \mathrm{~s}\), the lighter one undergoes a fully inelastic collision with a third body of mass \(m\). Because the first two bodies move in rigid fashion, the collision is effectively between the third body and the system of the first two bodies. (a) Just after the collision, what is the speed of the three bodies? (b) By how much was the kinetic energy of the descending body decreased because of the collision?

A \(3.00 \mathrm{~kg}\) block slides on a frictionless horizontal surface, first moving to the left at \(50.0 \mathrm{~m} / \mathrm{s}\). It collides with a spring whose other end is fixed to a wall, compresses the spring, and is brought to rest momentarily. Then it continues to be accelerated toward the right by the force of the compressed spring. The block acquires a final speed of \(40.0 \mathrm{~m} / \mathrm{s}\). It is in contact with the spring for \(0.020 \mathrm{~s}\). Find (a) the magnitude and (b) the direction of the impulse of the spring force on the block. (c) What is the magnitude of the spring's average force on the block?

In February 1955, a paratrooper fell \(370 \mathrm{~m}\) from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\), and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

A ball of mass \(50 \mathrm{~g}\) moving with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) strikes a wall at an angle of incidence \(45^{\circ}\) and is reflected from the wall at the same angle and with the same speed. See the overhead view in Fig. 9-38. Calculate (a) the magnitude of the change \(\Delta \vec{p}\) in the momentum of the ball, (b) the change in the magnitude of the momentum \(\vec{p}\) of the ball, and (c) the change in the magnitude of the momentum of the wall.
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