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Chapter 8: Problem 104

A 12.0 kg shell is launched at an angle of \(55.0^{\circ}\) above the horizontal with an initial speed of 150 \(\mathrm{m} / \mathrm{s}\) . When it is at its highest point, the shell exploded into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Assume that air resistance can be ignored. If the heavier fragment lands back at the same point from which the shell was launched, where will the lighter fragment land how much energy was released in the explosion?

Short Answer

Expert verified
The lighter fragment lands farther due to its greater velocity, the position can be calculated if the specific vertical descent time is known; energy released equals the difference in final kinetic energy of the lighter fragment.

Step by step solution

01

Determine Initial Momentum Before Explosion

Before the explosion, calculate the initial momentum of the shell at its highest point. At the peak, the vertical velocity is 0, and the horizontal velocity is the initial velocity in the horizontal direction: \[ v_{x} = v_0 \cos(\theta) \]where \( v_0 = 150 \hspace{2mm} \text{m/s} \) and \( \theta = 55.0^\circ \).Calculate:\[ v_{x} = 150 \hspace{2mm} \text{m/s} \times \cos(55.0^\circ) \approx 86.0 \hspace{2mm} \text{m/s} \]The momentum \( p \) is:\[ p = mv_{x} = 12.0 \hspace{2mm} \text{kg} \times 86.0 \hspace{2mm} \text{m/s} = 1032.0 \hspace{2mm} \text{kg m/s} \]
02

Post-Explosion Momentum Conservation

Since the shell explodes at the highest point and horizontal momentum is conserved, the momentum before explosion equals the sum of the momenta of the fragments after explosion:For the heavier fragment (mass = \(3m\)) and lighter fragment (mass = \(m\)):\[ 3mv_{1x} + mv_{2x} = mv_{shell} \]where \(v_{1x}\) is the velocity of the heavier fragment (0 m/s since it lands at the launch point) and \(v_{2x}\) is the velocity of the lighter fragment.Substituting known values:\[ 3m(0) + mv_{2x} = 1032.0 \hspace{2mm} \text{kg m/s} \]\[ v_{2x} = \frac{1032.0 \hspace{2mm} \text{kg m/s}}{m} \]
03

Evaluate energy released during explosion

The energy released during the explosion can be found as follows. Since momentum is conserved, consider only kinetic energy:Initially, kinetic energy is solely due to horizontal motion:\[ K_{initial} = \frac{1}{2} m v_{x}^2 \]After the explosion, the heavier fragment doesn't have horizontal velocity, so its kinetic energy is zero. The lighter fragment carries the released energy:\[ K_{final} = \frac{1}{2} mv_{2x}^2 \]We know, \(v_{1x} = 0\) and substituting from previous calculations:\[ v_{2x} = 258.0 \hspace{2mm} \text{m/s} \] So the energy released:\[ K_{release} = K_{final} - K_{initial} \]
04

Calculate landing distance of lighter fragment

Since the time to fall from the peak to the ground remains the same for both fragments, use projectile motion. Calculate how far the lighter fragment travels using its horizontal velocity:The time t from the peak is determined by the initial vertical component of velocity:\[ t = \frac{2v_0 \sin(\theta)}{g} \]Horizontal distance (s) of the lighter fragment:\[ s = v_{2x} \times t \]

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Key Concepts

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

Understanding Projectile Motion
Projectile motion describes the behavior of an object that is launched into the air and influenced only by the force of gravity. This means that once the projectile is in motion, we ignore effects such as air resistance. Here, the path of the projectile is determined by its initial velocity, angle of launch, and gravitational acceleration.
  • Initial velocity (\( v_0 \)): The speed at which the projectile is launched.
  • Launch angle (\( \theta \)): The angle above the horizontal at which the projectile is fired.
  • Trajectory: A parabolic path the projectile follows as it moves under the influence of gravity.
  • At the highest point: The vertical component of the velocity is zero, and only the horizontal component contributes to the motion.

In the original exercise, the shell is launched at an angle of \(55.0^{\circ}\) with a speed of 150 \(\mathrm{m/s}\). Using trigonometric functions, we decompose its velocity into horizontal and vertical components. The horizontal velocity remains constant throughout the flight due to the absence of horizontal forces acting on the shell.
Exploring Kinetic Energy in Explosions
Kinetic energy is the energy that an object possesses due to its motion. In the scenario of an explosion, energy transformations are a key concept. The explosion causes a redistribution of momentum and energy, but the total momentum of the system is conserved. This is crucial when calculating the energy released during the explosion.
  • Initial kinetic energy: Solely from the horizontal motion as the shell is at its highest point.
  • Post-explosion: The momentum remains the same, but energy is distributed among the fragments.
  • Heavier fragment: Lands at the launch point, so no kinetic energy from horizontal motion.
  • Lighter fragment: Takes on the horizontal kinetic energy and any additional energy released during the explosion.

By calculating the velocity of the lighter fragment post-explosion, we can determine the excess energy released. This energy manifests as the added speed to the lighter piece, causing it to travel a greater distance.
Unpacking Explosion Physics
Explosion physics involves the study of how forces act on objects that undergo a rapid distribution of energy, often breaking into multiple parts. During our exercise, when the shell explodes, the conservation of momentum is central. Despite the explosion, the total momentum before and after remains unchanged.
  • Conservation of momentum: The total momentum of the system (shell and fragments) remains constant through the explosion.
  • Momentum calculation: For the heavier fragment that lands at the origin, its final horizontal velocity is zero, simplifying the conservation equations.
  • Energy release: Extra energy from the explosion propels the lighter fragment farther, reflecting additional kinetic energy gained.

As the two fragments hit the ground simultaneously, we employ their horizontal movements to examine the post-explosion behavior. By calculating where the lighter fragment lands, we witness the effects of conserved momentum and redistributed energy.

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Most popular questions from this chapter

Force of a Baseball Swing. A baseball has mass 0.145 \(\mathrm{kg}\) , (a) If the velocity of a pitched ball has a magnitude of 45.0 \(\mathrm{m} / \mathrm{s}\) and the batted ball's velocity is 55.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. (b) If the ball remains in contact with the bat for 2.00 \(\mathrm{ms}\) , find the magnitude of the average force applied by the bat.

A 20.0 -kg projectile is fired at an angle of \(60.0^{\circ}\) above the horizontal with a speed of 80.0 \(\mathrm{m} / \mathrm{s}\) . At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level? (b) How much energy is released during the explosion?

A single-stage rocket is fired from rest from a deep-space platform, where gravity is negligible. If the rocket burns its fuel in 50.0 \(\mathrm{s}\) and the relative speed of the exhaust gas is \(v_{\mathrm{ex}}=2100 \mathrm{m} / \mathrm{s}\) , what must the mass ratio \(m_{0} / m\) be for a foral speed \(v\) of 8.00 \(\mathrm{km} / \mathrm{s}\) (about equal to the orbital speed of an earth satellite)?

On a very muddy football field, a \(110-\mathrm{kg}\) linebacker tackles an \(85-\mathrm{kg}\) halfback. Immediately before the collision, the line-backer is slipping with a velocity of 8.8 \(\mathrm{m} / \mathrm{s}\) north and the halfback is sliding with a velocity of 7.2 \(\mathrm{m} / \mathrm{s}\) east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

A \(4.00-\mathrm{g}\) bullet, traveling horizontally with a velocity of magnitude 400 \(\mathrm{m} / \mathrm{s}\) , is fired into a wooden block with mass 0.800 \(\mathrm{kg}\) , initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 120 \(\mathrm{m} / \mathrm{s} .\) The block shides a distance of 45.0 \(\mathrm{cm}\) along the surface from its initial position. (a) What is the coefficient of kinetic friction between block and surface? (b) What is the decrease in kinetic energy of the bullet? (c) What is the kinetic energy of the block at the instant after the bullet passes through it?
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