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(II) \(\mathrm{A} 28\) -g rifle bullet traveling 210 \(\mathrm{m} / \mathrm{s}\) buries itself in a 3.6 \(\mathrm{-kg}\) pendulum hanging on a 2.8 -long string, which makes the pendulum swing upward in an arc. Determine the vertical and horizontal components of the pendulum's maximum displacement.

Short Answer

Expert verified
Vertical: 0.134 m, Horizontal: 0.819 m.

Step by step solution

01

Understand the Situation

A bullet is fired into a pendulum, embedding itself. The bullet and pendulum then swing to a maximum height. We need to find the vertical and horizontal components of the pendulum's maximum displacement.
02

Calculate the Initial Momentum

The initial momentum of the system can be calculated using the mass of the bullet (0.028 kg) and its velocity (210 m/s). Apply the formula for linear momentum: \( p = mv \). The initial momentum is then \( p = 0.028 \times 210 = 5.88 \ \text{kg} \cdot \text{m/s} \).
03

Determine the Combined Mass

The mass of the combined system (bullet and pendulum) after the bullet embeds is \( m = 0.028 + 3.6 = 3.628 \ \text{kg} \).
04

Calculate the Velocity after Impact

Use conservation of momentum to find the velocity just after impact. Set the initial momentum equal to the final momentum: \( 5.88 = 3.628 \cdot v \). Solving for \( v \) gives \( v = \frac{5.88}{3.628} = 1.62 \ \text{m/s} \).
05

Calculate the Maximum Height

Use conservation of energy. Convert the kinetic energy at the lowest point into potential energy at the highest point: \( \frac{1}{2}mv^2 = mgh \). Simplify to find \( h \): \( \frac{1}{2}(1.62)^2 = 9.81h \). Solving for \( h \) gives \( h = \frac{1.62^2}{2 \times 9.81} = 0.134 \ \text{m} \).
06

Determine the Vertical Displacement

The vertical displacement is the maximum height the pendulum rises, which is \( h = 0.134 \ \text{m} \).
07

Calculate the Angular Displacement

Use trigonometry to find the angle \( \theta \) at the maximum height. \( \cos(\theta) = \frac{L - h}{L} = \frac{2.8 - 0.134}{2.8} = 0.9521 \). Solve for \( \theta \): \( \theta = \cos^{-1}(0.9521) = 17.03^\circ \).
08

Calculate the Horizontal Displacement

Using trigonometry and the angle calculated, \( \text{horizontal displacement} = L \sin(\theta) = 2.8 \sin(17.03^\circ) = 0.819 \ \text{m} \).

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Key Concepts

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

Linear Momentum
Linear momentum is a fundamental concept in physics, reflecting the quantity of motion an object has. It is expressed as the product of an object’s mass and its velocity. In mathematical terms, linear momentum is written as \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.
Linear momentum is a vector quantity, meaning it has both magnitude and direction. This is important when assessing the outcomes of collisions or interactions between objects, such as a bullet embedding into a pendulum.
In this example, a bullet traveling at high speed transfers its momentum to the pendulum upon impact. The initial momentum of the system, given by the bullet, is conserved through the collision, allowing us to find the velocity of the bullet-pendulum system post-impact. This demonstrated the principle of conservation of momentum, which asserts that the total linear momentum of an isolated system remains constant if no external forces act upon it.Key points about linear momentum involve:
  • Momentum is directional, given by both its speed and vector.
  • Conservation of momentum helps solve collision and explosion problems.
  • A system's total momentum before and after a interaction stays the same, provided no external forces intervene.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion, calculated with the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
Kinetic energy is scalar, meaning it only has magnitude and no direction, and it plays a vital role in understanding how energy transforms between different types in various physical systems.In our pendulum example, after the bullet becomes embedded in the pendulum, the system possesses kinetic energy due to its motion. This kinetic energy is exchanged for potential energy as the pendulum reaches its highest point in the swing, depicting conservation of energy principles.
Understanding this energy transformation:
  • Demonstrates that initial kinetic energy, upon impact, becomes potential energy at maximum height.
  • Illustrates the need to balance both kinetic and potential energies to solve the system’s behavior post-impact.
  • Shows how energy conservation allows for the computation of maximum displacement from speed post-impact.
This is crucial in predicting how much "swing" the pendulum will achieve after a collision, reflecting directly on the energy exchanged.
Pendulum Motion
Pendulum motion is a type of periodic motion and can be explained with concepts like amplitude, period, and frequency. However, in this scenario, we are more interested in how energy and momentum cause pendulum swings.
A pendulum consists of a mass (the pendulum bob) hanging from a string or rod of certain length, which swings back and forth under the influence of gravity and initial velocity input. When the bullet embeds in the pendulum bob, the system gains new momentum and kinetic energy, which control how high and far the pendulum can swing off to the side. The systematic conservation of energy from kinetic to potential and back informs us about the pendulum’s ability to reach a displacement vertically and horizontally. Key insights into pendulum motion:
  • Conservation of energy plays a significant role: transferring energy from kinetic (movement) to potential (height).
  • The maximum height (vertical displacement) reached by the bob signifies all kinetic energy has turned into potential energy.
  • Angular displacement, found using trigonometry, helps in calculating how far aside the pendulum swings horizontally.
These understandings illuminate the relationship between initial momentum and eventual swing, showing the effects of embedding momentum in pendulum behavior.

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Most popular questions from this chapter

(II) Rain is falling at the rate of \(5.0 \mathrm{~cm} / \mathrm{h}\) and accumulates in a pan. If the raindrops hit at \(8.0 \mathrm{~m} / \mathrm{s}\), estimate the force on the bottom of a \(1.0 \mathrm{~m}^{2}\) pan due to the impacting rain which does not rebound. Water has a mass of \(1.00 \times 10^{3} \mathrm{~kg}\) per \(\mathrm{m}^{3}\).

A hockey puck of mass \(4 m\) has been rigged to explode, as part of a practical joke. Initially the puck is at rest on a frictionless ice rink. Then it bursts into three pieces. One chunk, of mass \(m,\) slides across the ice at velocity \(v \hat{\mathbf{i}}\). Another chunk, of mass \(2 m\), slides across the ice at velocity \(2 v \hat{\mathbf{j}} .\) Determine the velocity of the third chunk.

(III) ( \(a\) ) Calculate the impulse experienced when a \(65-\mathrm{kg}\) person lands on firm ground after jumping from a height of \(3.0 \mathrm{~m} .\) (b) Estimate the average force exerted on the person's feet by the ground if the landing is stiff-legged, and again \((c)\) with bent legs. With stiff legs, assume the body moves \(1.0 \mathrm{~cm}\) during impact, and when the legs are bent, about \(50 \mathrm{~cm} .\) [Hint: The average net force on her which is related to impulse, is the vector sum of gravity and the force exerted by the ground.

Suppose two asteroids strike head on. Asteroid A \(\left(m_{\mathrm{A}}=7.5 \times 10^{12} \mathrm{~kg}\right)\) has velocity \(3.3 \mathrm{~km} / \mathrm{s}\) before the collision, and asteroid B \(\left(m_{\mathrm{B}}=1.45 \times 10^{13} \mathrm{~kg}\right)\) has velocity \(1.4 \mathrm{~km} / \mathrm{s}\) before the collision in the opposite direction. If the asteroids stick together, what is the velocity (magnitude and direction) of the new asteroid after the collision?

(III) A sled filled with sand slides without friction down a \(32^{\circ}\) slope. Sand leaks out a hole in the sled at a rate of \(2.0 \mathrm{~kg} / \mathrm{s} .\) If the sled starts from rest with an initial total mass of \(40.0 \mathrm{~kg},\) how long does it take the sled to travel \(120 \mathrm{~m}\) along the slope?

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