/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 A \(20.00-\mathrm{kg}\) lead sph... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(20.00-\mathrm{kg}\) lead sphere is hanging from a hook by a thin wire 3.50 \(\mathrm{m}\) long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a \(5.00-\mathrm{kg}\) steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

Short Answer

Expert verified
The dart must have an initial speed of at least 52.2 m/s.

Step by step solution

01

Understand the Problem

We need to find the minimum initial speed of the dart such that after embedding into the lead sphere, the system swings in a complete vertical circle. The problem involves conservation of linear momentum and concepts of circular motion.
02

Use Conservation of Momentum

Before the collision, only the dart has momentum, and immediately after, the dart and sphere move together. Using conservation of momentum: \[ m_d v_d = (m_d + m_s) v_f \]where \( m_d = 5.00 \text{ kg} \) (mass of the dart), \( m_s = 20.00 \text{ kg} \) (mass of the sphere), \( v_d \) is the speed of the dart, and \( v_f \) is the speed of the combination right after impact.
03

Calculate the Necessary Speed at the Top of the Loop

For the system to just make a complete loop, at the top of the circle the tension can be zero. Using physics of circular motion, the condition at the top of the loop is:\[ \frac{(m_d + m_s) v_{top}^2}{R} = (m_d + m_s)g \]where \( v_{top} \) is the speed at the top, \( R = 3.50 \text{ m} \), and \( g = 9.81 \text{ m/s}^2 \). Solving gives \( v_{top} = \sqrt{g R} = \sqrt{9.81 \times 3.50} \approx 5.86 \text{ m/s} \).
04

Apply Energy Conservation

Apply conservation of mechanical energy from just after the collision to the top of the swing:\[ \frac{1}{2}(m_d + m_s) v_f^2 = \frac{1}{2}(m_d + m_s) v_{top}^2 + (m_d + m_s)g \times 2R \]Solving for \( v_f \):\[ v_f^2 = v_{top}^2 + 4gR \]
05

Solve for the Initial Speed of the Dart

Combine results from Steps 2 and 4 to get the initial speed \( v_d \):- From Step 4, find \( v_f \) using \( v_f = \sqrt{v_{top}^2 + 4gR} \)- Use \[ v_d = \frac{(m_d + m_s)}{m_d} v_f \]Calculate \( v_f = \sqrt{5.86^2 + 4 \times 9.81 \times 3.50} \approx 10.44 \text{ m/s}\).- Find \( v_d \):\[ v_d = \frac{25}{5} \times 10.44 \approx 52.2 \text{ m/s} \]
06

Final Result

The minimum initial speed of the dart is approximately \( 52.2 \text{ m/s} \).

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Key Concepts

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

Conservation of Momentum
Before the collision occurs, we must consider the principles of conservation of momentum. Momentum is a crucial concept when objects interact. It is essentially the product of an object's mass and its velocity, representing how much "push" it has while moving. In isolated systems like our problem, momentum is conserved unless acted upon by an external force. This means the total momentum before and after the collision remains the same.
In this exercise, initially, only the dart has momentum. We calculate it using the formula \( p = m v \), where \( m \) is the mass and \( v \) is the velocity. After the dart hits and embeds into the lead sphere, both objects move together, combining their masses and sharing a new velocity, \( v_f \). Thus, immediately after the collision, the combined momentum (dart plus sphere) equals the initial momentum of the dart.
The equation for momentum conservation here is:
  • Before collision: \( m_d v_d \)
  • After collision: \((m_d + m_s) v_f \)
This ensures we can solve for the initial speed of the dart by comparing these two momentum values.
Circular Motion
Circular motion is the movement of an object along the circumference of a circle or rotation along a circular path. In this exercise, the lead sphere and dart, after collision, need to swing in a complete circle.
The critical condition here is that at the top of their swing, the sphere-dart system has just enough velocity to overcome gravitational pull without tension in the wire. This is where we use the concept of centripetal force, which is necessary for circular motion.
For the sphere to perform a complete circle, the centripetal force at the top must equal the gravitational force. This balance is given by the equation:
  • \( \frac{(m_d + m_s) v_{top}^2}{R} = (m_d + m_s)g \)
Here, \( v_{top} \) denotes the velocity at the top of the circle. The equation shows that this velocity depends on both the gravitational pull \( g \) and the radius \( R \) of the swing. By isolating and solving for \( v_{top} \), we can find the minimum speed necessary for a complete circular motion.
Energy Conservation
Energy conservation is a fundamental principle in physics, stating that energy cannot be created or destroyed, only transformed from one form to another. In the context of this problem, energy transforms between kinetic (motion) and potential (position) energy as the system oscillates.After the collision, we apply conservation of energy principles to understand how the motion transforms. At the lowest point of the swing (immediately post-collision), the system has maximum kinetic energy. As it ascends, kinetic energy converts to potential energy until, at the top, potential energy peaks while kinetic energy dips.
The conservation equation that captures this transformation as the system rises to complete the loop is:
  • \( \frac{1}{2}(m_d + m_s) v_f^2 = \frac{1}{2}(m_d + m_s) v_{top}^2 + (m_d + m_s)g \times 2R \)
This provides a relationship between initial velocity \( v_f \) and the velocity at the top \( v_{top} \). By rearranging and solving, we see how much kinetic energy must be retained or introduced to keep the system moving in a full loop. Understanding these energy transformations is crucial to solving the problem and predicting motion correctly.

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Most popular questions from this chapter

Three coupled railroad cars roll along and couple with a fourth car, which is initially at rest. These four cars roll along and couple with a fifth car initially at rest. This process continues until the speed of the final collection of railroad cars is one-fifth the speed of the initial three railroad cars. All the cars are identical. Ignoring friction, how many cars are in the final collection?

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