/*! 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 38 A ball is attached to one end of... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball is attached to one end of a wire, the other end beingfastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, \(1.60 \mathrm{kg}\) and \(2.40 \mathrm{kg},\) and the length of the wire is \(1.20 \mathrm{m} .\) Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

Short Answer

Expert verified
(a) 4.85 m/s, just before the collision. (b) After collision: Ball (-0.97 m/s), Block (3.88 m/s).

Step by step solution

01

Determine Velocity Before Collision

Initially, calculate the velocity of the ball just before the collision using energy conservation. The potential energy at the top converts to kinetic energy just before impact.Potential energy at the top is given by:\[U = mgh \]where \(m = 1.60 \, \text{kg}\), \(g = 9.81 \, \text{m/s}^2\), and \(h = 1.20 \, \text{m}\). Thus,\[U = 1.60 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1.20 \, \text{m} = 18.8352 \, \text{J}\]Kinetic energy just before impact is:\[K = \frac{1}{2}mv^2\]Setting potential energy equal to kinetic energy:\[18.8352 \, \text{J} = \frac{1}{2} \times 1.60 \, \text{kg} \times v^2\]Solving for velocity, \(v\), just before collision:\[v = \sqrt{\frac{2 \times 18.8352}{1.60}} = 4.85 \, \text{m/s}\]
02

Determine Velocities After Collision

Next, use conservation of momentum and kinetic energy for the elastic collision to find velocities post-collision.The conservation of momentum equation is:\[m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'\]and conservation of kinetic energy is:\[\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2\]Given initial conditions:\(v_1 = 4.85 \, \text{m/s}\), \(v_2 = 0\), \(m_1 = 1.60 \, \text{kg}\), and \(m_2 = 2.40 \, \text{kg}\).Simplifying equations:\[m_1v_1 = m_1v_1' + m_2v_2'\]\[4.85 \times 1.60 = 1.60v_1' + 2.40v_2'\]\[7.76 = 1.60v_1' + 2.40v_2'\]Using conservation of kinetic energy:\[\frac{1}{2} \times 1.60 \times (4.85)^2 = \frac{1}{2} \times 1.60 \times v_1'^2 + \frac{1}{2} \times 2.40 \times v_2'^2\]\[18.8352 = 0.8v_1'^2 + 1.2v_2'^2\]Solving these simultaneous equations, we find:\[v_1' = -0.97, v_2' = 3.88 \, \text{m/s}\]
03

Interpretation of Velocities

Interpret the results for the velocity directions. The negative velocity for the ball after collision \(v_1' = -0.97 \, \text{m/s}\) implies it's moving in the opposite direction to its initial movement. The block moves with a velocity of \(v_2' = 3.88 \, \text{m/s}\) in the same direction as the ball was initially moving.

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Key Concepts

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

Conservation of Momentum
In elastic collisions, the total momentum of the system before and after the collision remains constant. This principle is known as the conservation of momentum.
The momentum of an object is calculated using the formula:
\[ p = mv \]
where \( p \) represents momentum, \( m \) is the mass, and \( v \) is the velocity.
In the initial setup, the ball is in motion and has momentum, while the block is stationary with zero momentum. At the moment of collision, the momentum is transferred between the ball and the block. The conservation of momentum can be mathematically expressed as:
  • Before collision: \( m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2' \)
  • Where \( v_1 \) and \( v_2 \) are the velocities of the ball and block before collision, and \( v_1' \) and \( v_2' \) are their respective velocities after collision.
In this scenario, the momentum lost by the ball is equal to the momentum gained by the block, consistent with the law of conservation of momentum.
Conservation of Energy
The conservation of energy principle states that the total energy in a closed system remains constant over time. In the context of this problem, it plays a pivotal role in determining the ball's velocity just before colliding with the block.
The energy considered here includes both kinetic and potential energy. Initially, the ball is at rest, possessing potential energy due to its elevation. As it descends, the potential energy converts into kinetic energy.
Right before it hits the block, the ball's entire gravitational potential energy has transformed into kinetic energy, as expressed by the equation:
  • Potential Energy at the top, \( U = mgh \), becomes Kinetic Energy just before impact \( K = \frac{1}{2}mv^2 \)
  • Thus, \( mgh = \frac{1}{2}mv^2 \)
This conversion helps us calculate the velocity of the ball just before the collision, relying solely on the conservation of energy.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When the ball is released and moves downward, it transforms its initial potential energy into kinetic energy, which is crucial before the collision.
Calculating kinetic energy involves:
\[ K = \frac{1}{2}mv^2 \]
where \( K \) is kinetic energy, \( m \) is mass, and \( v \) is velocity.Post collision, the kinetic energy is distributed between the ball and block. In an elastic collision, such as this one, both momentum and kinetic energy are conserved. This means the system's total kinetic energy before and after the collision is equal.
The conservation can be set up as:
  • Before collision: \( \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \)
  • After collision: \( \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2 \)
Knowing what kinetic energy represents and how it balances before and after the collision is key to solving the velocities of the ball and block after impact.
Gravitational Potential Energy
Gravitational potential energy is an object's energy stored due to its height above a reference point, often associated with the earth's gravity. The ball held at the horizontal position initially has maximum potential energy.
The formula to calculate gravitational potential energy is:
\[ U = mgh \]
where \( U \) is potential energy, \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height.
As the ball swings downward, this potential energy gets converted to kinetic energy. At the lowest point of the descent, where the collision occurs, all the energy is kinetic. This perfect energy transformation is fundamental to understanding how the ball gains speed as it falls and hits the block.Gravitational potential energy gives insight into the energy changes at different points of the ball’s motion, showcasing the inherent link between potential and kinetic energy in this system.

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