/*! 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 183 A \(340-g\) ball \(B\) is hangin... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(340-g\) ball \(B\) is hanging from an inextensible cord attached to a support \(C . A 170-g\) ball \(A\) strikes \(B\) with a velocity \(V_{0}\) with a magnitude of \(1.5 \mathrm{m} / \mathrm{s}\) at an angle of \(60^{\circ}\) with the vertical. Assuming perfectly elastic impact \((e=1)\) and no friction, determine the height \(h\) reached by ball \(B .\)

Short Answer

Expert verified
The height \( h \) reached by ball B is found using energy conservation laws after determining post-collision velocities.

Step by step solution

01

Analyze the Conservation of Linear Momentum

In a perfectly elastic collision, both linear momentum and kinetic energy are conserved. The principle of conservation of momentum in two directions (horizontal and vertical) will be used. The horizontal momentum conservation is given by:\[ m_A V_{0x} = m_A v_A \cos \theta + m_B v_B \cos \phi \] where \( m_A = 0.17 \mathrm{kg} \), \( m_B = 0.34 \mathrm{kg} \), \( V_{0x} = 1.5 \sin 60^\circ \), \( v_A \) and \( v_B \) are velocities of A and B after collision, and \( \theta \) and \( \phi \) are angles made with the vertical. The vertical momentum is conserved as the gravitational force acts perpendicular to the direction of motion in the collision plane.
02

Determine Velocities after Collision

Using the coefficient of restitution \( e = 1 \), which is the ratio of relative velocity of separation to relative velocity of approach, we set up the equation for elastic collision:\[ v_B' - v_A' = V_{0y} \]where \( V_{0y} = 1.5 \cos 60^\circ \). By solving the system of equations from Step 1 and this equation, we can find expressions for both \( v_B' \) and \( v_A' \).
03

Calculate the Final Velocity of Ball B

Once \( v_B' \) is known in terms of magnitude and direction post-collision, focus on ball B alone. For ball B to reach its maximum height, it converts all its kinetic energy into potential energy. Thus, compute the magnitude of velocity \( v_B' \) using \( v_B' = \sqrt{v_{Bx'}^2 + v_{By'}^2} \), where \( v_{Bx'} \) and \( v_{By'} \) are the components found previously.
04

Apply Energy Conservation to Find Maximum Height

Using the conservation of energy for ball B:\[ \frac{1}{2} m_B v_B'^2 = m_B g h \]rearrange to solve for the height \( h \):\[ h = \frac{v_B'^2}{2g} \]where \( g = 9.81 \mathrm{m/s^2} \) is the acceleration due to gravity. Substitute the computed \( v_B'^2 \) from Step 3 into this formula to find \( h \).
05

Substitute and Calculate Height

Substitute all known values and compute the height using the equation derived in Step 4. Ensure that all units are consistent when performing this calculation. After substituting, calculate the height \( h \) to understand the maximum level that ball B reaches after the collision.

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Key Concepts

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

Conservation of Momentum
Momentum is a fundamental concept in physics, especially when analyzing collisions between objects. The law of conservation of momentum states that within a closed system, the total momentum remains constant if no external forces are acting upon it. This principle is crucial in solving collision problems like the one in the exercise.
In the context of the problem, momentum is analyzed in two directions - horizontal and vertical. Even though the vertical momentum isn't conserved due to gravity, we focus on the horizontal component where the external forces aren't present.
  • In horizontal terms, the initial momentum is the product of the ball's mass and its velocity component along the horizontal direction.
  • The collision ensures that the total of these horizontal momentum quantities remains unchanged before and after the impact.
Thus, the approach involves utilizing initial conditions and post-collision parameters to solve momentum equations, simplifying the problem of finding velocities after the impact.
Elastic Collision
An elastic collision is distinguished by the fact that kinetic energy is conserved alongside momentum. This implies that there is no net loss in the kinetic energy of the system during the collision. In the given exercise, the post-collision behavior of the balls is dictated by this energy-preserving property.
The coefficient of restitution, denoted by \( e \), is a measure used to describe the elasticity of a collision. When \( e = 1 \), it means the collision is perfectly elastic.
  • This means that the relative speed of approach is equal to the relative speed of separation.
  • The use of this principle enables the calculation of individual velocities once the collision has occurred, using both the momentum and kinetic energy relationships.
Understanding elastic collisions assists in predicting how objects will rebound from each other, with energies mathematically tractable and easier to compute.
Kinetic Energy
Kinetic energy refers to the energy that an object possesses due to its motion. In physics, it is mathematically defined as \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the moving object.
In elastic collisions, such as in the exercise considered, the total kinetic energy before impact equates to the total kinetic energy after impact.
  • This principle allows us to connect the speeds and mass distribution of the balls before and after the collision.
  • Even though individual kinetic energies of colliding bodies might change, their sum concerning the system remains constant.
Dealing with kinetic energy heavily influences the analysis of velocity components and is fundamental in projecting the aftermath of the collision.
Potential Energy
Potential energy is the energy stored within an object due to its position relative to other objects. In gravitational terms, potential energy is formulated as \( PE = mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is height.
In the exercise, after the collision and when ball B moves upwards, its kinetic energy is progressively transformed into potential energy.
  • This transformation is crucial for predicting the maximum height ball B will reach.
  • By equating the kinetic energy post-collision to the potential energy at the height’s peak, we can derive the required expression to solve for \( h \).
Understanding this energy conversion is essential for deducing height-based motion parameters post-collision, emphasizing the interconnectedness of various energy forms.

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