/*! 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 22 Figure 15-34 shows block 1 of ma... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 22

Figure 15-34 shows block 1 of mass \(0.200 \mathrm{~kg}\) sliding to the right over a frictionless elevated surface at a speed of \(8.00 \mathrm{~m} / \mathrm{s}\). The block undergoes an elastic collision with stationary block 2 , which is attached to a spring of spring constant \(1208.5 \mathrm{~N} / \mathrm{m}\). (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in SHM with a period of \(0.140 \mathrm{~s}\), and block 1 slides off the opposite end of the elevated surface, landing a distance \(d\) from the base of that surface after falling height \(h=4.90 \mathrm{~m} .\) What is the value of \(d ?\)

Short Answer

Expert verified
Use projectile motion and SHM principles: \(d = v_{1f} \times t_f\). Calculate each factor separately.

Step by step solution

01

Calculate Velocity of Block 2

Since this is an elastic collision, both momentum and kinetic energy are conserved. Let's denote the velocity of block 2 after the collision as \(v_2\) and the initial velocity of block 1 as \(v_1 = 8.00\, \text{m/s}\). Using the relationship for velocity in elastic collisions, we have:\[ v_2 = \frac{2m_1}{m_1 + m_2} \times v_1 \]But it is given that block 2 undergoes Simple Harmonic Motion (SHM) with a period. Therefore, first we find the mass of block 2, \(m_2\), using the period of SHM formula:\[ T = 2\pi \sqrt{\frac{m_2}{k}} \]Plugging in the values: \(T = 0.140\, \text{s}\), \(k = 1208.5\, \text{N/m}\).\[ m_2 = \left(\frac{T}{2\pi}\right)^2 \times k\] and we find \(m_2\). Once \(m_2\) is found, plug it into the formula for \(v_2\).
02

Calculate Time of Flight

After the collision, block 1 slides off the surface. When it leaves the elevated surface, it becomes a projectile and moves under the influence of gravity only. The time \( t_f \) taken to fall a height \(h = 4.90 \text{ m}\) can be calculated using:\[ h = \frac{1}{2}gt_f^2 \]Solving for \( t_f \) gives:\[ t_f = \sqrt{\frac{2h}{g}} \]Substituting \( g = 9.81 \text{ m/s}^2 \) and \( h = 4.90 \text{ m} \), compute \( t_f \).
03

Calculate Horizontal Distance

The horizontal distance \(d\) that block 1 travels before hitting the ground can be found from its horizontal velocity post-collision and the time of flight. Denote the velocity of block 1 post-collision as \(v_{1f}\):\[ v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} \times v_1 \]Then, the horizontal distance \(d\) is:\[ d = v_{1f} \times t_f \]Substitute the values obtained for \( v_{1f} \) from Step 1 and for \( t_f \) from Step 2 and solve for \(d\).
04

Solve for d

Using the results from the previous steps:1. Calculate the post-collision velocity \(v_{1f}\), using \( m_2 \) from Step 1.2. Substitute \(v_{1f}\) and \(t_f\) into the formula for \(d\).For example, if \(m_2\) and \(v_{1f}\) are calculated from their respective formulas\\[ d = v_{1f} \times t_f \]Calculate \(d\) using:- \(v_{1f} \) resulting from the conservation of momentum and after obtaining \(m_2\).- \(t_f \) as obtained in Step 2.

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Key Concepts

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

Momentum Conservation
In an elastic collision, one of the most crucial principles is the conservation of momentum. Momentum, a measure of motion for an object, is given by the product of its mass and velocity. During an elastic collision, the total momentum of the system before the collision is equal to the total momentum after the collision. This means:
  • Before collision:\[p_{initial} = m_1 \times v_1 + m_2 \times 0\]
  • After collision:\[p_{final} = m_1 \times v_{1f} + m_2 \times v_2\]
By equating these, we can solve for the velocities of the blocks after the collision. This conservation law is fundamental because it allows us to predict the outcome of collision events. Understanding this concept lets us analyze the motion of objects post-collision, such as the blocks in this exercise.
Kinetic Energy Conservation
Alongside momentum, kinetic energy is another key quantity conserved in elastic collisions. Kinetic energy is the energy that an object possesses due to its motion and is calculated using the formula \( KE = \frac{1}{2}mv^2 \). In an elastic collision, the total kinetic energy before the collision equals the total kinetic energy after. This helps in determining the velocities after collision, ensuring that no kinetic energy is lost to sound, heat, or deformation.
  • Initial kinetic energy:\[KE_{initial} = \frac{1}{2} m_1 v_1^2\]
  • Final kinetic energy:\[KE_{final} = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_2^2\]
Setting the kinetic energies equal to one another, we solve for the unknown velocities. This is crucial for problems involving elastic collisions because it provides another equation to find the variables needed.
Simple Harmonic Motion
After the collision, block 2 is linked to a spring and participates in simple harmonic motion (SHM), characterized by repetitive oscillations within a defined time period. SHM concerns the motion of block 2 once the collision has transferred energy to it. The period \(T\) of this oscillation provides valuable information about the system.
  • Period of SHM:\[T = 2\pi \sqrt{\frac{m_2}{k}}\]
  • The mass of block 2, \(m_2\), is crucial to determine from this formula, given \(k\), the spring constant.
Comprehending SHM allows us to understand how energy is stored and converted within a system, such as a spring-block system, with block 2 oscillating after receiving kinetic energy. Recognizing these oscillations helps in calculating the masses involved in the exercise.
Projectile Motion
After the collision, block 1 continues as a projectile when it slides off the elevated surface. Projectile motion describes the path of objects released in a gravitational field, neglecting air resistance. Block 1 falls from a height \(h = 4.90 \text{ m}\), under the influence of gravity, and its path can be described using basic kinematics.
  • Vertical motion is described by:\[h = \frac{1}{2}gt_f^2\]
  • Solving for time of flight, \(t_f\), gives:\[t_f = \sqrt{\frac{2h}{g}} \] using the known gravitational acceleration \(g = 9.81 \text{ m/s}^2\).
    • Ultimately, to find the horizontal distance \(d\), use:\[d = v_{1f} \times t_f\] , where \(v_{1f}\) is the horizontal velocity of block 1 after the collision.
Projectile motion principles allow the prediction of where the block will land, which is vital to solving for the horizontal distance in this problem.

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Most popular questions from this chapter

When a \(20 \mathrm{~N}\) can is hung from the bottom of a vertical spring, it causes the spring to stretch \(20 \mathrm{~cm} .\) (a) What is the spring constant? (b) This spring is now placed horizontally on a frictionless table. One end of it is held fixed, and the other end is attached to a \(5.0 \mathrm{~N}\) can. The can is then moved (stretching the spring) and released from rest. What is the period of the resulting oscillation?

A thin uniform rod \((\mathrm{mass}=0.50 \mathrm{~kg})\) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of \(1.5 \mathrm{~s}\) and an angular amplitude of \(10^{\circ} .\) (a) What is the length of the rod? (b) What is the maximum kinetic energy of the rod as it swings?

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of \(3.00 \mathrm{~Hz}\). (a) What is the spring constant of each spring if the mass of the car is \(1450 \mathrm{~kg}\) and the mass is evenly distributed over the springs? (b) What will be the oscillation frequency if five passengers, averaging \(73.0 \mathrm{~kg}\) each, ride in the car with an even distribution of mass?

A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position \(y_{i}\) such that the spring is at its rest length. The object is then released from \(y_{i}\) and oscillates up and down, with its lowest position being \(10 \mathrm{~cm}\) below \(y_{i}\). (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is \(8.0 \mathrm{~cm}\) below the initial position? (c) An object of mass \(300 \mathrm{~g}\) is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object? (d) How far below \(y_{i}\) is the new equilibrium (rest) position with both objects attached to the spring?

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. What is their phase difference?
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