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Chapter 9: Problem 79

A 124-g Frisbee is lodged on a tree branch \(7.25 \mathrm{~m}\) above the ground. To free it, you lob a \(240-\mathrm{g}\) dirt clod vertically upward. The dirt leaves your hand at a point \(1.23 \mathrm{~m}\) above the ground, moving at \(18.7 \mathrm{~m} / \mathrm{s}\). It sticks to the Frisbee. Find (a) the maximum height reached by the Frisbee-dirt combination and (b) the speed with which the combination hits the ground.

Short Answer

Expert verified
The maximum height reached by the Frisbee-dirt combination is calculated in Step 2, and the speed with which the combination hits the ground is calculated in Step 3. Both calculations depend on the initial velocity, which is calculated in Step 1 using the law of conservation of momentum.

Step by step solution

01

Calculate the initial velocity of the Frisbee-dirt combination

By the law of conservation of momentum, the initial velocity after the two bodies stick together is given by \((m_1v_1 + m_2v_2) / (m_1 + m_2)\), where \(m_1\) and \(m_2\) are masses and \(v_1\) and \(v_2\) are velocities. Here, \(m_1 = 0.240 \mathrm{~kg}\) , \(v_1 = 18.7 \mathrm{~m/s}\), \(m_2 = 0.124 \mathrm{~kg}\) , \(v_2 = 0 \mathrm{~m/s}\). Substituting these values will give the initial velocity of the Frisbee-dirt combination.
02

Calculate the maximum height reached

We can use the second equation of motion: \(v^2 = u^2 + 2a s\), where \(v\) is final velocity, \(u\) is initial velocity, \(a\) is acceleration, and \(s\) is distance. Here, \(v = 0 \mathrm{~m/s}\) (since at maximum height, velocity will be zero), \(u\) is the velocity calculated in step 1, \(a = -9.8 \mathrm{~m/s^2}\) (since the acceleration due to gravity acts downward), and \(s\) is the distance to be calculated. Solving the equation for \(s\) gives the maximum height achieved from the initial point of throw. To get the maximum height from the ground, add the height at which the object was initially thrown, i.e., \(1.23 \mathrm{~m}\).
03

Calculate the final speed

To find the final speed with which the combination hits the ground, we can reuse the second equation of motion. Now, \(u\) is equal to the velocity we found in Step 1, \(s\) is the height we found in Step 2, \(v\) is the final velocity that we're trying to find, and acceleration \(a\) is still \( -9.8 \mathrm{~m/s^2}\). Solving for \(v\) will give us the final speed at which the Frisbee and dirt combination hits the ground.

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Key Concepts

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

Physics Problem Solving
When tackling physics problems like the one involving a Frisbee and a dirt clod, a structured approach is crucial. It begins with understanding the problem: identifying given data, what needs to be found, and the physics principles that apply. Here, the central theme is the conservation of momentum, where two objects stick together. This scenario requires calculating quantities like mass, velocity, and height.
  • Start by writing down all known quantities and identifying unknowns.
  • Determine which physical laws or equations relate these variables. In this problem, momentum conservation aids in finding the initial velocity after collision.
  • Use logical steps to carry the calculation through, carefully substituting and manipulating equations.
By following this framework, you'll systematically navigate through to a solution that reflects the fundamental laws of physics.
Projectile Motion
In this problem, after the dirt clod and Frisbee stick together, they behave as a single object moving under gravity's influence. This results in projectile motion, a type of movement where only gravitational force acts.
Key characteristics of projectile motion include:
  • The vertical motion is independent of horizontal motion.
  • Acceleration due to gravity constantly affects the object, pulling it downwards.
  • The peak height of the motion is reached when the vertical component of velocity is zero.
Understanding projectile motion helps predict an object's trajectory by analyzing initial conditions like speed and angle of launch, which in this vertical launch, becomes simply upward speed.
Therefore, recognizing these aspects allows you to anticipate how the object behaves, using the equations of motion to calculate heights and speeds at various points.
Equations of Motion
Equations of motion are the mathematical description of an object's motion under the influence of forces. They provide powerful tools for solving problems, like determining heights and velocities during projectile motion.
In vertical motion problems, such as this exercise, the relevant equation is:\[ v^2 = u^2 + 2as \]Where:
  • \( v \): final velocity
  • \( u \): initial velocity
  • \( a \): acceleration (usually gravity, \( -9.8 \, \mathrm{m/s^2} \))
  • \( s \): distance traveled
This equation helps find variables like maximum height (when \( v = 0 \)) or final speed when returning to ground using values for \( u \) and \( s \).
Understanding these equations empowers you to calculate various aspects of motion practically and accurately.

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Most popular questions from this chapter

A car moving at speed \(v\) undergoes a one-dimensional collision with an identical car initially at rest. The collision is neither elastic nor fully inelastic; \(7 / 32\) of the initial kinetic energy is lost. Find the velocities of the two cars after the collision.

A block of mass \(M\) is moving at speed \(v_{0}\) on a frictionless surface that ends in a rigid wall, heading toward a stationary block of mass \(n M\), where \(n \geq 1\) (Fig. 9.29). Collisions between the two blocks or the left-hand block and the wall are elastic and one-dimensional. (a) Show that the blocks will undergo only one collision with each other if \(n \leq 3\). (b) Show that the blocks will undergo two collisions with each other if \(n=4 .\) (c) How many collisions will the blocks undergo if \(n=10\), and what will be their final speeds?

Two identical trucks have mass \(5500 \mathrm{~kg}\) when empty, and the maximum permissible load for each is \(8000 \mathrm{~kg}\). The first truck, carrying \(3800 \mathrm{~kg}\), is at rest. The second truck plows into it at \(65 \mathrm{~km} / \mathrm{h}\), and the pair moves away at \(37 \mathrm{~km} / \mathrm{h}\). As an expert witness, you're asked to determine whether the second truck was overloaded. What do you report?

Three \(200-\mathrm{g}\) objects have velocities given by \(\vec{v}_{1}=15.0 \hat{j} \mathrm{~m} / \mathrm{s}\), \(\vec{v}_{2}=6.7 \hat{i}-3.45 \hat{j} \mathrm{~m} / \mathrm{s}\), and \(\vec{v}_{3}=-6.7 \hat{i}-4.32 \hat{j} \mathrm{~m} / \mathrm{s}\). Find the kinetic energy of the center of mass and the internal kinetic energy of this system.

How is it possible to have a collision between objects that don't ever touch? Give an example of such a collision.
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