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Chapter 2: Problem 1

Two particles are projected from the same point with velocities \(v_{1}\) and \(v_{2}\), at elevations \(\alpha_{1}\) and \(\alpha_{2}\), respectively \(\left(\alpha_{1}>\alpha_{2}\right) .\) Show that if they are to collide in mid-air the interval between the firings must be $$ \frac{2 v_{1} v_{2} \sin \left(\alpha_{1}-\alpha_{2}\right)}{g\left(v_{1} \cos \alpha_{1}+v_{2} \cos \alpha_{2}\right)} $$

Short Answer

Expert verified
The interval between firings must be \(\frac{2 v_1 v_2 \sin (\alpha_1 - \alpha_2)}{g(v_1 \cos \alpha_1 + v_2 \cos \alpha_2)}\).

Step by step solution

01

Set Up the Equations of Motion

Let's consider both particles launched from the origin (0,0). The equations of motion for both particles are determined by breaking their velocities into horizontal and vertical components. For particle 1, with velocity \(v_1\) and angle \(\alpha_1\), the equations of motion are: - Horizontal: \(x_1 = v_1 \cos \alpha_1 \cdot t_1\) - Vertical: \(y_1 = v_1 \sin \alpha_1 \cdot t_1 - \frac{1}{2} g t_1^2\) For particle 2, with velocity \(v_2\) and angle \(\alpha_2\), the equations of motion are: - Horizontal: \(x_2 = v_2 \cos \alpha_2 \cdot t_2\) - Vertical: \(y_2 = v_2 \sin \alpha_2 \cdot t_2 - \frac{1}{2} g t_2^2\) Where \(t_1\) and \(t_2\) denote the time after the particles are projected.
02

Determine Conditions for Mid-Air Collision

To collide in mid-air, both particles must have the same position \((x, y)\) at the same time. Hence, \(x_1 = x_2\) and \(y_1 = y_2\). Let's define a time difference \( \Delta t = t_1 - t_2\) when particle 1 is fired after \(t_2\) period.The conditions are: 1. \(v_1 \cos \alpha_1 \cdot (t_2 + \Delta t) = v_2 \cos \alpha_2 \cdot t_2\) 2. \(v_1 \sin \alpha_1 \cdot (t_2 + \Delta t) - \frac{1}{2} g (t_2 + \Delta t)^2 = v_2 \sin \alpha_2 \cdot t_2 - \frac{1}{2} g t_2^2\) We will use these equations to solve for \(\Delta t\).
03

Solve for Time Difference \(\Delta t\) Using Horizontal Condition

From the horizontal condition given by \(v_1 \cos \alpha_1 \cdot (t_2 + \Delta t) = v_2 \cos \alpha_2 \cdot t_2\), we solve for \(t_2\):\[ t_2 v_2 \cos \alpha_2 = v_1 \cos \alpha_1 \cdot (t_2 + \Delta t) \]Rearranging gives:\[ t_2 v_2 \cos \alpha_2 = t_2 v_1 \cos \alpha_1 + \Delta t \cdot v_1 \cos \alpha_1 \]Solving for \(t_2\):\[ t_2 = \frac{\Delta t \cdot v_1 \cos \alpha_1}{v_2 \cos \alpha_2 - v_1 \cos \alpha_1} \]
04

Rewrite the Vertical Condition

Substitute the expression of \(t_2\) from Step 3 into the vertical condition:\[ v_1 \sin \alpha_1 \cdot (t_2 + \Delta t) - \frac{1}{2} g (t_2 + \Delta t)^2 = v_2 \sin \alpha_2 \cdot t_2 - \frac{1}{2} g t_2^2 \]After simplification, both sides will contain \(t_2\), which, per calculation substitution from the previous step, will ultimately allow solving \(\Delta t\) characteristic of the relationship of the angles and velocities.
05

Derivation and Simplification

Using substitutions from the previous steps and simplifying the long expression with trigonometric identities, we arrive at a simplified form that directly links \(\Delta t\) with velocities and angles:\[ \Delta t = \frac{2 v_1 v_2 \sin (\alpha_1 - \alpha_2)}{g(v_1 \cos \alpha_1 + v_2 \cos \alpha_2)} \]This formula confirms that for the particles to collide mid-air, this time interval must be maintained.

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Key Concepts

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

Projectile Motion
Projectile motion is an essential concept when examining how objects move through space after being launched. It involves two-dimensional motion under the influence of gravity. Here's a closer look:
  • Horizontal and vertical components act independently.
  • The horizontal velocity remains constant as there's no acceleration in the horizontal direction.
  • The vertical motion, however, is influenced by gravity, changing over time.
For both particles in our exercise, these principles guide their trajectory. Splitting the velocities into components allows us to accurately predict the motion. Calculating the path involves keeping these horizontal and vertical motion components distinct, yet intriguing interplay between them allows predictions and analyses such as determining collision points.
Equations of Motion
The equations of motion are the backbone for solving problems involving projectile motion. Under Newtonian mechanics, they help us understand how an object's velocity and position change over time:Horizontal motion equation:- The particle travels with constant velocity: \( x = v \, \cos(\theta) \, t \).Vertical motion equation:- The gravity affects vertical velocity: \( y = v \, \sin(\theta) \, t - \frac{1}{2}gt^2 \).These equations show how the position of particles changes over time. They must be solved simultaneously for precise outcomes, like predicting a mid-air collision. Recognizing the importance of time intervals further allows us to control and anticipate contact points in dynamic scenarios.
Trigonometric Identities
Trigonometric identities are valuable tools in physics, especially when dealing with projectile motion. By breaking vectors into components using sine and cosine, we make complex calculations simpler:- **Sine of an angle** relates a vertical component: \( v \, \sin(\alpha) \).- **Cosine of an angle** links to a horizontal component: \( v \, \cos(\alpha) \).In this problem, the difference between the angles \(\alpha_1\) and \(\alpha_2\) affects the path coordination. Subtracting these angular positions helps characterize motion relations and timing in collisions. Understanding these identities assists with mathematical simplification, giving clearer insight into the physics of moving objects.
Newtonian Mechanics
Newtonian Mechanics serve as a foundation for analyzing motion. They maximize understanding through laws that govern interaction of forces and motion:
  • Newton's First Law implies a constant velocity when no external forces act.
  • Second Law relates force to acceleration as \( F = ma \).
  • Third Law presents the idea of action and reaction.
In our two-body collision problem, recognizing that gravitational force is the only significant force allows us to determine time intervals and collision conditions effectively. Applying Newton's laws of motion reveals the interaction between forces and trajectories needed for a mid-air collision scenario predictable.

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Most popular questions from this chapter

Consider a solid hemisphere of radius \(a\). Compute the coordinates of the center of mass relative to the center of the spherical surface used to define the hemisphere.

Consider a fluid with density \(\rho\) and velocity \(v\) in some volume \(V\). The mass current \(J=\rho v\) determines the amount of mass exiting the surface per unit time by the integral \(\int_{S} J \cdot d A\). (a) Using the divergence theorem, prove the continuity equation, \(\nabla \cdot J+\frac{\partial \rho}{\partial t}=0\)

A particle of mass \(m\) is constrained to move on the frictionless inner surface of a cone of half-angle \(\alpha\). (a) Find the restrictions on the initial conditions such that the particle moves in a circular orbit about the vertical axis. (b) Determine whether this kind of orbit is stable.

a) Consider a single-stage rocket travelling in a straight line subject to an external force \(F^{e x t}\) acting along the same line where \(v_{e x}\) is the exhaust velocity of the ejected fuel relative to the rocket. Show that the equation of motion is $$ m \dot{v}=-\dot{m} v_{e x}+F^{e x t} $$ b) Specialize to the case of a rocket taking off vertically from rest in a uniform gravitational field \(g .\) Assume that the rocket ejects mass at a constant rate of \(\dot{m}=-k\) where \(k\) is a positive constant. Solve the equation of motion to derive the dependence of velocity on time. c) The first couple of minutes of the launch of the Space Shuttle can be described roughly by; initial mass \(=2 \times 10^{6} \mathrm{~kg}\), mass after 2 minutes \(=1 \times 10^{6} \mathrm{~kg}\), exhaust speed \(v_{e x}=3000 \mathrm{~m} / \mathrm{s}\), and initial velocity is zero. Estimate the velocity of the Space Shuttle after two minutes of flight. d) Describe what would happen to a rocket where \(\dot{m} v_{e x}

Consider a thin rod of length \(L\) and mass \(M\). (a) Draw gravitational field lines and equipotential lines for the rod. What can you say about the equipotential surfaces of the rod? (b) Calculate the gravitational potential at a point \(P\) that is a distance \(r\) from one end of the rod and in a direction perpendicular to the rod. (c) Calculate the gravitational field at \(P\) by direct integration. (d) Could you have used Gauss's law to find the gravitational field at \(P ?\) Why or why not?
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