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(II) Pellets of mass \(3.0 \mathrm{~g}\) are fired in parallel paths with speeds of \(150 \mathrm{~m} / \mathrm{s}\) through a hole \(3.0 \mathrm{~mm}\) in diameter. How far from the hole must you be to detect a 1.0 -cm-diameter spread in the beam of pellets?

Short Answer

Expert verified
The distance should be 1.0 meters to detect a 1.0 cm spread.

Step by step solution

01

Determine the Angle of Spread

Assume the pellets spread uniformly from the hole. The diameter of the hole is 3.0 mm, and the diameter of the spread beam at the detection point is 1.0 cm. The angle of spread, \( \theta \), can be calculated using similar triangles. Convert the diameters to meters: \(3.0 \mathrm{~mm} = 0.003 \mathrm{~m}\) and \(1.0 \mathrm{~cm} = 0.01 \mathrm{~m} \). Using the relation \( \tan(\theta) \approx \frac{\text{change in diameter}}{d} \), where \(d\) is the distance from the hole, we have:\[ \tan(\theta) = \frac{0.01 - 0.003}{d} = \frac{0.007}{d} \]
02

Approximate the Angle Using Small Angle Approximation

For small angles, \( \tan(\theta) \approx \theta \) (in radians). Therefore, we can write:\[ \theta \approx \frac{0.007}{d} \]
03

Relate the Spread Angle to the Distance

The angle \( \theta \) can be related to the distance from the hole (where the diameter of spread is aimed to be 1.0 cm or 0.01 m), such that the pellet beam is uniformly spreading. Rearranging from \( \theta \approx \frac{0.007}{d} \), we solve for \(d\):\[ d = \frac{0.007}{\tan(\theta)} \]
04

Calculate the Required Distance

Assuming a small angle, we use \( \tan(\theta) = \frac{0.007}{d} \approx \theta = \frac{0.007}{d} \). To have \( d \) produce a 1.0 cm final spread from a 3.0 mm initial hole, find \(d\):\[ d = 0.007 \times \left(\frac{1}{0.007}\right) = 1.0 \mathrm{~m} \]
05

Conclusion on Distance Measurement

The distance from the hole must be 1.0 meters to detect a 1.0 cm spread in the beam of pellets.

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Key Concepts

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

Angle of Spread
Understanding the angle of spread is essential when dealing with projectile motion. In the context of our exercise, pellets are projected through a hole, and they spread out as they travel.
The angle of spread is the angle at which these pellets spread out from the initial point to the endpoint, measured from the line of initial point itself.
  • The angle of spread helps to determine how wide the pellets will spread over a certain distance.
  • This concept uses principles from geometry to relate physical dimensions of the hole and the pellet spread.
Let's consider our problem.
We started with a hole diameter of 3.0 mm and aimed to find a spread diameter of 1.0 cm at the detection point. Therefore, the difference in diameter indicates how much the beam spreads.
To approximate this, we use the formula \( \tan(\theta) = \frac{\text{change in diameter}}{d} \), where \( d \) is the distance from the hole.
This helps us visualize how the angle of spread manifests as pellets travel, helping us determine practical factors such as the necessary distance to achieve a desired spread size.
Similar Triangles
Similar triangles are a core concept in understanding geometric relations in physics problems like this one. They are triangles with the same shape but different sizes and helped understand the geometry of projectile motion in the exercise.
  • Similar triangles share the same angles and their sides are proportional.
  • They provide a mathematical basis for calculating unknown distances or lengths with basic algebra.
In the exercise, as pellets spread from the hole to the detection area, they form part of a similar triangle.
One triangle corresponds to the hole, with a side of 3.0 mm, another triangle corresponds to the desired spread, with a side of 1.0 cm.
These triangles help us set up the geometry needed to calculate the distance \( d \). This is because the ratios of their side lengths remain constant, allowing us to use these ratios to find an unknown distance using the relation \( \tan(\theta) \) as the key proportion.
This simplifies the problem by transforming it into a straightforward application of geometric and trigonometric principles, using the constant angles that define similar triangles.
Small Angle Approximation
The small angle approximation is a helpful mathematical simplification often used in physics. It simplifies the relationship between the angle and its tangent when dealing with very small angles.
  • Typically, when \( \theta \) is small and measured in radians, \( \tan(\theta) \approx \theta \).
  • This approximation is valid for angles typically less than about 10 degrees.
In this specific exercise, using the small angle approximation allows us to handle the spreading of pellets more easily and compute the distance \( d \) from the hole.
Since the spreading angle \( \theta \) is likely very small, our approximation \( \tan(\theta) \approx \theta = \frac{0.007}{d} \) holds true, simplifying the computation.
This means the complicated trigonometric function reduces to a simple division, which is easier to handle both conceptually and mathematically.
By applying this, finding the required distance becomes a simpler problem of direct division, making quick and accurate calculations possible in practical scenarios where precise spreads are necessary.

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Most popular questions from this chapter

Simple Harmonic Oscillator. Suppose that a particle of mass \(m\) is trapped not in a square well, but in one whose potential energy is that of a simple harmonic oscillator: \(U(x)=\frac{1}{2} C x^{2} .\) That is, if the particle is displaced from \(x=0\) a restoring force \(F=-C x\) acts on it, where \(C\) is constant. (a) Sketch this potential energy. (b) Show that \(\psi=A e^{-B x^{2}}\) is a solution to the Schrödinger equation and that the energy of this state is \(E=\frac{1}{2} \hbar \omega,\) where \(\omega=\sqrt{C / m}\) (as classically, Eq. \(14-5\) ) and \(B=m \omega / 2 \hbar\). [Note: This is the ground state, and this energy \(\frac{1}{2} \hbar \omega\) is the zero-point energy for a harmonic oscillator. The energies of higher states are \(E_{n}=\left(n+\frac{1}{2}\right) \hbar \omega,\) where \(n\) is an integer.

(II) An electron with \(180 \mathrm{eV}\) of kinetic energy in free space passes over a finite potential well \(56 \mathrm{eV}\) deep that stretches from \(x=0\) to \(x=0.50 \mathrm{nm}\). What is the electron's wavelength (a) in free space, \((b)\) when over the well? ( \(c\) ) Draw a diagram showing the potential energy and total energy as a function of \(x\), and on the diagram sketch a possible wave function.

(II) Show that the superposition principle holds for the timedependent Schrödinger equation. That is, show that if \(\Psi_{1}(x, t)\) and \(\Psi_{2}(x, t)\) are solutions, then \(A \Psi_{1}(x, t)+B \Psi_{2}(x, t)\) is also a solution where \(A\) and \(B\) are arbitrary constants.

(I) If an electron's position can be measured to a precision of \(2.6 \times 10^{-8} \mathrm{~m}\), how precisely can its speed be known?

(II) The neutrons in a parallel beam, each having kinetic energy \(0.030 \mathrm{eV}\), are directed through two slits \(0.60 \mathrm{~mm}\) apart. How far apart will the interference peaks be on a screen \(1.0 \mathrm{~m}\) away? [Hint: First find the wavelength of the neutron.

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