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Chapter 2: Problem 25

Small steel balls fall from rest through the opening at \(A\) at the steady rate of two per second. Find the vertical separation \(h\) of two consecutive balls when the lower one has dropped 3 meters. Neglect air resistance.

Short Answer

Expert verified
The vertical separation is approximately 2.616 meters.

Step by step solution

01

Analyze the situation

Two steel balls are falling at a rate of two per second. We need to find the vertical separation between two consecutive balls when the first one has fallen 3 meters. They are dropping under gravity, starting from rest, so we'll use the kinematic equation for free fall.
02

Use kinematic equation for the first ball

The first ball falls from rest, so its initial velocity is 0. It has fallen 3 meters, with acceleration due to gravity \(g = 9.8 \text{ m/s}^2\). The kinematic equation for displacement \(s\) is given by \(s = ut + \frac{1}{2} g t^2\), where \(u\) is the initial velocity and \(t\) is the time. Since \(u = 0\), the equation simplifies to \(3 = \frac{1}{2} \times 9.8 \times t^2\). Solving for \(t\), we get \(t^2 = \frac{6}{9.8}\) and \(t = \sqrt{\frac{6}{9.8}}\).
03

Calculate time period and find time for first ball

Calculating, \(t = \sqrt{\frac{6}{9.8}} \approx 0.78 \text{ seconds}\). This is the time taken for the first ball to fall 3 meters.
04

Determine time gap between release of consecutive balls

Since the balls are released at a rate of two per second, there is a \(\frac{1}{2}\) second (0.5 seconds) time gap between the release of consecutive balls.
05

Calculate fall time of the second ball

During the 0.5 seconds before the second ball is released, the first ball has already fallen some distance. The second ball is released 0.5 seconds after the first, so when the first ball has been falling for 0.78 seconds, the second ball has only been falling for \(0.78 - 0.5 = 0.28 \text{ seconds}\).
06

Calculate distance fallen by the second ball

Using the kinematic equation \(s = \frac{1}{2} g t^2\) again for the second ball, with \(t = 0.28\) seconds, we calculate the distance fallen: \(s = \frac{1}{2} \times 9.8 \times (0.28)^2 = \frac{1}{2} \times 9.8 \times 0.0784 \approx 0.384 \text{ meters}\).
07

Find the vertical separation between the two balls

The first ball has fallen 3 meters, and the second ball has fallen 0.384 meters. Thus, the vertical separation \(h\) is the difference between these distances: \(h = 3 - 0.384 = 2.616 \text{ meters}\).

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Key Concepts

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

Free Fall
Free fall describes the motion of an object exclusively under the influence of gravity. When objects free fall in physics, air resistance is often ignored, simplifying calculations. Here, the balls are said to fall from rest (starting velocity is zero), so the only force acting on them is gravity.
The principle simplifies decision-making and calculations in physics by focusing solely on the gravitational acceleration, which remains constant for objects near Earth's surface. This simplification helps in using the kinematic equations effectively to solve real-world problems like our current exercise.
Acceleration Due to Gravity
The acceleration due to gravity, denoted by the symbol \( g \), is a constant value on the surface of the Earth. In our calculations, it is approximately 9.8 m/s虏. This value represents how quickly any free-falling object will speed up as it continues to fall.
In physics problems involving free fall, \( g = 9.8 ext{ m/s}^2 \) is typically used. This consistently helps in predicting how long it will take for objects to fall a certain distance and how fast they will be going at any point.
Because \( g \) is uniform, once you determine it, you can apply it to objects in free fall to model their movement predictably, making it a key component of kinematic equations.
Time of Flight
The time of flight is the duration an object spends in motion from the moment it begins its fall to when it reaches a specified point. Using kinematic equations, you can find this value by solving for time, \( t \).
For this exercise, once we know the distance (3 meters in the case of the lower ball), we use the rearranged kinematic equation: \( s = \frac{1}{2} g t^2\). Solving for \( t \) yields the time it takes for the ball to fall that distance.
The second ball's time of flight starts 0.5 seconds after the first ball, affecting its position relative to the first ball, which is critical in determining the vertical separation between the two.
Displacement Calculation
Displacement is a vector quantity that refers to the change in position of an object. It can be determined using the kinematic formula, which, for an object in free fall from rest, is: \( s = \frac{1}{2} g t^2\).
In this problem, to find the distance each ball has fallen, we plug the time values we calculated into this equation. For example, at 0.78 seconds, the first ball has fallen 3 meters. The second ball, starting its journey 0.5 seconds later, falls a shorter distance, calculated to be approximately 0.384 meters after 0.28 seconds.
The vertical separation, \( h \), between the two balls is simply the difference in their displacements. Knowing how to set up and solve these equations helps analyze motion accurately in physics.

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Most popular questions from this chapter

The acceleration of a particle is given by \(a=-k s^{2}\) where \(a\) is in meters per second squared, \(k\) is a constant, and \(s\) is in meters. Determine the velocity of the particle as a function of its position \(s\). Evaluate your expression for \(s=5 \mathrm{m}\) if \(k=0.1 \mathrm{m}^{-1} \mathrm{s}^{-2}\) and the initial conditions at time \(t=0\) are \(s_{0}=3 \mathrm{m}\) and \\[ v_{0}=10 \mathrm{m} / \mathrm{s} \\]

When the effect of aerodynamic drag is included, the \(y\) -acceleration of a baseball moving vertically upward is \(a_{u}=-g-k v^{2},\) while the acceleration when the ball is moving downward is \(a_{d}=-g+k v^{2},\) where \(k\) is a positive constant and \(v\) is the speed in meters per second. If the ball is thrown upward at \(30 \mathrm{m} / \mathrm{s}\) from essentially ground level, compute its maximum height \(h\) and its speed \(v_{f}\) upon impact with the ground. Take \(k\) to be \(0.006 \mathrm{m}^{-1}\) and assume that \(g\) is constant.

Train \(A\) is traveling at a constant speed \(v_{A}=\) \(35 \mathrm{mi} / \mathrm{hr}\) while car \(B\) travels in a straight line along the road as shown at a constant speed \(v_{B}\). A conductor \(C\) in the train begins to walk to the rear of the train car at a constant speed of \(4 \mathrm{ft} / \mathrm{sec}\) relative to the train. If the conductor perceives car \(B\) to move directly westward at \(16 \mathrm{ft} / \mathrm{sec},\) how fast is the car traveling?

At time \(t=0,\) the position vector of a particle moving in the \(x\) -y plane is \(\mathbf{r}=5 \mathbf{i} \mathrm{m} .\) By time \(t=0.02 \mathrm{s}\) its position vector has become \(5.1 \mathrm{i}+0.4 \mathrm{jm}\) Determine the magnitude \(v_{\mathrm{av}}\) of its average velocity during this interval and the angle \(\theta\) made by the average velocity with the positive \(x\) -axis.

A batter hits the baseball \(A\) with an initial velocity of \(v_{0}=100 \mathrm{ft} / \mathrm{sec}\) directly toward fielder \(B\) at an angle of \(30^{\circ}\) to the horizontal; the initial position of the ball is \(3 \mathrm{ft}\) above ground level. Fielder \(B\) requires \(\frac{1}{4}\) sec to judge where the ball should be caught and begins moving to that position with constant speed. Because of great experience, fielder \(B\) chooses his running speed so that he arrives at the "catch position" simultaneously with the baseball. The catch position is the field location at which the ball altitude is \(7 \mathrm{ft}\). Determine the velocity of the ball relative to the fielder at the instant the catch is made.
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