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Chapter 2: Problem 20

A girl rolls a ball up an incline and allows it to return to her. For the angle \(\theta\) and ball involved, the acceleration of the ball along the incline is constant at \(0.25 g\), directed down the incline. If the ball is released with a speed of \(4 \mathrm{m} / \mathrm{s}\), determine the distance \(s\) it moves up the incline before reversing its direction and the total time \(t\) required for the ball to return to the child's hand.

Short Answer

Expert verified
The distance moved up the incline is 3.27 meters and the total time for the round trip is 3.26 seconds.

Step by step solution

01

Identify the parameters

First, let's identify all the given parameters. The acceleration of the ball along the incline is constant at \(a = 0.25 g\), where \(g\) is the acceleration due to gravity \(9.8 \text{ m/s}^2\), hence, \(a = 0.25 \times 9.8 = 2.45 \text{ m/s}^2\). The initial velocity \(v_0\) is \(4 \text{ m/s}\).
02

Determine the distance moved up the incline

We use the equation of motion: \(v^2 = v_0^2 - 2as\), where \(v = 0\) because the ball stops momentarily before reversing direction. Plug in the values: \[0 = (4)^2 - 2(2.45)s\]Solve for \(s\):\[s = \frac{16}{4.9} = 3.27 \text{ meters}\]
03

Calculate the time to reach the highest point

Use the kinematic equation: \(v = v_0 - at\), where \(v = 0\).\[0 = 4 - (2.45)t_1\]Solve for \(t_1\), the time to reach the highest point:\[t_1 = \frac{4}{2.45} = 1.63 \text{ seconds}\]
04

Calculate the total time for the round trip

The time for the ball to return is the same as the time to reach the highest point because acceleration down the incline is constant. Thus, total time \(t = 2t_1\).\[t = 2 \times 1.63 = 3.26 \text{ seconds}\]

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Key Concepts

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

Inclined Plane
An inclined plane is simply a flat surface that is tilted at a certain angle, known as \(\theta\), compared to the horizontal ground. It's a common scenario in physics problems.
The real magic of an inclined plane comes into play when we analyze the forces acting on a body placed on it. When you have a ball rolling up an incline, like in our exercise, gravity works to pull the ball back down. This gravitational force can be divided into two components:
  • One acting perpendicular to the plane
  • One acting parallel to it
The parallel component is what causes the ball to accelerate down the slope. This acceleration is why the problem specifies a certain fraction of gravity, such as \(0.25g\). It is along this incline rather than vertically downward.Understanding these components is essential. It helps in working out how an object behaves as it moves along the tilted surface.
Constant Acceleration
Constant acceleration means that the acceleration of the object does not change over time. In simpler terms, if a ball is accelerating downhill on an incline, its speed increases at a steady rate.For the problem we have, the acceleration value is constant and equals \(0.25g\), which translates to \(2.45 \text{ m/s}^2\). A constant acceleration makes mathematical calculations straightforward, because most equations of motion assume such a condition. This characteristic simplifies the task of predicting future motion.The implications of constant acceleration are:
  • Uniform motion equations apply
  • Prediction of velocity at any given time becomes simple
  • The simplicity of calculations makes it easier to analyze motion
By dealing with constant acceleration, you don't have to worry about changing forces that could complicate calculations.
Equations of Motion
Equations of motion are fascinating tools in physics. They help us predict how an object will move, given initial conditions and constant acceleration. These equations are ideal for problems like the one with our rolling ball because they incorporate variables like initial velocity, time, distance, and acceleration.For example, in this exercise, we used:
  • \[v^2 = v_0^2 - 2as\]This equation helps find the distance \(s\) the ball rolls up the incline before stopping.
  • \[v = v_0 - at\]This equation is used for calculating the time \(t\) taken to stop momentarily.
The charm of these equations is that they work beautifully when you know certain values, like initial speed and acceleration, and need to find others such as distance or time. They're powerful because they bridge our understanding of an object's past motion to predict its future trajectory with precision.

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Most popular questions from this chapter

The car \(C\) increases its speed at the constant rate of \(1.5 \mathrm{m} / \mathrm{s}^{2}\) as it rounds the curve shown. If the magnitude of the total acceleration of the car is \(2.5 \mathrm{m} / \mathrm{s}^{2}\) at point \(A\) where the radius of curvature is \(200 \mathrm{m},\) compute the speed \(v\) of the car at this point.

A particle which moves with curvilinear motion has coordinates in meters which vary with time \(t\) in seconds according to \(x=2 t^{2}+3 t-1\) and \(y=5 t-2 .\) Determine the coordinates of the center of curvature \(C\) at time \(t=1\) s.

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?

A game requires that two children each throw a ball upward as high as possible from point \(O\) and then run horizontally in opposite directions away from O. The child who travels the greater distance before their thrown ball impacts the ground wins. If child \(A\) throws a ball upward with a speed of \(v_{1}=70 \mathrm{ft} / \mathrm{sec}\) and immediately runs leftward at a constant speed of \(v_{A}=16\) ft/sec while child \(B\) throws the ball upward with a speed of \(v_{2}=64 \mathrm{ft} / \mathrm{sec}\) and immediately runs rightward with a constant speed of \(v_{B}=18 \mathrm{ft} / \mathrm{sec},\) which child will win the game?

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} \\]
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