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Chapter 2: Problem 16

Two balls are dropped from rest from the same height. One of the balls is dropped 1.00 s after the other. What distance separates the two balls \(2.00 \mathrm{s}\) after the second ball is dropped?

Short Answer

Expert verified
24.5 meters

Step by step solution

01

- Drop time of the first ball

Define the time of travel for the first ball. Since it was dropped 1 second before the second, its total drop time is 2 seconds plus 1 second.
02

- Drop time of the second ball

For the second ball, it was dropped 1.00 s after the first ball and we need the distance after 2.00 s.
03

- Distance traveled by the first ball

Use the kinematic equation for the distance traveled under gravity. The equation: \( d = \frac{1}{2} g t^2 \) Calculate for the first ball (t = 3.00 seconds): \( d_1 = \frac{1}{2} \times 9.8 \times (3^2) = 44.1 \) meters.
04

- Distance traveled by the second ball

Use the same kinematic equation for the second ball, but t = 2.00 seconds: \( d_2 = \frac{1}{2} \times 9.8 \times (2^2) = 19.6 \) meters.
05

- Calculate the separation distance

Subtract the distance traveled by the second ball from that of the first ball to get the distance separating them: \( d_{sep} = d_1 - d_2 = 44.1 - 19.6 = 24.5 \) meters.

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Key Concepts

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

Distance Under Gravity
When objects fall freely under the influence of gravity, they accelerate at a constant rate until they hit the ground. This acceleration due to gravity is denoted by the symbol \( g \) and is approximately \( 9.8 \, m/s^2 \) on Earth's surface.
To calculate the distance an object falls under gravity, neglecting air resistance, we use the kinematic equation:
\[ d = \frac{1}{2} g t^2 \]
In this equation:
  • \( d \) represents the distance fallen
  • \( g \) is the acceleration due to gravity (\( 9.8 \, m/s^2 \)
  • \( t \) is the time the object has been falling
For example, in the exercise, the first ball falls for 3 seconds and the second ball falls for 2 seconds. Using this formula, we can calculate the distance each ball travels.
Kinematic Equations
Kinematic equations are essential tools for solving problems in classical mechanics, especially when dealing with motion under constant acceleration, such as free-fall. These equations relate the following quantities:
  • Displacement (\( d \))
  • Initial velocity (\( v_0 \))
  • Final velocity (\( v \))
  • Acceleration (\( a \))
  • Time (\( t \))
One of the key kinematic equations used in our problem is:
\[ d = \frac{1}{2} g t^2 \]
This equation is derived from integrating the acceleration due to gravity and is applicable when the object starts from rest (initial velocity equals zero). Using this equation simplifies the calculation of the falling distance for the balls in the problem.
Free-Fall Motion
Free-fall motion is a specific type of motion in which an object moves under the influence of gravity alone. In free-fall, we assume there is no air resistance, and the only force acting on the object is gravity. This means:
  • The object accelerates downwards at \( 9.8 \, m/s^2 \).
  • Objects in free-fall experience a constant acceleration irrespective of their mass.
  • A key aspect of free-fall is that two objects dropped from the same height will hit the ground at the same time if air resistance is negligible.
In the exercise, both balls experience free-fall, but the timing of their release causes a separation distance. The first ball falls for a longer time, resulting in it traveling a greater distance. By calculating the distances traveled by both balls, we find the separation after a specific time.

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Most popular questions from this chapter

A stone is thrown vertically upwards with an initial speed of \(10.0 \mathrm{m} \mathrm{s}^{-1}\) from a cliff that is \(50.0 \mathrm{m}\) high. (a) When does it reach the bottom of the cliff? (b) What speed does it have just before hitting the ground? (c) What is the total distance travelled by the stone? (Take the acceleration due to gravity to be \(\left.10.0 \mathrm{m} \mathrm{s}^{-2} .\right)\)

A car of mass 1200 kg starts from rest, accelerates uniformly to a speed of \(4.0 \mathrm{m} \mathrm{s}^{-1}\) in \(2.0 \mathrm{s}\) and continues moving at this constant speed in a horizontal straight line for an additional \(10 \mathrm{s}\). The brakes are then applied and the car is brought to rest in \(4.0 \mathrm{s}\). A constant resistance force of \(500 \mathrm{N}\) is acting on the car during its entire motion. (a) Calculate the force accelerating the car in the first \(2.0 \mathrm{s}\) of the motion. (b) Calculate the average power developed by the engine in the first \(2.0 \mathrm{s}\) of the motion. (c) Calculate the force pushing the car forward in the next \(10 \mathrm{s}\) (d) Calculate the power developed by the engine in those \(10 \mathrm{s}\) (e) Calculate the braking force in the last \(4.0 \mathrm{s}\) of the motion. (f) Describe the energy transformations that have taken place in the 16 s of the motion of this car.

A body of mass \(1.00 \mathrm{kg}\) is tied to a string and rotates on a horizontal, frictionless table. (a) If the length of the string is \(40.0 \mathrm{cm}\) and the speed of revolution is \(2.0 \mathrm{m} \mathrm{s}^{-1},\) find the tension in the string. (b) If the string breaks when the tension exceeds \(20.0 \mathrm{N},\) what is the largest speed the mass can rotate at? (c) If the breaking tension of the string is 20.0 N but you want the mass to rotate at \(4.00 \mathrm{m} \mathrm{s}^{-1},\) what is the shortest length string that can be used?

Show by applying Newton's law of gravitation and the second law of mechanics that a satellite (or planet) in a circular orbit of radius \(R\) around the earth (or the sun) has a period (i.e. time to complete one revolution) given by $$T^{2}=\frac{4 \pi^{2} R^{3}}{G M}$$ where \(M\) is the mass of the attracting body (earth or sun). This is Kepler's third law.

A spring of spring constant \(k=150 \mathrm{Nm}^{-1}\) is compressed by \(4.0 \mathrm{cm} .\) The spring is horizontal and a mass of \(1.0 \mathrm{kg}\) is held to the right end of the spring. If the mass is released, with what speed will it move away?
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