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Chapter 9: Problem 9

A stone is dropped at \(t=0 .\) A second stone, with twice the mass of the first, is dropped from the same point at \(t=100 \mathrm{~ms}\). (a) How far below the release point is the center of mass of the two stones at \(t=300 \mathrm{~ms}\) ? (Neither stone has yet reached the ground.) (b) How fast is the center of mass of the twostone system moving at that time?

Short Answer

Expert verified
(a) The center of mass is 0.278 meters below the release point. (b) Its speed is 2.287 m/s.

Step by step solution

01

Calculate Distance Fallen by First Stone

The first stone is in free fall for 300 ms (0.3 seconds). Using the formula for distance in free fall, \( s = \frac{1}{2} g t^2 \), where \( g = 9.8 \frac{m}{s^2} \) is the acceleration due to gravity, we calculate: \[ s_1 = \frac{1}{2} \times 9.8 \times (0.3)^2 \] \[= \frac{1}{2} \times 9.8 \times 0.09 \] \[= 0.441 \, \text{meters}\].Thus, the first stone has fallen 0.441 meters.
02

Calculate Distance Fallen by Second Stone

The second stone is in free fall for 200 ms (0.2 seconds), as it was dropped 100 ms after the first. Using the same equation, \[ s_2 = \frac{1}{2} \times 9.8 \times (0.2)^2 \] \[= \frac{1}{2} \times 9.8 \times 0.04 \] \[= 0.196 \, \text{meters}\].Thus, the second stone has fallen 0.196 meters.
03

Calculate the Center of Mass Position

Let the mass of the first stone be \( m \) and the second stone be \( 2m \). The center of mass \( x_{cm} \) is given by \[ x_{cm} = \frac{m \cdot s_1 + 2m \cdot s_2}{m + 2m} \] \[= \frac{m \times 0.441 + 2m \times 0.196}{3m} \] \[= \frac{0.441m + 0.392m}{3m} \] \[= \frac{0.833}{3} \]\[= 0.27766 \text{ meters}\].The center of mass is 0.27766 meters below the release point.
04

Calculate the Velocity of First Stone

To find the velocity of the first stone at 300 ms, we use the equation \( v = g t \). Thus: \[ v_1 = 9.8 \times 0.3 \] \[= 2.94 \text{ m/s} \].This is the velocity of the first stone.
05

Calculate the Velocity of Second Stone

To find the velocity of the second stone at 200 ms, use the same formula: \[ v_2 = 9.8 \times 0.2 \] \[= 1.96 \text{ m/s} \].This is the velocity of the second stone.
06

Calculate the Velocity of the Center of Mass

The velocity of the center of mass \( v_{cm} \) for the system is given by: \[ v_{cm} = \frac{m \cdot v_1 + 2m \cdot v_2}{m + 2m} \] \[= \frac{m \times 2.94 + 2m \times 1.96}{3m} \] \[= \frac{2.94m + 3.92m}{3m} \] \[= \frac{6.86}{3} \] \[= 2.28667 \text{ m/s} \].The center of mass is moving at 2.28667 m/s.

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Key Concepts

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

Free Fall
In free fall, an object is only influenced by gravity. This means there are no other forces, like air resistance, acting on it. Free-fall can be understood by imagining an object dropped from a height and accelerating downwards solely due to gravity. Understanding free fall is crucial for various applications in physics.
  • An object in free fall will continue to accelerate downwards until it impacts with the ground or another surface.
  • In our exercise, both stones are in free fall, meaning they accelerate towards the ground, governed only by gravity.
Free-fall formulas such as those for calculating distance and velocity provide a mathematical representation of this natural phenomenon. These equations help us predict how far and how fast an object will travel under the influence of Earth's gravitational pull.
Acceleration Due to Gravity
Gravity is the force that pulls objects toward the center of the Earth. On Earth, this force accelerates objects at a rate of approximately 9.8 meters per second squared (\(9.8 \, \frac{m}{s^2}\)). This acceleration is often denoted by the letter \(g\).
  • Every object, regardless of its mass, experiences this same gravitational pull when in free fall.
  • In our example, both stones are subjected to this constant acceleration as they fall.
Understanding acceleration due to gravity allows us to calculate how quickly an object's velocity will increase or how far it will move over time. This concept is pivotal in physics, enabling us to make predictions about free-falling bodies.
Velocity Calculation
Velocity describes how fast an object is moving and in which direction. It's important to remember that, unlike speed, velocity is a vector quantity. This means it has both magnitude and direction. In the context of free fall, we often focus on the magnitude, which increases steadily as the object falls.
  • The formula for calculating the velocity of an object in free fall is \( v = g \cdot t \), where \( v \) is the velocity, \( g \) is the acceleration due to gravity, and \( t \) is the time elapsed.
  • For the first stone, after 0.3 seconds, its velocity is calculated to be \(2.94 \; \text{m/s}\), whereas the second stone takes 0.2 seconds to reach \(1.96 \; \text{m/s}\).
The calculations for velocity are essential in understanding how quickly an object reaches a given speed while falling under gravity's influence.
Distance Calculation
Determining how far an object has fallen during free fall involves calculating the distance it has traveled over a given time. When an object is dropped, it doesn't move uniformly 鈥� it accelerates over time due to gravity.
  • The formula for calculating the distance an object falls under gravity is \( s = \frac{1}{2} g t^2 \).
  • This formula helps us understand that the distance fallen increases with the square of the time because of constant acceleration.
  • For the stones, the first falls 0.441 meters in 0.3 seconds, while the second falls 0.196 meters in 0.2 seconds.
Distance calculation thus enables us to quantify the effects of gravity on falling objects over time, providing insights into their motion during free fall.

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Most popular questions from this chapter

In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviation MVC. Suppose a \(1000 \mathrm{~kg}\) car slides into a stationary \(500 \mathrm{~kg}\) moose on a very slippery road, with the moose being thrown through the windshield (a common MVC result). (a) What percent of the original kinetic energy is lost in the collision to other forms of energy? A similar danger occurs in Saudi Arabia because of camel- vehicle collisions (CVC). (b) What percent of the original kinetic energy is lost if the car hits a \(300 \mathrm{~kg}\) camel? (c) Generally, does the percent loss increase or decrease if the animal mass decreases?

After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.

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