ilovejodeci
Answered

Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Discover solutions to your questions from experienced professionals across multiple fields on our comprehensive Q&A platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

Select the correct answer.

Two masses, each weighing [tex]$1.0 \times 10^3$[/tex] kilograms and moving with the same speed of 12.5 meters/second, are approaching each other. They have a head-on collision and bounce off away from each other. Assuming this is a perfectly elastic collision, what will be the approximate kinetic energy of the system after the collision?

A. [tex]$1.6 \times 10^5$[/tex] joules

B. [tex][tex]$2.5 \times 10^5$[/tex][/tex] joules

C. [tex]$1.2 \times 10^3$[/tex] joules

D. [tex]$2.5 \times 10^3$[/tex] joules

Sagot :

GinnyAnswer
To solve the problem, we need to understand the following concepts:

1. Kinetic Energy Formula: The kinetic energy (KE) of an object is given by the formula [tex]\( KE = \frac{1}{2}mv^2 \)[/tex], where [tex]\( m \)[/tex] is the mass and [tex]\( v \)[/tex] is the speed of the object.
2. Conservation of Kinetic Energy in Perfectly Elastic Collisions: In perfectly elastic collisions, the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision.

Let's break this problem down into steps:

### Step 1: Calculate the kinetic energy of each mass before the collision

- Mass of each object, [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex], is [tex]\( 1.0 \times 10^3 \)[/tex] kilograms.
- Speed of each object, [tex]\( v_1 \)[/tex] and [tex]\( v_2 \)[/tex], is 12.5 meters per second.

Using the kinetic energy formula for each object:

[tex]\[ KE_1 = \frac{1}{2} m_1 v_1^2 \][/tex]
[tex]\[ KE_2 = \frac{1}{2} m_2 v_2^2 \][/tex]

Substituting the given values:

[tex]\[ KE_1 = \frac{1}{2} \times 1.0 \times 10^3 \, \text{kg} \times (12.5 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_1 = \frac{1}{2} \times 1.0 \times 10^3 \, \text{kg} \times 156.25 \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ KE_1 = 0.5 \times 1.0 \times 10^3 \, \text{kg} \times 156.25 \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ KE_1 = 78.125 \times 10^3 \, \text{J} \][/tex]
[tex]\[ KE_1 = 78,125 \, \text{J} \][/tex]

Similarly, for the second mass:

[tex]\[ KE_2 = \frac{1}{2} \times 1.0 \times 10^3 \, \text{kg} \times (12.5 \, \text{m/s})^2 \][/tex]
[tex]\[ KE_2 = 78,125 \, \text{J} \][/tex]

### Step 2: Calculate the total kinetic energy before the collision

The total kinetic energy before the collision is the sum of the kinetic energies of both masses:

[tex]\[ KE_{\text{initial}} = KE_1 + KE_2 \][/tex]
[tex]\[ KE_{\text{initial}} = 78,125 \, \text{J} + 78,125 \, \text{J} \][/tex]
[tex]\[ KE_{\text{initial}} = 156,250 \, \text{J} \][/tex]

### Step 3: Apply the conservation of kinetic energy for a perfectly elastic collision

In a perfectly elastic collision, the total kinetic energy is conserved. Therefore, the total kinetic energy of the system after the collision will be the same as it was before the collision:

[tex]\[ KE_{\text{final}} = KE_{\text{initial}} \][/tex]
[tex]\[ KE_{\text{final}} = 156,250 \, \text{J} \][/tex]

### Step 4: Select the correct answer

Based on our calculations, the approximate kinetic energy of the system after the collision is [tex]\( 156,250 \, \text{J} \)[/tex], which is closest to option:

A. [tex]\( 1.6 \times 10^5 \, \text{Joules} \)[/tex]

Therefore, the correct answer is:

A. [tex]\( 1.6 \times 10^5 \, \text{Joules} \)[/tex]

Answer:Since this is a perfectly elastic collision, the total kinetic energy of the system will be conserved.

Here's how to solve this problem:

Find the total mass:  Since there are two identical masses, the total mass (m) will be the sum of the individual masses: m = 2m_individual

Find the total initial kinetic energy (KE_initial):

We can use the formula for kinetic energy: KE = 1/2 * m * v^2

where:

m is the mass (individual mass for now)

v is the velocity (12.5 m/s)

BUT we need the total kinetic energy, so:

KE_initial = 1/2 * (2 * m_individual) * (12.5 m/s)^2

Perfectly elastic collision: In a perfectly elastic collision, the total kinetic energy is conserved. This means the total kinetic energy after the collision (KE_final) will be equal to the total kinetic energy before the collision (KE_initial).

KE_final = KE_initial

We don't need to find the individual masses: Notice that the individual mass (m_individual) cancels out in both the KE_initial and KE_final equations. Therefore, we don't need the specific value of the individual mass to solve for the final kinetic energy.

Answer:

Since the total kinetic energy is conserved, option B (which represents the initial kinetic energy we calculated) is the approximate kinetic energy of the system after the collision.

B.  1/2 * (2 * m_individual) * (12.5 m/s)^2 joules

pen_spark

tune

share

more_vert

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.