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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



Answer :

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

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