Answer :
To solve this problem, we need to determine the total kinetic energy of the system after an elastic collision. In an elastic collision, the total kinetic energy of the system is conserved. This means that the total kinetic energy before the collision will be equal to the total kinetic energy after the collision.
Let's start by calculating the initial kinetic energy of both objects.
1. Calculate the initial kinetic energy of Object A:
[tex]\[ \text{KE}_A = \frac{1}{2} m_A v_A^2 \][/tex]
Given:
- Mass of object [tex]\(A\)[/tex]: [tex]\(m_A = 25\)[/tex] kilograms
- Velocity of object [tex]\(A\)[/tex]: [tex]\(v_A = 5.98\)[/tex] meters/second
Plugging the values into the formula:
[tex]\[ \text{KE}_A = \frac{1}{2} \times 25 \times (5.98)^2 \][/tex]
2. Calculate the initial kinetic energy of Object B:
[tex]\[ \text{KE}_B = \frac{1}{2} m_B v_B^2 \][/tex]
Given:
- Mass of object [tex]\(B\)[/tex]: [tex]\(m_B = 25\)[/tex] kilograms
- Velocity of object [tex]\(B\)[/tex]: [tex]\(v_B = 0\)[/tex] meters/second (stationary)
Since object [tex]\(B\)[/tex] is stationary, its initial kinetic energy will be:
[tex]\[ \text{KE}_B = \frac{1}{2} \times 25 \times 0^2 = 0 \][/tex]
3. Calculate the total initial kinetic energy of the system:
[tex]\[ \text{Total Initial KE} = \text{KE}_A + \text{KE}_B \][/tex]
[tex]\[ \text{Total Initial KE} = \frac{1}{2} \times 25 \times (5.98)^2 + 0 \][/tex]
Performing the calculation:
[tex]\[ \text{Total Initial KE} \approx 447.005 \ \text{joules} \][/tex]
Since the collision is elastic, the total kinetic energy after the collision will be the same as the total initial kinetic energy.
Therefore, the total kinetic energy of the system after the collision is approximately [tex]\(447.005\)[/tex] joules.
Now, let's match this value to the given options:
- A. [tex]\(1.2 \times 10^2\)[/tex] joules (120 joules)
- B. [tex]\(4.5 \times 10^2\)[/tex] joules (450 joules)
- C. [tex]\(5.0 \times 10^2\)[/tex] joules (500 joules)
- D. [tex]\(9.5 \times 10^2\)[/tex] joules (950 joules)
- E. [tex]\(1.1 \times 10^3\)[/tex] joules (1100 joules)
The closest answer to [tex]\(447.005\)[/tex] joules is B. [tex]\(4.5 \times 10^2\)[/tex] joules.
Thus, the correct answer is:
[tex]\[ \boxed{4.5 \times 10^2 \text{ joules}} \][/tex]
Let's start by calculating the initial kinetic energy of both objects.
1. Calculate the initial kinetic energy of Object A:
[tex]\[ \text{KE}_A = \frac{1}{2} m_A v_A^2 \][/tex]
Given:
- Mass of object [tex]\(A\)[/tex]: [tex]\(m_A = 25\)[/tex] kilograms
- Velocity of object [tex]\(A\)[/tex]: [tex]\(v_A = 5.98\)[/tex] meters/second
Plugging the values into the formula:
[tex]\[ \text{KE}_A = \frac{1}{2} \times 25 \times (5.98)^2 \][/tex]
2. Calculate the initial kinetic energy of Object B:
[tex]\[ \text{KE}_B = \frac{1}{2} m_B v_B^2 \][/tex]
Given:
- Mass of object [tex]\(B\)[/tex]: [tex]\(m_B = 25\)[/tex] kilograms
- Velocity of object [tex]\(B\)[/tex]: [tex]\(v_B = 0\)[/tex] meters/second (stationary)
Since object [tex]\(B\)[/tex] is stationary, its initial kinetic energy will be:
[tex]\[ \text{KE}_B = \frac{1}{2} \times 25 \times 0^2 = 0 \][/tex]
3. Calculate the total initial kinetic energy of the system:
[tex]\[ \text{Total Initial KE} = \text{KE}_A + \text{KE}_B \][/tex]
[tex]\[ \text{Total Initial KE} = \frac{1}{2} \times 25 \times (5.98)^2 + 0 \][/tex]
Performing the calculation:
[tex]\[ \text{Total Initial KE} \approx 447.005 \ \text{joules} \][/tex]
Since the collision is elastic, the total kinetic energy after the collision will be the same as the total initial kinetic energy.
Therefore, the total kinetic energy of the system after the collision is approximately [tex]\(447.005\)[/tex] joules.
Now, let's match this value to the given options:
- A. [tex]\(1.2 \times 10^2\)[/tex] joules (120 joules)
- B. [tex]\(4.5 \times 10^2\)[/tex] joules (450 joules)
- C. [tex]\(5.0 \times 10^2\)[/tex] joules (500 joules)
- D. [tex]\(9.5 \times 10^2\)[/tex] joules (950 joules)
- E. [tex]\(1.1 \times 10^3\)[/tex] joules (1100 joules)
The closest answer to [tex]\(447.005\)[/tex] joules is B. [tex]\(4.5 \times 10^2\)[/tex] joules.
Thus, the correct answer is:
[tex]\[ \boxed{4.5 \times 10^2 \text{ joules}} \][/tex]
