To determine the total kinetic energy of the system after the collision, we need to follow these steps:
1. Calculate the kinetic energy of object [tex]\( A \)[/tex] before the collision.
The kinetic energy ([tex]\( KE \)[/tex]) of an object is given by the formula:
[tex]\[
KE = \frac{1}{2} m v^2
\][/tex]
where [tex]\( m \)[/tex] is the mass and [tex]\( v \)[/tex] is the velocity.
For object [tex]\( A \)[/tex]:
[tex]\[
m_A = 25 \text{ kg}, \quad v_A = 5.98 \text{ m/s}
\][/tex]
[tex]\[
KE_A = \frac{1}{2} \times 25 \times (5.98)^2
\][/tex]
The result is approximately 447.005 joules.
2. Calculate the kinetic energy of object [tex]\( B \)[/tex] before the collision.
Since object [tex]\( B \)[/tex] is stationary, it has no kinetic energy.
[tex]\[
m_B = 25 \text{ kg}, \quad v_B = 0 \text{ m/s}
\][/tex]
[tex]\[
KE_B = \frac{1}{2} \times 25 \text{ kg} \times (0)^2 = 0 \text{ joules}
\][/tex]
3. Determine the total kinetic energy before the collision.
[tex]\[
KE_{\text{total before}} = KE_A + KE_B = 447.005 \text{ joules} + 0 \text{ joules} = 447.005 \text{ joules}
\][/tex]
4. Use conservation of kinetic energy for a perfectly elastic collision.
In a perfectly elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
[tex]\[
KE_{\text{total after}} = KE_{\text{total before}} = 447.005 \text{ joules}
\][/tex]
Given the above steps, the total kinetic energy of the system after the collision is approximately:
[tex]\[
447.005 \text{ joules}
\][/tex]
Comparing this result with the provided options, the closest match is:
[tex]\[
B. 4.5 \times 10^2 \text{ joules}
\][/tex]
So, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
