91Ó°ÊÓ

Translational momentum refers to the quantity of motion that an object has while moving in a given direction. It is a product of an object's mass and its velocity, expressed as \( \textbf{p} = m \times v \). In our exercise, the space probe has a mass (\(m\)) of 2500 kg and an initial velocity (\(v\)) of 300 m/s. This makes its initial translational momentum \( \textbf{p} = 2500 \text{ kg} \times 300 \text{ m/s} = 750,000 \text{ kg m/s} \). When a force acts on the probe, its translational momentum changes according to the direction and duration of the force applied. This change is guided by the Impulse-Momentum Theorem, illustrating how translational momentum is a vector quantity, meaning that both magnitude and direction matter.

The Impulse-Momentum Theorem states that the change in an object's momentum is equal to the impulse applied to it. Mathematically, this is represented as \( \textbf{Δp} = \textbf{J} \) or \( F \times t \), where \( F \) is the force and \( t \) is the time period over which the force acts. In our space probe scenario, the given thrust force is 3000 N, and the duration is 65 seconds. Using the formula, the resulting impulse is \( \textbf{J} = 3000 \text{ N} \times 65 \text{ s} = 195,000 \text{ Ns} \). This impulse causes a change in the probe's momentum by 195,000 Ns. Depending on the direction of the thrust, the magnitude of the change differs. If forward, the change is positive; backward, it is negative, and if thrust is sideways, the direction changes, but the magnitude remains 195,000 Ns.

Kinetic energy (KE) quantifies the energy of motion and is calculated using the formula \( KE = \frac{1}{2} m v^2 \). For the space probe, the initial kinetic energy is \( KE_i = \frac{1}{2} \times 2500 \text{ kg} \times (300 \text{ m/s})^2 = 112,500,000 \text{ J} \). When the thrust is forward, the change in velocity is calculated to be 78 m/s (from \( \textbf{Δv} = \frac{195,000 \text{ Ns}}{2500 \text{ kg}} \)), resulting in a final velocity of 378 m/s. This changes the kinetic energy to \( KE_f = \frac{1}{2} \times 2500 \text{ kg} \times (378 \text{ m/s})^2 = 178,290,000 \text{ J} \), showing an increase of 65,790,000 J in KE. When the thrust is backward, \( \textbf{Δv} = -78 \text{ m/s} \), resulting in a final velocity of 222 m/s and a KE_f of \( 61,695,000 \text{ J} \), indicating a reduction of 50,805,000 J in KE. When the thrust is sideways, the speed remains 300 m/s because the thrust direction is perpendicular to the motion, so the kinetic energy remains unchanged, and \( \text{ΔKE} = 0 \text{ J} \).