91Ó°ÊÓ

Momentum components break down an object's overall momentum into specific directions. In a two-dimensional setting as in this exercise, we consider both the x and y directions. This simplified approach makes it easier to analyze the momentum of each object separately. - Momentum is calculated by multiplying the mass of an object by its velocity. For instance, the pickup truck's momentum in the x-direction is given as: \[ p_{1x} = m_1 \cdot v_{1x} = 2500 \, \text{kg} \cdot (-14.0 \, \text{m/s}) = -35000 \, \text{kg} \, \text{m/s} \]- Similarly, for the sedan in the y-direction: \[ p_{2y} = m_2 \cdot v_{2y} = 1500 \, \text{kg} \cdot 23.0 \, \text{m/s} = 34500 \, \text{kg} \, \text{m/s} \]By focusing on momentum components, you can easily handle more complex scenarios of multidirectional motion.

The net momentum is the combined momentum from all objects involved, found by adding the corresponding components together. In this exercise:- **Net momentum in the x-direction:** The pickup contributes with no additional x-momentum from the sedan. Therefore, \[ p_{x, \text{net}} = -35000 \, \text{kg} \, \text{m/s} \]- **Net momentum in the y-direction:** Only the sedan contributes, hence, \[ p_{y, \text{net}} = 34500 \, \text{kg} \, \text{m/s} \]These values show straightforward addition, where the directions of motion are orthogonal, making calculations easier. Net momentum offers a complete overview of how the momentum distributes across the system.

The Pythagorean theorem helps find the magnitude of the net momentum when we have perpendicular momentum components. This is due to the property of right triangles:- We have the net momentum components: \[ p_{x, \text{net}} = -35000 \, \text{kg} \, \text{m/s} \] \[ p_{y, \text{net}} = 34500 \, \text{kg} \, \text{m/s} \]- The magnitude is: \[ \sqrt{p_{x, \text{net}}^2 + p_{y, \text{net}}^2} = \sqrt{(-35000)^2 + (34500)^2} \] \[ = \sqrt{1225000000 + 1190250000} \, \text{kg} \, \text{m/s} = 49142.19 \, \text{kg} \, \text{m/s} \]This method is vital for converting component momentums into a single usable value representing an object's total momentum.

Determining the angle of momentum allows us to understand the direction of the resultant momentum vector with respect to a reference axis.- The angle \( \theta \) from the negative x-axis can be calculated using the inverse tangent function: \[ \tan(\theta) = \frac{p_{y, \text{net}}}{|p_{x, \text{net}}|} = \frac{34500}{35000} = 0.9857 \]- Solving for \( \theta \), we find: \[ \theta = \tan^{-1}(0.9857) = 44.64^\circ \]This angle is measured from the x-axis in a counter-clockwise direction, which in our scenario, translates to north of west. Calculating this angle allows us to fully understand the direction in which the combined momentum is acting.