91Ó°ÊÓ

Vectors are central in physics because they provide direction and magnitude. In this exercise, we deal with velocity vectors, which show both how fast an object moves and its direction. To handle vectors effectively, we often break them into components along the x and y axes.

For example, Part A’s velocity is given as \( (10 \, \hat{i} + 12 \, \hat{j}) \, \mathrm{m/s} \). Here:

• 10 \( \hat{i} \) represents 10 m/s in the x direction.
• 12 \( \hat{j} \) represents 12 m/s in the y direction.

When dealing with angles, like Part B’s velocity at 110° counterclockwise:

• The x-component is found using \( v_{Bx} = 14.0 \cos(110^\text{o}) \).
• The y-component is found using \( v_{By} = 14.0 \sin(110^\text{o}) \).

This breakdown simplifies the calculations involving vectors, making it easier to combine or resolve them.

Momentum is a measure of an object's motion, calculated as the product of mass and velocity (\( \mathbf{p} = m \mathbf{v} \)). It is a vector quantity, meaning it has both magnitude and direction.

In our exercise, a package explodes into three parts. Since the total initial momentum was zero (the package was stationary), the total final momentum must also be zero due to conservation of momentum.

To address this, we sum the momenta of Parts A and B:

• Part A’s momentum: \( \mathbf{P}_A = 0.500 \times (10 \hat{i} + 12 \hat{j}) = (5 \hat{i} + 6 \hat{j}) \mathrm{kg} \mathrm{m/s} \)
• Part B’s momentum is broken down into components and then summed: \( \mathbf{P}_B = 0.750 \times (-4.788 \hat{i} + 13.156 \hat{j}) = (-3.591 \hat{i} + 9.867 \hat{j}) \mathrm{kg} \mathrm{m/s} \).

This gives us the momentum of Part C by ensuring their sum equates to zero:

• \( \mathbf{P}_C = -(\mathbf{P}_A + \mathbf{P}_B) \)
• \( \mathbf{P}_A + \mathbf{P}_B = (5 \hat{i} + 6 \hat{j}) + (-3.591 \hat{i} + 9.867 \hat{j}) = (1.409 \hat{i} + 15.867 \hat{j}) \mathrm{kg} \mathrm{m/s} \)

So the momentum of Part C is \( \mathbf{P}_C = -(1.409 \hat{i} + 15.867 \hat{j}) = -1.409 \hat{i} - 15.867 \hat{j} \mathrm{kg} \mathrm{m/s} \).

Determining an object's travel direction in physical problems often requires converting between angles and vector coordinates.

For Part B, traveling at 110° counterclockwise from the positive x-axis, we convert this angle to vector form using trigonometric functions:

• The x-component: \( v_{Bx} = 14.0 \cos(110^\text{o}) \)
• The y-component: \( v_{By} = 14.0 \sin(110^\text{o}) \)

This conversion helps integrate the vectors with others in the problem.

Once we find the momentum of Part C, we also calculate its velocity components:

• \( \mathbf{v}_C = \frac{-1.409 \hat{i} - 15.867 \hat{j}}{1.400} = (-1.006 \hat{i} - 11.334 \hat{j}) \mathrm{m/s} \)

Finally, to understand Part C’s overall motion, we calculate the magnitude of its velocity using the Pythagorean theorem:

• \( v_C = \sqrt{(-1.006)^2 + (-11.334)^2} = 11.378 \mathrm{m/s} \)

The direction angle, or heading, is found using the inverse tangent function:

\( \theta = \tan^{-1} \left( \frac{-11.334}{-1.006} \right) \approx 85^\text{o} \) which means it travels almost vertically downward, just slightly to the left.