Problem 80 - Fundamentals of Physics

Question

A \(2.50 \mathrm{~kg}\) particle that is moving horizontally over a floor with velocity \((-3.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) undergoes a completely inelastic collision with a \(4.00 \mathrm{~kg}\) particle that is moving horizontally over the floor with velocity \((4.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}} .\) The collision occurs at \(x y\) coordinates \((-0.500 \mathrm{~m},-0.100 \mathrm{~m}) .\) After the collision and in unit- vector notation, what is the angular momentum of the stuck-together particles with respect to the origin?

Step-by-Step Solution

Answer

The angular momentum is \(5.54 \hat{\mathbf{k}}\,\mathrm{kg\,m^2/s}\).

Step 1: Calculate Initial Momenta of Particles

Before the collision, the momentum of the first particle (\(m_1 = 2.50\,\mathrm{kg}\), \(\vec{v}_{1} = (-3.00\,\mathrm{m/s}) \hat{\mathbf{j}}\)) is \(\vec{p}_1 = m_1 \vec{v}_{1} = (2.50\,\mathrm{kg})\times(-3.00\,\mathrm{m/s}) \hat{\mathbf{j}} = (-7.50\,\mathrm{kg\, m/s}) \hat{\mathbf{j}}\).Similarly, for the second particle (\(m_2 = 4.00\,\mathrm{kg}\), \(\vec{v}_{2} = (4.50\,\mathrm{m/s}) \hat{\mathbf{i}}\)), \(\vec{p}_2 = m_2 \vec{v}_{2} = (4.00\,\mathrm{kg})\times(4.50\,\mathrm{m/s}) \hat{\mathbf{i}} = (18.00\,\mathrm{kg\, m/s}) \hat{\mathbf{i}}\).Both momenta need to be combined after the collision which is inelastic.

Step 2: Apply Conservation of Momentum

In a completely inelastic collision, the total momentum before the collision equals the total momentum after the collision.The combined momentum is\( \vec{p}_{\text{total}} = \vec{p}_1 + \vec{p}_2 = (-7.50\,\mathrm{kg\, m/s}) \hat{\mathbf{j}} + (18.00\,\mathrm{kg\, m/s}) \hat{\mathbf{i}}\).Thus,\( \vec{p}_{\text{total}} = (18.00\,\mathrm{kg\, m/s}) \hat{\mathbf{i}} - (7.50\,\mathrm{kg\, m/s}) \hat{\mathbf{j}}\).This momentum is of the combined mass \(m_1 + m_2 = (2.50 + 4.00)\,\mathrm{kg} = 6.50\,\mathrm{kg}\).

Step 3: Find the Velocity of the Combined Mass

The velocity \(\vec{v}\) of the combined object can be calculated using the relation\[ \vec{v} = \frac{\vec{p}_{\text{total}}}{m_1 + m_2} = \frac{(18.00 \hat{\mathbf{i}} - 7.50 \hat{\mathbf{j}})}{6.50}\].Calculating gives\( \vec{v} = (2.77\,\mathrm{m/s}) \hat{\mathbf{i}} - (1.15\,\mathrm{m/s}) \hat{\mathbf{j}}\).

Step 4: Calculate Angular Momentum about the Origin

The position vector \(\vec{r}\) where the collision happens is given as \((-0.500, -0.100)\,\mathrm{m}\), or\(\vec{r} = -0.500 \hat{\mathbf{i}} - 0.100 \hat{\mathbf{j}}\,\mathrm{m}\).The angular momentum \(\vec{L}\) is calculated by the cross product:\( \vec{L} = m \cdot (\vec{r} \times \vec{v})\),where \(m\) is the combined mass \(6.50\,\mathrm{kg}\).\(\vec{L} = 6.50 \left[\begin{array}{|ccc|} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \ -0.500 & -0.100 & 0 \ 2.77 & -1.15 & 0 \end{array}\right]\).

Step 5: Compute the Determinant for Cross Product

Set up and calculate the determinant:- The determinant of the matrix involving cross-product is\[\vec{L} = 6.50 \Big( (0 \times (-1.15) - 0.0) \hat{\mathbf{i}} - (0 \times 2.77 - 0.0) \hat{\mathbf{j}} + ((-0.500) \times (-1.15) - (-0.100) \times 2.77) \hat{\mathbf{k}}) \Big)\].After calculation, the non-zero component\( \vec{L} = 6.50 \Big( 0.575 + 0.277 \Big) \hat{\mathbf{k}}\)\( \vec{L} = 6.50 \times 0.852 \hat{\mathbf{k}}\)\( \vec{L} = 5.54 \hat{\mathbf{k}}\,\mathrm{kg\,m^2/s}\).

Step 6: Final Result for Angular Momentum

The angular momentum of the combined object relative to the origin is\( \vec{L} = 5.54 \hat{\mathbf{k}}\,\mathrm{kg\,m^2/s}\). This represents a vector directed out of the plane defined by the \(\hat{\mathbf{i}}, \hat{\mathbf{j}}\) directions, with magnitude \(5.54\,\mathrm{kg\,m^2/s}\).

Key Concepts

Inelastic Collision

In an inelastic collision, objects collide and stick together, moving with a common velocity after the impact. This scenario is an example of a perfectly inelastic collision, where the maximum amount of kinetic energy is lost. However, it's important to note that while kinetic energy is not conserved, the total momentum of the system is conserved.

In this specific exercise, a particle with mass 2.50 kg moving in the negative j-direction collides with a 4.00 kg particle moving in the positive i-direction. After the collision, these two particles merge and move together as one entity.

Key characteristics of inelastic collisions include:

• The system loses kinetic energy due to deformation or other factors, such as sound or heat.
• The final velocity of the combined mass is in the direction of the resultant momentum vector, not necessarily along the line of their original paths.
• Calculations often require considering both individual momenta before collision and the joint system's momentum afterward.

Understanding these characteristics helps in analyzing and solving collision problems effectively.

Conservation of Momentum

Conservation of momentum is a fundamental principle applicable to all types of collisions, including inelastic. It states that the total momentum of a closed system remains constant if no external forces act on it. This principle is crucial when analyzing the motion of objects before and after a collision.

In the problem given, the combined momentum before collision is calculated by adding the momentum vectors of both particles:
- Particle 1 has a momentum: \(\vec{p}_1 = m_1 \vec{v}_1 \ = (2.50 \ \text{kg}) \times (-3.00 \ \text{m/s}) \hat{\mathbf{j}} \ = (-7.50 \ \text{kg}\, \text{m/s}) \hat{\mathbf{j}}\).- Particle 2 has a momentum: \(\vec{p}_2 = m_2 \vec{v}_2 \ = (4.00 \ \text{kg}) \times (4.50 \ \text{m/s}) \hat{\mathbf{i}} \ = (18.00 \ \text{kg}\, \text{m/s}) \hat{\mathbf{i}}\).
Once these are summed, the system's momentum after the collision remains the same at \(\vec{p}_{\text{total}} = (18.00 \ \text{kg}\, \text{m/s}) \hat{\mathbf{i}} - (7.50 \ \text{kg}\, \text{m/s}) \hat{\mathbf{j}}\). This total momentum allows us to determine the velocity of the combined mass by dividing the momentum by the total mass. Understanding conservation of momentum is key in predicting motion changes in collisions, providing insights into velocity directions and magnitude after impact.

Cross Product

The cross product is a vector operation used to find a vector perpendicular to two given vectors in three-dimensional space. It is particularly useful in physics when calculating angular quantities, like angular momentum.

In this exercise, once the new velocity of the combined mass is found, the angular momentum \(\vec{L}\) is calculated using the position vector \(\vec{r}\) and the velocity \(\vec{v}\) as follows:\[\vec{L} = m \cdot (\vec{r} \times \vec{v})\]Here, \(\vec{r} = -0.500 \hat{\mathbf{i}} - 0.100 \hat{\mathbf{j}}\) is the position vector from the origin to the point where the collision occurs.

The cross product of \(\vec{r}\) and \(\vec{v}\) is evaluated using the determinant formula:\[\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \-0.500 & -0.100 & 0 \2.77 & -1.15 & 0 \\end{array}\right|\]
The non-zero component of this operation results in \(\vec{L} = 6.50 \times 0.852 \hat{\mathbf{k}} = 5.54 \hat{\mathbf{k}}\,\text{kg\,m}^2/\text{s}\). This result tells us that the angular momentum vector is directed along the third dimension (\(\hat{\mathbf{k}}\)), showing its perpendicular characteristic relative to \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\). Understanding the process of calculating cross products forms the basis for determining angular momentum and resolving more complex three-dimensional motion problems.