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Chapter 7: Problem 36

@ Object A is moving due east, while object \(\mathrm{B}\) is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of \(m_{\mathrm{A}}=17.0 \mathrm{kg}\) and an initial velocity of \(\vec{v}_{0}=8.00 \mathrm{m} / \mathrm{s}\) , due east. Object \(\mathrm{B}\) , however, has a mass of \(m_{\mathrm{B}}=29.0 \mathrm{kg}\) and an initial velocity of \(\overrightarrow{\mathrm{v}}_{\mathrm{oB}}=5.00 \mathrm{m} / \mathrm{s},\) due north. Find the magnitude and direction of the total momentum of the two-object system after the collision.

Short Answer

Expert verified
Magnitude is 198.80 kg路m/s; direction is 46.01掳 from east towards north.

Step by step solution

01

Understand the Problem

We are dealing with a completely inelastic collision where two objects stick together after impact. We must conserve momentum to find the magnitude and direction of their total momentum post-collision.
02

Identify Given Information

Object A has a mass of \( m_A = 17.0 \; \text{kg} \) and velocity \( \vec{v}_{0A} = 8.00 \; \text{m/s east}. \) Object B has a mass of \( m_B = 29.0 \; \text{kg} \) and velocity \( \vec{v}_{0B} = 5.00 \; \text{m/s north}. \)
03

Calculate Total Initial Momentum

We calculate momentum in both x (east-west) and y (north-south) directions. For Object A: \( p_{xA} = m_A \times v_{0A} = 17.0 \times 8.00 = 136 \; \text{kg.m/s}. \) For Object B: \( p_{yB} = m_B \times v_{0B} = 29.0 \times 5.00 = 145 \; \text{kg.m/s}. \)
04

Use Pythagorean Theorem for Total Momentum

Total momentum \( \vec{p} \) is a vector with components from the previous step. The magnitude \( p \) is found by: \[ p = \sqrt{p_{xA}^2 + p_{yB}^2} = \sqrt{136^2 + 145^2} \].
05

Compute Magnitude

Calculate the above expression: \[ p = \sqrt{136^2 + 145^2} = \sqrt{18496 + 21025} = \sqrt{39521} = 198.80 \; \text{kg.m/s} \].
06

Calculate Direction Angle

Use the tangent function to find the direction of total momentum \( \theta \): \[ \theta = \tan^{-1}\left(\frac{p_{yB}}{p_{xA}}\right) = \tan^{-1}\left(\frac{145}{136}\right) \].
07

Calculate Direction Angle Result

Compute \( \theta \): \[ \theta = \tan^{-1}\left(\frac{145}{136}\right) = 46.01^\circ \]. This angle is measured from east towards north.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Momentum Conservation
In physics, the concept of momentum conservation is fundamental, especially in collisions. According to the law of conservation of momentum, the total momentum of a system remains constant if no external forces act upon it. In an inelastic collision, like the one described in this problem, the two objects stick together post-collision, and their combined momentum is conserved. This means:
  • The total momentum before the collision equals the total momentum after.
  • In our exercise, both objects have linear momentum in perpendicular directions, one moving east and the other north.
These independent momentum components in two dimensions (x and y) need to be considered separately and then combined to find the total system momentum after the collision.
Vector Components
When dealing with momentum or any vector quantity, it's crucial to break down vectors into their components. Here, the task is to analyze momentum in two perpendicular directions: east-west (x-axis) and north-south (y-axis).

For object A (moving east), its momentum is purely in the x-direction:
  • \( p_{xA} = m_A \times v_{0A} \)
  • Given as 136 kg路m/s, since object A is moving only east.
For object B (moving north), its momentum is purely in the y-direction:
  • \( p_{yB} = m_B \times v_{0B} \)
  • Calculated as 145 kg路m/s, as object B solely travels north.
These components help us in later calculations, providing a foundation for determining the resultant vector's overall magnitude and direction.
Pythagorean Theorem
The Pythagorean Theorem is a mathematical principle used to find the magnitude of a vector when its components are known. In the realm of this collision problem, it helps us find the magnitude of the total momentum vector from its perpendicular x (east) and y (north) components.
  • We use: \[p = \sqrt{p_{xA}^2 + p_{yB}^2} \]
  • This equation computes the hypotenuse of the right triangle formed in the component space.
  • Performing the calculation gives \( p \approx 198.80 \; \text{kg.m/s} \).
This step is crucial for understanding how individual movements in their respective axes combine into a single movement or momentum in space.
Angle Calculation
Once the magnitude of the total momentum vector is known, the next step is to determine its direction or angle. For vectors in two dimensions, this angle provides clarity on how the momentum vector physically applies in space.

To find the angle \( \theta \), tan inverse (arctan) is often used:
  • \( \theta = \tan^{-1}\left(\frac{p_{yB}}{p_{xA}}\right) \)
  • Given the components \( p_{xA} = 136 \; \text{kg.m/s} \) and \( p_{yB} = 145 \; \text{kg.m/s} \), the resultant angle \( \theta \) is \( 46.01^\circ \).
  • This angle is from the east towards the north, indicating the momentum direction post-collision.
Understanding the angle helps in predicting the trajectory of objects post-impact, giving a complete picture of motion and direction.

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