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Chapter 7: Problem 48

A collision occurs between a \(2.00 \mathrm{~kg}\) particle traveling with velocity \(\vec{v}_{A 1}=(-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(-5.00 \mathrm{~m} / \mathrm{s}) \mathrm{j}\) and a \(4.00 \mathrm{~kg}\) particle traveling with velocity \(\vec{v}_{B 1}=(6.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+\) \((-2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). The collision connects the two particles. What then is their velocity in (a) unit-vector notation and (b) magnitude-angle notation?

Short Answer

Expert verified
The final velocity in unit-vector notation is \( (2.67 \mathrm{~m/s}) \hat{i} + (-3.00 \mathrm{~m/s}) \hat{j} \. The magnitude is \( 4.01 \mathrm{~m/s} \) and the angle is -48.4 degrees.

Step by step solution

01

- Find the initial momentum of each particle

The momentum of a particle is given by \(\textbf{p} = m \textbf{v}\). Calculate the momentum for particle A and particle B.\( \textbf{p}_{A1} = m_A \vec{v}_{A1} = 2.00 \mathrm{~kg} \times \left[ (-4.00 \mathrm{~m/s}) \hat{i} + (-5.00 \mathrm{~m/s}) \hat{j} \right] = (-8.00 \mathrm{~kg~m/s}) \hat{i} + (-10.00 \mathrm{~kg~m/s}) \hat{j}\)\( \textbf{p}_{B1} = m_B \vec{v}_{B1} = 4.00 \mathrm{~kg} \times \left[ (6.00 \mathrm{~m/s}) \hat{i} + (-2.00 \mathrm{~m/s}) \hat{j} \right] = (24.00 \mathrm{~kg~m/s}) \hat{i} + (-8.00 \mathrm{~kg/m/s}) \hat{j}\)
02

- Calculate the total initial momentum

Add the momentum vectors of particle A and particle B to obtain the total initial momentum.\( \textbf{p}_{\text{total1}} = \textbf{p}_{A1} + \textbf{p}_{B1} = \left[ (-8.00 \mathrm{~kg~m/s}) \hat{i} + (-10.00 \mathrm{~kg~m/s}) \hat{j} \right] + \left[ (24.00 \mathrm{~kg~m/s}) \hat{i} + (-8.00 \mathrm{~kg~m/s}) \hat{j} \right] = (16.00 \mathrm{~kg~m/s}) \hat{i} + (-18.00 \mathrm{~kg~m/s}) \hat{j}\)
03

- Calculate the total mass

Add the masses of the two particles to get the total mass of the system.\( m_{\text{total}} = m_A + m_B = 2.00 \mathrm{~kg} + 4.00 \mathrm{~kg} = 6.00 \mathrm{~kg}\)
04

- Calculate the final velocity in unit-vector notation

Using the conservation of momentum, find the final velocity after the collision. The final velocity \(\textbf{v}_f\) is given by the total momentum divided by the total mass.\( \textbf{v}_f = \frac{\textbf{p}_{\text{total1}}}{m_{\text{total}}} = \frac{(16.00 \mathrm{~kg~m/s}) \hat{i} + (-18.00 \mathrm{~kg~m/s}) \hat{j}}{6.00 \mathrm{~kg}} = (2.67 \mathrm{~m/s}) \hat{i} + (-3.00 \mathrm{~m/s}) \hat{j}\)
05

- Calculate the magnitude of the final velocity

The magnitude of the velocity vector is given by the Pythagorean theorem.\( v_f = \sqrt{(2.67 \mathrm{~m/s})^2 + (-3.00 \mathrm{~m/s})^2} = \sqrt{7.13 + 9.00} = \sqrt{16.13} = 4.01 \mathrm{~m/s} \)
06

- Calculate the angle of the final velocity

The angle \(\theta\) can be found using the arctangent function.\( \theta = \tan^{-1} \left(\frac{-3.00 \mathrm{~m/s}}{2.67 \mathrm{~m/s}}\right) = \tan^{-1}(-1.124) = -48.4^\text{o} \)

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Key Concepts

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

Collision Physics

Understanding collision physics is crucial in analyzing the behavior of particles when they come into contact.
Two primary types of collisions exist: elastic and inelastic. In this exercise, we deal with an inelastic collision because the two particles stick together after the impact.
During an inelastic collision, momentum is conserved, but kinetic energy is not necessarily conserved.
This means the total momentum before the collision will equal the total momentum after the collision, regardless of potential energy losses due to heat, sound, or deformation.
Momentum Calculation
Momentum is a vector quantity defined by the product of mass and velocity.
The formula for momentum is \( \textbf{p} = m \textbf{v} \). For this exercise, we have two particles with different masses and velocities.
The steps to calculate initial momenta are:
  • Identify mass (m) and velocity (\( \textbf{v} \)) for each particle.
  • Compute individual momenta using \( \textbf{p}_{A} = m_{A} \textbf{v}_{A} \) and \( \textbf{p}_{B} = m_{B} \textbf{v}_{B} \).
  • Sum up the individual momenta vectors to get the total initial momentum, \( \textbf{p}_{\mathrm{total1}} = \textbf{p}_{A1} + \textbf{p}_{B1} \).
Vector Notation
Vector notation is essential to describe quantities with both magnitude and direction. Here鈥檚 a breakdown:
  • Vectors are represented as \( \textbf{v} = v_{i} \hat{i} + v_{j} \hat{j} \), where \( \hat{i} \) and \( \hat{j} \) are unit vectors along the x and y axes.
  • In the given exercise, the velocities are expressed as vectors, denoted by their components in terms of \( \hat{i} \) (x-direction) and \( \hat{j} \) (y-direction).
  • When dealing with vector quantities, addition and subtraction must consider both magnitude and direction.
Understanding how to manipulate vectors allows us to solve for the system's resultant momentum and identify the final velocity vector.
Velocity Magnitude
The velocity magnitude is the speed of the moving object, considering both x and y components. To calculate it:
  • Square the components of the final velocity vector.
  • Sum the squared components.
  • Take the square root of the sum.

Using the given example:

Final velocity vector: \( \textbf{v}_{f} = (2.67 \mathrm{~m/s}) \hat{i} + (-3.00 \mathrm{~m/s}) \hat{j} \).
Velocity magnitude, \( v_{f} = \sqrt{(2.67 \mathrm{~m/s})^{2} + (- 3.00 \mathrm{~m/s})^{2}} = 4.01 \mathrm{~m/s} \).
This magnitude represents the overall speed of the combined system after the collision.
Angle Calculation
To describe the direction of the resulting velocity vector, we calculate its angle relative to a given axis:
  • Use the arctangent function of the ratio of the y-component to the x-component of the velocity vector.

The formula is: \( \theta = \arctan \left(\dfrac{-3.00 \mathrm{~m/s}}{2.67 \mathrm{~m/s}} \right) = -48.4 ^{\circ} \).This negative angle indicates the direction of the velocity vector in the standard coordinate system, measured counterclockwise from the positive x-axis.
Therefore, understanding how to calculate the direction allows us to complete the description of the velocity in magnitude-angle notation.

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Most popular questions from this chapter

A rocket is moving away from the solar system at a speed of \(6.0 \times 10^{3} \mathrm{~m} / \mathrm{s}\). It fires its engine, which ejects exhaust with a speed of \(3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}\) relative to the rocket. The mass of the rocket at this time is \(4.0 \times 10^{4} \mathrm{~kg}\), and its acceleration is \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the thrust of the engine? (b) At what rate, in kilograms per second is exhaust ejected during the firing?

In Edmund Rostand's famous play, Cyrano de Bergerac, Cyrano, in an attempt to distract a suitor from visiting Roxanne, claims to have descended to Earth from the Moon and proclaims to have invented six novel and fantastical methods for traveling to the Moon. One is as follows. Sitting on an iron platform- thence To throw a magnet in the air. This is A method well conceived - the magnet flown, Infallibly the iron will pursue: Then quick! relaunch your magnet, and you thus Can mount and mount unmeasured distances! \(^{*}\) In an old cartoon, there is another version of this method. A character in the old West is on a hand-pumped, two-person rail car. After getting tired of pumping the handle up and down to make the car move along the rails, he takes out a magnet, hangs it from a fishing pole, and holds it in front of the cart. The magnet pulls the cart toward it, which pushes the magnet forward, and so on, so the cart moves forward continually. What do you think of these methods? Can some version of them work? Discuss in terms of the physics you have learned.

A \(150 \mathrm{~g}\) baseball pitched at a speed of \(40 \mathrm{~m} / \mathrm{s}\) is hit straight back to the pitcher at a speed of \(60 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the average force on the ball from the bat if the bat is in contact with the ball for \(5.0 \mathrm{~ms}\) ?

A \(60 \mathrm{~kg}\) man is ice-skating due north with a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) when he collides with a \(38 \mathrm{~kg}\) child. The man and child stay together and have a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(35^{\circ}\) north of east immediately after the collision. What are the magnitude and direction of the velocity of the child just before the collision?v

A projectile body of mass \(m_{A}\) and initial \(x\) component velocity \(v_{A x}\left(t_{1}\right)=10.0 \mathrm{~m} / \mathrm{s}\) collides with an initially stationary target body of mass \(m_{B}=2.00 m_{A}\) in a one-dimensional collision. What is the velocity of \(m_{B}\) following the collision if the two masses stick together?
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