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(II) Billiard ball A of mass \(m_{\mathrm{A}}=0.120 \mathrm{kg}\) moving with speed \(v_{\mathrm{A}}=2.80 \mathrm{m} / \mathrm{s}\) strikes ball \(\mathrm{B}\) , initially at rest, of mass \(m_{\mathrm{B}}=0.140 \mathrm{kg} .\) As a result of the collision, ball \(\mathrm{A}\) is deflected off at an angle of \(30.0^{\circ}\) with a speed \(v_{\mathrm{A}}^{\prime}=2.10 \mathrm{m} / \mathrm{s}\) (a) Taking the \(x\) axis to be the original direction of motion of ball \(A,\) write down the equations expressing the conservation of momentum for the components in the \(x\) and \(y\) directions separately. (b) Solve these equations for the speed, \(v_{\mathrm{B}}^{\prime},\) and angle, \(\theta_{\mathrm{B}}^{\prime},\) of ball B. Do not assume the collision is elastic.

Short Answer

Expert verified
The speed of ball B is approximately 1.23 m/s, and the angle is about 47.6 degrees.

Step by step solution

01

Define Conservation of Momentum in the x-direction

In the x-direction, the momentum conservation equation is given by:\[ m_A \cdot v_A = m_A \cdot v_A' \cdot \cos(30^{\circ}) + m_B \cdot v_B' \cdot \cos(\theta_B') \]Substituting known values, it becomes:\[ 0.120 \times 2.80 = 0.120 \times 2.10 \times \cos(30^{\circ}) + 0.140 \times v_B' \cdot \cos(\theta_B') \].
02

Define Conservation of Momentum in the y-direction

In the y-direction, the momentum conservation equation is expressed as:\[ 0 = m_A \cdot v_A' \cdot \sin(30^{\circ}) - m_B \cdot v_B' \cdot \sin(\theta_B') \]Here, the momentum of ball A in the y-direction after collision is equal to the momentum of ball B in the negative y-direction, resulting in the equation:\[ 0.120 \times 2.10 \times \sin(30^{\circ}) = 0.140 \times v_B' \cdot \sin(\theta_B') \].
03

Solve the y-direction Equation for v_B' \sin(\theta_B')

From the y-direction momentum equation, solve for \(v_B' \sin(\theta_B')\):\[ 0.126 = 0.140 \times v_B' \cdot \sin(\theta_B') \]\[ v_B' \cdot \sin(\theta_B') = \frac{0.126}{0.140} \approx 0.90 \].
04

Solve the x-direction Equation for v_B' \cos(\theta_B')

From the x-direction momentum equation, isolate \(v_B' \cdot \cos(\theta_B')\):\[ 0.336 = 0.218 + 0.140 \times v_B' \cdot \cos(\theta_B') \]\[ 0.140 \times v_B' \cdot \cos(\theta_B') = 0.336 - 0.218 = 0.118 \]\[ v_B' \cdot \cos(\theta_B') = \frac{0.118}{0.140} \approx 0.843 \].
05

Combine Expressions to Solve for v_B'

Use the identity \((v_B')^2 = (v_B' \sin(\theta_B'))^2 + (v_B' \cos(\theta_B'))^2\) to solve for \(v_B'\):\[ (v_B')^2 = (0.90)^2 + (0.843)^2 \]\[ (v_B')^2 = 0.81 + 0.710 \approx 1.52 \]\[ v_B' = \sqrt{1.52} \approx 1.23 \text{ m/s} \].
06

Calculate Angle θ_B’

Find \(\theta_B'\) using \(\tan(\theta_B') = \frac{v_B' \sin(\theta_B')}{v_B' \cos(\theta_B')}\):\[ \tan(\theta_B') = \frac{0.90}{0.843} \approx 1.067 \]\[ \theta_B' = \tan^{-1}(1.067) \approx 47.6^{\circ} \].

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Key Concepts

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

Collision Mechanics
In the world of physics, collision mechanics is the study of what happens when two objects strike each other. This topic is commonly encountered in different forms, **elastic** or **inelastic** collisions, which describe how objects behave post-collision. In simple terms, when two objects collide, their movement before and after impact is studied through principles like momentum and energy. The key factor is understanding how these quantities are transferred or transformed.

During a collision, objects exert forces on each other for a brief period. Depending on the type and circumstances of the collision, the manner in which these forces affect the velocities of the objects can differ. One of the primary concepts used to study collisions is the conservation of momentum, which asserts that the total momentum of a system remains constant if no external forces act on the system. This becomes especially important when analyzing systems where friction and other forces are negligible.

Collision mechanics ultimately aims to predict the post-collision characteristics of the involved objects, like velocity and direction. These predictions help us understand real-world applications such as traffic accidents, sports, and even in structural engineering.
Inelastic Collisions
Inelastic collisions are a type of collision in which the kinetic energy is not conserved, though momentum is. This contrasts with **elastic collisions**, where both kinetic energy and momentum are conserved. Inelastic collisions are common in the real world because real-world surfaces usually don't perfectly bounce off each other; they often deform or generate heat, leading to a loss of kinetic energy.

In the case of our exercise, the collision between the two billiard balls is inelastic. This is evident as it allows for the transformation of kinetic energy into other forms, such as sound or heat. However, the total momentum before and after the impact remains the same. **Momentum conservation** is used to determine the velocity and angle of ball B after collision. This demonstrates how energy transition doesn't hinder the ability to predict momentum outcomes.

The key takeaway is that while kinetic energy may not be the same after a collision, using momentum conservation equations still enables predictions about the resulting velocities and angles of the objects. This highlights the importance of separating energy and momentum conservation, as they independently govern different aspects of the collision dynamics.
Momentum Components
Understanding momentum components is essential to solving collision problems in physics. Momentum is a vector quantity, which means it has both magnitude and direction. When analyzing problems involving momentum, it is crucial to break down the momentum into components along each axis – typically the x-axis and y-axis.

By decomposing the momentum into its components, it is easier to apply the conservation of momentum laws in each direction separately. This is seen in our exercise, where the initial momentum of the billiard balls is split into two components; one parallel to the original direction (x-direction), and the other perpendicular (y-direction).

For the billiard balls collision problem, conservation laws are applied separately:
- In the x-direction, the equation considers the cosine of the angles. - In the y-direction, the equation involves the sine of the angles.

This split helps in addressing each directional momentum uniquely and obtaining separate equations for both directions. Combining the results from these calculations provides the tools necessary to find the precise speed and directional angle of the balls post-collision.

By utilizing these momentum components, students can gain clarity in breaking down complex vector problems into simpler, more manageable tasks, ensuring a clear path from problem to solution.

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Most popular questions from this chapter

The space shuttle launches an \(850-\mathrm{kg}\) satellite by ejecting it from the cargo bay. The ejection mechanism is activated and is in contact with the satellite for 4.0 s to give it a velocity of \(0.30 \mathrm{~m} / \mathrm{s}\) in the \(z\) -direction relative to the shuttle. The mass of the shuttle is \(92,000 \mathrm{~kg} .\) ( \(a\) ) Determine the component of velocity \(v_{\mathrm{f}}\) of the shuttle in the minus \(z\) -direction resulting from the ejection. (b) Find the average force that the shuttle exerts on the satellite during the ejection.

(II) A ball of mass \(0.220 \mathrm{~kg}\) that is moving with a speed of \(7.5 \mathrm{~m} / \mathrm{s}\) collides head-on and elastically with another ball initially at rest. Immediately after the collision, the incoming ball bounces backward with a speed of \(3.8 \mathrm{~m} / \mathrm{s}\). Calculate (a) the velocity of the target ball after the collision, and (b) the mass of the target ball.

(II) A uniform thin wire is bent into a semicircle of radius \(r\). Determine the coordinates of its center of mass with respect to an origin of coordinates at the center of the "full" circle.

(II) Show that, in general, for any head-on one-dimensional elastic collision, the speeds after collision are $$ v_{\mathrm{B}}^{\prime}=v_{\mathrm{A}}\left(\frac{2 m_{\mathrm{A}}}{m_{\mathrm{A}}+m_{\mathrm{B}}}\right)+v_{\mathrm{B}}\left(\frac{m_{\mathrm{B}}-m_{\mathrm{A}}}{m_{\mathrm{A}}+m_{\mathrm{B}}}\right) $$ and $$ v_{\mathrm{A}}^{\prime}=v_{\mathrm{A}}\left(\frac{m_{\mathrm{A}}-m_{\mathrm{B}}}{m_{\mathrm{A}}+m_{\mathrm{B}}}\right)+v_{\mathrm{B}}\left(\frac{2 m_{\mathrm{B}}}{m_{\mathrm{A}}+m_{\mathrm{B}}}\right) $$ where \(v_{\mathrm{A}}\) and \(v_{\mathrm{B}}\) are the initial speeds of the two objects of mass \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}}\)

Astronomers estimate that a \(2.0-\mathrm{km}\) -wide asteroid collides with the Earth once every million years. The collision could pose a threat to life on Earth. ( \(a\) ) Assume a spherical asteroid has a mass of \(3200 \mathrm{~kg}\) for each cubic meter of volume and moves toward the Earth at \(15 \mathrm{~km} / \mathrm{s}\). How much destructive energy could be released when it embeds itself in the Earth? (b) For comparison, a nuclear bomb could release about \(4.0 \times 10^{16} \mathrm{~J}\). How many such bombs would have to explode simultaneously to release the destructive energy of the asteroid collision with the Earth?

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