91Ó°ÊÓ

A 16 -g mass is moving in the \(+x\) -direction at \(30 \mathrm{~cm} / \mathrm{s}\), while a \(4.0\) -g mass is moving in the \(-x\) -direction at \(50 \mathrm{~cm} / \mathrm{s}\). They collide headon and stick together. Find their velocity after the collision. Assume negligible friction. This is a completely inelastic collision for which KE is not conserved, although momentum is. Let the 16 -g mass be \(m_{1}\) and the 4.0-g mass be \(m_{2}\). Take the \(+x\) -direction to be positive. That means that the velocity of the \(4.0\) -g mass has a scalar value of \(u_{2 x}\) \(=-50 \mathrm{~cm} / \mathrm{s}\). We apply the law of conservation of momentum to the system consisting of the two masses: $$ \begin{array}{l} \text { Momentum before impact = Momentum after impact }\\\ \begin{aligned} m_{1} v_{1 x}+m_{2} v_{2 x} &=\left(m_{1}+m_{2}\right) v_{x} \\ (0.016 \mathrm{~kg})(0.30 \mathrm{~m} / \mathrm{s})+(0.0040 \mathrm{~kg})(-0.50 \mathrm{~m} / \mathrm{s}) &=(0.020 \mathrm{~kg}) v_{x} \\ v_{x} &=+0.14 \mathrm{~m} / \mathrm{s} \end{aligned} \end{array} $$ (Notice that the \(4.0\) -g mass has negative momentum.) Hence, \(\overrightarrow{\mathbf{v}}=\) \(0.14 \mathrm{~m} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION

Step by step solution

01

Understand the Conservation of Momentum

Momentum is always conserved in a closed system. This means that the total momentum before the collision equals the total momentum after the collision. For two masses, this is expressed as: \[ m_1 v_{1x} + m_2 v_{2x} = (m_1 + m_2) v_x \] where \(v_x\) is the velocity after collision.

02

Convert Mass and Velocity into Compatible Units

For calculations, mass should be in kilograms and velocity in meters per second. Convert: - 16 g to 0.016 kg - 30 cm/s to 0.30 m/s - 4 g to 0.004 kg - 50 cm/s to 0.50 m/s.

03

Substitute Given Values into the Momentum Equation

Plug the values of mass and velocity into the conservation of momentum formula: \[ (0.016 \, \text{kg})(0.30 \, \text{m/s}) + (0.0040 \, \text{kg})(-0.50 \, \text{m/s}) = (0.020 \, \text{kg}) v_x \].

04

Calculate the Momentum Before Collision

Calculate the momentum for each object: - Momentum of the 16-g mass is \(0.016 \times 0.30 = 0.0048 \, \text{kg m/s}\).- Momentum of the 4-g mass is \(0.004 \times (-0.50) = -0.002 \, \text{kg m/s}\). - Total momentum before is \(0.0048 - 0.002 = 0.0028 \, \text{kg m/s}\).

05

Solve for the Final Velocity \(v_x\)

Set the total initial momentum equal to the momentum after collision and solve for \(v_x\): \[ 0.0028 = (0.020) v_x \]Rearrange to find \(v_x\): \[ v_x = \frac{0.0028}{0.020} = 0.14 \, \text{m/s} \].

06

Determine the Direction of the Velocity

Since the final velocity \(v_x = 0.14 \, \text{m/s}\) is positive, the direction is in the +x direction.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Conservation of Momentum

When it comes to understanding physics, the principle of conservation of momentum is foundational. Momentum, which is the product of an object's mass and velocity, is conserved in an isolated system. This means that just before and just after any collision, the total momentum remains unchanged. For our specific problem, the sum of the initial momenta of both masses before they collide is equal to the combined momentum of both masses after they stick together. This is represented by the equation: \[ m_1 v_{1x} + m_2 v_{2x} = (m_1 + m_2) v_x \].Key points about conservation of momentum:

• It applies to all kinds of collisions, whether elastic or inelastic.
• Momentum is a vector quantity, which means it has both magnitude and direction.
• In our problem, even though the individual momenta may change, the total system momentum stays constant.

Understanding this principle helps in solving many collision-related problems and is crucial for calculating post-collision velocities.

Inelastic Collision

In the realm of physics, collisions can be classified as elastic or inelastic, based on whether kinetic energy is conserved. In the scenario presented, we deal with a completely inelastic collision. Here, the two colliding masses stick together after they crash, making them move as one unit thereafter. Important points to note about inelastic collisions are:

• Kinetic energy is not conserved, but momentum is.
• After the collision, the objects do not bounce off each other but instead cling together, sharing a common velocity.
• This change in system behavior affects how velocities and forces are modified post-collision.
• In this enabled problem, the final merged mass moves with a common velocity that can be determined using the conservation of momentum.

Recognizing the type of collision is critical for selecting the right physics principles to apply.

Unit Conversion

An essential part of solving physics problems is ensuring that all units are compatible for calculation. This process, known as unit conversion, requires careful attention. Variables like mass and velocity often need adjustments to maintain consistency across equations. For the given exercise, mass in grams must be converted to kilograms, and velocities in centimeters per second converted to meters per second: - Mass: Convert 16 g to 0.016 kg and 4 g to 0.004 kg. - Velocity: Convert 30 cm/s to 0.30 m/s and 50 cm/s to 0.50 m/s. Key considerations when performing unit conversions:

• Use familiar conversion factors (e.g., 1 kg = 1000 g, 1 m = 100 cm) to simplify the process.
• Ensure all related terms in equations reflect the same unit type (e.g., meters, kilograms) before solving.
• Regular consistency checks can prevent errors that might arise from incorrect unit application.

Mastering unit conversion is vital to obtaining accurate and meaningful results from calculations in physics.