Problem 6 - Applied Physics

Question

A \(75.0\) -kg paint cart with rubber bumpers is rolling \(0.965 \mathrm{~m} / \mathrm{s}\) to the right and strikes a second cart of mass \(85.0 \mathrm{~kg}\) moving \(1.30 \mathrm{~m} / \mathrm{s}\) to the left. After the collision, the heavier cart is traveling \(0.823 \mathrm{~m} / \mathrm{s}\) to the right. What is the velocity of the lighter cart after the collision?

Step-by-Step Solution

Answer

The velocity of the lighter cart after the collision is \(-1.4412 \, \text{m/s}\) to the left.

Step 1: Identify the Components of the System

Let's set up our problem. We have two carts involved in a collision. Cart A (the paint cart) has a mass of \( m_A = 75.0 \, \text{kg} \) and is moving to the right (positive direction) with a velocity of \( v_{A1} = 0.965 \, \text{m/s} \). The second cart, Cart B, has a mass of \( m_B = 85.0 \, \text{kg} \) and is moving to the left (negative direction) with a velocity of \( v_{B1} = -1.30 \, \text{m/s} \). After the collision, Cart B is moving to the right with a velocity of \( v_{B2} = 0.823 \, \text{m/s} \).

Step 2: Apply the Law of Conservation of Momentum

The law of conservation of momentum states that the total momentum of a system before a collision is equal to the total momentum after the collision, for isolated systems. The equation is given as:\[m_A v_{A1} + m_B v_{B1} = m_A v_{A2} + m_B v_{B2}\]Substitute the known values:\[75.0 \, \text{kg} \times 0.965 \, \text{m/s} + 85.0 \, \text{kg} \times (-1.30 \, \text{m/s}) = 75.0 \, \text{kg} \times v_{A2} + 85.0 \, \text{kg} \times 0.823 \, \text{m/s}\]

Step 3: Solve for the Unknown Velocity

Now we solve for \( v_{A2} \), the velocity of the lighter cart after the collision. Calculate the left side of the equation:\[75.0 \, \text{kg} \times 0.965 \, \text{m/s} = 72.375 \, \text{kg} \cdot \text{m/s}\]\[85.0 \, \text{kg} \times (-1.30 \, \text{m/s}) = -110.5 \, \text{kg} \cdot \text{m/s}\]Thus, the total initial momentum is:\[72.375 \, \text{kg} \cdot \text{m/s} - 110.5 \, \text{kg} \cdot \text{m/s} = -38.125 \, \text{kg} \cdot \text{m/s}\]Calculate the momentum for Cart B after the collision:\[85.0 \, \text{kg} \times 0.823 \, \text{m/s} = 69.965 \, \text{kg} \cdot \text{m/s}\]Now set up the equation and solve for \( v_{A2} \):\[-38.125 = 75.0 \, \text{kg} \times v_{A2} + 69.965\]\[75.0 \, \text{kg} \times v_{A2} = -38.125 - 69.965 = -108.09\]\[v_{A2} = \frac{-108.09}{75}\]\[v_{A2} = -1.4412 \, \text{m/s}\]

Step 4: Interpret the Solution

The negative sign in \( v_{A2} = -1.4412 \, \text{m/s} \) means that the lighter cart is moving to the left after the collision.

Key Concepts

Collisions

Collisions in physics refer to an event where two or more objects exert forces on each other for a short duration. These interactions can tell us a lot about the behaviors of systems under the influence of forces.
In this scenario, the carts' collision demonstrates the key principles governing momentum conservation.

• A specific collision can be classified as elastic or inelastic, depending on whether the kinetic energy is conserved. In elastic collisions, both momentum and kinetic energy remain constant, while in inelastic collisions, only momentum is conserved.
• Collisions can also be one-dimensional or multi-dimensional. Here, we have a straight-line path, making it a one-dimensional problem.

Understanding the momentum exchange during these interactions helps us predict the subsequent movement of involved objects. For the given carts, though the question does not specify, it's likely slightly inelastic since momentum—but not necessarily kinetic energy—is conserved.

Physics Problems

Solving physics problems involves identifying the known and unknown variables, applying the appropriate principles or laws, and performing calculations to find a solution.
In this particular exercise, some major steps need to be undertaken:

• Understanding the scenario: recognizing that we are dealing with carts involved in a collision requiring momentum principles.
• Listing known values: masses, initial velocities, and post-collision velocity of one cart.
• Writing down the conservation of momentum equation to account for all momentum before and after the collision.
• Simplified logical steps to isolate unknowns and solve the equation.

When tackling problems, such as the one involving these carts, it is critical to analyze the situation thoroughly and apply equations consistently to resolve for unknowns.

System of Particles

In physics, a system of particles refers to a collection of two or more objects considered as a single entity.
This allows us to analyze their collective behavior under certain physical laws. For this exercise:

• The system comprises two carts, Cart A and Cart B, colliding with one another.
• Considering them as a system simplifies analyzing their interactions due to momentum conservation, enabling us to write a single equation governing their total momentum.
• By treating the carts collectively, it’s easier to apply the law of conservation of momentum with the pre-determined boundary: no external forces acting, which assumes it was an isolated system during the interaction.

Assessing the collective momentum, instead of each individually, allows for streamlined calculations and is fundamental to simplifying complex interactions.

Velocity Calculation

Velocity calculation, especially after a collision, requires careful application of physics principles to determine the resulting speed and direction of movement.
To find the velocity of the lighter cart after impact:

• Apply the formula for momentum conservation: \[ m_A v_{A1} + m_B v_{B1} = m_A v_{A2} + m_B v_{B2} \]
• Reverse engineer the equation to solve for the unknown, in this case, \( v_{A2} \).
• Execute algebraic manipulation - isolating the term \( v_{A2} \) by substituting known values and rearranging the equation appropriately.
• The result \( v_{A2} = -1.4412 \, \text{m/s} \) indicates the direction (negative as left post-collision).

Understanding how to derive a velocity after interactions like collisions enables predictions on object dynamics, crucial for everything from simple physics projects to complex engineering applications.