Problem 10 - Applied Physics

Question

One cart of mass \(15.0 \mathrm{~kg}\) is moving \(5.00 \mathrm{~m} / \mathrm{s}\) to the right on a frictionless track and collides with a cart of mass \(3.00 \mathrm{~kg}\). The final velocity of the carts that become stuck together after the collision is \(1.50 \mathrm{~m} / \mathrm{s}\) to the right. Find the velocity of the second cart before the collision.

Step-by-Step Solution

Answer

The velocity of the second cart before the collision was \(-16.0 \, \mathrm{m/s}\), indicating it was moving to the left.

Step 1: Understand the Conservation of Momentum Principle

In this problem, we apply the principle of conservation of momentum. For any isolated system, momentum before collision equals momentum after collision. This means: \[ m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \] where \( m_1 \) and \( m_2 \) are the masses of the two carts, \( v_1 \) and \( v_2 \) are their velocities before the collision, and \( v_f \) is their common velocity after collision.

Step 2: Identify Known and Unknown Variables

We know: - \( m_1 = 15.0 \, \mathrm{kg} \)- \( v_1 = 5.00 \, \mathrm{m/s} \)- \( m_2 = 3.00 \, \mathrm{kg} \)- \( v_f = 1.50 \, \mathrm{m/s} \)We need to find: \( v_2 \), the velocity of the second cart before the collision.

Step 3: Plug the Values into the Momentum Equation

Substitute the known values into the momentum equation:\[ 15.0 \times 5.00 + 3.00 \times v_2 = (15.0 + 3.00) \times 1.50 \] This simplifies to:\[ 75.0 + 3.00v_2 = 27.0 \]

Step 4: Solve for the Unknown Velocity \( v_2 \)

Isolate \( v_2 \):\[ 75.0 + 3.00v_2 = 27.0 \] Subtract 75.0 from both sides:\[ 3.00v_2 = 27.0 - 75.0 \] \[ 3.00v_2 = -48.0 \] Now, divide both sides by 3.00:\[ v_2 = \frac{-48.0}{3.00} \]\[ v_2 = -16.0 \, \mathrm{m/s} \]

Step 5: Interpret the Solution

The solution, \( v_2 = -16.0 \, \mathrm{m/s} \), indicates that the second cart was moving to the left (opposite direction to the initial movement of the first cart) with a velocity of 16.0 m/s before the collision.

Key Concepts

Collisions in Physics

In physics, collisions are events where two or more bodies exert forces on each other for a short period. These interactions can be exciting to explore because they involve exchanges of energy and momentum. There are two main types of collisions — elastic and inelastic. In elastic collisions, the total kinetic energy and momentum are conserved, meaning objects bounce off each other without losing energy. In contrast, during inelastic collisions, such as the one described in our exercise, kinetic energy is not conserved, though momentum still is.

• Elastic collisions: Both momentum and kinetic energy are conserved.
• Inelastic collisions: Only momentum is conserved.
• Perfectly inelastic collisions: The colliding objects stick together.

In our exercise, it was a perfectly inelastic collision, as the two carts stuck together post-collision, sharing a common velocity. Understanding these distinctions helps in analyzing scenarios and predicting outcomes of physical interactions.

Momentum Equation

The momentum equation is a fundamental principle used to solve collision problems. Momentum (\(p\)) is defined as the product of an object's mass and velocity: \(p = mv\). The principle of the conservation of momentum states that in a closed system with no external forces, the total momentum before an event must equal the total momentum after. When looking at the momentum equation for our cart collision:\[m_1v_1 + m_2v_2 = (m_1 + m_2)v_f\]

• Variables:
  • \(m_1\) and \(m_2\) are the masses of the carts.
  • \(v_1\) and \(v_2\) are the initial velocities.
  • \(v_f\) is the final velocity of the combined carts.
• The left side represents the total momentum before the collision.
• The right side represents the total momentum after the collision.

By plugging known values into this equation, you can solve for an unknown, as seen by calculating \(v_2\), the initial velocity of the second cart before the collision.

Velocity Calculations

Calculating velocities is a core part of analyzing collisions and understanding how systems move. To solve problems involving velocities, it's crucial to clearly define the directions and magnitudes involved, as velocity is a vector quantity. In the given problem, we determined the velocity of the second cart (\(v_2\)) before the collision by rearranging the momentum equation. This process involved:

• Substituting all known quantities from the problem statement into the equation.
• Isolating the variable of interest (\(v_2\)) using algebraic manipulation.
• Maintaining awareness of the sign of the velocity, which indicates direction relative to the reference point.

The computed \(v_2 = -16.0 \, \mathrm{m/s}\) signifies that the second cart was initially moving to the left, opposite to the first cart's direction.
It is important to use consistent units throughout the calculation: kilograms for mass and meters per second for velocity in standard SI units. By ensuring good practice in velocity calculations, students can accurately predict and interpret the dynamics of physical systems.