Problem 61 - College Physics

Question

In a high-speed chase, a policeman's car bumps a criminal's car directly from behind to get his attention. The policeman's car is moving at \(40.0 \mathrm{~m} / \mathrm{s}\) to the right and has a total mass of \(1800 \mathrm{~kg}\). The criminal's car is initially moving in the same direction at \(38.0 \mathrm{~m} / \mathrm{s}\). His car has a total mass of \(1500 \mathrm{~kg}\). Assuming an elastic collision, determine their two velocities immediately after the bump.

Step-by-Step Solution

Answer

After the bump, the policeman's car moves at \(38.2 \mathrm{~m/s}\) and the criminal's car at \(39.8 \mathrm{~m/s}\).

Step 1: Identify Given Information

Start by understanding and writing down all given data of the problem. The policeman's car has a mass of \(1800 \mathrm{~kg}\) and is moving at a velocity of \(40.0 \mathrm{~m/s}\). The criminal's car has a mass of \(1500 \mathrm{~kg}\) and is moving at \(38.0 \mathrm{~m/s}\). The collision is elastic.

Step 2: Apply Conservation of Momentum

In an elastic collision, momentum is conserved. The total momentum before the collision equals the total momentum after the collision. Set up the equation: \[m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'\] where \(m_1 = 1800 \mathrm{~kg}\), \(v_1 = 40.0 \mathrm{~m/s}\), \(m_2 = 1500 \mathrm{~kg}\), \(v_2 = 38.0 \mathrm{~m/s}\), and \(v_1'\) and \(v_2'\) are the velocities we need to find.

Step 3: Apply Conservation of Kinetic Energy

Since the collision is elastic, kinetic energy is also conserved. Set up the kinetic energy conservation equation: \[\frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_1 v_1'^2 + \frac{1}{2}m_2 v_2'^2\] Insert the known values to form a second equation.

Step 4: Solve the System of Equations

You now have two equations:1. \(1800 \cdot 40.0 + 1500 \cdot 38.0 = 1800 \cdot v_1' + 1500 \cdot v_2'\)2. \(\frac{1}{2} \cdot 1800 \cdot 40.0^2 + \frac{1}{2} \cdot 1500 \cdot 38.0^2 = \frac{1}{2} \cdot 1800 \cdot (v_1'^2) + \frac{1}{2} \cdot 1500 \cdot (v_2'^2)\)Solve this system of equations to find \(v_1'\) and \(v_2'\).

Step 5: Calculate Final Velocities

Using the conservation equations, solve them simultaneously (manually or using a calculator) to find the values:\(v_1' = 38.2 \mathrm{~m/s}\) and \(v_2' = 39.8 \mathrm{~m/s}\).

Key Concepts

Conservation of Momentum

The conservation of momentum is a key principle in understanding collisions, especially elastic ones. In any closed system not subject to external forces, like our cars on a frictionless road, the total momentum before a collision is equal to the total momentum after the collision. This principle can be expressed with the equation:
\[ m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2' \]where:

• \( m_1 \) and \( m_2 \) are the masses of the two colliding bodies.
• \( v_1 \) and \( v_2 \) are their velocities before the collision.
• \( v_1' \) and \( v_2' \) are the velocities after the collision.

In the exercise, the policeman's car (\(1800 \,\mathrm{kg}\)) traveling at \(40.0 \,\mathrm{m/s}\) collides with the criminal's car (\(1500 \,\mathrm{kg}\)) moving at \(38.0 \,\mathrm{m/s}\). Applying the conservation of momentum directly provides one of the two necessary equations to solve for the final velocities after the collision.

Conservation of Kinetic Energy

Unlike a regular collision, in an elastic collision, the total kinetic energy is also conserved. This means that the kinetic energy before and after the collision remains the same. The formula for kinetic energy conservation is:
\[ \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_1 v_1'^2 + \frac{1}{2}m_2 v_2'^2 \]This equation ensures that the power of motion (kinetic energy) remains unchanged, maintaining the speed form of the physics. In our context, both cars together possess certain kinetic energy before they hit each other. Post-collision, this energy remains the same, though distributed differently due to changes in each car’s velocity. By combining this equation with the conservation of momentum, we can more accurately deduce the accurate outcomes of both velocity changes during the event of collision.

Velocity Calculation

The goal of solving the exercise is to determine the final velocities of the two cars after the collision. By setting up the momentum and kinetic energy equations, we developed a system of equations. Here's how we calculate:

• After identifying these equations, simplify and manipulate them using algebraic methods to isolate and solve for the unknown velocities \( v_1' \) and \( v_2' \).
• Typically, this involves substituting one equation into the other to eliminate a variable and solving the resulting equations step by step.

In this particular problem, solving provides us with values: \( v_1' = 38.2\,\mathrm{m/s} \) and \( v_2' = 39.8\,\mathrm{m/s} \). The velocities have effectively swapped to depict the elastic nature of their interaction in the head-on bump. Understanding velocity changes in this way is fundamental in collision physics, highlighting the conservation ideas applied efficiently.

System of Equations in Physics

Physics problems like these often involve a system - a set of equations that need to be simultaneously solved. In elastic collision problems, the system consists of the momentum and kinetic energy conservation equations. These systems are essential because:

• They allow us to solve multiple variables at once, underpinning many practical physics problems.
• Manipulating equations methodically allows deriving relationships between their properties.

To solve such a system, typically:
1. Rearrange each equation to express one variable in terms of the others. 2. Substitute equations into one another to eliminate variables step by step. 3. Solve the equations, often resulting in roots or zero points indicating the variable values. In our collision example, the derived system lets us solve for the post-collision velocities, ensuring both momentum and kinetic energy conservation criteria are met. Becoming proficient in these methods not only simplifies complex problems but hones analytical skills broadly applicable across physics contexts.