Problem 34 - College Physics

Question

S This is a symbolic version of Problem 33. A railroad car of mass \(M\) moving at a speed \(v_{1}\) collides and couples with two coupled railroad cars, each of the same mass \(M\) and moving in the same direction at a speed \(t_{2}\). (a) What is the speed \(v_{y}\) of the three coupled cars after the collision in terms of \(v_{1}\) and \(v_{2}\) ? (b) How much kinetic energy is lost in the collision? Answer in terms of \(M\), \(v_{1}\), and \(v_{2}\).

Step-by-Step Solution

Answer

The speed of the three coupled cars after the collision is \(v_{f}\) = \(\frac{v_{1} + 2v_{2}}{3}\) and the loss in kinetic energy can be found by subtracting the final kinetic energy from the initial kinetic energy.

Step 1: Applying the Principle of Conservation of Momentum

According to the principle of conservation of momentum, total momentum before the collision equals total momentum after it. Before collision, there are two different systems: a single railroad car with mass \(M\) and velocity \(v_{1}\), and two coupled railroad cars with individual masses \(M\) moving at velocity \(v_{2}\). So total momentum can be calculated as \(Mv_{1} + 2Mv_{2} = (M + 2M)v_{f}\), where \(v_{f}\) is the final velocity of three coupled cars.

Step 2: Solving for the Final Velocity

On solving the equation from the first step, we find the final velocity \(v_{f}\) = \(\frac{v_{1} + 2v_{2}}{3}\).

Step 3: Calculate Initial and Final Kinetic Energy

Next, we calculate the initial kinetic energy (before collision) and the final kinetic energy (after collision). The initial kinetic energy can be calculated as \(\frac{1}{2}Mv_{1}^2 + 2*\frac{1}{2}*M*v_{2}^2\). The final kinetic energy is \(\frac{1}{2}(3M)v_{f}^2\).

Step 4: Calculate the Loss in Kinetic Energy

The kinetic energy lost during the collision is the difference between the initial and final kinetic energy, calculated as \(\Delta KE = KE_{initial} - KE_{final}\).

Key Concepts

Kinetic Energy

Kinetic energy is an essential concept in physics, describing the energy an object possesses due to its motion. It is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In this problem, we have two systems we're considering before they collide:

• A single railroad car moving with velocity \( v_1 \) and mass \( M \).
• Two coupled railroad cars, each having the same mass \( M \) and moving with velocity \( v_2 \).

Thus, the initial kinetic energy is calculated as the sum of the kinetic energies of these two systems. After the collision, the three cars move together with a common velocity \( v_f \), which changes the distribution of kinetic energy. Any change in kinetic energy during a collision indicates that energy has not been conserved as mechanical energy, due to factors like deformation or heat. The challenge lies in calculating both initial and final kinetic energies to compute the energy lost.

Inelastic Collision

An inelastic collision is one of several types of collisions where kinetic energy is not conserved. However, unlike perfectly elastic collisions, the objects in an inelastic collision stick together post-collision, lending itself to problems involving coupled objects, such as our railroad cars. In the given problem, we observe this phenomenon with a railroad car colliding with and coupling into a set of two previously coupled cars. Because they stick together, the situation is a prime example of an inelastic collision. Key characteristics to note:

• The objects stick together, resulting in a decrease in kinetic energy.
• This type of collision conserves momentum but not kinetic energy.

Understanding inelastic collisions involves being able to evaluate how the speed and energy of a combined mass system differ from separate mass systems both before and after the collision.

Momentum Conservation

The principle of momentum conservation is fundamental in analyzing collisions. It states that the total momentum of a closed system remains constant if no external forces act on it. Momentum \( p \) is the product of mass and velocity, given by \( p = mv \).For the recall case in the exercise:

• The total initial momentum consists of the momentum of one car with mass \( M \) moving at \( v_1 \) and the two coupled cars, each with mass \( M \), moving at \( v_2 \).
• The total momentum before the collision is: \( Mv_1 + 2Mv_2 \).
• After the collision, the cars move together, so we have a combined mass of \( 3M \) moving at a new velocity \( v_f \).

The equation \( Mv_1 + 2Mv_2 = 3Mv_f \) allows us to solve for \( v_f \). This direct application of momentum conservation explains how interacting objects behave when forces only affect one another.