Problem 69 - College Physics

Question

A freight car with a mass of \(25000 \mathrm{~kg}\) rolls down an inclined track through a vertical distance of \(1.5 \mathrm{~m}\). At the bottom of the incline, on a level track, the car collides and couples with an identical freight car that was at rest. What percentage of the initial kinetic energy is lost in the collision

Step-by-Step Solution

Answer

20% of the kinetic energy is lost in the collision.

Step 1: Calculate Initial Potential Energy

The potential energy at the top of the incline is calculated using the formula for gravitational potential energy: \[ PE = mgh \]where \( m = 25000 \text{ kg} \) is the mass of the freight car,\( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity,and \( h = 1.5 \text{ m} \) is the height.Plug these values in to find the initial potential energy:\[ PE = 25000 \times 9.8 \times 1.5 = 367500 \text{ J} \]

Step 2: Determine Initial Kinetic Energy at the Bottom

Since the freight car rolls down without friction, all potential energy is converted to kinetic energy at the bottom of the incline.Thus, the initial kinetic energy at the bottom is:\[ KE_{initial} = 367500 \text{ J} \]

Step 3: Calculate Final Kinetic Energy After Collision

After the collision, the two freight cars move together. Using conservation of momentum, first find the final speed:Initial momentum:\[ p_{initial} = m_1 v_1 = 25000 \cdot v \]Final momentum (combined mass):\[ p_{final} = (m_1 + m_2)v_f = 50000 \cdot v_f \]Equating initial and final momentum:\[ 25000v = 50000v_f \]\[ v_f = \frac{v}{2} \]Thus, the final kinetic energy:\[ KE_{final} = \frac{1}{2}(50000)v_f^2 \]Since \( KE_{initial} = \frac{1}{2} (25000) v^2 \), using \( v_f = \frac{v}{2} \),\[ KE_{final} = \frac{1}{2} (50000) \left( \frac{v}{2} \right)^2 = \frac{1}{2} (50000) \cdot \frac{v^2}{4} = \frac{25000}{4} \cdot v^2 = \frac{6250 v^2}{2} \]

Step 4: Calculate Percentage of Kinetic Energy Lost

Initial kinetic energy:\[ KE_{initial} = \frac{1}{2} (25000) v^2 \]Final kinetic energy:\[ KE_{final} = \frac{6250v^2}{2} \]Percentage of energy lost:\[ \text{Percent lost} = \frac{367500v^2 - 6250v^2}{367500v^2} \times 100t = \frac{0.5}{0.5 + 0.125} \times 100 = 20\%\]

Key Concepts

Potential Energy

Potential energy is the stored energy of an object because of its position or state. In this scenario, we explore gravitational potential energy, which depends on the height of the freight car above the ground. The formula to calculate gravitational potential energy is:\[ PE = mgh \]where:

• \(m\) is the mass of the object,
• \(g\) is the acceleration due to gravity (\(9.8 \text{ m/s}^2\)),
• \(h\) is the height of the object above the ground.

As the freight car sits at the top of the incline, it holds potential energy because it can move downhill. This energy will later change forms as it moves. In our exercise, by knowing the mass (\(25000 \text{ kg}\)) and height (\(1.5 \text{ m}\)), we calculated that the potential energy is initially \(367500 \text{ J}\). This value expresses how much energy the freight car can exert as it descends.

Kinetic Energy

Kinetic energy is the energy of motion, which an object possesses because of its velocity. As the freight car rolls down the track, it speeds up, and its potential energy is transformed into kinetic energy. The kinetic energy at any given moment can be calculated using the formula:\[ KE = \frac{1}{2} mv^2 \]In our example, as the freight car reaches the bottom of the incline, all its potential energy has converted into kinetic energy due to its descent, assuming no energy loss to friction. This means the kinetic energy at this point is also \(367500 \text{ J}\). This conversion exemplifies how energy is conserved when switching from one form to another. The speed of the freight car at the bottom determines this energy when plugged into the equation above.

Momentum Conservation

Momentum conservation is a principle stating that in a closed system, the total momentum before any interaction is the same as the total momentum after that interaction. In the context of our freight car problem, momentum is conserved during the collision between the moving freight car and the stationary one.Initially, only one freight car has momentum, which is given by:\[ p_{initial} = m_1v \]After colliding and coupling with the rest freight car, the combined system's total momentum becomes:\[ p_{final} = (m_1 + m_2)v_f \]where \(m_1\) and \(m_2\) are the masses of the two freight cars and \(v_f\) is the final speed after collision.By equating initial and final momentum to solve for \(v_f\), we ensure that the total momentum remains unchanged. This balance holds even as the moving car strikes and combines with another freight car, demonstrating the robustness of the momentum conservation law outside the influence of external forces.

Energy Transformation

Energy transformation refers to the change of energy from one form to another, as seen in this freight car exercise. When the freight car descends the incline, its gravitational potential energy shifts entirely into kinetic energy—until it collides with the second car.During the collision, kinetic energy doesn't get entirely conserved as momentum does—some of it transforms into other forms like sound, heat, or deformation energy, which are not recoverable in kinetic terms. So, after the collision, when the two cars move together, we use the conservation of momentum to find their new kinetic energy:\[ KE_{final} = \frac{1}{2} (m_1 + m_2)v_f^2 \]We determined that the final kinetic energy after collision is reduced to \( \frac{6250v^2}{2} \), leading to a loss of 20% of the initial kinetic energy.This energy loss points to non-elastic collisions where energy dissipates into various non-mechanical energy forms, showcasing the practical nuances of real-world energy transformations.