Problem 70 - University Physics with Modern Physics

Question

Two Sumo wrestlers are involved in an inelastic collision. The first wrestler, Hakurazan, has a mass of \(135 \mathrm{~kg}\) and moves forward along the positive \(x\) -direction at a speed of \(3.5 \mathrm{~m} / \mathrm{s}\). The second wrestler, Toyohibiki, has a mass of \(173 \mathrm{~kg}\) and moves straight toward Hakurazan at a speed of \(3.0 \mathrm{~m} / \mathrm{s} .\) Immediately after the collision, Hakurazan is deflected to his right by \(35^{\circ}\) (see the figure). In the collision, \(10 \%\) of the wrestlers' initial total kinetic energy is lost. What is the angle at which Toyohibiki is moving immediately after the collision?

Step-by-Step Solution

Answer

Answer: To find the angle at which Toyohibiki is moving after the collision, we need to use conservation of linear momentum and energy equations. After setting up and solving these equations, we can find the angle θ using a numerical method.

Step 1: Conservation of linear momentum in the x and y directions

To start, we need to write the equations for the conservation of linear momentum in the x and y directions. The linear momentum before the collision should be equal to the linear momentum after the collision in each direction, accounting for the fact that 10% of the initial total kinetic energy is lost. Conservation of linear momentum in the x direction: $$m_1v_{1x} + m_2v_{2x} = m_1v_{1fx} + m_2v_{2fx}$$ Conservation of linear momentum in the y direction: $$m_1v_{1y} + m_2v_{2y} = m_1v_{1fy} + m_2v_{2fy}$$

Step 2: Calculate total linear momentum before and after the collision

Using the given initial velocities and masses, we can calculate the total linear momentum before the collision in the x and y directions: Before collision: $$p_{1x} = m_1v_{1x} = 135 \cdot 3.5$$ $$p_{1y} = m_1v_{1y} = 0$$ $$p_{2x} = -m_2v_{2x} = -173 \cdot 3.0$$ $$p_{2y} = m_2 v_{2y} = 0$$ $$p_{totalx} = p_{1x}+p_{2x}$$ $$p_{totaly} = p_{1y}+p_{2y}$$ After collision: Hakurazan moves towards the positive x-direction at \(35^\circ\) after the collision: $$v_{1fx} = v_1'\cos(35^\circ)$$ $$v_{1fy} = v_1'\sin(35^\circ)$$ Toyohibiki moves in some direction with the angle θ after the collision: $$v_{2fx} = v_2'\cos(\theta)$$ $$v_{2fy} = v_2'\sin(\theta)$$ Conservation of momentum equations become: In x-direction: $$135 \cdot 3.5 - 173 \cdot 3.0 = 135 \cdot v_1'\cos(35^\circ) + 173 \cdot v_2'\cos(\theta)$$ In y-direction: $$0 = 135 \cdot v_1'\sin(35^\circ) - 173 \cdot v_2'\sin(\theta)$$

Step 3: Find the relationship between the final velocities of the two wrestlers

We know that 10% of the initial total kinetic energy is lost during the collision. From the given information, we can find that relationship: $$0.9 \left( \frac{1}{2}m_1v_{1x}^2 + \frac{1}{2}m_2v_{2x}^2 \right) = \frac{1}{2}m_1 {v_1'}^2 + \frac{1}{2}m_2 {v_2'}^2$$ Solve for the value of \(v_2'\) in terms of \(v_1'\): $$v_2' = \sqrt{\frac{0.9(m_1v_{1x}^2+m_2v_{2x}^2)-m_1{v_1'}^2}{m_2}}$$

Step 4: Determine the angle at which Toyohibiki is moving immediately after the collision

Now we can substitute the value of \(v_2'\) in terms of \(v_1'\) into the conservation of momentum equations. Using the conservation of momentum equation in the y direction: $$0 = 135 \cdot v_1'\sin(35^\circ) - 173 \cdot v_2'\sin(\theta)$$ Substitute \(v_2'\) and solve for \(\theta\): $$0 = 135 \cdot v_1'\sin(35^\circ) - 173 \cdot \sqrt{\frac{0.9(m_1v_{1x}^2+m_2v_{2x}^2)-m_1{v_1'}^2}{m_2}}\sin(\theta)$$ Using a numerical method, we can solve this equation for \(\theta\). After finding the value of \(\theta\), we have the angle at which Toyohibiki is moving immediately after the collision.

Key Concepts

Conservation of Linear Momentum

The principle of the conservation of linear momentum is one of the cornerstones of classical mechanics, asserting that the momentum of an isolated system remains constant if no external forces are acting upon it. This concept is especially important in the analysis of collision problems, such as the collision between Hakurazan and Toyohibiki, our hefty sumo wrestlers.

During an inelastic collision, the objects involved stick together or merge post-collision. Although the collision is inelastic and kinetic energy is not conserved, the conservation of linear momentum holds true. Intuitively, you can think of momentum as the 'oomph' an object has due to its motion; it's the product of mass and velocity. In the case of our sumo wrestlers, despite kinetic energy being lost during the collision (since a percentage converts into other forms of energy like sound or heat), the total momentum remains the same before and after the impact.

It's essential to understand that this conservation applies to both the horizontal (x) and vertical (y) components separately. That's why we separate the momentum into these components to solve for the unknowns in the problem, allowing us to untangle the wrestlers' post-collision velocities and directions.

Kinetic Energy Loss

Kinetic energy is the energy of motion. In our sumo wrestling scenario, both wrestlers are moving, thus they possess kinetic energy. In an inelastic collision, some of this kinetic energy is transformed into other energy forms, resulting in a 'loss' from the perspective of the system's kinetic energy pool. However, energy cannot be created or destroyed; this is a fundamental law of physics known as the conservation of energy.

What exactly happens in a 10% kinetic energy loss, like in our wrestlers' encounter? It means that if we were to calculate the total kinetic energy of both wrestlers before the collision, we'd find that, after the collision, only 90% of that energy remains as kinetic energy. The remaining 10% has been transformed due to forces during the impact, potentially into sound, heat, or even internal energy, like potential energy within deformed materials.

When we apply this knowledge to solving the collision problem, it allows us to create a relationship between the speeds of the wrestlers post-collision, since we have an equation that tells us the final kinetic energy is 90% of the initial kinetic energy. This information is crucial because it helps us narrow down the possible outcomes of the collision and find the specific answer to our problem.

Momentum Conservation Equations

The momentum conservation equations are the mathematical expressions that embody the concept of momentum conservation, and they come into play heavily when dealing with collision problems. To reiterate, they are based on the premise that the total linear momentum of a closed system (one without external forces) is conserved. These equations are vector equations and thus can be broken down into components, which is helpful in collisions that are not one-dimensional.

In the sumo wrestler problem, there are equations for both the x (horizontal) and y (vertical) directions. For instance, the equation for the conservation of momentum along the x-axis is \(m_1 v_{1x} + m_2 v_{2x} = m_1 v_{1fx} + m_2 v_{2fx}\), where \(m_1\) and \(m_2\) are the masses, \(v_{1x}\) and \(v_{2x}\) are the initial velocities in the x-direction, and \(v_{1fx}\) and \(v_{2fx}\) are the final velocities after collision in the x-direction.

To solve our exercise, we used these momentum conservation equations for both directions, treating the x and y components separately. By inserting the known values, we managed to tie the equations to the given 10% kinetic energy loss, and ultimately we could calculate the unknown angle at which Toyohibiki moves post-collision. These equations are powerful tools for predicting the outcome of collisions, given adequate information about the system's initial conditions.