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 \(3.50 \mathrm{~m} / \mathrm{s}\). The second wrestler, Toyohibiki, has a mass of \(173 \mathrm{~kg}\) and moves straight toward Hakurazan at a speed of \(3.00 \mathrm{~m} / \mathrm{s} .\) Immediately after the collision, Hakurazan is deflected to his right by \(35.0^{\circ}\) (see the figure). In the collision, \(10.0 \%\) of the wrestlers' initial total kinetic energy is lost. What is the angle at which Toyohibiki is moving immediately after the collision?

Short answer

Answer: To find the angle at which Toyohibiki is moving immediately after the collision, first obtain the final velocities (\(v_{1f}\), \(v_{2f}\)) by solving the three equations mentioned in the previous steps. Then, use the formula \(\tan(\theta) = \frac{v_{1f}\sin(35^{\circ})}{v_{2i} + v_{2xf}}\) to calculate the angle \(\theta\).

Step-by-step solution

Conservation of Momentum along the x-axis

First, let's apply the conservation of momentum along the x-axis. The equation for conservation of momentum is: \( m_1v_{1i} + m_2v_{2i} = m_1v_{1xf} + m_2v_{2xf} \) Where \(m_1=135\mathrm{~kg}\) and \(m_2=173\mathrm{~kg}\); \(v_{1i}=3.50\mathrm{~m/s}\) and \(v_{2i}=-3.00\mathrm{~m/s}\) as Toyohibiki is moving toward Hakurazan. Let \(v_{1xf}\) and \(v_{2xf}\) be the final velocities of Hakurazan and Toyohibiki along the x-axis, respectively. Plugging in the values: \( 135(3.50) + 173(-3.00) = 135(v_{1xf}) + 173(v_{2xf}) \) This simplifies to: \( 126(v_{1xf}) - 105(v_{2xf}) = -31.5 \)

Conservation of Momentum along the y-axis

Now, let's apply the conservation of momentum along the y-axis. In this case, the equation for conservation of momentum is: \( m_1v_{1yf} + m_2v_{2yf} = 0 \) Let \(v_{1yf}\) and \(v_{2yf}\) be the final velocities of Hakurazan and Toyohibiki along the y-axis, respectively. We know Hakurazan is deflected by \(35.0^{\circ}\) to his right, so we can determine his final y-momentum using trigonometry. \(v_{1yf}=\frac{m_1v_{1yf}}{m_1}=-v_{1f}\sin(35^{\circ})\) Substituting this into the conservation of momentum equation for the y-axis, we get: \( -v_{1f}\sin(35^{\circ}) + v_{2yf} = 0 \) This simplifies to: \( v_{2yf} = v_{1f}\sin(35^{\circ}) \)

Use Kinetic Energy Relationship

Since \(10.0 \%\) of the wrestlers' initial total kinetic energy is lost, we can write the equation for the energy relationship as: \( (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) \times 0.9 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2 \) Plugging in the values and simplifying: \( 0.9(3.50^2 \times 135 + 3.00^2 \times 173) = 67.5v_{1f}^2 + 86.5v_{2f}^2 \)

Solve for final velocities and the angle

Now we have three equations and three unknowns: \(v_{1xf}\), \(v_{1f}\), and \(v_{2f}\). We can solve for these using any method, such as substitution or matrices. Once we have found the final velocities, we can find the angle \(\theta\) between Toyohibiki's initial and final paths using the following formula: \( \tan(\theta) = \frac{v_{2yf}}{v_{2i} + v_{2xf}} \) Substitute the known values: \( \tan(\theta) = \frac{v_{1f}\sin(35^{\circ})}{v_{2i} + v_{2xf}} \) Now, calculate the angle \(\theta\) to find the angle at which Toyohibiki is moving immediately after the collision.

Key concepts

Conservation of Momentum

When a sumo wrestler meets another in the ring, we witness a textbook example of conservation of momentum. In physics, this principle dictates that the total momentum of a closed system remains constant if it's not affected by external forces. Momentum, which is the product of mass and velocity, becomes a crucial aspect in analyzing such collisions.

Applying this to our sumo wrestlers, we can say that the initial combined momentum of Hakurazan and Toyohibiki before the collision is equal to their combined momentum afterward. Regardless of whether the collision is elastic or inelastic, this conservation law holds true. This is why understanding momentum is key in predicting the outcome of collisions. For example, knowing that no external forces are at play, if a lighter wrestler moves at a higher speed and a heavier wrestler at a slower speed, the post-collision scenario will reflect the balance of momentum they brought to the encounter.

Kinetic Energy Loss

Now, not all collisions are made equal. In an inelastic collision like the one between our hefty athletes, some kinetic energy—energy of motion—is transformed or lost. This energy isn't destroyed but converted into other forms of energy, like sound, heat, or potential energy within deformed materials.

Inelastic collisions are characterized by this loss of kinetic energy. While total kinetic energy isn't preserved, something remarkable happens: a new formation is created post-collision that retains the system's momentum. In the sumo wrestlers' case, 10% of their initial kinetic energy is lost, likely heard as a thunderous stomp and felt as a tremor on the dohyō (sumo ring). Calculating this loss is essential to understand the post-collision speed and direction of the wrestlers.

Momentum Conservation Equations

To navigate through the dynamics of collisions, we use momentum conservation equations. These equations set the stage for solving any collision problem.

For horizontal motion, the conservation of momentum equation simplifies to the sum of the initial momenta equaling the sum of the final momenta. In a vertical sense, especially when no external vertical forces act upon the system (like our sumo encounter), the sum of the initial and final vertical momenta must also be equal.

By using these equations alongside the kinetic energy loss expression, we can solve for unknowns such as final velocities and deflection angles. This mathematical approach transforms the complex dance of mass and movement into a problem that, with careful calculation, reveals the aftermath of a collision as clearly as the numbers on a page.