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A projectile proton with a speed of \(500 \mathrm{~m} / \mathrm{s}\) collides elastically with a target proton initially at rest. The two protons then move along perpendicular paths, with the projectile path at \(60^{\circ}\) from the original direction. After the collision, what are the speeds of (a) the target proton and (b) the projectile proton?

Short Answer

Expert verified
Target proton speed: \(250\sqrt{3} \ \mathrm{m/s}\); Projectile proton speed: \(250 \ \mathrm{m/s}\).

Step by step solution

01

Understand the Initial Conditions

Before the collision, the projectile proton has a velocity of \(v_1 = 500 \ \mathrm{m/s}\), and the target proton is at rest with \(v_2 = 0\). Initially, the projectile moves along the x-axis, and the target is stationary.
02

Apply Conservation of Momentum

In an elastic collision, momentum is conserved. The momentum in the x and y directions must be separately conserved. This gives two equations:1. \( m_1 v_{1i} = m_1 v_{1f} \cos(60°) + m_2 v_{2f} \cos(30°) \) 2. \( 0 = m_1 v_{1f} \sin(60°) - m_2 v_{2f} \sin(30°) \) Since both particles are protons with the same mass (\(m_1 = m_2\), which cancels out in the equations), they simplify to:\[ v_{1i} = v_{1f} \cos(60°) + v_{2f} \cos(30°) \]\[ 0 = v_{1f} \sin(60°) - v_{2f} \sin(30°) \]
03

Apply Conservation of Kinetic Energy

In elastic collisions, kinetic energy is also conserved. Thus,\[ \frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \]Simplifying, since masses are equal and cancel out:\[ v_{1i}^2 = v_{1f}^2 + v_{2f}^2 \]
04

Solve the Equations

Using the equations from Steps 2 and 3:1. \( 500 = v_{1f} \cdot 0.5 + v_{2f} \cdot \sqrt{3}/2 \)2. \( 0 = v_{1f} \cdot \sqrt{3}/2 - v_{2f}/2 \)3. \( 500^2 = v_{1f}^2 + v_{2f}^2 \)From the second equation, solve for \(v_{2f}\) in terms of \(v_{1f}\):\[ v_{2f} = \sqrt{3}v_{1f} \]Substitute into the first equation:\[ 500 = 0.5v_{1f} + \sqrt{3}/2 \cdot \sqrt{3}v_{1f} \]\[ 500 = 0.5v_{1f} + 1.5v_{1f} \]\[ 500 = 2v_{1f} \]\[ v_{1f} = 250 \ \mathrm{m/s} \]Now solve for \(v_{2f}\):\[ v_{2f} = \sqrt{3} \cdot 250 = 250\sqrt{3} \ \mathrm{m/s} \]
05

Verify with Kinetic Energy Conservation

Substitute \(v_{1f} = 250\) and \(v_{2f} = 250\sqrt{3}\) into the kinetic energy equation:\[ 500^2 = 250^2 + (250\sqrt{3})^2 \]\[ 250000 = 62500 + 62500 \cdot 3 \]\[ 250000 = 250000 \]The equality verifies that our speeds satisfy the conservation of kinetic energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum
Momentum is a fundamental concept in physics, expressing the quantity of motion an object possesses. In any closed system free from external forces, momentum persists through processes like collisions. During an elastic collision, such as a proton-proton collision, the total momentum of the system before the impact equals the total momentum afterward.

The exercise involves an initial condition where a moving projectile proton collides with a stationary target proton. Here, the momentum must be conserved separately in both the x and y directions:
  • In the x-direction: \(v_{1i} = v_{1f} \cos(60°) + v_{2f} \cos(30°)\)
  • In the y-direction: \(0 = v_{1f} \sin(60°) - v_{2f} \sin(30°)\)
Understanding these momentum equations aids in predicting how the individual velocities of the protons change post-collision.

These equations consider mass equality, thereby allowing simplification by canceling the masses, further indicating that the motion's effect is entirely a function of the speed and angle changes.
Conservation of Kinetic Energy
In elastic collisions, not only is momentum conserved, but kinetic energy also remains constant. Kinetic energy describes the energy an object possesses due to its motion. The exercise's collision exhibits this principle as the protons undergo an exchange of energy, maintaining the total kinetic energy of the system.

Mathematically, this is expressed as: \(v_{1i}^2 = v_{1f}^2 + v_{2f}^2\), implying that the initial kinetic energy before collision equals the sum of kinetic energies following impact. Simplified for the case of identical proton masses, this expression highlights the distribution of speeds following their interaction.

By ensuring kinetic energy conservation, predictions of proton velocity (250 m/s for the projectile and \(250\sqrt{3} \ \mathrm{m/s}\) for the target) become reliable. This verification step confirms that calculated speeds satisfy the energy conservation principle and the outcome aligns with physical observations of elastic collisions.
Projectile Motion
Projectile motion is the natural movement of an object in motion that is governed primarily by gravity, usually displaying a parabolic trajectory. However, in this exercise, the proton moves through space as if part of projectile motion analysis.

Before collision, the projectile proton's path is a straightforward linear motion along the x-axis at 500 m/s. Post-impact, its path diverts significantly due to the collision at a 60-degree angle.

Although primarily used in problems involving gravitational force fields, analyzing the proton's movement through projectile concepts helps in understanding how its vector components transform due to the elastic collision and the interplay of direction and speed.
Proton-Proton Collision
A proton-proton collision represents a fundamental interaction where two protons, devoted particles constituting atomic nuclei, engage. Analyzing such collisions—a cornerstone of particle physics—provides insights about forces and energy transformations at subatomic levels.

The exercise addressed frames an elastic proton-proton collision: initiated by one proton at 500 m/s while the other remains at rest. The collision results in deflection along perpendicular paths, a hallmark of such interactions where the intrinsic symmetry and conservation laws make predicting the aftermath apparent.

Such proton interactions often illuminate foundational principles laid out through the intersecting domains of classical mechanics and quantum physics, highlighting deeper subatomic phenomena, otherwise invisible in macroscopic views.

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Most popular questions from this chapter

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