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Ch 11: Impulse and Momentum
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 11: Impulse and MomentumProblem 22
Chapter 11, Problem 22

A proton is traveling to the right at 2.0 x 107 m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speed and direction of each after the collision?

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Step 1: Recognize that this is a perfectly elastic collision, meaning both momentum and kinetic energy are conserved. Let the mass of the proton be mₚ and the mass of the carbon atom be mₐ = 12mₚ. Let the initial velocity of the proton be vₚ₁ = 2.0 × 10⁷ m/s, and the initial velocity of the carbon atom be vₐ₁ = 0 m/s (since it is initially at rest).
Step 2: Write the equation for conservation of momentum: mₚvₚ₁ + mₐvₐ₁ = mₚvₚ₂ + mₐvₐ₂, where vₚ₂ and vₐ₂ are the final velocities of the proton and carbon atom, respectively. Substitute the known values: mₚ(2.0 × 10⁷) + 12mₚ(0) = mₚvₚ₂ + 12mₚvₐ₂.
Step 3: Simplify the momentum equation by canceling out mₚ (since it is common to all terms): 2.0 × 10⁷ = vₚ₂ + 12vₐ₂. This is the first equation relating vₚ₂ and vₐ₂.
Step 4: Write the equation for conservation of kinetic energy: (1/2)mₚvₚ₁² + (1/2)mₐvₐ₁² = (1/2)mₚvₚ₂² + (1/2)mₐvₐ₂². Substitute the known values: (1/2)mₚ(2.0 × 10⁷)² + (1/2)(12mₚ)(0)² = (1/2)mₚvₚ₂² + (1/2)(12mₚ)vₐ₂².
Step 5: Simplify the kinetic energy equation by canceling out (1/2)mₚ: (2.0 × 10⁷)² = vₚ₂² + 12vₐ₂². This is the second equation relating vₚ₂ and vₐ₂. Solve the system of equations from Step 3 and Step 5 to find the final velocities vₚ₂ and vₐ₂. The direction of each particle can be determined from the signs of their velocities.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Conservation of Momentum

In a collision, the total momentum of a system remains constant if no external forces act on it. This principle allows us to relate the velocities of the colliding objects before and after the collision. For a perfectly elastic collision, the momentum before the collision equals the momentum after the collision, which can be expressed mathematically as m1*v1 + m2*v2 = m1*v1' + m2*v2', where m is mass and v is velocity.
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Elastic Collision

An elastic collision is one in which both momentum and kinetic energy are conserved. In such collisions, the objects bounce off each other without any loss of kinetic energy. This is crucial for solving the problem, as it allows us to use both conservation laws to find the final velocities of the proton and the carbon atom after their interaction.
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Mass Ratio and Velocity Relationship

The mass ratio between colliding objects significantly affects their post-collision velocities. In this scenario, the carbon atom's mass is 12 times that of the proton, which means it will have a much smaller change in velocity compared to the proton. Understanding how mass influences the final speeds and directions of the objects is essential for accurately calculating their outcomes after the collision.
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