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Chapter 6: Problem 46

A billiard ball rolling across a table at \(1.50 \mathrm{~m} / \mathrm{s}\) makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision (a) when the second ball is initially at rest, (b) when the second ball is moving toward the first at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\), and (c) when the second ball is moving away from the first at a speed of \(1.00 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
For the first scenario, both ball speeds will be \(1.50 \mathrm{~m} / \mathrm{s}\). For the second, the first ball moves at \(0.50 \mathrm{~m} / \mathrm{s}\) and the second at \(2.50 \mathrm{~m} / \mathrm{s}\). For the final part, both balls move at \(0.50 \mathrm{~m} / \mathrm{s}\).

Step by step solution

01

Collision with Second Ball at Rest

Start with the equation for conservation of momentum before and after the collision for this scenario, thus: \(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\). Where the \(m_i\) and \(v_i\) are the masses and initial speeds of the balls respectively, and the \(v_f\) are their final speeds. For this scenario, the second ball starts at rest, meaning \(v_{2i}=0\). Solving the equation will give \(v_{2f} = 2v_{1i}\).
02

Second Ball Moving Towards First Ball

The second part of the problem is solved similarly to the first. The initial speed of the second ball, however, is no longer \(0\), but \(1.00 \mathrm{~m} / \mathrm{s}\) in the opposite direction. Thus, the equation for conservation of momentum becomes \(v_{2f} = v_{1i} - v_{2i}\). So, \(v_{2f} = 1.50 \mathrm{~m} / \mathrm{s} - (-1.00 \mathrm{~m} / \mathrm{s}) = 2.50 \mathrm{~m} / \mathrm{s}\), and \(v_{1f} = v_{1i} - v_{2i} = 1.50 \mathrm{~m} / \mathrm{s} - 1.00 \mathrm{~m} / \mathrm{s} = 0.50 \mathrm{~m} / \mathrm{s}\).
03

Second Ball Moving Away from First Ball

The final scenario has the second ball moving away from the first initially. Again, the equation for conservation of momentum will be used. Now \(v_{2i} = 1.00 \mathrm{~m} / \mathrm{s}\), but in the same direction as the first ball. Therefore, \(v_{2f} = v_{1i} - v_{2i} = 1.50 \mathrm{~m} / \mathrm{s} - 1.00 \mathrm{~m} / \mathrm{s} = 0.50 \mathrm{~m} / \mathrm{s}\), and \(v_{1f} = v_{1i} - v_{2i} = 1.50 \mathrm{~m} / \mathrm{s} - 1.00 \mathrm{~m} / \mathrm{s} = 0.50 \mathrm{~m} / \mathrm{s}\).

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Key Concepts

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

Conservation of Momentum
In an elastic collision, the principle of conservation of momentum plays a crucial role. This principle states that the total momentum of two colliding objects is the same before and after the collision.
For billiard balls, momentum can be expressed in the equation:
  • \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \)
Here, \( m_i \) represents the mass and \( v_i \) is the velocity of each ball (initial and final).
Given that both billiard balls are identical, their masses cancel out, simplifying calculations. Using the principle of conservation of momentum allows us to find post-collision velocities even in various scenarios of initial motion.
Billiard Ball Physics
The physics of billiard ball collisions is intriguing, particularly because billiard ball collisions are typically elastic.
An elastic collision means that not only is momentum conserved, but kinetic energy is also conserved. Kinetic energy helps determine the movement of these balls post-collision. Billiard balls are nearly ideal for studying elastic collisions because they are hard, smooth, and hardly deform on impact, minimizing energy loss.
  • An elastic collision can be represented with both momentum and kinetic energy to predict the post-collision outcomes.
  • The nature of billiard ball physics offers insight into real-world applications wherever elastic collisions occur.
Head-on Collision
A head-on collision refers to two objects colliding along a single line of motion. In our scenarios, this happens when billiard balls meet directly face-to-face along their path on the billiard table.
To visualize such collisions:
  • Consider the first billiard ball moving directly towards the second.
  • The movement leads to an exchange of momentum along the line of impact.
In classical physics, analyzing head-on collisions teaches us about how forces and motions transfer between objects.
The analysis involves vector calculations, but for aligned motions like billiard balls, it simplifies to scalar calculation, using direction only as sign (positive or negative).
Initial and Final Speed Analysis
Understanding the initial and final speeds of interacting objects enhances our analysis of collisions.
In the given scenarios:
  • When the second ball is at rest, the initial momentum is carried forward, resulting in the first ball coming to rest and the second assuming the initial speed of the first.
  • For opposing initial directions, momentum exchange results in reversing and modifying speeds of both balls, balancing as per conservation laws.
  • In the case where both balls move in the same direction, speeds adjust less violently but still conserve momentum and energy.
This analysis requires careful attention to vectors and orientation to accurately solve for each speed in different scenarios.

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Most popular questions from this chapter

An object has a kinetic energy of \(275 \mathrm{~J}\) and a momentum of magnitude \(25.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). Find the speed and mass of the object.

ecp An astronaut in her space suit has a total mass of \(87.0 \mathrm{~kg}\), including suit and oxygen tank. Her tether line loses its attachment to her spacecraft while she's on a spacewalk. Initially at rest with respect to her spacecraft, she throws her \(12.0\) -kg oxygen tank away from her spacecraft with a speed of \(8.00 \mathrm{~m} / \mathrm{s}\) to propel herself back toward it (Fig. P6.25). (a) Determine the maximum distance she can be from the craft and still return within \(2.00 \mathrm{~min}\) (the amount of time the air in her helmet remains breathable). (b) Explain in terms of Newton's laws of motion why this strategy works.

A pitcher claims he can throw a \(0.145\) -kg baseball with as much momentum as a \(3.00-\mathrm{g}\) bullet moving with a speed of \(1.50 \times 10^{3} \mathrm{~m} / \mathrm{s}\). (a) What must the baseball's speed be if the pitcher's claim is valid? (b) Which has greater kinetic energy, the ball or the bullet?

An archer shoots an arrow toward a \(300-\mathrm{g}\) target that is sliding in her direction at a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) on a smooth, slippery surface. The \(22.5-\mathrm{g}\) arrow is shot with a speed of \(35.0 \mathrm{~m} / \mathrm{s}\) and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target?

ecp Two ice skaters are holding hands at the center of a frozen pond when an argument ensues. Skater A shoves skater B along a horizontal direction. Identify (a) the horizontal forces acting on \(\mathrm{A}\) and (b) those acting on \(\mathrm{B}\). (c) Which force is greater, the force on \(\mathrm{A}\) or the force on B? (d) Can conservation of momentum be used for the system of \(A\) and \(B\) ? Defend your answer. (e) If \(A\) has a mass of \(0.900\) times that of \(\mathrm{B}\), and \(\mathrm{B}\) begins to move away with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\), find the speed of \(\mathrm{A}\).
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