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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

High-speed stroboscopic photographs show that the head of a \(200-\mathrm{g}\) golf club is traveling at \(55 \mathrm{~m} / \mathrm{s}\) just before it strikes a 46 -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at \(40 \mathrm{~m} / \mathrm{s}\). Find the speed of the golf ball just after impact.

Gayle runs at a speed of \(4.00 \mathrm{~m} / \mathrm{s}\) and dives on a sled, initially at rest on the top of a frictionless, snow-covered hill. After she has descended a vertical distance of \(5.00 \mathrm{~m}\), her brother, who is initially at rest, hops on her back, and they continue down the hill together. What is their speed at the bottom of the hill if the total vertical drop is \(15.0 \mathrm{~m}\) ? Gayle's mass is \(50.0 \mathrm{~kg}\), the sled has a mass of \(5.00 \mathrm{~kg}\), and her brother has a mass of \(30.0 \mathrm{~kg}\).

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?

A \(65.0\) -kg person throws a \(0.0450\) -kg snowball forward with a ground speed of \(30.0 \mathrm{~m} / \mathrm{s}\). A second person, with a mass of \(60.0 \mathrm{~kg}\), catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of \(2.50 \mathrm{~m} / \mathrm{s}\), and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard friction between the skates and the ice.

A \(1200-\mathrm{kg}\) car traveling initially with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) in an easterly direction crashes into the rear end of a 9000 -kg truck moving in the same direction at \(20.0 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{P} 6.42\) ). The velocity of the car right after the collision is \(18.0 \mathrm{~m} / \mathrm{s}\) to the east. (a) What is the velocity of the truck right after the collision? (b) How much mechanical energy is lost in the collision? Account for this loss in energy.
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