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Chapter 6: Problem 46

A billiand 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 mowing 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
The speeds of the balls after the collision are: (a) first ball: 0 m/s, second ball: 1.50 m/s, (b) first ball: 0.50 m/s, second ball: -0.50 m/s, (c) first ball: 1.00 m/s, second ball: 1.50 m/s.

Step by step solution

01

Understanding the Principles of Momentum and Energy Conservation

Firstly, it's important to know two key principles. The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces act on it. In other words, the sum of the momenta of the two balls before the collision is equal to the sum of their momenta after the collision. For an elastic collision, not only is momentum conserved but kinetic energy is also conserved. This means, the sum of the kinetic energy of the two balls before the collision is equal to the sum of their kinetic energy after the collision.
02

Case (a): When the Second Ball is Initially at Rest

Given that the first ball is moving at 1.50 m/s and collides with the second ball which is at rest (velocity = 0), we apply the principle of conservation of momentum and energy conservation. We set up the equations based on these principles and solve for the unknowns, the final velocities of each ball. The calculations show that since the balls are identical and the second ball is initially at rest, it takes over the entire velocity of the first ball. Therefore, after the collision, the first ball stops (velocity = 0) and the second ball moves with 1.50 m/s.
03

Case (b): Second Ball Moving Towards the First

Next, we consider the scenario where the second ball is moving towards the first at a speed of 1.00 m/s. The calculations for this scenario, using the conservation laws, yield that after the collision, both balls are moving in the same direction as the initially moving ball, but the first one moves with a speed of 0.50 m/s and the second one moves with -0.50 m/s (negative indicating direction opposite to initial direction). Thus, the second billiard ball is moving at a slower speed than before the collision.
04

Case (c): Second Ball Moving Away from the First

In this situation, when the second ball is moving away from the first at a speed of 1.00 m/s. The calculations, again based on the conservation laws of momentum and kinetic energy, yield that after the collision, both balls keep moving in their initial directions. The first ball keeps moving with a velocity of 1.00 m/s and the second ball with 1.50 m/s. So, in this case, the collision makes the first ball slow down while the second ball speeds up.

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Key Concepts

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

Conservation of Momentum
In the world of physics, the conservation of momentum is a fundamental principle. Imagine two billiard balls engaging in a collision. The total momentum before and after the collision remains constant, provided no external forces interfere. Think of momentum as the product of an object's mass and velocity.
  • In an isolated system like two colliding billiard balls, conservation of momentum tells us that the sum of their momenta before the collision must equal the sum of their momenta post-collision.
  • This principle helps predict the final velocities of the balls after they collide.
Understanding how momentum is preserved enables us to solve many complex collision problems with clarity and precision.
Kinetic Energy Conservation
Kinetic energy conservation is another pillar of collision physics, especially in elastic collisions. An elastic collision is one where both momentum and kinetic energy are conserved. The kinetic energy of an object is given by the equation \( KE = \frac{1}{2}mv^2 \), where \( m \) is mass and \( v \) is velocity.
  • During an elastic collision, not only does the sum of the momenta remain constant, but the total kinetic energy also stays the same before and after impact.
  • This is unlike inelastic collisions where kinetic energy is transformed into other forms of energy, such as heat or deformation.
In our billiard ball example, these conservation laws allow us to calculate the final velocities post-collision.
Billiard Ball Physics
Billiard ball physics is a great hands-on example of both momentum and kinetic energy conservation. When a billiard ball strikes another, the collision is almost perfectly elastic due to their hard surface and minimal energy loss.
  • Consider a billiard ball moving at 1.50 m/s hitting a stationary identical ball; after the impact, the second ball moves away with the speed the first ball had, evidencing conservation of momentum and kinetic energy.
  • If the second ball approaches the first at 1.00 m/s, they end up sharing the velocities, moving in opposite directions due to the elastic collision dynamics.
  • When the second ball initially moves away, the first ball's speed reduces slightly after impact, while the second ball speeds up.
Observing these collisions gives insights into how complex motion follows simple physics principles.
Physics Problem Solving
Solving physics problems often involves breaking down scenarios into fundamental principles and clearly understanding the sequence of events. Applying conservation laws helps simplify complex interactions like those seen in billiard ball collisions.
  • Identify known quantities and set up equations for momentum and energy conservation.
  • Clearly define initial and final states: what speeds objects have pre- and post-collision.
  • Construction of these equations allows for the calculation of unknown values.
Using structured approaches helps in systematically tackling and solving even the most perplexing physics problems.

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Most popular questions from this chapter

8 A car of mass \(m\) moving at a speed \(v_{1}\) collides and couples with the back of a truck of mass 2 m moving initially in the same direction as the car at a lower speed \(v_{2}\) - (a) What is the speed \(y_{f}\) of the two vehicles immediately after the collision? (b) What is the change in kinetic energy of the car-truck system in the collision?

\(\mathrm{Q} \mathrm{C}\) A pitcher throws a 0.14-\textrm{kg baseball toward the } batter so that it crosses home plate horizontally and has a speed of \(42 \mathrm{~m} / \mathrm{s}\) just before it makes contact with the bat. The batter then hits the ball straight back at the pitcher with a speed of \(48 \mathrm{~m} / \mathrm{s}\). Assume the ball travels along the same line leaving the bat as it followed before contacting the bat. (a) What is the magnitude of the impulse delivered by the bat to the baseball? (b) If the ball is in contact with the bat for \(0.00508\), what is the magnitude of the average force exerted by the bat on the ball? (c) How does your answer to part (b) compare to the weight of the ball?

A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass \(m_{1}=48.0 \mathrm{~kg}\) travels in the positive \(x\)-direction at \(12.0 \mathrm{~m} / \mathrm{s}\), and a second p盲ece of mass \(m_{2}=62.0 \mathrm{~kg}\) travels in the \(x y\)-plane at an angle of \(105^{\circ}\) at \(15.0 \mathrm{~m} / \mathrm{s}\). The third piece has mass \(\mathrm{m}_{3}=112 \mathrm{~kg}\). (a) Sketch a diagram of the situation, labeling the different masses and their velocities. (b) Write the general expression for conservation of momentum in the \(x\)-and \(y\)-directions in terms of \(m_{1}, m_{2}, m_{3}, v_{1}, \tau_{2}\), and \(\tau_{3}\) and the sines and cosines of the angles, taking \(\theta\) to be the unknown angle. (c) Calculate the final \(x\)-componens of the momenta of \(m_{1}\) and \(m_{2}\). (d) Calculate the final \(y\)-components of the momenta of \(m_{1}\) and \(m_{2}\). (e) Substitute the known momentum components into the general equations of momentum for the \(x\)-and \(y\)-directions, along with the known mass \(m_{3}\). (1) Solve the two momentum equations for \(x_{3} \cos \theta\) and \(v_{3} \sin \theta\), respectively, and use the identity \(\cos ^{2} \theta+\sin ^{2} \theta=1\) to obtain \(x_{3}\) (g) Divide the equation for \(v_{3} \sin \theta\) by that for \(v_{3} \cos \theta\) to obtain tan \(\theta\), then obtain the angle by taking the inverse tangent of both sides. (h) In general, would three such pieces necessarily have to move in the same plane? Why?

Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that exerted on the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false; Newton's third law tells us that both objects are acted upon by forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at \(8.00 \mathrm{~m} / \mathrm{s}\) and that they undergo a perfectly inelastic head-on collision. Each driver has mass \(80.0 \mathrm{~kg}\). Inchuding the masses of the drivers, the total masses of the vehicles are \(800 \mathrm{~kg}\) for the car and \(4000 \mathrm{~kg}\) for the truck. If the collision time is \(0.120 \mathrm{~s}\), what force does the seat belt exert on each driver?

A \(45.0-\mathrm{kg}\) girl is standing on a \(150-\mathrm{kg}\) plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of \(1.50 \mathrm{~m} / \mathrm{s}\) to the right relative to the plank. (a) What is her velocity relative to the surface of the ice? (b) What is the velocity of the plank relative to the surface of the ice?
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