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Chapter 9: Problem 72

\(\|\) At the county fair, Chris throws a \(0.15 \mathrm{kg}\) baseball at a \(2.0 \mathrm{kg}\) wooden milk bottle, hoping to knock it off its stand and win a prize. The ball bounces straight back at \(20 \%\) of its incoming speed, knocking the bottle straight forward. What is the bottle's speed, as a percentage of the ball's incoming speed?

Short Answer

Expert verified
The final speed of the wooden milk bottle is 3% of the incoming speed of the baseball.

Step by step solution

01

Understand the problem

The ball has an initial mass \(m_1 = 0.15kg\) and an unknown initial speed \(v_{1i}\). After hitting the bottle, it bounces back with \(20\%\) of its initial speed, so the final speed \(v_{1f} = -0.2v_{1i}\) is negative because it鈥檚 going in the opposite direction. The wooden milk bottle has a mass of \(m_2 = 2kg\), its initial speed \(v_{2i} = 0\) (since it's stationary) and an unknown final velocity \(v_{2f}\). The question asks for the value of \(v_{2f}\) as a percentage of the initial speed of the ball \(v_{1i}\).
02

Using the principle of conservation of momentum

The principle of conservation of momentum states that the total momentum of a system of objects is constant if no external forces are acting upon it. In this case, we have no external forces acting on the ball and the bottle. So, we have:\[m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}\]Substituting the known values we have: \(0.15v_{1i} = 0.15(-0.2v_{1i}) + 2v_{2f}\). Solving the equation for \(v_{2f}\) (final speed of the bottle), we get: \(v_{2f} = 0.03v_{1i}\).
03

Find the final speed as a percentage

To find the final speed of the bottle as a percentage of the ball鈥檚 initial speed, just multiply the ratio by 100:\[Percentage = 100 \times v_{2f} / v_{1i} = 100 \times 0.03 = 3\%\]This means that the final speed of the bottle is 3% of the ball's initial speed.

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Key Concepts

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

Momentum
The concept of momentum in physics is fundamental to understanding various phenomena, especially when it comes to motion and collisions. Momentum, symbolized by the letter 'p', is the product of an object's mass (m) and its velocity (v). Hence, the momentum (\( p \) is given by the equation \( p = m \times v \)em>. In the context of our exercise, both the baseball and the wooden milk bottle possess momentum. The key feature of momentum is that in an isolated system (where no external forces act on the system), the total momentum before and after an event, like a collision, remains conserved.

When Chris throws the baseball at the bottle, the system consists of the ball and bottle. Initially, the bottle is at rest, carrying no momentum, while the moving ball has a certain amount of momentum. After collision, the ball reverses its direction, and the bottle starts moving, indicating a transfer of momentum has occurred. Applying this principle of conservation of momentum allows us to predict the motion of objects post collision, as seen in the step-by-step solution provided.
Inelastic Collision
In an inelastic collision, two objects collide and do not maintain their individual shapes or energy; instead, they lose some kinetic energy during the process. This kind of collision contrasts with a perfectly elastic collision, where there's no net loss in kinetic energy within the system. Although the ball in the problem doesn't stick to the bottle, which would make it a perfectly inelastic collision, it is still considered inelastic because the ball loses speed after hitting the bottle.

Despite the energy loss, the principle of conservation of momentum still holds true. This exercise exemplifies inelastic collision by showing that the total momentum before and after the collision remains the same but does not consider the kinetic energy transferred to sound, heat, and deformation. The question challenges students to determine the change in momentum by comparing velocities, providing practice in associating inelastic collisions with momentum conservation.
Physics Problem-Solving
Physics problems often require a structured approach to arrive at the correct solution. The problem featuring Chris and the baseball employs a sequential method to problem-solving. This method involves:
  • Understand the Problem: Identifying known variables and what needs to be solved.
  • Implement Physics Principles: Applying laws, such as the conservation of momentum.
  • Mathematical Solution: Manipulating equations to reach a numerical solution.

In this scenario, understanding the problem entails recognizing that the 'collision' event is key and that the 'speed' of each object will change as a result. The use of conservation of momentum is critical to formulating an equation with the known and unknown variables. Finally, algebraic manipulation leads to finding the unknown speed as a percentage of the ball's initial speed, which cultivates the students鈥� ability to transition from conceptual understanding to practical application.

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Most popular questions from this chapter

One billiard ball is shot east at \(2.00 \mathrm{m} / \mathrm{s}\). A second, identical billiard ball is shot west at \(1.00 \mathrm{m} / \mathrm{s}\). The balls have a glancing collision, not a head-on collision, deflecting the second ball by \(90^{\circ}\) and sending it north at \(1.41 \mathrm{m} / \mathrm{s}\). What are the speed and direction of the first ball after the collision?

A tennis player swings her 1000 g racket with a speed of \(10 \mathrm{m} / \mathrm{s} .\) She hits a \(60 \mathrm{g}\) tennis ball that was approaching her at a speed of \(20 \mathrm{m} / \mathrm{s} .\) The ball rebounds at \(40 \mathrm{m} / \mathrm{s} .\) a. How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision. b. If the tennis ball and racket are in contact for \(10 \mathrm{ms}\), what is the average force that the racket exerts on the ball?

A kid at the junior high cafeteria wants to propel an empty milk carton along a lunch table by hitting it with a \(3.0 \mathrm{g}\) spit ball. If he wants the speed of the 20 g carton just after the spit ball hits it to be \(0.30 \mathrm{m} / \mathrm{s},\) at what speed should his spit ball hit the carton?

I Immediately after the collision, the momentum of the club \(+\) ball system will be A. Less than before the collision. B. The same as before the collision. C. Greater than before the collision.

II The saut茅 demi pli茅 begins with a phase in which the net force on the dancer is negative. During this phase of the jump, A. The normal force of the floor on her is zero. B. The normal force of the floor on her is less than her weight but greater than zero. C. The normal force of the floor on her is equal to her weight. D. The normal force of the floor on her is greater than her weight.
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