/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 A golf ball strikes a hard, smoo... [FREE SOLUTION] | 91Ó°ÊÓ

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A golf ball strikes a hard, smooth floor at an angle of \(30.0^{\circ}\) and, as the drawing shows, rebounds at the same angle. The mass of the ball is \(0.047 \mathrm{kg},\) and its speed is \(45 \mathrm{m} / \mathrm{s}\) just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the fl oor, and ignore the weight of the ball.)

Short Answer

Expert verified
The impulse is 2.115 kg·m/s.

Step by step solution

01

Understanding the Problem

The golf ball hits the floor at an angle of 30.0° with the same speed before and after the impact. We need to determine the impulse imparted to the ball by the floor. This requires examining the change in vertical momentum, as the horizontal component remains constant.
02

Calculate the Vertical Component of Velocity

The ball's velocity can be decomposed into horizontal and vertical components using trigonometry. The vertical component of velocity is given by \( v_{y} = v \times \sin(\theta) \), where \( v = 45 \) m/s and \( \theta = 30.0^{\circ} \). Thus, \( v_{y} = 45 \times \sin(30.0^{\circ}) = 22.5 \) m/s.
03

Determine Change in Vertical Momentum

Momentum is given by \( p = m \times v \). Initially, the vertical momentum is \( p_{i} = m \times v_{y} = 0.047 \times 22.5 \) kg·m/s. After the bounce, the vertical momentum is \( p_{f} = 0.047 \times (-22.5) \), where the negative sign indicates a change in direction. The change in momentum \( \Delta p = p_{f} - p_{i} \).
04

Calculate the Impulse

Impulse \( J \) is the change in momentum, \( J = \Delta p = 0.047 \times (-22.5) - 0.047 \times 22.5 \). Simplifying gives \( J = 2 \times 0.047 \times 22.5 = 2.115 \) kg·m/s, since the change is essentially a reversal in direction.

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics. It's the product of an object's mass and its velocity. The formula for momentum is \( p = m \times v \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.
For a golf ball impacting a floor, as in our exercise, momentum is crucial. Initially, the ball's velocity is split into horizontal and vertical components. When the ball hits the floor, its vertical momentum changes direction, while the horizontal momentum remains constant. This principle helps simplify the problem because we only need to consider changes in vertical momentum.
Understanding how momentum changes occur, particularly during collisions, is important. The concept of impulse is tied closely to momentum. Impulse measures the change in momentum; it's calculated as \( J = \Delta p \), meaning the impulse \( J \) equals the change in momentum \( \Delta p \). This allows us to determine forces acting over time, providing insight into motion dynamics.
Velocity Components
When analyzing motion, breaking down velocity into components is essential. Velocity can be split into horizontal and vertical components based on the direction of movement.
For our golf ball example, we use trigonometry to find these components. The vertical velocity component \( v_{y} \) is calculated using \( v_{y} = v \times \sin(\theta) \), where \( v \) is the total velocity and \( \theta \) is the angle of incidence. For an angle of \( 30.0^{\circ} \), the ball's vertical velocity becomes \( 22.5 \mathrm{m/s} \).
Horizontal velocity, \( v_{x} = v \times \cos(\theta) \), remains unchanged during our scenario because the problem specifies that only the vertical momentum changes. By understanding these components, we can more easily analyze different motion vectors separately.
Trigonometry in Physics
Trigonometry is a vital tool in physics for resolving vector components and analyzing forces and motions. Using trigonometric functions like sine and cosine helps break down complex movements into simpler parts.
In the context of this exercise, the use of sine (\( \sin \)) is pivotal. It allows us to calculate the vertical component of the ball's velocity. The angle \( 30.0^{\circ} \) is plugged into the formula \( v_{y} = v \times \sin(30.0^{\circ}) \) to yield a value of \( 22.5 \mathrm{m/s} \) for the vertical velocity component.
Similarly, the cosine function would apply for horizontal components with the formula \( v_{x} = v \times \cos(\theta) \). Trigonometry simplifies the handling of angle-based problems, making complex calculations manageable and providing valuable insights into physical behaviors in diverse situations.

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Most popular questions from this chapter

A car (mass \(=1100 \mathrm{kg}\) ) is traveling at \(32 \mathrm{m} / \mathrm{s}\) when it collides head on with a sport utility vehicle (mass \(=2500 \mathrm{kg}\) ) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?

A \(40.0-\mathrm{kg}\) boy, riding a \(2.50-\mathrm{kg}\) skateboard at a velocity of \(+5.30 \mathrm{m} / \mathrm{s}\) across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is \(6.00 \mathrm{m} / \mathrm{s}, 9.50^{\circ}\) above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's veloc- ity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

A \(50.0-\mathrm{kg}\) skater is traveling due east at a speed of \(3.00 \mathrm{m} / \mathrm{s} .\) A \(70.0-\mathrm{kg}\) skater is moving due south at a speed of \(7.00 \mathrm{m} / \mathrm{s}\). They collide and hold on to each other after the collision, managing to move off at an angle \(\theta\) south of east, with a speed of \(v_{\mathrm{r}}\). Find (a) the angle \(\theta\) and (b) the speed \(v_{f},\) assuming that friction can be ignored.

Multiple-Concept Example 7 presents a model for solving problems such as this one. A \(1055-\mathrm{kg}\) van, stopped at a traffic light, is hit directly in the rear by a 715 -kg car traveling with a velocity of \(+2.25 \mathrm{m} / \mathrm{s}\). Assume that the transmission of the van is in neutral, the brakes are not being applied, and the collision is elastic. What are the final velocities of (a) the car and (b) the van?

A ball is dropped from rest from the top of a \(6.10-\mathrm{m}\) -tall building, falls straight downward, collides inelastically with the ground, and bounces back. The ball loses \(10.0 \%\) of its kinetic energy every time it collides with the ground. How many bounces can the ball make and still reach a windowsill that is \(2.44 \mathrm{m}\) above the ground?

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