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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 floor, and ignore the weight of the ball.)

Short Answer

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

Step by step solution

01

Understand the Situation

The golf ball strikes the floor at an angle of \(30.0^{\circ}\) and rebounds at the same angle. The speed before and after impact is the same at \(45 \text{ m/s}\). We need to find the change in momentum, considering only the vertical component affected by the floor.
02

Determine Initial and Final Vertical Velocities

Calculate the initial and final vertical velocities using trigonometry.\[v_{y_i} = v \cdot \sin(30.0^{\circ}) = 45 \cdot \sin(30.0^{\circ}) = 22.5 \text{ m/s}\ \]\[v_{y_f} = -v \cdot \sin(30.0^{\circ}) = -45 \cdot \sin(30.0^{\circ}) = -22.5 \text{ m/s}\\] The negative sign on \(v_{y_f}\) indicates the change in direction.
03

Calculate Change in Vertical Velocity

The change in vertical velocity is given by: \[\Delta v_y = v_{y_f} - v_{y_i} = -22.5 - (22.5) = -45 \text{ m/s}\]
04

Calculate Change in Vertical Momentum

Change in momentum \(\Delta p_y\) is the mass times the change in vertical velocity:\[\Delta p_y = m \cdot \Delta v_y = 0.047 \text{ kg} \cdot (-45 \text{ m/s}) = -2.115 \text{ kg} \cdot \text{m/s} \]
05

Find the Magnitude of Impulse

Impulse is the magnitude of change in momentum, so we take the absolute value: \[|\Delta p_y| = 2.115 \text{ kg} \cdot \text{m/s}\]

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Key Concepts

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

Momentum
In physics, momentum is a key concept used to describe the motion of an object. It is the product of an object's mass and velocity. The momentum of an object can be represented by the formula:\[ p = m \cdot v \]where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity of the object.

When studying impulse, which is the change in momentum, it's crucial to understand how momentum varies in different situations. In the case of the golf ball hitting the floor, only the vertical component of momentum changes. This is because the horizontal components of velocity before and after the impact are equal in magnitude and direction.
  • The vertical momentum before the impact, \( p_{y_i} \), is calculated using the vertical velocity.
  • The vertical momentum after the impact, \( p_{y_f} \), considers the change in direction of the vertical velocity.
  • The impulse applied to the golf ball is the change in its momentum in the vertical direction.
The magnitude of this impulse is the absolute value of the change in the ball's vertical momentum.
Vertical Velocity
Vertical velocity is a component of an object's total velocity that is directed along a vertical axis. It plays a crucial role in scenarios involving projectiles or objects rebounding, such as the golf ball situation mentioned above.

To determine the vertical velocity of a projectile, we use trigonometric functions. For an angle \( \theta \) of incidence:\[v_{y} = v \cdot \sin(\theta)\]where \( v \) is the speed of the object.
  • In the golf ball example, the initial vertical velocity \( v_{y_i} = 22.5 \text{ m/s} \) was upward.
  • After rebounding, the vertical velocity \( v_{y_f} = -22.5 \text{ m/s} \) was downward, indicating direction change.
  • This change in direction is critical to calculating the change in momentum.
The change in vertical velocity, specifically, yields the key to find impulse, as impulse involves the force over time that causes the momentum change.
Angle of Incidence
The angle of incidence is the angle formed between an incoming object or wave and a surface it encounters. In many physics problems, this is important for understanding the behavior of objects after they strike a surface.

In this golf ball scenario, the angle of incidence is \(30^{\circ}\). This angle plays a pivotal role in dictating how the ball behaves post-impact.
  • Given the angle of incidence, you can determine the vertical and horizontal components of velocity using trigonometry.
  • The angle's symmetrical rebound nature means it left at the same \(30^{\circ}\) but in the opposite direction.
  • The consistent speed before and after hitting the floor signifies no energy loss along the impact plane, which often simplifies calculations.
Understanding the angle of incidence helps predict how a projectile will react upon hitting a surface, aiding in problems' solutions involving momentum and impulse.

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Most popular questions from this chapter

A model rocket is constructed with a motor that can provide a total impulse of \(29.0 \mathrm{N} \cdot\) s. The mass of the rocket is \(0.175 \mathrm{kg} .\) What is the speed that this rocket achieves when launched from rest? Neglect the effects of gravity and air resistance.

An \(85-\mathrm{kg}\) jogger is heading due east at a speed of \(2.0 \mathrm{m} / \mathrm{s}\). A \(55-\mathrm{kg}\) jogger is heading \(32^{\circ}\) north of east at a speed of \(3.0 \mathrm{m} / \mathrm{s} .\) Find the magnitude and direction of the sum of the momenta of the two joggers.

Two stars in a binary system orbit around their center of mass. The centers of the two stars are \(7.17 \times 10^{11} \mathrm{m}\) apart. The larger of the two stars has a mass of \(3.70 \times 10^{30} \mathrm{kg},\) and its center is \(2.08 \times 10^{11} \mathrm{m}\) from the system's center of mass. What is the mass of the smaller star?

A mine car (mass \(=440 \mathrm{kg}\) ) rolls at a speed of \(0.50 \mathrm{m} / \mathrm{s}\) on a horizontal track, as the drawing shows. A \(150-\mathrm{kg}\) chunk of coal has a speed of \(0.80 \mathrm{m} / \mathrm{s}\) when it leaves the chute. Determine the speed of the car-coal system after the coal has come to rest in the car.

A \(2.3-\mathrm{kg}\) cart is rolling across a frictionless, horizontal track toward a \(1.5-\mathrm{kg}\) cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is \(+4.5 \mathrm{m} / \mathrm{s},\) and the second cart's velocity is \(-1.9 \mathrm{m} / \mathrm{s}\) (a) What is the total momentum of the system of the two carts at this instant? (b) What was the velocity of the first cart when the second cart was still at rest?

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