/*! 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 10 A bat strikes a 0.145-kg basebal... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 40.0 m/s; when it leaves the bat, the ball is traveling to the left at an angle of 30\(^\circ\) above horizontal with a speed of 52.0 m/s. If the ball and bat are in contact for 1.75 ms, find the horizontal and vertical components of the average force on the ball.

Short Answer

Expert verified
The average horizontal force is \(-7043.0 \text{ N}\) and vertical force is \(2154.3 \text{ N}\).

Step by step solution

01

Determine Initial and Final Velocity Components

First, let's break down the initial and final velocities of the baseball into their horizontal and vertical components. Initially, the baseball is moving to the right with a velocity of 40.0 m/s, meaning its horizontal velocity is \( v_{ix} = 40.0 \) m/s and its vertical velocity \( v_{iy} = 0 \) m/s as it has no vertical motion. After striking, the baseball is traveling to the left at a 30° angle above the horizontal with a speed of 52.0 m/s. The horizontal component of this velocity is \( v_{fx} = -52.0 \cdot \cos(30^\circ) \approx -45.0 \) m/s (used negative because it moves to the left), and the vertical component is \( v_{fy} = 52.0 \cdot \sin(30^\circ) = 26.0 \) m/s.
02

Apply Impulse-Momentum Theorem

Impulse-momentum theorem states that the change in momentum equals the impulse applied. The change in horizontal momentum is \( m(v_{fx} - v_{ix}) \) and in vertical momentum is \( m(v_{fy} - v_{iy}) \). Calculate these changes: for horizontal momentum, \[\Delta p_{x} = 0.145 \times (-45.0 - 40.0) = -12.325 \text{ kg.m/s} \], and for vertical momentum, \[\Delta p_{y} = 0.145 \times (26.0 - 0) = 3.77 \text{ kg.m/s} \].
03

Calculate Average Force Components

The impulse experienced by the ball equals the average force times the time of contact \( F_{avg} \Delta t = \Delta p \). Using this formula, calculate the average horizontal force component: \( F_{avg,x} = \frac{\Delta p_{x}}{\Delta t} = \frac{-12.325}{1.75 \times 10^{-3}} = -7043.0 \text{ N} \), and the average vertical force component: \( F_{avg,y} = \frac{\Delta p_{y}}{\Delta t} = \frac{3.77}{1.75 \times 10^{-3}} = 2154.3 \text{ N} \). Remember that the negative sign in the horizontal force indicates that the force is in the opposite direction of the initial motion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Average Force
When discussing the average force in the context of the Impulse-Momentum Theorem, the aim is to understand how force is applied over the time during which an object is in contact with another. In this example, the interaction between a baseball and a bat is key to understanding the average force.
The average force is not constant, but it gives a simplified picture of the complex forces during the contact time of 1.75 milliseconds (very brief!). To find this, we use the impulse, which is the product of the average force and the contact time, and relate it to momentum change.
  • The impulse experienced by the object is calculated using: \[ F_{\text{avg}} \Delta t = \Delta p \] where \( F_{\text{avg}} \) is the average force and \( \Delta p \) is the change in momentum.
  • In the exercise, the average horizontal and vertical force components were calculated by rearranging the above equation to solve for \( F_{\text{avg}} \).
Understanding the average force helps model real-world dynamic interactions by simplifying the calculations while still capturing the essence of the event.
Velocity Components
Breaking down velocity into components is crucial for solving problems where angles are involved, such as when the baseball changes direction after being struck by the bat. Velocity components help visualize and calculate individual directions of an object's motion.
In physics, velocity is often split into horizontal and vertical components. These components form the legs of a right triangle, with the resultant velocity being the hypotenuse.
  • The horizontal component of velocity (\( v_x \)) is found using \( v_x = v \cdot \cos(\theta) \), where \( \theta \) is the angle of motion relative to the horizontal.
  • The vertical component (\( v_y \)) is found with \( v_y = v \cdot \sin(\theta) \).
For our baseball: after impact, the initial horizontal velocity was positive (to the right), while the final horizontal velocity became negative (to the left), indicating a direction change. The vertical velocity, initially zero, increased, reflecting the upward motion after the hit. Decomposing velocity into these components helps in accurately computing the momentum changes, which are essential to determining the forces involved.
Momentum Change
Momentum change is a central concept in the Impulse-Momentum Theorem, where momentum of an object is altered due to applied force within a period. Momentum is defined as the product of an object's mass and its velocity, \( p = mv \).
In the given exercise, the change in momentum is used to calculate the average force experienced by the baseball.
  • The change in momentum in a specific direction is given by: \( \Delta p = m(v_{f} - v_{i}) \), where \( m \) is the mass, \( v_{f} \) is the final velocity, and \( v_{i} \) is the initial velocity in that direction.
  • For horizontal movement, this involves a switch from a rightward 40.0 m/s to a leftward 52.0 m/s, resulting in significant momentum change.
  • Vertically, the baseball starts at 0 m/s and moves upwards at a speed influenced by the angle and speed change.
Understanding momentum change is crucial for evaluating how forces exerted over time can alter an object’s motion. This provides insights into not just the magnitude of forces but also their directional impact on the motion involved.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Jonathan and Jane are sitting in a sleigh that is at rest on frictionless ice. Jonathan's weight is 800 N, Jane's weight is 600 N, and that of the sleigh is 1000 N. They see a poisonous spider on the floor of the sleigh and immediately jump off. Jonathan jumps to the left with a velocity of 5.00 m/s at 30.0\(^\circ\) above the horizontal (relative to the ice), and Jane jumps to the right at 7.00 m/s at 36.9\(^\circ\) above the horizontal (relative to the ice). Calculate the sleigh's horizontal velocity (magnitude and direction) after they jump out.

A small rocket burns 0.0500 kg of fuel per second, ejecting it as a gas with a velocity relative to the rocket of magnitude 1600 m/s. (a) What is the thrust of the rocket? (b) Would the rocket operate in outer space where there is no atmosphere? If so, how would you steer it? Could you brake it?

You are standing on a concrete slab that in turn is resting on a frozen lake. Assume there is no friction between the slab and the ice. The slab has a weight five times your weight. If you begin walking forward at 2.00 m/s relative to the ice, with what speed, relative to the ice, does the slab move?

A 70-kg astronaut floating in space in a 110-kg MMU (manned maneuvering unit) experiences an acceleration of 0.029 m/s\(^2\) when he fires one of the MMU's thrusters. (a) If the speed of the escaping N\(_2\) gas relative to the astronaut is 490 m/s, how much gas is used by the thruster in 5.0 s? (b) What is the thrust of the thruster?

Jack (mass 55.0 kg) is sliding due east with speed 8.00 m/s on the surface of a frozen pond. He collides with Jill (mass 48.0 kg), who is initially at rest. After the collision, Jack is traveling at 5.00 m/s in a direction 34.0\(^\circ\) north of east. What is Jill's velocity (magnitude and direction) after the collision? Ignore friction.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.