/*! 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 54 A person stands in a stationary ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 54

A person stands in a stationary canoe and throws a \(5.00-\mathrm{kg}\) stone with a velocity of 8.00 \(\mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. The person and canoe have a combined mass of \(105 \mathrm{~kg}\). Ignoring air resistance and effects of the water, find the horizontal recoil velocity (magnitude and direction) of the canoe.

Short Answer

Expert verified
The horizontal recoil velocity of the canoe is approximately -0.377 m/s.

Step by step solution

01

Understand the Problem

A stone is thrown from a canoe, and we need to find the horizontal recoil velocity of the canoe due to the throw. Since no external horizontal forces are acting, the horizontal momentum of the system must be conserved.
02

Identify Given Values

The mass of the stone is given as 5.00 kg, the velocity of the stone is 8.00 m/s at an angle of 30.0 degrees above the horizontal, and the combined mass of the person and canoe is 105 kg.
03

Calculate Horizontal Velocity of the Stone

The horizontal component of the stone's velocity is given by \(v_{x} = v \cos(\theta)\), where \(v = 8.00\, \mathrm{m/s}\) is the velocity of the stone and \(\theta = 30.0^{\circ}\). Therefore, \(v_{x} = 8.00 \cos(30.0^{\circ})\).
04

Compute Horizontal Component of Stone's Momentum

The horizontal momentum of the stone is \(p_x = m_s \cdot v_x\), where \(m_s = 5.00\, \mathrm{kg}\). Substitute the value from the previous step: \(p_x = 5.00 \cdot (8.00 \cdot \cos(30.0^{\circ}))\).
05

Apply Conservation of Momentum

According to the conservation of momentum, the initial horizontal momentum (0, as everything is stationary) equals the final horizontal momentum. Thus, \(0 = -p_x + (m_c \cdot v_r)\), where \(m_c = 105\, \mathrm{kg}\) is the mass of the canoe and \(v_r\) is the recoil velocity of the canoe.
06

Solve for Canoe's Recoil Velocity

Rearrange the previous equation to find \(v_r\). We have \(v_r = \frac{p_x}{m_c}\). Substitute the value of \(p_x\) and \(m_c\) to calculate \(v_r\).
07

Calculate Numerical Result

Substitute the calculated momentum value to find \(v_r = \frac{5.00 \cdot 8.00 \cdot \cos(30.0^{\circ})}{105}\). Calculate this to get the final horizontal recoil velocity.

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 91影视!

Key Concepts

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

Momentum Conservation
Momentum conservation is a fundamental principle in physics. It states that the total momentum of a closed system remains constant when no external forces act upon it. In simpler terms, what goes in must come out, at least in terms of momentum.
  • In our exercise, the system consists of a person, a canoe, and a stone. Initially, everything is at rest, so the total momentum is zero.
  • When the stone is thrown, it gains momentum in the direction of its motion. To keep the total system momentum zero, the canoe must move in the opposite direction.
Think of it like a tiny explosion. You push the stone forward, and the canoe is pushed backward equally and oppositely. This backward motion of the canoe is what we call recoil.
Horizontal Velocity
Horizontal velocity refers to the component of velocity parallel to the ground. In projectile motion, it's essential to distinguish between the horizontal and vertical components.
  • When the stone is thrown at an angle of 30 degrees, its velocity has both horizontal and vertical parts.
  • To find just the horizontal velocity, trigonometry is used. The formula is:\[v_{x} = v \, \cos(\theta)\]where \(v\) is the stone's speed and \(\theta\) is the angle.
Grasping this distinction helps in understanding how the stone moves across the water versus how it rises and falls.
Recoil Velocity
Recoil velocity describes the speed and direction in which the canoe moves after the stone is thrown. It's akin to the kick you feel when firing a gun.
  • According to momentum conservation, the initial momentum (zero when everything is stationary) must be the same as the final momentum.
  • When solving for recoil velocity, one isolates the variable representing the canoe's velocity and rearranges the equation:
\[v_r = \frac{p_x}{m_c}\]where \(p_x\) is the horizontal momentum of the stone and \(m_c\) is the combined mass of the person and canoe.
This calculation reveals how fast and in what direction the canoe moves after the stone is thrown.
Vector Components
Vectors are quantities that have both magnitude and direction, making them suitable for describing velocities in physics. In our problem, the stone's velocity is a vector that can be divided into horizontal and vertical parts.
  • These parts are called components, derived using trigonometric functions based on the angle of the throw.
  • For instance, breaking down the stone's 8 m/s velocity at a 30-degree angle gives us a clearer understanding of how it affects the system's momentum.
By examining each component separately, we simplify complex scenarios into manageable calculations.
This approach helps make the effects of forces and motions in different directions more understandable.

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

Two ice skaters have masses \(m_{1}\) and \(m_{2}\) and are initially stationary. Their skates are identical. They push against one another, as in Figure \(7-11,\) and move in opposite directions with different speeds. While they are pushing against each other, any kinetic frictional forces acting on their skates can be ignored. However, once the skaters separate, kinetic frictional forces eventually bring them to a halt. As they glide to a halt, the magnitudes of their accelerations are equal, and skater 1 glides twice as far as skater 2\. What is the ratio \(m_{1} / m_{2}\) of their masses?

ssm Starting with an initial speed of \(5.00 \mathrm{~m} / \mathrm{s}\) at a height of \(0.300 \mathrm{~m}\), a \(1.50-\mathrm{kg}\) ball swings downward and strikes a \(4.60\) -kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the \(1.50-\mathrm{kg}\) ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?

An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision?

Kevin has a mass of \(87 \mathrm{~kg}\) and is skating with in-line skates. He sees his \(22-\mathrm{kg}\) younger brother up ahead standing on the sidewalk, with his back turned. Coming up from behind, he grabs his brother and rolls off at a speed of \(2.4 \mathrm{~m} / \mathrm{s}\). Ignoring friction, find Kevin's speed just before he grabbed his brother.

A student \((m=63 \mathrm{~kg})\) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of \(0.010 \mathrm{~s}\). The average force exerted on him by the ground is \(+18000 \mathrm{~N}\), where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Thermodynamics

Read Explanation

Magnetism

Read Explanation

Scientific Method Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
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.