/*! 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 69 \(\bullet\) Jonathan and Jane ar... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
0.098 m/s to the right.

Step by step solution

01

Understand the Problem

We need to calculate the sleigh's horizontal velocity after Jonathan and Jane jump off. Since the problem occurs on a frictionless surface, we apply the conservation of momentum in the horizontal direction.
02

Calculate Total Mass

Convert the weights to masses by dividing by gravity \( m_{Jonathan} = \frac{800}{9.8} \, \text{kg} = 81.63 \, \text{kg} \), \( m_{Jane} = \frac{600}{9.8} \, \text{kg} = 61.22 \, \text{kg} \), and \( m_{sleigh} = \frac{1000}{9.8} \, \text{kg} = 102.04 \, \text{kg} \).
03

Calculate Initial Momentum

Initially, the total system (Jonathan, Jane, and the sleigh) was at rest, so the total momentum was zero.
04

Calculate Jonathan's Momentum Components

Jonathan's horizontal momentum is \(m_{Jonathan} \times 5.00 \times \cos(30.0^{\circ}) = 81.63 \times 5.00 \times 0.866 = 353.18 \, \text{kg}\textrm{m/s}\). This is directed to the left, thus is negative: \(-353.18 \text{ kg}\cdot \text{m/s}\).
05

Calculate Jane's Momentum Components

Jane's horizontal momentum is \(m_{Jane} \times 7.00 \times \cos(36.9^{\circ}) = 61.22 \times 7.00 \times 0.8 = 343.19 \, \text{kg}\text{m/s}\). This is directed to the right, so it is positive.
06

Apply Conservation of Momentum

By the conservation of momentum, the initial momentum of Jonathan and Jane, plus the momentum of the sleigh, equals zero:\(-353.18 + 343.19 + m_{sleigh} \cdot v_{sleigh} = 0\). Solve for \(v_{sleigh}\): \(v_{sleigh} = \frac{353.18 - 343.19}{102.04} = \frac{9.99}{102.04} = 0.098 \, \text{m/s}\) to the right.

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.

Physics Problem Solving
Physics often involves understanding complex scenarios by breaking them down into simpler parts. In this exercise, Jonathan and Jane jumping from the sleigh involves momentum conservation concepts on a frictionless surface. By identifying the essential components like velocity, angle, and direction of both people, we can calculate the sleigh's resulting motion.
Start by assessing what you know about the system before and after the event. Initially, the sleigh and its occupants are at rest. After they jump off, the system must conserve momentum. This means any change in their motion must result in the sleigh moving in a way that balances the total momentum to zero.
For effective physics problem-solving:
  • Break down the problem into clear steps.
  • Identify known and unknown variables.
  • Apply relevant physics principles, like the conservation of momentum.
Once you have a clear path, solve using mathematical expressions.
Frictionless Surface
A frictionless surface, like the ice on which the sleigh rests, simplifies calculations by removing forces that could complicate momentum conservation. With no friction, the movement of Jonathan, Jane, and the sleigh are only affected by their interactions and gravity.
This lack of friction means that there is no force opposing the movement caused by Jonathan and Jane jumping, allowing us to directly calculate the sleigh's movement using the momentum principles. Consider friction in general terms as the force hindering motion between surfaces. Here, it's absent, causing no interference and making momentum calculations straightforward.
Horizontal Velocity
Horizontal velocity refers to the component of velocity that runs parallel to the ground, and it is crucial when analyzing motion on surfaces. In this problem, both Jonathan's and Jane's jumps have components of velocity affecting this direction.
Understanding how to find these components involves using trigonometry. We use the given angles to calculate each person's horizontal velocity using cosine functions:
- Jonathan's horizontal velocity is calculated as: \[ 5.00 \times \cos(30.0^{\circ}) \]
- Jane's horizontal velocity is determined similarly: \[ 7.00 \times \cos(36.9^{\circ}) \]
With these, we determine each jump's contribution to the sleigh's horizontal velocity. The sleigh's velocity is then found by ensuring the total horizontal momentum remains zero, accounting for direction.
Weight to Mass Conversion
Weight and mass are distinct, though often confused, concepts in physics. Weight is the gravitational force on an object, typically measured in newtons (N), while mass is the amount of matter in an object, measured in kilograms (kg). Understanding the difference is crucial for correctly applying physics formulas.
To convert weight to mass, use the gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \). The formula is:
\[ m = \frac{W}{g} \]
where \( m \) is mass and \( W \) is weight.
For this exercise, we use it to find:
  • Jonathan's mass: \( \frac{800}{9.8} \approx 81.63 \, \text{kg} \)
  • Jane's mass: \( \frac{600}{9.8} \approx 61.22 \, \text{kg} \)
  • Sleigh's mass: \( \frac{1000}{9.8} \approx 102.04 \, \text{kg} \)
These values are foundational for calculating momentum.

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

\(\bullet\) To warm up for a match, a tennis player hits the 57.0 g ball vertically with her racket. If the ball is stationary just before it is hit and goes 5.50 \(\mathrm{m}\) high, what impulse did she impart to it?

\(\bullet\) Detecting planets around other stars. Roughly 500 planets have so far been detected beyond our solar system. This is accomplished by looking for the effect the planet has on the star. The star is not truly stationary; instead, it and its planets orbit around the center of mass of the system. Astronomers can measure this wobble in the position of a star.(a) For a star with the mass and size of our sun and having a planet with five times the mass of Jupiter, where would the center of mass of this system be located, relative to the center of the star, if the distance from the star to the planet was the same as the distance from Jupiter to our sun? (Consult Appendix E.) (b) If the planet had earth's mass, where would the center of mass of the system be located if the planet was just as far from the star as the earth is from the sun? (c) In view of your results in parts (a) and (b), why is it much easier to detect stars having large planets rather than small ones?

Your little sister (mass 25.0 \(\mathrm{kg}\) ) is sitting in her little red wagon (mass 8.50 \(\mathrm{kg} )\) at rest. You begin pulling her forward and continue accelerating her with a constant force for 2.35 \(\mathrm{s}\) at the end of which time she's moving at a speed of 1.80 \(\mathrm{m} / \mathrm{s}\) . (a) Calculate the impulse you imparted to the wagon and its passenger. (b) With what force did you pull on the wagon?

. Biomechanics. The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 ms. (This number can also vary, depending on the racket and swing.) We shall assume a 30.0 \(\mathrm{ms}\) contact time throughout this problem. The fastest-known served tennis ball was served by "Big Bill" Tilden in \(1931,\) and its speed was measured to be 73.14 \(\mathrm{m} / \mathrm{s}\) .(a) What impulse and what force did Big Bill exert on the ten- nis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of \(55 \mathrm{m} / \mathrm{s},\) what force and what impulse did he exert on the ball, assuming only horizontal motion?

\(\bullet\) On a very muddy football field, a \(110-\) kg linebacker tack- les an 85 -kg halfback. Immediately before the collision, the linebacker is slipping with a velocity of 8.8 \(\mathrm{m} / \mathrm{s}\) north and the halfback is sliding with a velocity of 7.2 \(\mathrm{m} / \mathrm{s}\) east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

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.