/*! 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 104 Jonathan and Jane are sitting in... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 of 5.00 \(\mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal (relative to the ice), 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
The sleigh's velocity is approximately 0.104 m/s to the right.

Step by step solution

01

Calculate Masses of Jonathan, Jane, and Sleigh

First, convert the weights of Jonathan, Jane, and the sleigh into masses by using the formula: \( \text{mass} = \frac{\text{weight}}{g} \), where \( g = 9.8 \, \text{m/s}^2 \). For Jonathan:\[ m_J = \frac{800 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 81.63 \, \text{kg} \] For Jane:\[ m_j = \frac{600 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 61.22 \, \text{kg} \]For the sleigh:\[ m_s = \frac{1000 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 102.04 \, \text{kg} \]
02

Resolve Velocities into Horizontal Components

Resolve the velocities of Jonathan and Jane into their horizontal components using trigonometric functions:For Jonathan:\[ v_{Jx} = 5.00 \, \text{m/s} \cdot \cos(30.0^{\circ}) = 5.00 \, \text{m/s} \cdot \frac{\sqrt{3}}{2} = 4.33 \, \text{m/s} \] (Note: This direction is to the left, so it is negative relative to the sleigh)For Jane:\[ v_{jx} = 7.00 \, \text{m/s} \cdot \cos(36.9^{\circ}) = 7.00 \, \text{m/s} \cdot 0.8 = 5.60 \, \text{m/s} \] (Note: This direction is to the right, so it is positive relative to the sleigh.)
03

Apply Conservation of Momentum

Use the conservation of momentum principle, where the total momentum before the jump equals the total momentum after the jump. Initially, the system is at rest, so the initial total momentum is zero:Let \( v_s \) be the sleigh's velocity:\[ m_J \cdot v_{Jx} + m_j \cdot v_{jx} + m_s \cdot v_s = 0 \]Substitute the known values:\[ (81.63 \, \text{kg})(-4.33 \, \text{m/s}) + (61.22 \, \text{kg})(5.60 \, \text{m/s}) + (102.04 \, \text{kg})(v_s) = 0 \]
04

Solve for Sleigh's Velocity

Simplify and solve the equation obtained from the momentum conservation:\[ -353.46 \, \text{kg m/s} + 342.83 \, \text{kg m/s} + 102.04 \, \text{kg} \cdot v_s = 0 \]Combine the constant terms:\[ -10.63 \, \text{kg m/s} + 102.04 \, \text{kg} \cdot v_s = 0 \]Solve for \( v_s \):\[ v_s = \frac{10.63 \, \text{kg m/s}}{102.04 \, \text{kg}} \approx 0.104 \, \text{m/s} \]The sleigh moves to the right due to the positive 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 91Ó°ÊÓ!

Key Concepts

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

Physics Problem Solving
In physics, tackling problems relies on understanding fundamental laws and principles. One such principle is the conservation of momentum, playing a key role in dynamics scenarios like Jonathan and Jane's sleigh situation. When two individuals jump in opposite directions from a frictionless surface, we must consider the conservation of momentum principle.

The law states that in the absence of external forces, the total momentum of a system remains constant. This principle is pivotal in solving the problem, especially since our system—the sleigh and the jumpers initially at rest—must have a net zero momentum. This implies that any change in momentum from the jumpers must be balanced by the sleigh's motion.

Effective physics problem solving involves breaking down the problem into more manageable parts—calculating masses from weights, resolving velocities into components, and applying known laws like momentum conservation. This structured approach helps in formulating solutions logically and effectively.
Horizontal Velocity Calculation
Calculating horizontal velocity involves decomposing apparent velocities that occur at angles. In our scenario, both Jonathan and Jane jump at specific angles above the horizontal. To find the sleigh's velocity, we first resolve the velocities of Jonathan and Jane into their horizontal components using trigonometric functions.

For Jonathan, jumping at a 30.0° angle, we use the cosine function:
  • Horizontal velocity: \( v_{Jx} = v_J \cos(30.0^{\circ}) = 5.00 \text{ m/s} \times \frac{\sqrt{3}}{2} = 4.33 \text{ m/s} \)
For Jane, with an angle of 36.9°:
  • Horizontal velocity: \( v_{jx} = v_j \cos(36.9^{\circ}) = 7.00 \text{ m/s} \times 0.8 = 5.60 \text{ m/s} \)
The distinct directions (left and right) impact sign convention—negative for Jonathan and positive for Jane, reflecting the horizontal motion direction relative to the sleigh.

This calculation provides the necessary components to apply momentum principles and solve for the desired sleigh velocity.
Frictionless Surface Dynamics
Dynamics on a frictionless surface create unique conditions where no energy is lost to friction, allowing us to directly apply principles like conservation of momentum. When on such a surface, objects exerting forces on each other do so without experiencing resistance from friction.

In the sleigh problem, the ice ensures a truly closed system where our calculations can proceed without accounting for energy loss. Jonathan and Jane's jumps propel them and serve as the internal forces that determine the motion of the sleigh. Given that they jump in opposite directions, their actions are directly compensated by the sleigh moving to the right.

Solving the motion requires summing all horizontal forces and setting them equal to zero, keeping total momentum conserved as initially, there was no movement. Recognizing these conditions helps simplify the problem, as actions like Jonathan's and Jane's don't face other opposition except their own mass' reaction on the sleigh, thus making mathematical predictions precise based on pure dynamics.

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

An atomic nucleus suddenly bursts apart (fissions) into two pieces. Piece \(A,\) of mass \(m_{A},\) travels off to the left with speed \(v_{A} .\) Piece \(B,\) of mass \(m_{B},\) travels off to the right with speed \(v_{B}\). (a) Use conservation of momentum to solve for \(v_{B}\) in terms of \(m_{A}\) , \(m_{B},\) and \(v_{A}\) . (b) Use the results of part (a) to show that \(K_{A} / K_{B}=m_{B} / m_{A},\) where \(K_{A}\) and \(K_{B}\) are the kinetic energies of the two pieces.

On a frictionless, horizontal air table, puck \(A\) (with mass 0.250 \(\mathrm{kg}\) is moving toward puck \(B\) (with mass \(0.350 \mathrm{kg} ),\) which is initially at rest. After the collision, puck \(A\) has a velocity of 0.120 \(\mathrm{m} / \mathrm{s}\) to the left, and puck \(B\) has a velocity of 0.650 \(\mathrm{m} / \mathrm{s}\) to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

Three identical pucks on a horizontal air table have repelling magnets. They are held together and then released simultaneously. Each has the same speed at any instant. One puck moves due west. What is the direction of the velocity of each of the other two pucks?

BIO Changing Your Center of Mass. To keep the calculations fairly simple, but still reasonable, we shall model a human leg that is 92.0 \(\mathrm{cm}\) long (measured from the hip joint) by assuming that the upper leg and the lower leg (which includes the foot) have equal lengths and that each of them is uniform. For a 70.0 -kg person, the mass of the upper leg would be \(8.60 \mathrm{kg},\) while that of the lower leg (including the foot) would be 5.25 \(\mathrm{kg} .\) Find the location of the center of mass of this leg, relative to the hip joint, if it is (a) stretched out horizontally and (b) bent at the knee to form a right angle with the upper leg remaining horizontal.

Energy Sharing in Elastic Collisions. A stationary object with mass \(m_{B}\) is struck head-on by an object with mass \(m_{A}\) that is moving initially at speed \(v_{0} .\) (a) If the collision is elastic, what percentage of the original energy does each object have after the collision? (b) What does your answer in part (a) give for the special cases (i) \(m_{A}=m_{B}\) and (ii) \(m_{A}=5 m_{B} ?(\mathrm{c})\) For what values, if any, of the mass ratio \(m_{A} / m_{B}\) is the original kinetic energy shared equally by the two objects 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.