Problem 104 - University Physics with Modern Physics

Question

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.

Step-by-Step Solution

Answer

The sleigh's velocity is approximately 0.104 m/s to the right.

Step 1: 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:\[ m_s = \frac{1000 \, \text{N}}{9.8 \, \text{m/s}^2} \approx 102.04 \, \text{kg} \]

Step 2: 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.)

Step 3: 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:\[ (81.63 \, \text{kg})(-4.33 \, \text{m/s}) + (61.22 \, \text{kg})(5.60 \, \text{m/s}) + (102.04 \, \text{kg})(v_s) = 0 \]

Step 4: 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.

Key Concepts

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.