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Two skaters, one with mass \(65 \mathrm{~kg}\) and the other with mass 40 \(\mathrm{kg}\), stand on an ice rink holding a pole of length \(10 \mathrm{~m}\) and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the \(40 \mathrm{~kg}\) skater move?

Short Answer

Expert verified
The 40 kg skater moves 6.19 m.

Step by step solution

01

Understand the Problem

We need to determine the distance moved by the 40 kg skater when two skaters pull themselves along a pole and meet. The center of mass remains constant as there are no external forces.
02

Define the System's Center of Mass

Calculate the initial position of the system's center of mass. Let the 65 kg skater be at position 0 m and the 40 kg skater be at 10 m. The formula for the center of mass is: \\[ x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \] \Substitute the values: \\[ x_{cm} = \frac{65 \times 0 + 40 \times 10}{65 + 40} \]
03

Calculate the Initial Center of Mass

Compute the center of mass: \\[ x_{cm} = \frac{65 \times 0 + 40 \times 10}{65 + 40} = \frac{400}{105} \approx 3.81 \text{ m} \] \This is the initial center of mass, which remains unchanged during the movement.
04

Determine the Meeting Point

Since the center of mass doesn't move, both skaters meet at the center of mass location, which is 3.81 m from the position of the 65 kg skater.
05

Calculate the Distance Moved by the 40 kg Skater

The initial position of the 40 kg skater is 10 m. The final meeting position is 3.81 m. Calculate the distance moved: \\[ \text{Distance moved} = 10 \text{ m} - 3.81 \text{ m} = 6.19 \text{ m} \]

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Key Concepts

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

Conservation of Momentum
When two skaters on a frictionless ice rink pull themselves towards each other using a pole, it's a perfect illustration of the conservation of momentum principle. Momentum deals with the quantity of motion an object has, calculated as the product of mass and velocity. In this scenario, the system is isolated, and there are no external forces acting on the skaters. Hence, the total momentum before they start pulling will equal the total momentum when they meet.

Since both skaters are initially at rest, their combined momentum is zero. Because the ice rink is frictionless, this zero initial momentum must also be maintained throughout their movement.

  • Both skaters move towards the center of mass of the system.
  • The center of mass is a point where the weighted relative position of the skaters can be considered.
  • As no external forces act on the skaters, their movement keeps the center of mass stationary.
Hence, conservation of momentum ensures they meet exactly at the center of mass. It perfectly demonstrates this crucial physics principle, and why the path each skater takes results in them meeting at that exact point.
Skater Dynamics
In the study of skater dynamics, the problem with two skaters on an ice rink shows how forces and motion interact. Skaters convert their internal energy to movement by pulling the pole. This dynamic interaction showcases Newton's third law: "For every action, there is an equal and opposite reaction."

  • The 65 kg skater applies a force on the pole, pulling themselves towards the lighter skater.
  • Simultaneously, the 40 kg skater exerts a force in the opposite direction.

Because of the mass difference, the skaters move different distances. The lighter one, experiencing less resistance to the opposing force, travels a greater distance.

Both skaters will meet at a point where balancing forces and masses results in the center of mass position as calculated in the original solution. This interaction of forces and differing motions illustrates the fascinating dynamics that occur on frictionless surfaces.
Mass Distribution
Mass distribution plays a key role when determining how the skaters move. It affects the location of the center of mass, where the skaters eventually meet. The center of mass is computed based on the weights and positions of the skaters along the ice rink.

  • The 65 kg skater initially is at one end of the pole (0 m), and the 40 kg skater is at the other end (10 m).
  • The calculation for the center of mass is performed using \( x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \).

Solving gives us a center of mass at approximately 3.81 m from the position of the 65 kg skater. This point does not shift when they move, because it's determined solely by their relative masses and positions at the start. Understanding mass distribution helps to predict how the skaters will move and meet, revealing insights into physics beyond the superficial motion.

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Most popular questions from this chapter

Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of \(36 \mathrm{~m}\). During the collision at the bottom of the elevator shaft, a \(90 \mathrm{~kg}\) passenger is stopped in \(5.0 \mathrm{~ms}\). (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the passenger were to jump upward with a speed of \(7.0\) \(\mathrm{m} / \mathrm{s}\) relative to the cab floor just before the cab hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force (assuming the same stopping time)?

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with the cue ball's original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don't consider the rotation) conserved?

A rocket that is in deep space and initially at rest relative to an inertial reference frame has a mass of \(2.55 \times 10^{5} \mathrm{~kg}\), of which \(1.81 \times 10^{5} \mathrm{~kg}\) is fuel. The rocket engine is then fired for \(250 \mathrm{~s}\) while fuel is consumed at the rate of \(480 \mathrm{~kg} / \mathrm{s}\). The speed of the exhaust products relative to the rocket is \(3.27 \mathrm{~km} / \mathrm{s}\). (a) What is the rocket's thrust? After the \(250 \mathrm{~s}\) firing, what are (b) the mass and (c) the speed of the rocket?

Two \(2.0 \mathrm{~kg}\) bodies, \(A\) and \(B\), collide. The velocities before the collision are \(\vec{v}_{A}=(15 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\) and \(\vec{v}_{B}=(-10 \hat{\mathrm{i}}+5.0 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\). After the collision, \(\vec{v}_{A}^{\prime}=(-5.0 \hat{\mathrm{i}}+20 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}\). What are (a) the final velocity of \(B\) and (b) the change in the total kinetic energy (including sign )?

At time \(t=0\), force \(\vec{F}_{1}=(-4.00 \hat{\mathrm{i}}+5.00 \hat{\mathrm{j}}) \mathrm{N}\) acts on an initially stationary particle of mass \(2.00 \times 10^{-3} \mathrm{~kg}\) and force \(\vec{F}_{2}=(2.00 \hat{\mathrm{i}}-4.00 \hat{\mathrm{j}}) \mathrm{N}\) acts on an initially stationary particle of mass \(4.00 \times 10^{-3} \mathrm{~kg}\). From time \(t=0\) to \(t=2.00 \mathrm{~ms}\), what are the (a) magnitude and (b) angle (relative to the positive direction of the \(x\) axis) of the displacement of the center of mass of the twoparticle system? (c) What is the kinetic energy of the center of mass at \(t=2.00 \mathrm{~ms} ?\)

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