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Sam \((61 \mathrm{~kg})\) and Alice \((44 \mathrm{~kg})\) stand on an ice rink, providing them with a nearly frictionless surface to slide on. Sam gives Alice a push, causing her to slide away at a speed (with respect to the rink) of \(1.20 \mathrm{~m} / \mathrm{s}\). a) With what speed does Sam recoil? b) Calculate the change in the kinetic energy of the SamAlice system. c) Energy cannot be created or destroyed. What is the source of the final kinetic energy of this system?

Short Answer

Expert verified
Answer: The source of the final kinetic energy of Sam and Alice is the work done by Sam's push on Alice and the equal and opposite force exerted by Alice on Sam. The change in the kinetic energy of the system is 38.19 Joules.

Step by step solution

01

Part (a) - Find Sam's Recoil Speed

To solve this part, we need to apply the law of conservation of momentum. The momentum of the system before Sam's push is equal to the momentum of the system after the push. Let's consider the total momentum of Sam and Alice be \(p_{initial}\) and \(p_{final}\) \(p_{initial} = p_{final}\) Initially, Sam and Alice are at rest so: \(p_{initial} = 0\) After the push, Sam recoils with a speed \(v_s\) and Alice slides with a speed of \(1.20 \mathrm{~m} \mathrm{s}^{-1}\). So, the final momentum of Sam and Alice is given by: \(p_{final} = m_sv_s + m_av_a\) Applying the conservation of momentum: \(m_sv_s + m_av_a = 0\) We're given that \(m_s = 61 \mathrm{~kg}, m_a = 44 \mathrm{~kg}\) and \(v_a = 1.20 \mathrm{~m} \mathrm{s}^{-1}\). Now, we can solve for Sam's recoil speed, \(v_s\): \(61v_s + 44(1.20) = 0\) \(v_s = -\frac{44(1.20)}{61} \approx -0.87 \mathrm{~m} \mathrm{s}^{-1}\) Sam's recoil speed is \(0.87 \mathrm{~m} \mathrm{s}^{-1}\) in the opposite direction to the push.
02

Part (b) - Change in the Kinetic Energy

Now we'll calculate the change in kinetic energy of the system. Let the initial kinetic energy be \(K_i\) and the final kinetic energy be \(K_f\). The initial kinetic energy is zero as both Sam and Alice are at rest. \(K_i = 0\) The final kinetic energy is given by: \(K_f = \frac{1}{2} m_s v_s^2 + \frac{1}{2} m_a v_a^2\) Substituting the known values: \(K_f = \frac{1}{2} (61)(-0.87)^2 + \frac{1}{2} (44)(1.20)^2 \approx 38.19 \mathrm{~Joules} \) The change in kinetic energy is: \(\Delta K = K_f - K_i = 38.19 - 0 = 38.19 \mathrm{~Joules}\)
03

Part (c) - Source of the Final Kinetic Energy

As mentioned in the question, energy cannot be created or destroyed. In this scenario, the final kinetic energy of the system comes from the work done by Sam while pushing Alice. When Sam pushes Alice, he exerts a force on her, and in turn, she exerts an equal and opposite force on Sam (according to Newton's third law). As Sam applies the force on Alice, he does positive work on her, which increases her kinetic energy. At the same time, the force exerted by Alice on Sam does negative work on him, which decreases his potential energy and increases his kinetic energy as he recoils. Therefore, the source of the final kinetic energy of the system is the work done by Sam's push on Alice and the equal and opposite force exerted by Alice on Sam.

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Key Concepts

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

Understanding Kinetic Energy
In the exercise, both Sam and Alice initially had no kinetic energy because they were not moving. However, after Sam pushes Alice, they both begin to slide on the ice, converting energy from the push into kinetic energy. The futures of the people who have been pushed will be very unpredictable, with outcomes ranging from simple acceleration to complex orbital maneuvers. To calculate the change in kinetic energy, we simply find the difference between the final kinetic energy (after they start moving) and the initial kinetic energy (when they were at rest). As seen in the solution, once you have the mass and velocities of both objects, you can determine the overall kinetic energy of the system.
Newton's Third Law in Action
Newton's third law of motion states that for every action, there is an equal and opposite reaction. This can sometimes be counterintuitive, but it's a fundamental principle of mechanics. When Sam pushes Alice, he exerts a force on her; according to Newton's third law, Alice exerts an equal force in the opposite direction on Sam. While this might seem like it would cause a deadlock, with neither being able to move, the forces act on different objects, causing them each to move in opposite directions. In the context of the exercise, the force of the push made Alice skate forward, while the equal and opposite reaction force made Sam recoil backward. Both forces are internal to the system, meaning they don't change the system's overall momentum, which stays zero, just as it was initially.
The Law of Conservation of Energy
Energy cannot be created or destroyed; this is the principle behind the law of conservation of energy. In our physical exercise, the kinetic energy held by Sam and Alice after the push had to come from somewhere—specifically, the chemical potential energy stored in Sam's muscles. When he pushes Alice, Sam does work by applying force over a distance, which transforms his stored chemical energy into mechanical work. This energy transfer results in the increase in kinetic energy for both Alice and Sam. The increase in their motion doesn't violate the conservation of energy because the total energy of the system, including the energy of the push (work done), remains constant; it has merely been converted from one form to another.

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Most popular questions from this chapter

An artillery shell is moving on a parabolic trajectory when it explodes in midair. The shell shatters into a very large number of fragments. Which of the following statements is true (select all that apply)? a) The force of the explosion will increase the momentum of the system of fragments, and so the momentum of the shell is not conserved during the explosion. b) The force of the explosion is an internal force and thus cannot alter the total momentum of the system. c) The center of mass of the system of fragments will continue to move on the initial parabolic trajectory until the last fragment touches the ground. d) The center of mass of the system of fragments will continue to move on the initial parabolic trajectory until the first fragment touches the ground. e) The center of mass of the system of fragments will have a trajectory that depends on the number of fragments and their velocities right after the explosion.

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