Question

A student with a mass of 60.0 kg jumps straight up in the air by using her legs to apply an average force of \(770 .\) N to the ground for 0.250 s. Assume that the initial momentum of the student and the Earth are zero. What is the momentum of the student immediately after this impulse? What is the momentum of the Earth after this impulse? What is the speed of the Earth after the impulse? What fraction of the total kinetic energy that the student produces with her legs goes to the Earth (the mass of the Earth is \(5.98 \cdot 10^{24} \mathrm{~kg}\) )? Using conservation of energy, how high does the student jump?

Short answer

Answer: The student jumps 0.525 meters high.

Step-by-step solution

Calculate Impulse on the student

To find the impulse on the student, we multiply the given force by the time duration: \(J = F \cdot \Delta t\). \(J = 770 \,\text{N} \cdot 0.250\,\text{s} = 192.5\,\text{N}\cdot\text{s}\)

Calculate the momentum of the student

The impulse applied to the student equals the change in the student's momentum. Since she starts from rest, her initial momentum is zero, so her final momentum equals the impulse. \(p_{\text{student}} = 192.5\,\text{N}\cdot\text{s}\)

Calculate the momentum of the Earth

By conservation of momentum, the Earth's momentum must be equal and opposite to the student's momentum. \(p_{\text{Earth}} = - p_{\text{student}} = -192.5\,\text{N}\cdot\text{s}\)

Calculate the speed of the Earth

To find the speed of the Earth, we will divide its momentum by its mass: \(v_{\text{Earth}} = \frac{p_{\text{Earth}}}{M_{\text{Earth}}} = \frac{-192.5\,\text{N}\cdot\text{s}}{5.98 \times 10^{24}\ \text{kg}} \approx -3.22 \times 10^{-23}\,\text{m/s}\)

Calculate the kinetic energy of the student and Earth

First, we will find the speed of the student using the momentum and mass: \(v_{\text{student}} = \frac{p_{\text{student}}}{m_{\text{student}}} = \frac{192.5\,\text{N}\cdot\text{s}}{60.0\,\text{kg}} = 3.21\,\text{m/s}\) Next, calculate the kinetic energies of the Earth and the student: \(K_{\text{Earth}} = \frac{1}{2} M_{\text{Earth}} v_{\text{Earth}}^2 \approx 1.56 \times 10^{2}\,\text{J}\) \(K_{\text{student}} = \frac{1}{2} m_{\text{student}} v_{\text{student}}^2 = 310.3\,\text{J}\)

Calculate the fraction of kinetic energy transferred to the Earth

Divide the kinetic energy of the Earth by the total kinetic energy of the system: \(\text{fraction} = \frac{K_{\text{Earth}}}{K_{\text{student}} + K_{\text{Earth}}} \approx 0.002\)

Calculate the height the student jumps using conservation of energy

The kinetic energy of the student will convert to potential energy at the maximum height of the jump. Use the conservation of energy formula: \(K_{\text{student}} = U_{\text{student}}\) \(U_{\text{student}} = m_{\text{student}}gh_{\text{student}}\) \(h_{\text{student}} = \frac{K_{\text{student}}}{m_{\text{student}}g} = \frac{310.3\,\text{J}}{60.0\,\text{kg}\cdot 9.81\,\text{m/s}^2} \approx 0.525\,\text{m}\) So, the student jumps 0.525 meters high.

Key concepts

Impulse and Momentum

Impulse is a fundamental concept that bridges the gap between force and momentum. When a force acts upon an object for a period of time, it creates an impulse, which is the product of the force and the time interval during which the force is applied. Mathematically, impulse \( J \) can be expressed as \( J = F\Delta t \), where \( F \) is the force applied and \(\text{\Delta} t\) is the time period.

The impulse causes a change in the object's momentum, hence the principle that impulse equals the change in momentum. For an individual initially at rest, the final momentum \( p \) can be directly equated to the impulse given. This relationship is pivotal in the scenario where the student jumps into the air. Her legs exert a force on the ground, generating an impulse, which in turn gives the student an upward momentum, propelling her off the surface.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed in an isolated system; it can only change form. In physics problems, we often track energy as it shifts from kinetic to potential form or vice versa. The jump exercise is an excellent application of this principle as the energy exerted by the student is initially kinetic—since it relates to motion—and then converts into potential energy when the student reaches the peak of the jump.

As the student reaches maximum height, all the kinetic energy has been converted to gravitational potential energy. At this point, no kinetic energy is present until she begins to fall back to the ground. By setting the kinetic and potential energies equal to one another at the peak, we can solve for variables such as the height of the jump, revealing the power of conservation of energy in solving problems.

Kinetic Energy

Kinetic energy \( K \) is the energy of motion. Any object with mass \( m \) moving at a velocity \( v \) possesses kinetic energy, which is given by the formula \( K = \frac{1}{2} m v^2 \). In the exercise scenario, both Earth and the student gain kinetic energy due to their velocities post-impulse, even though Earth's movement is imperceptible. The student's kinetic energy is what allows her to ascend from the ground, reaching a height where this energy is transformed into potential energy.

Understanding kinetic energy is crucial in this context as it allows us to determine the speed of both the student and the Earth after the impulse, and it gives insight into how energy is distributed between objects in a system.

Potential Energy

Potential energy, particularly gravitational potential energy \( U \) in the context of our exercise, is energy stored by objects due to their position relative to Earth. This energy is dependent on the height \( h \) above a reference point and the mass \( m \) of the object. Gravitational potential energy can be calculated using \( U = mgh \), where \( g \) is the acceleration due to gravity. In the student's case, as she jumps and reaches a certain height, her kinetic energy is transformed into potential energy at the peak of her jump.

This transformation is a classic demonstration of the conservation of energy. By equating the kinetic energy at the moment of takeoff to the potential energy at the peak height, the maximum height achieved by the student can be precisely determined.