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Astronauts are playing baseball on the International Space Station. One astronaut with a mass of \(50.0 \mathrm{~kg}\), initially at rest, hits a baseball with a bat. The baseball was initially moving toward the astronaut at \(35.0 \mathrm{~m} / \mathrm{s},\) and after being hit, travels back in the same direction with a speed of \(45.0 \mathrm{~m} / \mathrm{s}\). The mass of a baseball is \(0.14 \mathrm{~kg}\). What is the recoil velocity of the astronaut?

Short Answer

Expert verified
Answer: The recoil velocity of the astronaut is 0.028 m/s.

Step by step solution

01

Establish the conservation of linear momentum equation

According to the conservation of linear momentum, the total momentum before the hit is equal to the total momentum after the hit. Mathematically, this can be expressed as: $$(m_1u_1 + m_2u_2) = (m_1v_1 + m_2v_2)$$ where: \(m_1\) = mass of astronaut (50.0 kg) \(u_1\) = initial velocity of astronaut (0 m/s) \(v_1\) = final velocity of astronaut (unknown) \(m_2\) = mass of baseball (0.14 kg) \(u_2\) = initial velocity of baseball (-35.0 m/s, negative since it's in the opposite direction) \(v_2\) = final velocity of the baseball (-45.0 m/s, negative since it's traveling back in the same direction) Now, we will substitute the known values and calculate the final velocity of the astronaut, \(v_1\).
02

Substitute the known values into the formula

Substituting the known values into the equation results in: $$(50.0 \mathrm{~kg} \cdot 0 \mathrm{~m/s} + 0.14 \mathrm{~kg} \cdot -35.0 \mathrm{~m/s}) = (50.0 \mathrm{~kg} \cdot v_1 + 0.14 \mathrm{~kg} \cdot -45.0 \mathrm{~m/s})$$ Solving for \(v_1\) using algebra, we get: $$50.0 \mathrm{~kg} \cdot v_1 = 0.14 \mathrm{~kg} (45.0 \mathrm{~m/s} - 35.0 \mathrm{~m/s})$$
03

Calculate the final velocity of the astronaut

Now, we can solve for the final velocity of the astronaut: $$v_1 = \frac{0.14 \mathrm{~kg} (45.0 \mathrm{~m/s} - 35.0 \mathrm{~m/s})}{50.0 \mathrm{~kg}}$$ $$v_1 = \frac{0.14 \mathrm{~kg} \cdot 10.0 \mathrm{~m/s}}{50.0 \mathrm{~kg}}$$ $$v_1 = \frac{1.4 \mathrm{~kg~m/s}}{50.0 \mathrm{~kg}}$$ $$v_1 = 0.028 \mathrm{~m/s}$$ Thus, the recoil velocity of the astronaut is \(0.028~\mathrm{m/s}\).

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Key Concepts

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

Momentum Conservation
When we talk about momentum conservation in physics, we're referring to one of the most fundamental principles in classical mechanics. In a closed system, where no external forces are acting, the total momentum of all the objects remains constant. This principle is crucial for understanding interactions like collisions or separations.

To illustrate, imagine two ice skaters pushing off from one another; they move in opposite directions after the push. This is because the total momentum before and after the push remains unchanged; it's simply redistributed between the skaters. In more technical terms, if you have two objects (like our astronaut and baseball), the total momentum before an event is equal to the total momentum after the event, assuming no external forces interfere.

Applying Momentum Conservation
In the context of our space baseball game, we can see momentum conservation in action. Before the astronaut hits the baseball, the momentum of the system is the sum of the astronaut's and baseball's momenta. Since the astronaut is initially at rest, only the baseball has momentum. After the hit, the baseball moves in the opposite direction, and the astronaut moves backward. The sum of their momenta post-impact is equal to the initial momentum of the ball — beautifully demonstrating the conservation of linear momentum.
Recoil Velocity
Recoil velocity is the speed at which an object moves back after exerting a force on another object. It's commonly observed with guns; when a bullet is fired, the gun kicks back. That's recoil. In space, where there's no gravity or friction to affect motion, recoil effects become much more noticeable.

In our astronaut and baseball scenario, when the astronaut hits the ball, they experience recoil. The astronaut moves in the opposite direction to the baseball's new trajectory. The magnitude of the recoil velocity is directly related to the mass and velocity of both the astronaut and the baseball. Since the astronaut has a more significant mass compared to the baseball, the recoil velocity is relatively small but still present.

The Calculation of Recoil Velocity
Mathematically, we calculate recoil velocity using the conservation of momentum. The astronaut's recoil velocity is found by arranging the conservation of momentum formula to solve for the unknown velocity. The key point is to note that the directions are important – velocities in the opposite direction from the initial motion are negative. This understanding guides us correctly through the algebraic solution.
Physics in Space
The laws of physics, including the conservation of momentum, don't change just because we're in space — but the outcomes might look a little different due to the absence of gravity and atmospheric resistance. In space, objects in motion will stay in motion at a constant velocity unless acted on by another force, according to Newton's First Law of Motion.

For astronauts, every action they take from walking, to pushing off a wall, or hitting a baseball, causes a reaction. There are no external forces to quickly dampen these movements as there are on Earth, like friction or air resistance. This is why a simple act of hitting a baseball can propel an astronaut backward, and without any external forces to stop them, they'll continue to drift until they interact with another object (like the wall of the space station).

Implications in Space
Because of this persistence of motion, all movements must be carefully considered in space. Astronauts must secure themselves when performing tasks that require force to avoid unwanted travel. They even use their own mass and inertia for maneuvering, which is beautifully exemplified in the baseball and astronaut example.
Astronaut Mass and Velocity
In our example with the astronaut playing baseball, the concept of 'astronaut mass and velocity' comes into play, literally. Here, the mass of the astronaut (and the baseball) is a key component in determining the outcome post-impact. This concept ties into Newton's Second Law of Motion, which states that the acceleration of an object depends on the net force acting upon it and the mass of the object.

The larger the mass of the object (in this case, our astronaut), the less it will accelerate when a force is applied. That's why, despite hitting the ball hard, the astronaut only recoils with a small velocity. Consequently, when dealing with mass and velocity in momentum problems, it's crucial to account for the mass of both objects involved in the collision or interaction to predict each object's velocity post-event.

Effect of Mass on Recoil Velocity
In the given solution, we've used the mass of both the astronaut and the baseball to calculate the astronaut's recoil velocity. This involves setting up an equation based on the principle of momentum conservation, entering the known masses and velocities, and solving for the unknown value. The astronaut's larger mass results in a much smaller velocity change compared to the baseball after the interaction, a great example of how mass can influence motion.

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Most popular questions from this chapter

The value of the momentum for a system is the same at a later time as at an earlier time if there are no a) collisions between particles within the system. b) inelastic collisions between particles within the system. c) changes of momentum of individual particles within the system. d) internal forces acting between particles within the system. e) external forces acting on particles of the system.

7.57 A 1439 -kg railroad car traveling at a speed of \(12 \mathrm{~m} / \mathrm{s}\) strikes an identical car at rest. If the cars lock together as a result of the collision, what is their common speed (in \(\mathrm{m} / \mathrm{s})\) afterward?

Some kids are playing a dangerous game with fireworks. They strap several firecrackers to a toy rocket and launch it into the air at an angle of \(60^{\circ}\) with respect to the ground. At the top of its trajectory, the contraption explodes, and the rocket breaks into two equal pieces. One of the pieces has half the speed that the rocket had before it exploded and travels straight upward with respect to the ground. Determine the speed and direction of the second piece.

An uncovered hopper car from a freight train rolls without friction or air resistance along a level track at a constant speed of \(6.70 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction. The mass of the car is \(1.18 \cdot 10^{5} \mathrm{~kg}\). a) As the car rolls, a monsoon rainstorm begins, and the car begins to collect water in its hopper (see the figure). What is the speed of the car after \(1.62 \cdot 10^{4} \mathrm{~kg}\) of water collects in the car's hopper? Assume that the rain is falling vertically in the negative \(y\) -direction. b) The rain stops, and a valve at the bottom of the hopper is opened to release the water. The speed of the car when the valve is opened is again \(6.70 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) -direction (see the figure). The water drains out vertically in the negative \(y\) -direction. What is the speed of the car after all the water has drained out?

An automobile with a mass of \(1450 \mathrm{~kg}\) is parked on a moving flatbed railcar; the flatbed is \(1.5 \mathrm{~m}\) above the ground. The railcar has a mass of \(38,500 \mathrm{~kg}\) and is moving to the right at a constant speed of \(8.7 \mathrm{~m} / \mathrm{s}\) on a frictionless rail. The automobile then accelerates to the left, leaving the railcar at a speed of \(22 \mathrm{~m} / \mathrm{s}\) with respect to the ground. When the automobile lands, what is the distance \(D\) between it and the left end of the railcar? See the figure.

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