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Chapter 7: Problem 9

A space probe is traveling in outer space with a momentum that has a magnitude of \(7.5 \times 10^{7} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) A retrorocket is fired to slow down the probe. It applies a force to the probe that has a magnitude of \(2.0 \times 10^{6} \mathrm{N}\) and a direction opposite to the probe's motion. It fires for a period of 12 s. Determine the momentum of the probe after the retrorocket ceases to fire.

Short Answer

Expert verified
The final momentum is \(5.1 \times 10^{7} \text{ kg} \cdot \text{m/s} \).

Step by step solution

01

Understanding the Problem

We are given the initial momentum of the probe as \(7.5 \times 10^{7} \text{ kg} \cdot \text{m/s} \), the force applied by the retro-rocket as \(2.0 \times 10^{6} \text{ N} \), and the time duration of the force as 12 seconds. We need to find the final momentum of the probe.
02

Identify the Relationship

Momentum is the product of mass and velocity, and the change in momentum ( \(\Delta p\) ) equals the impulse applied to the system. Impulse is the product of force and time, so \( \Delta p = F \times t \).
03

Calculate Impulse

Calculate the impulse using the formula: \[ \Delta p = F \cdot t \]Substitute the given values: \[ \Delta p = (2.0 \times 10^{6} \text{ N}) \times (12 \text{ s}) = 2.4 \times 10^{7} \text{ kg} \cdot \text{m/s} \]
04

Determine Final Momentum

Since the force is opposite to the direction of motion, it will reduce the momentum. Hence, the change in momentum is subtracted from the initial momentum:\[ \text{Final momentum} = \text{Initial momentum} - \Delta p = 7.5 \times 10^{7} - 2.4 \times 10^{7} \]\[ = 5.1 \times 10^{7} \text{ kg} \cdot \text{m/s} \]
05

Conclude the Solution

The final momentum of the probe after the retrorocket ceases to fire is \(5.1 \times 10^{7} \text{ kg} \cdot \text{m/s} \).

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Key Concepts

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

Understanding Impulse
Impulse is a fundamental concept when dealing with momentum. It represents the change in momentum of an object when a force is applied over a certain period of time. Mathematically, impulse is calculated as the product of force and time. In our case:
  • The force is given as \(2.0 \times 10^{6} \text{ N}\).
  • The time for which the force is applied is 12 seconds.
Thus, the impulse exerted on the space probe is \[ \Delta p = F \cdot t = (2.0 \times 10^{6} \text{ N}) \times (12 \text{ s}) = 2.4 \times 10^{7} \text{ kg} \cdot \text{m/s} \] The impulse tells us how much the momentum of the probe will change due to the retrorocket. Since the rocket's force opposes the probe's motion, this impulse decreases the probe's momentum.
Role of Force
Force is any interaction that, when unopposed, changes the motion of an object. In this exercise, the central role of force is to alter the speed of the space probe by acting in the opposite direction to its motion. Force is measured in newtons (N) and is vectorial, meaning it has both magnitude and direction. Here we have:
  • A force of \(2.0 \times 10^{6} \text{ N}\).
  • The direction opposite to the probe's initial motion.
When this force acts over time, it creates an impulse which causes a change in momentum. Understanding force helps us see how actions, like firing a retrorocket, can significantly alter the path and speed of objects in space by changing their momentum.
Change in Momentum
Momentum, often denoted as \(p\), is a measure of an object's motion, dependent on mass and velocity. The change in momentum (\(\Delta p\)), is key to grasp in scenarios like this where the motion of an object has been altered. Initially, the space probe had a large momentum due to its speed, given by \(7.5 \times 10^{7} \text{ kg} \cdot \text{m/s}\). After the force provided by the retrorocket acts on it:
  • Impulse on the probe changes its momentum by \(2.4 \times 10^{7} \text{ kg} \cdot \text{m/s}\).
  • The force and its opposing direction reduce the probe's momentum thereafter.
The final momentum of the probe is calculated by subtracting the impulse from the initial momentum, using:\[ \text{Final momentum} = \text{Initial momentum} - \Delta p = 7.5 \times 10^{7} - 2.4 \times 10^{7} \]\[ = 5.1 \times 10^{7} \text{ kg} \cdot \text{m/s} \]Thus, the probe moves slower after the retrorocket ceases, demonstrating how momentum changes in response to external forces.

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Most popular questions from this chapter

A ball is attached to one end of a wire, the other end beingfastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, \(1.60 \mathrm{kg}\) and \(2.40 \mathrm{kg},\) and the length of the wire is \(1.20 \mathrm{m} .\) Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

A \(46-\mathrm{kg}\) skater is standing still in front of a wall. By pushing against the wall she propels herself backward with a velocity of \(-1.2 \mathrm{m} / \mathrm{s}\). Her hands are in contact with the wall for 0.80 s. Ignore friction and wind resistance. Find the magnitude and direction of the average force she exerts on the wall (which has the same magnitude as, but opposite direction to, the force that the wall applies to her).

Two ice skaters have masses \(m_{1}\) and \(m_{2}\) and are initially stationary. Their skates are identical. They push against one another, as in Interactive Figure \(7.9,\) and move in opposite directions with different speeds. While they are pushing against each other, any kinetic frictional forces acting on their skates can be ignored. However, once the skaters separate, kinetic frictional forces eventually bring them to a halt. As they glide to a halt, the magnitudes of their accelerations are equal, and skater 1 glides twice as far as skater \(2 .\) What is the ratio \(m_{1} / m_{2}\) of their masses?

You and your crew must dock your \(25000 \mathrm{kg}\) spaceship at Spaceport Alpha, which is orbiting Mars. In the process, Alpha's control tower has requested that you ram another vessel, a freight ship of mass \(16500 \mathrm{kg},\) latch onto it, and use your combined momentum to bring it into dock. The freight ship is not moving with respect to the colossal Spaceport Alpha, which has a mass of \(1.85 \times 10^{7} \mathrm{kg} .\) Alpha's automated system that guides incoming spacecraft into dock requires that the incoming speed is less than \(2.0 \mathrm{m} / \mathrm{s}\). (a) Assuming a perfectly linear alignment of your ship's velocity vector with the freight ship (which is stationary with respect to Alpha) and Alpha's docking port, what must be your ship's speed (before colliding with the freight ship) so that the combination of the freight ship and your ship arrives at Alpha's docking port with a speed of \(1.50 \mathrm{m} / \mathrm{s} ?\) (b) How does the velocity of Spaceport Alpha change when the combination of your vessel and the freight ship successfully docks with it? (c) Suppose you made a mistake while maneuvering your vessel in an attempt to ram the freight ship and, rather than latching on to it and making a perfectly inelastic collision, you strike it and knock it in the direction of the spaceport with a perfectly elastic collision. What is the speed of the freight ship in that case (assuming your ship had the same initial velocity as that calculated in part (a))?

A 60.0-kg person, running horizontally with a velocity of \(+3.80 \mathrm{m} / \mathrm{s}\), jumps onto a \(12.0-\mathrm{kg}\) sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast \(30.0 \mathrm{m}\) on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?
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