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At time $t=0,$ a $2150-\mathrm{kg}$ rocket in outer space fires an engine that exerts an increasing force on it in the $+x$ -direction. This force obeys the equation $F_{x}=A t^{2},$ where $t$ is time, and has a magnitude of 781.25 $\mathrm{N}$ when $t=1.25 \mathrm{s}$ . (a) Find the SI value of the constant $A,$ including its units. $(\mathrm{b})$ What impulse does the engine exert on the rocket during the 1.50 -s interval starting 2.00 s after the engine is fired? (c) By how much does the rocket's velocity change during this interval?

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(a) $A=500 \mathrm{N} / \mathrm{s}^{2}$(b) $\vec{J}=5812.5 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$(c) $\Delta \overrightarrow{\mathbf{v}}=2.7 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

Cornell University

Hope College

University of Sheffield

Lectures

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

03:30

In physics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Given a force, F, applied for a time, t, the resulting change in momentum, p, is equal to the impulse, I. Impulse applied to a mass, m, is also equal to the change in the object's kinetic energy, T, as a result of the force acting on it.

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in this question. We have a rocket with a mass off 2000 and 150 kg, and some force is being exerted by its engines on that force. Depends on time according to this relation, we also know that when time is 1.25 seconds, the force has this magnitude in the X direction. In the first item off this question, we have to better mind What is the value off this constant? A. For that, we can do the following. We know that F X is equals to eight times t squared. So we know that F X is 781.25. So F X is 708 1.25 and this is equals to a Times T squared. But T is 1.25. Therefore, we have 1.25 squared and this means that a is given by 708. 1.25 divided by 1.25 squared on. This results in 500 for the units we can do the following. We know that the expression tells us that something that has units off force is equals to the units off a times something that has units off time. In this case, we're measuring time in seconds and that thing is squared. So this is f equals to a T squared. But this is written in terms off its units. Therefore, we can better mind the units off a as follows. The units off a are given by Newton's by seconds squared on. These are the units off that number. So a is equals to 500 newtons per second squared. And this is the answer to the first item off this question. In the next item, we have to determine what is the impulse exerted by the engine. Starting from T equals to 41.5 seconds. Therefore, we want to deter mined. What is the impulse exerted from T equals two up two t equals to 3.5, which is two plus 1.5. The impulse is given by the following. It is the integral from the initial time 2.0 up to the final time 3.50 off the force which is given by eight times t squared the T. And of course, this is the impulse in the X direction since we are taking the integral off a force that is being exerted in the X direction, then that integral results in the following we can take that a out off the integral because it's a constant and then we integrate t squared d t. The integral FTI squared is t tow the third divided by three and we evaluated from 2.0 up to 3.50. This results in 500 which is a divided by three, which we can factor out times 3.52 the third minus 2.0 to the third and this results in 5812.5. Note that we have to approximate the result 23 significant digits as this is the smallest number off significant figures involved in these calculation. By doing that, we approximate this result to 5.81 times 10 to the third. On the units are units off impulse which are the same as units off momentum. So kilograms meters per second on this is the answer to the second item In the third item, we have to dr mind What is the change in the velocity that is produced by that impulse. So we know that the momentum is given by the mass times the velocity therefore a change in the momentum is given by the final momentum, minus the initial momentum. And this is the mass times the final velocity minus the mass times the initial velocity where I'm neglecting any change in the mass off that rocket. This results in Delta P being equals to the mass times the variation in the velocity, the reform Ah, variation in the velocity is given by the variation of the momentum divided by the mass. But we know that the variation of the momentum is equal to the impulse. So the variation in the velocity is the impulse divided by the mass. Now we plug in the values that we have. The impulse is 5.81 times, 10 to the third on the mass. Off the rocket is 2150 kg. On This results in approximately 2.70 m per second. Most that I approximated to three significant figures because we only had three significant figures for the impulse on. This is the answer to this question.

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