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At time $t$ = 0 a 2150-kg rocket in outer space fires an engine that exerts an increasing force on it in the $+x$-direction. This force obeys the equation $F_x = At^2$, where $t$ is time, and has a magnitude of 781.25 N when $t$ = 1.25 s. (a) Find the SI value of the constant $A$, including its units. (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? Assume constant mass.

(a) $A=\frac{F_{x}}{t^{2}}=\frac{781.25 \mathrm{N}}{(1.25 \mathrm{s})^{2}}=500 \mathrm{N} / \mathrm{s}^{2}$(b) $J_{x}=\int_{t_{1}}^{t_{2}} A t^{2} d t=\frac{1}{3} A\left(t_{2}^{3}-t_{1}^{3}\right)=\frac{1}{3}\left(500 \mathrm{N} / \mathrm{s}^{2}\right)\left([3.50 \mathrm{s}]^{3}-[2.00 \mathrm{s}]^{3}\right)=5.81 \times 10^{3} \mathrm{N} \cdot \mathrm{s}$(c) $\Delta v_{x}=v_{2 x}-v_{1 x}=\frac{J_{x}}{m}=\frac{5.81 \times 10^{3} \mathrm{N} \cdot \mathrm{s}}{2150 \mathrm{kg}}=2.70 \mathrm{m} / \mathrm{s}$. The $x$ -component of the velocity of the rocketincreases by $2.70 \mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Section 1

Momentum and Impulse

Moment, Impulse, and Collisions

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problem. 8.11. We have a poorly drawn rocket in outer space with a massive 2150 kilograms. And at time T equals zero. It starts firing of an engine that produces a force that's proportional to time squared. We're told that the value of this force at one and 1/4 seconds after it starts firing is 781.25 moons. So we have several things were supposed to find, the first of which is what is the numerical value with units of A. So if we know that Oh, if we know T and F, then you know, we just solved this for a it's going to be 50. I didn't buy squared. So this is 700 81 a quarter Britons, divided by one and 1/4 seconds squared. And so this works out P 500. Newton's her second square. You can already tell what the units need to be. Uh, just this, you know, this has to have be have units of Newton's. This is gonna have units of seconds squared. So this needs to impudence of Newtons per second squared. So this this the unit's work out what you would hope And this is this is the number you get apart me is a little bit more interesting. It asks us to find what is the impulse that the this force produces? Does it exert on the rocket during the 1.5 2nd interval starting at two seconds after the T equals zero with the engine starts firing. So we know the J is equal to the after girl between two times of the force in this case to you, one is two seconds and he too will be three and 1/2 seconds. And so, putting in 80 spared here A doesn't end on times that comes out of the integral. That's medication Don't change. She's very t and just from Diggle, rules for polynomial is we know that the integral of T squared is 1 30 cubed. This has to be evaluated between are too times me and this ends up being 5008 110 Newton seconds, which is your recall. The impulse is also equal to the change in momentum. And so, you know, Newton seconds has the same units. If you expand out what Newton is a kilogram meters per second squared. It ends up having units of kilogram meters per second. So this is the same. It's the units for a moment now are at sea. What we want is to find what is the change in the rocket speed? Uh, because all of this is about the change in the momentum and everything. The actual initial speed of the rocket doesn't really matter. Before, there were no forces acting on it. So it was in, you know, an inertial reference frame. And, you know, you could look at it from whatever inertial reference for have you like and it wouldn't have changed anything, So I don't care what the speed was initially. Uh, admit that zero. It could have been whatever. So we know, Like I just said that the impulse is equal to the change in momentum. And now the problem says to assume that the mass, his constant of the rocket so it doesn't change from this 2150 kilograms, which is not how real world rockets work, but it doesn't make our lives a little bit. He's here here because now M doesn't change. And so this is going to become him Delta V and this is what we're looking for. So change in the speed is going to be equal to the change in momentum or or the impulse which you've already found divided by the mass which patrol. So you know, you have 5000 810 in seconds here, 2150 kilograms down here and this works out to be 2.7 meters per second. And you can see that even though it's, you know, a fair amount of force that this is exerting over quite a while. Um, because the mass of the rocket is so large, it doesn't speed up a lot in this one and 1/2 2nd interval, and that makes you know that makes intuitively.

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