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(II) The force on a bullet is given by the formula $F=\left[740-\left(2.3 \times 10^{5} \mathrm{s}^{-1}\right) t\right] \mathrm{N}$ over the time interval $t=0$ to $t=3.0 \times 10^{-3} \mathrm{s}$ . $(a)$ Plot a graph of $F$ versus $t$ for $t=0$ to $t=3.0 \mathrm{ms}$ . $(b)$ Use the graph to estimate the impulse given the bullet. (c) Determine the impulse by integration. (d) If the bullet achieves a speed of 260 $\mathrm{m} / \mathrm{s}$ as a result of this impulse, given to it in the barrel of a gun, what must the bullet's mass be? (e) What is the recoil speed of the 4.5 -kg gun?

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a) See graphb) 1.2 $\mathrm{N} \cdot \mathrm{s}$c) 1.2 $\mathrm{N} \cdot \mathrm{s}$d) 4.6 $\mathrm{g}$e) 0.26 $\mathrm{m} / \mathrm{s}$

Physics 101 Mechanics

Chapter 9

Linear Momentum

Motion Along a Straight Line

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

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of this problem comes, Wants to wants us to product crack. So given this equation for force in terms of time, we want a plot, a graph of force versus time and so on, reporting it from zero seconds from zero seconds. 2.3 seconds. So the simplest thing to do here will be to plug in F at equal zero and f T equals or plug in T equals zero for often than two equals. Ah, 20.3 for 1/2 and And get the four. Get those two extreme ends on. Then you notice that this is a linear relation S O. So we don't have any curvy, uh, many curves going on here. So there's ah constant this business constant slope from the zero point to this end point and you plot that there. So you don't really need thio. Plug in values every you know, every few steps or whatever. Just plot the importance and, uh, use linear relation anywhere you plotted that. Ah, and part B. We want to use graphical methods to estimate impulse. So impulses you'll remember is just f times down to tea. And so ah, you want in this portion. The area under the graph right on, then air. It's our area. Looking of the craft would be, um, approximated as area of a trapeze iam and so think of so the trapeze. Um, is this a four sided figure and whose area is given by 1/2? Uh um, 1/2 times the base. Uh uh, Or height or whatever. It's called height, Times a plus B. So, uh, this so, um, the sum of these of this side and that side, and so the trapeze, um, we're interested in here. Um, great. That's so, um, the trippy and the trapeze and we're interested in here will just be so J equals area will be 1/2 times height, which is, uh, basically, just ah, time, uh, at the end. So 0.3 seconds, Uh, times a plus B. And so that would be 7 40 here, here at 7 40 and then here to hear it's 50. So time 7 40 plus 50. Uh, Newton's okay. And so the area will be one point to Newton seconds. All right. Um, that's and that's my graphical method part. See, we use integration, and so, uh, here We integrate f over deep with respect to d t over the interval. T equals zero seconds to t V equals point. 03 seconds. Right. So, uh, we integrate 7 40 minus 2.3 times 10 to the five has to to respect to do t from zero equals zero to tickles, point or three. And so this ending the result of this integration a 7 40 t minus 2.2 25 divided by two. So 1.15 times 10 to the five times t to the one place once a T squared. Ah, and that is evaluated from T equals zero to t equals 00.0, or three. Ah, and what you get from that is 1.185 which is approximately one point to Newton seconds. Um, and this tells us that this was a more accurate method. But mum, bless The answers are approximately the same, and b and C okay, and part d, we want to find the massive those bullets so the impulse is equal to mass times change in velocity. So mass is just impulse over changing velocity on dso that's impulsive. One point to Newton seconds over. Uh, change in velocity is to 60 meters per second. And so we get 4.6 times 10 to the negative three kilograms. Um, which is also 4.6 grams for the bullets mess. And finally, in part E, we invoke momentum conservation. Soapy initial is equal to P final. Initially, you have no velocity for the bullet or the gun. So zero momentum. That is what the final momentum has to be. So that is a massive bullet. Tens mass of the velocity minus master the gun. Tons of velocity of the gun. Ah, and so? So that means that velocity of the gun is the momentum off the bullet over the mass of the gun. So that sums. So that's point. Oh Oh, for six times to 60 um, which will be 1.2 over a 4.5, giving us, um, 0.26 meters per second for the velocity of the guns and the

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