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Problem 21 Easy Difficulty

A hunter on a frozen, essentially frictionless pond uses a rifle that shoots $4.20-\mathrm{g}$ bullets at 965 $\mathrm{m} / \mathrm{s}$ . The mass of the hunter (including his gun) is 72.5 $\mathrm{kg}$ , and the hunter holds tight to the gun after firing it. Find the recoil velocity of the hunter if he fires the
rifle (a) horizontally and $(b)$ at $56.0^{\circ}$ above the horizontal.

Answer

(a) $v_{A 2 x}=0.056 \mathrm{m} / \mathrm{s}$ in the opposite direction of the bullet
(b) $v_{A 2 x}=0.0313 \mathrm{m} / \mathrm{s}$ in the opposite direction to the horizontal component of th

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Video Transcript

{'transcript': "it's a problem. 8.25. We have a hunter on a frozen pond that is basically frictionless right there. You have to wonder why the hunter isn't wearing shoes that have some sort of traction with the frozen pond. Easy. But perhaps he moves himself around by firing his gun, which just raises more questions. But anyway, let's take the problem. Face value s so he told his mass. The mass of a bullet and speed at which the bullet leaves his gun. He's holding the gun tightly. So he had the gun. We'll move together. Um, and their first asked, What is his velocity from the recoil from firing the bullet when he's holding the gun horizontally? And then and he's holding it a 56 degree angle. So now, in both of these cases, the horizontal component of the momentum is going to be concerned. So the initial momentum zero, of course. And so his momentum backwards has to be equal to the horizontal component of bullets. Momentum going forwards, which horizontal is the entire momentum in here like to do a little bit trigonometry toe, Get the cracked, uh, components. The reason that only the horizontal is conserved is that the the ice has a normal force that acts on the system, so it causes the momentum to not be conserved. If you're only considering the hunter of the bullet, If you're considering the hunter and the bullet and the ice and then the rest of the earth below that, then the momentum would still be conserved. But you'd still be doing this where you have to, you know, take into account all of that. You could find out how much you know what the what the speed of the hunter and the rest of the earth downward are. But yeah, it's gonna be quite small because they're very fast. So anyway, let's get down to it. Uh, like I said, the momentum of the hunter is going to be negative. The momentum of the bullet and the bullets momentum. Of course, it's mass 4.2 grams speed, 65 meters. So that for do what I should've done in the first place, which is, uh, actually solve this for the hundreds of Roscoe. So excuse that, please. So solving the hunters velocity, negative fines, the ratio of the masses and then this works out CEO point Bye meters. So if he is using is the request from his gun to get around the lake, he's He's not getting anywhere very fast, but I guess so. Here we have the same story, of course, but now we have to be careful only to take the X component of the velocity of the bullet and, of course, X component because of measuring the angle from the ex direction. Just the coastline of it's bloods. And don't know that this is this going to be this answer multiplied by the coastline of 56 which comes out to zero place here."}