a-soldier-loads-a-10-mathrmkg-cannonball-into-a-300-kg-cannon-that-is-initially-at-rest-on-the-groun

Question

A soldier loads a 10 $\mathrm{kg}$ cannonball into a 300 kg cannon that is initially at rest on the ground. What is the recoil speed of the cannon if the cannonball is fired with a horizontal velocity of 300 $\mathrm{m} / \mathrm{s} ?$
(A) 5 $\mathrm{m} / \mathrm{s}$
(B) 7.5 $\mathrm{m} / \mathrm{s}$
(C) 10 $\mathrm{m} / \mathrm{s}$
(D) 15 $\mathrm{m} / \mathrm{s}$

Salamat Ali · Accepted Answer

the initial speed off again is the cannonball is there also? The initial momentum is zero with the final moment of off cannon is Mars off a cannon times Dave lost your cannon, plus Ah, the mass of a cannonball plus velocity ofa cannonball can move, then solving for not plus here. Sorry. So this is we lost pieces off Cannonball. Then we console for her E c b. So we see this accuse us A minus M. C. We see divided by myself again in both substituting the values which is a minus 10 does 300 300 divided by 300 gives us so they&#x27;re lost your cannonball minus 10 meters per second. So our option C is the right answer.

Shital Rijal · Answer

All right. So we have a cannon here of this cannon Fire off Fire a shell off, Frank. Easy. It angle off 20 degrees. So this is the direction off the shelf. And this is from angle off French degrees. No, this angle here he&#x27;s 20 degree, though, given that mass off the cam is 300 kissy started 2000 kissing on DDE. What we need to come late is the required velocity with the mask off. Still, when the massive still lives the cannon Somare&#x27;s upset is given, as you can kiss it on, it leaves diversity. We this is the drinks must be. And the manager is 480 meters per second. No, the question is, what is the required velocity off the cannon? Now, the question say is that it can record on Lee in the horizontal direction, right? So only their hearts into component off the moment. Um, off the shelf. That is the cost. 20. It would be used for the moment of all for the required off the cannon. So using the condemned person off a mere moment. Um, initially, the moment, um what zero? Because both the ghannam silver crest, then consider the required velocity off can be in D. C. And this is concerned. We are only looking at the hearts and the motion. Right? So this should be called to muscle cell the V s on That should be in our general direction so that its cost rent. Right. So this is a long time in the horizontal direction. No, What we need to calculate is selected, elected as we s because this is Mr Shell. So we need to cover the velocity off cannon. So that is a culture negative. Off um, s V s. The momentum of shell in the horizontal direction cost 20. Boom and see. So you have the veil of M S, d S, m C. And we know the value of cost 22 this chance out to be fuel point 26 me just for a second. So this is about the question. No, this will want to part B. What is the kind of big energy off? Come on, this is pretty easy, right? So energy off cannon is huh? Massive can sometimes let us see the Squire. So this turns out to be 7.66 Villagers. Similarly for our part. What happens to the vertical component of moment? Um, so this here? I&#x27;m sure you hear. So this is the vertical component of moment. Off the shell on this reacts downward. Strike to the vertical moment reacts to us the art. So what we got right is who are taking a moment is ultimately he blocked it. Two Who blocked ID two? The cannon. Sorry to the earth. To Canada through camp. All right, so we have answered the big question.

Eric Mockensturm · Answer

we have a can monitor on the back of a ship, and it fires a 50 kilogram cannonball in the horizontal direction at the speed of 150 meters per second. Um, the ship and the cannon have a combined mass of 40,000 kilograms, and I initially arrest what is the speed of the ship just after shooting a cannon and assumptions may. So we have again are massive our cannonball mess of our ship, the Velocity cannonball used canon. The initial mo mentum in the system is zero because Cannonball isn&#x27;t moving the final moment, Um, in the system is the momentum of the cannonball. Yeah, and then the momentum of the ship. So what that tells us is that we know the change in momentum, um, zero for their system as a whole, because there&#x27;s no other impulse acting on it. And we&#x27;re assuming that there is no force acting from the water, which clearly there&#x27;s going to be. But we have no idea what that is. No, um so if we have a neck impulsive zero than the change in momentum of the system is zero. And so we can solve for V two x two and that becomes out to be 0.1875 meters per second. Now again, we&#x27;re assuming that there is no other forces acting on the boat. Well, it fires cannon, which is really a very poor assumption because there&#x27;s obviously going to be water resisting motion of this of the boat.

Lisa Tarman · Answer

number 91. There&#x27;s a cannon that shooting a cannibal straight upward with the velocity of initial velocity of 46 meters per second. And then you&#x27;re standing 105 meters away. And the question ask How fast is the cannonball going when you hear it? Well, we just need to find how it want. It takes you to hear it. So for the sound vehicles do pretty we know the speed of sound is 3 43 The distance you are way is one of five we can find him which time that takes. So it takes tweets 306 seconds for you to hear it. So now for the cannonball that&#x27;s in free fall So we know the initial speed Waas 46 three. We want to know what the final spirit is. It&#x27;s in freefall, so acceleration will be negative. 9.8 and we know it has a time of 0.3 or six seconds that looks like the acceleration equation multiply. And then I&#x27;ll add 46 to both sides. I get that final velocity of that will be 43 meters per second

Joseph Petrullo · Answer

for this problem, we can use conservation of momentum to find the recoil velocity of rifle. The system is initially at rest, and so there&#x27;s zero total initial momentum. And finally, after the bullet is shot will have the some of the momentum of the rifle and the momentum of the bullet. So what we get when doing some algebra is that PR is simply equal to minus m bvb for m R. And when we plug in the values from the problem, this gives us minus 0.1 times 820 divided by 4.5. This gives us minus 1.8 meters per second. So this means that the velocity is to the left. The recall velocity will be opposite of the bullet velocity.

Suzanne W. · Answer

cannon. Located 60 m from the base of a vertical 25 m tall clip, Shoot a 15 kg shall, at 43° above the horizontal towards the cliff. What must the minimum muzzle velocity be for the shell to clear the top of the cliff And part b. The ground at the top of the cliff is level with the constant elevation of 25 m. Under the conditions of party, how far does the shell land past the edge of the cliff? Okay, so it&#x27;s a projectile, so positive X is going to be pointed up. Sorry, positive Y. Positive X is going to be towards the hill. So to clear the top of the hill, X minus X. Man is equal to you 60 m. The shell has to reach 25 m high. We know the acceleration in the X. direction is equal to zero since it&#x27;s falling down, the acceleration in the Y direction is negative G. So we also need the velocity in the X. Direction and the velocity in the Y direction. So this is we not co sign 43° and this is the not Sign of 43° for part A. We know that x minus X amount is equal to abusive, not X. T. So we know x minus x. Not. So this gives us T. is equal to 60 m over. You said not co sign of 43°. Looking at the vertical motion, why minus? Why not? Is visa not Y. T. Plus one half 80 squared. So this is negative G. She squared. So plugging in This time into this equation and knowing that why minus why not is 25 m we can solve for the only unknown which is the velocity. So V. Sub not. So we can say that this is Piece of not sign of 43° T. This 1/2 negative g. C squared, solving for abusive nights. Using Algebra, we get 32.6 meters per second, and now, knowing this velocity, we find that the time is 2.51 seconds. So for part B we have that velocity in the Y direction is the initial velocity Plus that other velocity. So, plugging in values, we get negative 2.4 meters per second, so it is traveling downwards.

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