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Topic: **Act it out problem solving****Question:**
*The figure is just a picture of a cannon firing a cannonball.
Assume that every projectile fired by the toy cannon shown above experiences a constant net force (F) along the entire length of the barrel. If a projectile of mass (m) leaves the barrel of the cannon with a speed (v), at what speed will a projectile of mass (2m) leave the barrel?
A) v/2
B) v/(2^1/2)
C) v
D) 2v
E) 4v
Could you please explain how you got the answer?
This is an SAT question and I don't know what level it is.
Technot: Why do both equations have the same amount of work done to them?

June 21, 2019 / By Jackie

This is work/energy problem. If you got a constant force on an object with a distance moved, you get work done right? Work, in SI units, is measured in jules. Likewise, energy is too. So what you have is a conservation of energy problem with a mass that had no initial velocity, a force appplied, then a mass with a velocity once it leaves the barrel with no additional force acting on it. Well, there is always gravity, but that is not important in this question. To start, you have work done, W. This work is equal to the kenitic energy of the mass when the mass leaves the barrel, so W=1/2m(v^2) Since you have the same work done on the larger mass, (2m), the second equation is W= 1/2(2m)(v^2) Now since these to equations both equal the same work, set them equal to eachother and solve for the second velocity in terms of the first. To show the math, I will represent the velocity of the smaller mass with the letter Q. 1/2m(Q^2)=1/2(2m)(v^2) the "halves" and the "m's" cancle out Q^2 = 2(v^2) Divide both sides by 2 (Q^2)/2 = v^2 Now that the square root of both sides sqrt[(Q^2)/2] = v or Q/(2^(1/2)) = v answer B Additional info: The reason why both cannonballs have the same amount of work done on them is because work only depends on is the force applied and the distance traveled of a object. Since the force is constant for any size projectile, as stated in the question, and the distance that the force is applied to the object, a.k.a the lenght of the barrel of the cannon, is constant, the work done will be the same. Work = Force x Distance ("Mass" is not in the equation) Since both the cannonballs have the same amount of work applied to them, they will have the same kinetic energy once they leave the barrel. The difference between them is their velocities. Think about it, if you put the same amount of work, or energy, into throwing two different sized objects, the one with the smaller mass will go faster right? Think about the difference between a baseball and a bowling ball.

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Did you like the answer? We found more questions related to the topic: **Act it out problem solving**

This is work/energy problem. If you got a constant force on an object with a distance moved, you get work done right? Work, in SI units, is measured in jules. Likewise, energy is too. So what you have is a conservation of energy problem with a mass that had no initial velocity, a force appplied, then a mass with a velocity once it leaves the barrel with no additional force acting on it. Well, there is always gravity, but that is not important in this question. To start, you have work done, W. This work is equal to the kenitic energy of the mass when the mass leaves the barrel, so W=1/2m(v^2) Since you have the same work done on the larger mass, (2m), the second equation is W= 1/2(2m)(v^2) Now since these to equations both equal the same work, set them equal to eachother and solve for the second velocity in terms of the first. To show the math, I will represent the velocity of the smaller mass with the letter Q. 1/2m(Q^2)=1/2(2m)(v^2) the "halves" and the "m's" cancle out Q^2 = 2(v^2) Divide both sides by 2 (Q^2)/2 = v^2 Now that the square root of both sides sqrt[(Q^2)/2] = v or Q/(2^(1/2)) = v answer B Additional info: The reason why both cannonballs have the same amount of work done on them is because work only depends on is the force applied and the distance traveled of a object. Since the force is constant for any size projectile, as stated in the question, and the distance that the force is applied to the object, a.k.a the lenght of the barrel of the cannon, is constant, the work done will be the same. Work = Force x Distance ("Mass" is not in the equation) Since both the cannonballs have the same amount of work applied to them, they will have the same kinetic energy once they leave the barrel. The difference between them is their velocities. Think about it, if you put the same amount of work, or energy, into throwing two different sized objects, the one with the smaller mass will go faster right? Think about the difference between a baseball and a bowling ball.

"How do you know what formula to use?" Sometimes you don't, but if the first thing you do is try to find the right formula, then GUARANTEED you are doing the problem wrong. Physics is not the process of looking up formulas in textbooks, nor is it plug-and-chug algebra. Every equation describes a *relationship among quantities*, for instance, F=ma tells you how force, mass, and acceleration are all interrelated. When you study physics, don't look at the equations. Listen to your professor and read the *text* in the *text*book. Visualize in your mind which quantities are related to which other quantities and make up sentences to explain them, as in "More force means more acceleration, but more mass means less acceleration." In a particular problem, first of all ALWAYS draw a picture, label everything and write down what you know. Then imagine where the relationships can get you from the starting point. DO NOT try to sit there and figure out every step in advance. Just use the relationships between quantities to clue you in to equations that you can use to obtain new quantities that are hopefully helpful, even if they don't seem immediately useful. Eventually you'll see your way through to the end of the problem.

👍 80 | 👎 5

Well you do not have a picture here but you left out a few details. How much is the constant force is apply to the canon ball You did not give the length of the barrel the diameter of the barrel without these factors it is hard to determine.

👍 75 | 👎 4

I'm am at the moment taking this sort of stuff. I just looked in my High School Physics book at as I see it it seems the answer is (D). The reason for this is because 2v= 2 x V = speed. So the mass times the velocity which equals the speed.

👍 70 | 👎 3

F = mA A = F/(m) but if the mass is 2m then A =F/(2m) So clearly the acceleration with the increased mass is 1/2A But v = at +v(0) since v(0) = 0 v = at but if the mass is 2m then v = (a/2)t The new velocity when the mass is 2m is 1/2 the velocity when the mass is m.

👍 65 | 👎 2

I did b) first and then a) b) vf^2 - vi^2 = 2ax At t1, the maximum height is .225 m At t2, the maximum height is .9 To find out its increasing rate, in other words, how to find the constant of proportionality, multuply what you originally found with a proportionality x = .9/.225 = 4. Exactly a) t = sqrt(2h/x) t1 = .21428 t2 = .4287 x = 2

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