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Question

In the diagram above, a mass $m$ starting at point $A$ is projected with the same initial horizontal velocity $v_{0}$ along each of the three tracks shown here (with negligible friction) sufficient in each case to allow the mass to reach the end of the track at point $B .$ (Path 1 is directed up, path 2 is directed horizontal, and path 3 is directed down.) The masses remain in contact with the tracks throughout their motions. The displacement $A$ to $B$ is the same in each case, and the total path length of path 1 and 3 are equal. If $t_{1}, t_{2},$ and $t_{3}$ are the total travel times between $A$ and $B$ for paths $1,2,$ and $3,$ respectively, what is the relation among these times?
(A) $t_{3} < t_{2} < t_{1}$
(B) $t_{2} < t_{3} < t_{1}$
(C) $t_{2} < t_{1}=t_{3}$
(D) $t_{2}=t_{3} < t_{1}$

Vishal Gupta · Accepted Answer

Hi in the given problem there are three possible parts between two point E and be the first part curved upward, vertically, straight, horizontal. And the 3rd part garb like this girl vertically downward. An object is projected from point here and it has to reach up to POINT B. So we have to compare the time taken by this object to cover All these three parts No and long but one august to climb up against gravity. And then after reaching up to the maximum height then it will come down again to recheck point B. But its overall average velocity will reduce and long but two being horizontal. The velocity average velocity well, really insane while at last a long but three the object goes down first with gravity so its average speed average velocity will increase. Hence we conclude that we win means the average velocity along Part three. V three will be greatest. It will be more than that a long path to and it will be more than that a long path, one, So the time will be reversed. Hence, time taken by the particle along. Part three will be shortest, then intermediate along path to, and then maximum time will be taken along. But one so you can say here. Our option E. Is correct. Thank you.

Dading Chen · Answer

according to energy conservation are the initial total Energy Shwe go to really find out of the energy. So therefore, have such energy equation here we just the initial gravitational potential Energy is you go to Lee my Nokia Angie last e r no ah, gravitational potential Energy in the loop. So we know when Ah, a block star rotating in loop the gravity who you go to? The centrifugal force there will have m G&#x27;s is equal to Embry Square over our And as you can tell, mass m can cancel out And it took about three square is eulogy are so now he&#x27;s gr subsidy evil we square to have ah, MGH is a good 1/2 mgr uh, plus energy to our as you can tell, uh, mg Jharkand because all and it will give us the high h is you go to won&#x27;t have our plus two are, which is equally true if I over to our so for part B. So now let&#x27;s take a look at the situation and a partner of the loop will still apply the sand energy equation. But now the energy question is a study difference since at a bottle over a loop. The highs is as your point. So there&#x27;s no, uh, gravitational potential energy in a loop right now. Therefore, have the, um, energy equation is I m Jewish. You won&#x27;t have every square since Now the height is increased due to age. Remember ages ago? Fire which you are. So if we use file with who are substitute palatial have m g two times flowers Who are you with? The Why have every square as you can sell to? Can we cancel out mess semcken Because a lot and this will give us by grs Could do want happy score So therefore we square you see little tens er and the net force off the block in the loop should be go to the normal for us. Ah, minus the gravity off the block. And this is equal centrifugal force because let me draw that We&#x27;re here. So where blocking is an A button off the loop. Poison was right here The number of forces pointing upward directly to the center of the circle by the gravity He&#x27;s point down where? Okay, disease, this is grab penises number force And we know centrifugal force is always pulling toward the center. So therefore have m minus MGI is with memories Grow our and we know Um please, we&#x27;re can be good too. 10 z are therefore have m times tens er over our as you can see how are can because all and that&#x27;s why we have, um minus m g is the little 10 mg Eventually have the normal force on the block At bottom of the loop is you go to 11 MGI. So for Harbi, I&#x27;m sorry, posse should be Parsi year. So now the block is at the top over the loop. So this can be our energy question for this case, which is m g to H is equal to 1/2 and we score plus energy to our things. A nice l travel loop. That means, uh is at a maximum height in a loop, which is two times radius. So remember the highs issue is you go to ah, fire were to has are so we will have. Ah, And if i m g r is you don&#x27;t want happened. We square plus two mdrv substitute violent to our voyage. It was not run from here and eventually you have to be we square is you could do 60 are So now have the net force in a loop is you go to numerous past grab the things So this is the look So now the block is at the top of the look. As you can tell in the movie yours is pointing downward and a gravity is always point Our, therefore haven&#x27;t class M g is equal duties and tropical force which is MB square of our and when you re squares grew 60 are so if we used six years have stable, we square eventually we&#x27;ll have I&#x27;m past mg Zulu six energy. Therefore the normal force should be able to buy energy so for party. So now the block thief the loop enters the class action. So when the block enters the class action you know more force on the block Should we go to the gravity of the block? And these are the answers for this question

Abhishek Jana · Answer

so we can use conservation off energy to figure out the velocity when the block riches in the level ground. So in the top the block starts with dress, so that means there&#x27;s no velocity. Let&#x27;s call this point I our initial point. And when the block reaches at the bottom, let&#x27;s call that final position. So the total energy a financial position is basically kind idiot energy plus potential energy. Similarly, we have by kind of the engine final position. That&#x27;s potential energy. So the total energy is conserved? No, at the initial position, there is no velocity. So we can ignored the kind of energy and that final position. There is no potential energy because we&#x27;re setting our origin as this flat ground. So this is, you know, no, this height. Let&#x27;s call it eight. So that means, um, let&#x27;s call this height are so m times g times are should be equal. Do ah, half envies grade. And the reason for that is because ah, the height is basically the radius off this part of the circle. So from here we know velocity is equal to the square root off price gr so that&#x27;s two times 10 meters per second. Squared times are with his 0.8 meter, which gives us for meetings for a second. So option bees the correct choice. Thank you.

Farhanul Hasan · Answer

All right. We start with a free body diagram for the such for the case where the block is on the top of the curve loop. Um And so, uh, in this case, noble force and gravity of gravitational force a bar back in downwards. Eso you can equate this part es eso You confined the net radio force, which is normal force plus force of gravity. And that&#x27;s be called to n v squared over r centripetal force and V will be visa to meaning velocity of the top of the velocity of the top. Uh, and so for the case that the block shot is on the verge of falling off, the normal force will be zero. Giving us velocity of the top squared is just gr okay, then we use energy conservation initially. Ah, MGH is a gravitational potential energy that is converted to kinetic energy. One half em last year, the top squared plus potential idea, the top end you times the diameter of a circle or to our um And so, uh, we put plug in Vt Square and you get G h equals ah, 1/2 and she are plus two gr Ah, the G&#x27;s cancel and we&#x27;re left with H on left hand side and 2.5 are in the right hand side. So that&#x27;s the minimum release height. Ah, and part Devi. We&#x27;re looking at the second scenario where the block is on the bottom of the loop. And so what happens is that, uh, in this case, normal force acting upwards and gravitational forces acting downwards. And so the net radio force is normal force minus gravitational force. Is he called to centripetal force and B squared over r V be doing velocity of the bottom. All right, So, um and then and so you have that normal force. Is MQ plus M v d squared over r uh, And so next you do energy conservation and it starts initially, it&#x27;s that mg times twice the height. Right at the bottom. You have twice the r you&#x27;re starting a twice I on That&#x27;s equal to 1/2 times velocity of and Time&#x27;s lost a lawsuit of the bottom squared of EMS cancel. And so you&#x27;re left with V B squared is two times g times h. And remember h is 2.5 are So this is five. Oops. I should say V B squared us four. Um, do you, H senses five times two, uh, are four times 2.5. Meaning 10 times g r gets a lot of you b squared. Therefore, in here we have normal force is equal to mg plus m times tens. You are over our the arse. Cancel, um, arse cancel and you get, um, tandem G plus Angie saw normal forest in this case is 11 mg. Okay, Ben part. See, it&#x27;s identical. Depart A With the exception, that normal force is not zero. So, in this case, normal force is just, um m v t squared over r minus mg. Therefore, hams, mom and and then we use energy conservation mg. Times three are in the face, double equal to 1/2 the t squared, right? Uh, and so the EMS cancel, so luscious v t squared equals six g r. Therefore, normal force is ah m times six g r. Those are minus and cheese. Six minus warning calls. Five, haven&#x27;t she, um, And finally, in party, you have zero centripetal force on the flat part, Uh, meaning that this term goes arrest. So the magnitude of normal forces, just you and that&#x27;s it.

Averell Hause · Answer

So this would be the free body diagram for the system that we have here and for we&#x27;re gonna apply Newton&#x27;s second law to both masses. So fair, massive A we have The sum, of course, is in the ex direction. This is going to be equal to force tension minus m sub A g sign of fate US of a. Uh this is equaling M sub eight times A for the master B. We have the sum of forces in the X direction equaling AMSA be times g times sign of fate us that be this is gonna be minus force tension. This will equal m sub b times a And now we&#x27;re gonna add these to a crate to equations Together we can say forced tension minus amps of a G sign of data sub A plus. I&#x27;m so be G sign of status of B minus force of tension. This would equal and some eight times a plus. M c b times a force. Tension, of course, cancels out and we find that the acceleration is gonna be equal to M C B time. Sign of status of B minus mass of a time sign of fate us up A This would all be divided by the total mess. So this would be our answer for part A Now, for the, uh except for part B for the acceleration to be at rest acceleration. So at rest this means that the acceleration must equal zero meters per second squared. Ah, therefore the acceleration would equal. I&#x27;m so be again. Sign of fate us A B minus mass of a sign of fate us of a divided by massive A plus massive p and I forgot for this part. Um, there&#x27;s a factor of G. My apologies Just realized so. So this entire term would be times the acceleration due to gravity. Okay, so here and there will be times G. This would equal zero at this point if this equal zero This means that Massa be time sign of status that be I would be equal two massive a time sign of fate s of a therefore massive B would be equal to massive eh Times Sign of fate us of a divided by sign of fate a sub B This is gonna equal 5.0 kilograms. Uh, most applied by sign of 30 two degrees divided by sign of 23 degrees, and this is equaling 6.8 kilograms. So if the massive block A is equal in five kilograms and the massive block B&#x27;s equaling 6.8 kilograms, the system will be at rest and acceleration would be equal to zero meters per second. If we wanted to find attention here, we can say that for the massive, eh? The force tension minus mass of a times G sign of status of A. This is equaling zero. And so the force tension would simply be equal to the massive A 5.0 kilograms multiplied by the acceleration due to gravity 9.80 meters per second squared times sign of 32 degrees and we find that the forest tension is gonna be 26 Newton&#x27;s if the if the system is at rest and four part the sorry poor for for part c uh, just like in part be, the acceleration will be zero for a constant velocity. So we can say for constant velocity, the acceleration is again zero meters per second squared. Ah, therefore the ah, the same exact relationship applies. However, we can say that the massive a divided by the massive be then equals G sign of Fada, so be divided by G sign of fate. US of a, the G&#x27;s cancel out and we have sign of 23 degrees divided by sign of 32 degrees. This is equal in 0.74 So that would be your answer for part C. That is the end of the solution. Thank you for watching.

Robert Daine · Answer

and we have three cuts moving, colliding and then moving together. No, we have two cars moving to the right on one car, moving to the left. Our situation is on an elastic collision. So the equation for an elastic collision will bay and one V one plus to be to plus and three V three. No, because they stick. We just have to add the three masses together, multiplied by a final common speed. From here, it&#x27;s just a case of plugging in the numbers and figuring out what our final speed has got. A big so four times five plus 10 times three plus three times negative for because it&#x27;s moving to the left, we have to make sure we include that negative. Yeah, and then the three matters together for plus 10 plus three. Multiply it by the final speech. So four times five is 20 10 times three is 30 three times four is 12 on its months because of that negative, and that is equal to 17 via so dividing both sides. Now, by 17 we can see that the is equal to 38 over 17 which is 2.2 full. Meat is per second. Now, our second part of the question asked those what happens if they collide in a different order? Well, it actually doesn&#x27;t matter what order they collide in because momentum is indiscriminate, won&#x27;t save all collided. They will be moving at the same speed and in the same direction regardless. So I will just right here. Collision order does not matter.

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An object of mass $m_{1}$ experiences a linear, e…

Problem 23 Easy Difficulty

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