For Physics RN Chapter 3 quiz, the last two questions :
For 12 :
What is the acceleration of a 20-kg box if a force of 100 N is applied, as depicted in the ramp and pulley system shown?If there's a force of 100 being exerted, why would it double because of the two lines coming from the pulley? I was thinking that each of the two lines would do 50 N of work, but of course I'm thinking about it wrong, so what concept am I mixing up?
Also for 13:Two identical molecules, A and B, move with the same horizontal velocities but opposite vertical velocities. Which of the following statements may be true after the two molecules collide?
I got the question correct, but it was more that I eliminated B and D and then randomly chose an answer. I also don't really understand the explanation that the quiz solution gives. Can you help me out with why it's A?
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For question 12:
This is the explanation they use in the book:
We know that work is conserved. If that is the case, the work done on the bottom pulley (which is moving) is the same as the work done on the block. Since the pulley allows the string to length on both sides, it goes to reason that if the block moves, say, 2 meters, then the pulley moves down 1 meter (since each string will be 1 meter longer, generating the 2 meters the block moved). Setting up conservation of work:
W = Fd
If d in the strings is half of what it is for the block, F must twice in the strings of what it is in the block.
Thus, F on the block is 200 N.
The problem with assuming that each string is going to have 50 N of tension is that that implies translational equilibrium. If we have 100 N acting down on the pulley, and 100 N acting up, then the pulley will not be accelerating. However, we know that bottom pulley moves. Admittedly, they don't show that directly in the figure -- know that, on the MCAT, they would explicitly say that the bottom pulley is not fixed to avoid ambiguity.
For question 13:
Let's talk about this collision.
So, momentum has to be conserved. In this case, the particles have exactly the same momentum in the horizontal direction, but opposite in the vertical direction. What will happen when they collide? Well, these two objects are exactly the same mass, so after the collision, they'll have to continue traveling with the exact same horizontal velocity (so horizontal momentum is conserved), and they'll have exactly opposite vertical velocities (since the sum of the momenta in the vertical direction before the collision is 0, it will have to be 0 after also).
What happens with the energy? Well, if it is an elastic collision, then the kinetic energy will be conserved. If it is inelastic, then some kinetic energy will be lost. In other words, it's possible in this scenario (since they didn't tell us if it is or is not elastic) that the energy is either conserved or reduced after the collision.
Let's look at the answers...
A states that the kinetic energy will go down after the collision. Sounds good. As long as momentum is conserved (and it can be -- keep the horizontal component of velocity the same throughout, and just reduce the vertical velocity as long as it's opposite-but-equal-magnitude for the two molecules), energy can go down in an inelastic collision.
B is wrong because we know the overall vertical momentum is zero. The two molecules have equal masses, so their momenta after the collision will again have to cancel in the vertical direction.
C is wrong because it implies that energy is gained in the collision. Energy can only be conserved. In a collision, this means that the kinetic energy is conserved (if it's elastic), or some kinetic energy is lost (if it's inelastic), and the remainder of the energy in the inelastic collision is converted to heat, sound, and sometimes light (sparks). But never in a collision will you gain kinetic energy.
D is wrong because, like in B, we know that the vertical component of the velocities for the two molecules will have to be equal but opposite direction.
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For question 12:
This is the explanation they use in the book:
We know that work is conserved. If that is the case, the work done on the bottom pulley (which is moving) is the same as the work done on the block. Since the pulley allows the string to length on both sides, it goes to reason that if the block moves, say, 2 meters, then the pulley moves down 1 meter (since each string will be 1 meter longer, generating the 2 meters the block moved). Setting up conservation of work:
W = Fd
If d in the strings is half of what it is for the block, F must twice in the strings of what it is in the block.
Thus, F on the block is 200 N.
The problem with assuming that each string is going to have 50 N of tension is that that implies translational equilibrium. If we have 100 N acting down on the pulley, and 100 N acting up, then the pulley will not be accelerating. However, we know that bottom pulley moves. Admittedly, they don't show that directly in the figure -- know that, on the MCAT, they would explicitly say that the bottom pulley is not fixed to avoid ambiguity.
For question 13:
Let's talk about this collision.
So, momentum has to be conserved. In this case, the particles have exactly the same momentum in the horizontal direction, but opposite in the vertical direction. What will happen when they collide? Well, these two objects are exactly the same mass, so after the collision, they'll have to continue traveling with the exact same horizontal velocity (so horizontal momentum is conserved), and they'll have exactly opposite vertical velocities (since the sum of the momenta in the vertical direction before the collision is 0, it will have to be 0 after also).
What happens with the energy? Well, if it is an elastic collision, then the kinetic energy will be conserved. If it is inelastic, then some kinetic energy will be lost. In other words, it's possible in this scenario (since they didn't tell us if it is or is not elastic) that the energy is either conserved or reduced after the collision.
Let's look at the answers...
A states that the kinetic energy will go down after the collision. Sounds good. As long as momentum is conserved (and it can be -- keep the horizontal component of velocity the same throughout, and just reduce the vertical velocity as long as it's opposite-but-equal-magnitude for the two molecules), energy can go down in an inelastic collision.
B is wrong because we know the overall vertical momentum is zero. The two molecules have equal masses, so their momenta after the collision will again have to cancel in the vertical direction.
C is wrong because it implies that energy is gained in the collision. Energy can only be conserved. In a collision, this means that the kinetic energy is conserved (if it's elastic), or some kinetic energy is lost (if it's inelastic), and the remainder of the energy in the inelastic collision is converted to heat, sound, and sometimes light (sparks). But never in a collision will you gain kinetic energy.
D is wrong because, like in B, we know that the vertical component of the velocities for the two molecules will have to be equal but opposite direction.
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