When I took the 2018 exam, part b of this question was the one problem that stumped me, so I've been looking forward to your video on this question. I was so focused on the oscillation that I didn't even think of using conservation of momentum. Once you use it to derive Vpq = (1/3)Vp, the final answer is very clear. Thank you!
I think a lot of students completely missed the collision. The problem basically makes you think it is a simple harmonic motion question, when it actually is not. Glad to know I helped clear that up for you!
The equations are not essential to a paragraph length response. Referencing the decrease in velocity, instead of the exact 1/3 value, would still meet the goals of the problem. This could be argued conceptually without any calculations (just referencing the relationships); this approach would make just using a keyboard for this year's exam significantly easier. With equations, you now have to cut and paste from an equation editor, or write it all out by hand and upload a picture. Still a great video!
My angle on this was that we need to show the kinetic energy of PQ is decreased from just P. Because the mass of PQ is greater than that of P, just showing the speed of PQ is less than the speed of P is not enough for showing that the kinetic energy of PQ is decreased. I know that is not quite what the scoring guidelines say, however, I do not always agree with the scoring guidelines. 😳
Great video Mr. P, thanks again. I was wondering if you mind covering FRQ #1 from the 2019 exam, I ask because that one is QQT and I was having some trouble understanding the conceptual ideas of Center Of Mass and how the speed of COM changes.
great video! Could this also work? I used conservation of momentum and found that the final velocity was one third the original. Then I use Conservation of energy, kinetic equal elastic potential, used the variables and solved for x or the amplitude.x=root mv^2/k. Thus I concluded the amplitude decrease because the velocity was less and was in the numerator of the root/equation. Is this okay
Great video! I have a question: can you tell me in what all equations do we only use the magnitudes of the vectors? Like in W = Fdcos(theta) , we only use the magnitudes of F and D. Similarly, do we only use the magnitudes of r and F in torque = rFsin(theta)
Any time we introduce the trig term in the formula for the dot product and cross product, we only use the magnitudes of the vectors. For the cross product, we need to assign a direction with what we get for the product of magnitudes, via the right hand rule. For the dot product, there is no direction, and all we care about is the sign and magnitude. If you want to use the vectors instead of the magnitudes of the vectors, you use the component-by-component calculations for the dot and cross product. For the dot product, multiply each pair of corresponding components, and add them up. Just like when you calculate the total on a grocery receipt, by multiplying the quantity by the unit cost for each line item, and then add them up to get the total. For the cross product, construct the 3x3 matrix with the determinant notation. Unit vectors of coordinate axes go in the top row, first vector goes in the second row, and second vector goes in the final row. Take the determinant of the 3x3 matrix. My preferred method is to multiply along the diagonals, and assign positive signs to the positive diagonals, and negative signs to the negative diagonals, and then add up the total of the 6 products. Down and to the right, are the positive diagonals. This works for 2x2 and 3x3 matrices, but a different method for calculating determinants is needed for 4x4 and greater. Fortunately for us, we only have a 3-D world and a cross product that is only defined in up to 3 dimensions.
Great video! Can you tell me when do we use the magnitudes of vectors in equations? Like we only use the magnitudes of F and d in W = Fdcos(theta). Similarly, do we use only the magnitude of F in torque = rFsin(theta), etc. In What all equations do we use only the magnitudes?
Whenever u can give direction by your common sense then u can write magnitude only and give the direction by your common sense rather than using cross product
Momentum is only conserved in the horizontal axis, because that's the only direction where external forces are negligible, due to the collision happening quickly, and in a time period that the spring force doesn't apply a significant impulse. In the vertical direction, which ultimately isn't of interest to us in this problem, the normal force will be as large as necessary to prevent the blocks from sinking through the surface, and as a result, it provides an opposing impulse to slow down the blocks to rest in the vertical axis. Ultimately, the momentum goes in to the Earth, which has significantly greater mass than the blocks, and doesn't even move by a distance of the size of one atom.
![ARGENTINA vs. BOLIVIA [6-0] | RESUMEN | ELIMINATORIAS SUDAMERICANAS | FECHA 10](http://i.ytimg.com/vi/C3wPtKNbAOA/mqdefault.jpg)
When I took the 2018 exam, part b of this question was the one problem that stumped me, so I've been looking forward to your video on this question. I was so focused on the oscillation that I didn't even think of using conservation of momentum. Once you use it to derive Vpq = (1/3)Vp, the final answer is very clear. Thank you!
I think a lot of students completely missed the collision. The problem basically makes you think it is a simple harmonic motion question, when it actually is not. Glad to know I helped clear that up for you!
The equations are not essential to a paragraph length response. Referencing the decrease in velocity, instead of the exact 1/3 value, would still meet the goals of the problem. This could be argued conceptually without any calculations (just referencing the relationships); this approach would make just using a keyboard for this year's exam significantly easier. With equations, you now have to cut and paste from an equation editor, or write it all out by hand and upload a picture.
Still a great video!
My angle on this was that we need to show the kinetic energy of PQ is decreased from just P. Because the mass of PQ is greater than that of P, just showing the speed of PQ is less than the speed of P is not enough for showing that the kinetic energy of PQ is decreased. I know that is not quite what the scoring guidelines say, however, I do not always agree with the scoring guidelines. 😳
Great video Mr. P, thanks again. I was wondering if you mind covering FRQ #1 from the 2019 exam, I ask because that one is QQT and I was having some trouble understanding the conceptual ideas of Center Of Mass and how the speed of COM changes.
Currently editing that. Like right now. Wish I could attach a picture... Probably post on Sunday or Monday.
great video! Could this also work? I used conservation of momentum and found that the final velocity was one third the original. Then I use Conservation of energy, kinetic equal elastic potential, used the variables and solved for x or the amplitude.x=root mv^2/k. Thus I concluded the amplitude decrease because the velocity was less and was in the numerator of the root/equation. Is this okay
Great video! I have a question: can you tell me in what all equations do we only use the magnitudes of the vectors? Like in W = Fdcos(theta) , we only use the magnitudes of F and D. Similarly, do we only use the magnitudes of r and F in torque = rFsin(theta)
Any time we introduce the trig term in the formula for the dot product and cross product, we only use the magnitudes of the vectors. For the cross product, we need to assign a direction with what we get for the product of magnitudes, via the right hand rule. For the dot product, there is no direction, and all we care about is the sign and magnitude.
If you want to use the vectors instead of the magnitudes of the vectors, you use the component-by-component calculations for the dot and cross product.
For the dot product, multiply each pair of corresponding components, and add them up. Just like when you calculate the total on a grocery receipt, by multiplying the quantity by the unit cost for each line item, and then add them up to get the total.
For the cross product, construct the 3x3 matrix with the determinant notation. Unit vectors of coordinate axes go in the top row, first vector goes in the second row, and second vector goes in the final row. Take the determinant of the 3x3 matrix. My preferred method is to multiply along the diagonals, and assign positive signs to the positive diagonals, and negative signs to the negative diagonals, and then add up the total of the 6 products. Down and to the right, are the positive diagonals. This works for 2x2 and 3x3 matrices, but a different method for calculating determinants is needed for 4x4 and greater. Fortunately for us, we only have a 3-D world and a cross product that is only defined in up to 3 dimensions.
Great video! Can you tell me when do we use the magnitudes of vectors in equations? Like we only use the magnitudes of F and d in W = Fdcos(theta). Similarly, do we use only the magnitude of F in torque = rFsin(theta), etc. In What all equations do we use only the magnitudes?
Whenever u can give direction by your common sense then u can write magnitude only and give the direction by your common sense rather than using cross product
Another fantastic video, came out really great! Thanks again!
Dude. Thanks for all our help. Your alacrity is amazing!
How Can momentum be conserved if the block q has some initial momentum downwards and after there is none?
Momentum is only conserved in the horizontal axis, because that's the only direction where external forces are negligible, due to the collision happening quickly, and in a time period that the spring force doesn't apply a significant impulse.
In the vertical direction, which ultimately isn't of interest to us in this problem, the normal force will be as large as necessary to prevent the blocks from sinking through the surface, and as a result, it provides an opposing impulse to slow down the blocks to rest in the vertical axis. Ultimately, the momentum goes in to the Earth, which has significantly greater mass than the blocks, and doesn't even move by a distance of the size of one atom.
TME is NOT conserved in inelastic collisions, so I believe that your foundation of part 2 was false.
Vaibhav I think they mean it's conserved after the collision, as in the energy at post-collision equilibrium and post-collision amplitude is the same