For the circuit and motor problem it seems that you could also check the answer that the resistor should be placed in parallel with the motor, which would accomplish the same goal as described.
I actually disagree with this, but I may not be right. If a resistor is in parallel with the motor, it is true that equivalence resistance decreases which would increase equivalence current. However, I believe that the current in the path with the motor will decrease despite this because a new alternative path would be opened, so more current would go through that path. This would decease power in the motor because P=I^2R, which would increase ∆t, not decrease it. Another way to think about this is that the voltage across the motor is equal to the voltage across the battery minus the voltage across R1. If the new resistor is put in parallel with the motor, the equivalence current would increase as said before. This would increase the voltage across R1 because the current across R1 would increase as well, so voltage across R1 would increase because V=IR. If the voltage across R1 increase, the voltage across the motor would decrease. Because P=V^2/R, this would decrease power and increase ∆t.
Whereisamikz as the block moves up its potential energy increases, so the final value is going to be greater than the initial value, and delta U should be positive. If you are familiar with there being a minus sign there I’m guessing that’s another convention that ultimately will probably not influence the amount of points earned, but from what I know the energy is increasing so it makes sense that delta E is positive. If a minus sign is there it is probably a convention for talking about the “energy given to the motor” vs “energy dissipated by the motor” which I don’t think is what that question is testing. If you put -MgH you would probably still earn credit for the correct equation.
I am still sort of lost when block 1 hits block 2 yet the velocity stays the same. Shouldn't it be an inelastic collision where m1v1+m2v2=(m1+m2)vf. m2v2=0, so techincally m1v1=(m1+m2)vf.
Lohit Murali Nice are you prepping for this year’s online exam too? The reason the velocity stays the same is because you are looking in perspective with the center of mass for the two block system. The definition states that the velocity for the center of mass only changes if there is an external force applied on the system. However, in this case, the two blocks colliding is considered an internal interaction, which does not change the position nor velocity for the center of mass of the two-block system.
I don't think so -- I believe that is going to be a very common mistake on this question (I made the same mistake in fact) -- since the problem asks us to graph the velocity of the center of mass of the two block system the entire time, it only changes due to an external force like from the plunger or the force from the friction. Since only the first block is moving initially the graph seems to be a graph of its velocity alone, which is misleading and not true. This is inherent in the idea that the velocity of the center of mass of a system is constant unless an external force acts on that system. An external force does not act on the system when block 1 collides with block 2. Does that make sense?
What about this way of thinking of it... The center of mass was actually moving that slower velocity before the collision. I mean to say that the velocity of the center of mass was already half the velocity of m1 when m1 approached the resting m2 during segment DE. I think it helps to actually calculate the velocity of the center of mass to see that it doesn't change. formula of v center of mass= (m1v1 +m2v2)/(m1+m2) written for two objects (m1vi1 + m2 vi2)/ (m1+m2) before collision (mvi+0)/(2m) vi/2 (m1vf +m2vf)/(m1+m2) after collision and both vf are same (2mvf/2m) vf (obviously) you can get vf of the two blocks=(vi/2) from balancing momentum for the inelastic collision. m1(vi) +m2(0)= (m1+m2)vf mvi/(m+m)=vf (vi/2)=vf Although, you're not supposed to have to calculate it for the AP physics 1 test, it helps to do it to see what's going on. Otherwise, internal forces to the system don't change the velocity center of mass of the system is the way to solve it.
Charlie Koepp I think that you could have STARTED the problem by talking about Hooke’s law, but ultimately it doesn’t work because in the next part it specifically says you have to launch the ball. I can’t think of a way to only use Hooke’s law while also launching the ball. Let me know if you think of a way of doing this that fits with the subsequent parts of the problem.
I think you may get points off if you used kinematics because it only works with constant acceleration. Springs don't provide constant acceleration. If you however measured initial acceleration directly rather than final velocity, you should be fine I think.
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this was incredibly helpful, thanks
Thank you so much this was very helpful
Top tier educational content🤩
For the circuit and motor problem it seems that you could also check the answer that the resistor should be placed in parallel with the motor, which would accomplish the same goal as described.
I actually disagree with this, but I may not be right. If a resistor is in parallel with the motor, it is true that equivalence resistance decreases which would increase equivalence current. However, I believe that the current in the path with the motor will decrease despite this because a new alternative path would be opened, so more current would go through that path. This would decease power in the motor because P=I^2R, which would increase ∆t, not decrease it. Another way to think about this is that the voltage across the motor is equal to the voltage across the battery minus the voltage across R1. If the new resistor is put in parallel with the motor, the equivalence current would increase as said before. This would increase the voltage across R1 because the current across R1 would increase as well, so voltage across R1 would increase because V=IR. If the voltage across R1 increase, the voltage across the motor would decrease. Because P=V^2/R, this would decrease power and increase ∆t.
Good point! I think you’re right. I forgot that a lot of the current wouldn’t even be delivered to the motor.
bro this so helpful
@Crabtree I final gonna pass one of your tests !
at around 35:10 - Power does equal Delta E, however Delta E equals -delta U, Therefore negative Delta Ug, and -mgH.
Please correct me if I am wrong
Whereisamikz as the block moves up its potential energy increases, so the final value is going to be greater than the initial value, and delta U should be positive. If you are familiar with there being a minus sign there I’m guessing that’s another convention that ultimately will probably not influence the amount of points earned, but from what I know the energy is increasing so it makes sense that delta E is positive. If a minus sign is there it is probably a convention for talking about the “energy given to the motor” vs “energy dissipated by the motor” which I don’t think is what that question is testing. If you put -MgH you would probably still earn credit for the correct equation.
I am still sort of lost when block 1 hits block 2 yet the velocity stays the same. Shouldn't it be an inelastic collision where m1v1+m2v2=(m1+m2)vf. m2v2=0, so techincally m1v1=(m1+m2)vf.
Lohit Murali Nice are you prepping for this year’s online exam too? The reason the velocity stays the same is because you are looking in perspective with the center of mass for the two block system. The definition states that the velocity for the center of mass only changes if there is an external force applied on the system. However, in this case, the two blocks colliding is considered an internal interaction, which does not change the position nor velocity for the center of mass of the two-block system.
For problem 3 is it okay if I had Us=KE instead of Us=Ug?
Matthew Dwyer I think so, as long as you list a device for measuring the speed as the sphere exits the launcher
For the first one i thought the velocity of the center of mass would slow down as block 1 reaches block 2.. does anyone else see what im talking about
I don't think so -- I believe that is going to be a very common mistake on this question (I made the same mistake in fact) -- since the problem asks us to graph the velocity of the center of mass of the two block system the entire time, it only changes due to an external force like from the plunger or the force from the friction. Since only the first block is moving initially the graph seems to be a graph of its velocity alone, which is misleading and not true. This is inherent in the idea that the velocity of the center of mass of a system is constant unless an external force acts on that system. An external force does not act on the system when block 1 collides with block 2. Does that make sense?
What about this way of thinking of it...
The center of mass was actually moving that slower velocity before the collision. I mean to say that the velocity of the center of mass was already half the velocity of m1 when m1 approached the resting m2 during segment DE. I think it helps to actually calculate the velocity of the center of mass to see that it doesn't change.
formula of v center of mass= (m1v1 +m2v2)/(m1+m2) written for two objects
(m1vi1 + m2 vi2)/ (m1+m2) before collision
(mvi+0)/(2m)
vi/2
(m1vf +m2vf)/(m1+m2) after collision and both vf are same
(2mvf/2m)
vf (obviously)
you can get vf of the two blocks=(vi/2) from balancing momentum for the inelastic collision.
m1(vi) +m2(0)= (m1+m2)vf
mvi/(m+m)=vf
(vi/2)=vf
Although, you're not supposed to have to calculate it for the AP physics 1 test, it helps to do it to see what's going on. Otherwise, internal forces to the system don't change the velocity center of mass of the system is the way to solve it.
Could you have used hookes law for problem 3
Charlie Koepp I think that you could have STARTED the problem by talking about Hooke’s law, but ultimately it doesn’t work because in the next part it specifically says you have to launch the ball. I can’t think of a way to only use Hooke’s law while also launching the ball. Let me know if you think of a way of doing this that fits with the subsequent parts of the problem.
Azulon yeah that’s what I did
Azulon you know initial velocity is 0, so you can use a photo gate to determine the launch velocity and acceleration
I think you may get points off if you used kinematics because it only works with constant acceleration. Springs don't provide constant acceleration. If you however measured initial acceleration directly rather than final velocity, you should be fine I think.
Bob Jenkins let’s hope 😬 haha
i feel it would've been easier to do the experiment horizontal and converted Es to Ek right off the bat
but still a good video!
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