I did a Mechanics A2 maths exam and I think I did fairly well on it.
After the exam all of my friends got together and we kind of discussed it. One particular question which they all didn't get but I did is bugging me now. They said that I did it wrong but any way I look at it, I'm right.
Thats why I want to give you the question and method of answering so you can decide wether I'm right or wrong...
First some values;
mass of P = 0.5 kg
Modulus of Elasticity = 20 N
Natural Length = 1.25 m
A light elastic string is connected to point O at one end and partice P to another.
When velocity of the particle = V;
i) Show that V^2= -32X^2+20X+25
ii) Find the maximum velocity
iii) Find the acceleration when the particle is at the lowest point.
i) For this question I just equate the energy at O right before the particle is dropped. This is only Potential energy because there isn't any extension or velocity when P is at O. This means mgh = (0.5)(10)(1.25+x).
Then I equate the energy at any point during the drop. Because this new position is taken as the lowest level, there is no Potential energy. Only Kinetic and Elastic Potential energy(if the particle dropped lower than 1.25 m).
This means that KE+EPE= 0.25V^2 + 8X^2. Initial energy = Final energy -> and the rest is all algebra.
ii) I assumed that the maximum velocity is at the equilibrium position.
I assume this because, when velocity is maximum, acceleration is zero.
And according to Newtons law; Forces with the motion - forces against the motion = mass x acceleration.
So this means that T=MG since when its moving, the resisting force is the tension and the driving force is the weight.
I just worked out the extension at the equilibrium position by substituting T = (Modulus of Elasticity x Extension)/Natural Length
That way I can find the extension at this point and just substitute it into the equation found earlier.
iii) This was done by equating energies like described before but now there isn't any kinetic energy since the particle is at the lowest point and therefore has stopped moving. I find the extension at this point by just equating energies again.
Then I use Newtons law, which in this case looks like this; T - MG = MA. Then I just find the accelleration.
This one question bothered me really badly so please tell me I did this right
...
My friend said that the maximum speed isn't at the equilibrium position so if he's right, this messed up most of the question since part ii had the most marks.
If I got this right, I'm pretty sure that I've aced it.
Btw, this isn't the final question. The final question was a simple question to do with motion under variable forces. Lots of integration/differentiation (though not a lot when compared to the pure) and you get a shit load of marks for only about 5 lines of work.
After the exam all of my friends got together and we kind of discussed it. One particular question which they all didn't get but I did is bugging me now. They said that I did it wrong but any way I look at it, I'm right.
Thats why I want to give you the question and method of answering so you can decide wether I'm right or wrong...
First some values;
mass of P = 0.5 kg
Modulus of Elasticity = 20 N
Natural Length = 1.25 m
A light elastic string is connected to point O at one end and partice P to another.
When velocity of the particle = V;
i) Show that V^2= -32X^2+20X+25
ii) Find the maximum velocity
iii) Find the acceleration when the particle is at the lowest point.
i) For this question I just equate the energy at O right before the particle is dropped. This is only Potential energy because there isn't any extension or velocity when P is at O. This means mgh = (0.5)(10)(1.25+x).
Then I equate the energy at any point during the drop. Because this new position is taken as the lowest level, there is no Potential energy. Only Kinetic and Elastic Potential energy(if the particle dropped lower than 1.25 m).
This means that KE+EPE= 0.25V^2 + 8X^2. Initial energy = Final energy -> and the rest is all algebra.
ii) I assumed that the maximum velocity is at the equilibrium position.
I assume this because, when velocity is maximum, acceleration is zero.
And according to Newtons law; Forces with the motion - forces against the motion = mass x acceleration.
So this means that T=MG since when its moving, the resisting force is the tension and the driving force is the weight.
I just worked out the extension at the equilibrium position by substituting T = (Modulus of Elasticity x Extension)/Natural Length
That way I can find the extension at this point and just substitute it into the equation found earlier.
iii) This was done by equating energies like described before but now there isn't any kinetic energy since the particle is at the lowest point and therefore has stopped moving. I find the extension at this point by just equating energies again.
Then I use Newtons law, which in this case looks like this; T - MG = MA. Then I just find the accelleration.
This one question bothered me really badly so please tell me I did this right
My friend said that the maximum speed isn't at the equilibrium position so if he's right, this messed up most of the question since part ii had the most marks.
If I got this right, I'm pretty sure that I've aced it.
Btw, this isn't the final question. The final question was a simple question to do with motion under variable forces. Lots of integration/differentiation (though not a lot when compared to the pure) and you get a shit load of marks for only about 5 lines of work.
