@LoopBack Yes. Potential energy is always relative to some reference point. It is only differences in PE which matter. You can choose whichever reference point is convenient and assign that point as zero or any other figure. It will make no difference to the result of your calculation. The only time it matters is when you are asked for the absolute potential (or potential energy).
@sammygerbill but this ain't true in my case, wait I'll write what I did
Let the spring potential energy as well as gravitational potential energy to be zero at equilibrium position. If a sudden blow is given to the block it's total mechanical energy just after the blow will be ½mv². When the block reach the extreme position the total mechanical energy will be ½kx²-mgx.
As no external force is acting total mechanical energy is conserved (here the spring and earth are considered as the system therefore gravitational force is not external force) equating both I'll get a quadratic in x. ½mv²=½kx²-mgx.
Next case: if I take potential energy of the spring to be zero in it's natural length and gravitational potential energy to be zero at equilibrium position. In equilibrium position mg=ky. Or y=mg/k. Just after a sharp blow total mechanical energy will be ½mv²+½ky²=½mv²+m²g²/2k. At the extreme position total mechanical energy will be ½k(mg/k+x)²-mgx. Equating both I will get ½mv²=½kx². So where have I gone wrong
@sammygerbil just the paperwork and big calculations. The question is how much kinetic energy of the particles is lost after collision. Assuming both had initial masses $m_1$ and $m_2$ and non zero velocities $v_1$ and $v_2$. And coefficient of restitution is $\epsilon$ .
I am doing mistakes in calculation.
The answer they are giving is $\frac{m_1 m_2(1-\epsilon^2)(v_1-v_2)^2}{2(m_1+ m_2)}$
@Nobodyrecognizeable if you don't mind I have moved this question for you. I can't type it here but I'll a snapshot of the solution. I have take initial speeds to be u1 and u2
And if you are thinking what is k in the first paper see this
And I meant to say I have solved this question not moved it(damn autocorrect)
@LoopBack Although elastic PE can be given the value of zero at the equilibrium position, it does not mean that the difference in elastic PE can be calculated using $\frac12 k x^2$ with $x$ as difference in position.
In this formula $x$ refers to extension from natural length.
This sounds confusing, but what you were intending to ask was not "Can I define elastic energy as zero at any point?" which is true, but "Can I use the formula $E=\frac12 kx^2$ using $x$ relative to any point?" which is not true.
Let's suppose I chose elastic pe at equilibrium position 0. So the elastic potential energy at the extreme should be working where x is extension from natural length
Which will be ½k(mg/y+x)²
So it will be same no matter what 0 reference I chose for elastic pe
Correction in the first statement --> should be ½kx²
@Abcd The formula you need to use for resultant intensity when waves of unequal intensity interfere is $I=I_1+I_2+2\sqrt{I_1}{I_2}\cos{\phi}$ where $\phi$ is phase difference.
@Abcd Not yet. I'm getting tired. I may have to leave this for John Rennie to look at. I guess he might be starting soon. I think I need to sleep a while, and I have something else I need to do when I get up.
@Abcd I couldn't sleep so I had another try and got it using the formula I quoted along with your value of $\phi=\frac23\pi$. If the incident intensity is $I_s$ then the intensities on the screen from each slit are $I_1=0.5I_s$ and $I_2=2I_s$ (slit 2 is twice as wide as slit 1). The maximum intensity on the screen is $I=4.5I_s$ while the intensity at O is $I_O=1.5I_s$. Therefore $I_O=\frac{1.5}{4.5}I=\frac13I$.
@Abcd Your answer of $-10.2eV$ is the difference between the ground state and 1st state energy which is $-3.4eV$. (I recall we had this value in a question before I went on my vacation.)
@Abcd I don't know why are you wasting you time writing such a big derivation. Total energy of hydrogen atom is -13.6/n² eV. And the potential energy by -27.2/n² eV. The question says potential energy is take as zero and not total energy.
Potential energy in the second excited state is -6.8eV so if this is zero means you will have to add 6.8eV to total energy at the ground state. So energy will become -6.8eV
Would someone please help me with this question? The answer is A-p, B-q, C-t, D-s. The only way I can think of is using the result pressure at any point in the box at a depth h fromupper surface and distance l from left surface is hpg + lpa. But how do I integrate this? For instance I can't divide the upper face into strips since pressure will vary throughout each strip. This is confusing me. Or is there another way to do this?
I if you want to solve the question within a minute you can follow my steps.
Now the answer is clear R3 should be ma+something so it will be t, which means R4 should be q. Similarly R1 should be mg+something which means R1 is s, so R2 should be p. These are the reaction forces the liquid. Liquid will exert equal and opposite force on the container
@Hema This ain't the proper solution but we have to solve questions quickly.
@Hema a simple answer reaction is a pushing force. You cannot lift a block just by placing your hand over it. You can can lift it only when you place your hand beneath it
@AvnishKabaj according to my understanding the net force on upper face would be P0A and on lower face would be (P0 + lpg) A where l is side or P0A+mg,is that right?
@LoopBack Find the range of the Projectile on the inclined plane which is projected perpendicular to the inclined plane with the velocity 20m/s Angle of incline = 37 deg
Quantum computing is computing using quantum-mechanical phenomena, such as superposition and entanglement. A quantum computer is a device that performs quantum computing. Such a computer is different from binary digital electronic computers based on transistors. Whereas common digital computing requires that the data be encoded into binary digits (bits), each of which is always in one of two definite states (0 or 1), quantum computation uses quantum bits or qubits, which can be in superpositions of states. A quantum Turing machine is a theoretical model of such a computer, and is also known as...
@AbhasKumarSinha Hey, AITS starting this week. Are you in FIITJEE?
@AbhasKumarSinha AITS JEE Advanced Physics Questions.......Diagrams are enough to scare you away from the question...We had mock tests of previous year papers and I experienced that ... At least thats the case in Mechanics.
I seem to have forgot this Consider Rutherford's experiment and alpha particle at the other end and some already very close to nucleus,, in which case will P.E be maximum and in which case K.E maximum? I am confused because of sign
I need to urgently know this please someone help..