Problem statement: 
Model a passenger aircraft as spring-mass-damper system and find suitable values of spring constant and damping ratio for the system. 

Theory:
An aircraft can be taken as a lumped mass attached to a spring and a damper. If the aircraft has a mass `m`, net spring constant of suspension system be `k` and net damping coefficients of the damper system is `C`, the damping ratio can be defined as:

`\xi = \frac(C) (C_c)`

`C` = Actual damping
`C_c` = Critical damping

`C_c` can be defined as:

`C_c = 2\sqrt(km)`

Thus in terms of `\xi`, `C` can be written as:

`C = 2\xi\sqrt(km)`

The benefit of using this conversion is that `\xi` is a better known term in damping systems rather than `C`.

Thus differential equation of the aircraft system can be written as:


`m\ddotu + 2\xi\sqrt(km)\dotu + ku = 0`

Maximum landing weight of a passenger aircraft is around 260,000 kg and maximum take off weight is around 360,000 kg.
Assuming worst case scenario when aircraft needs to land right after take off, we have taken 360,000 kg as the mass for all calculations.

Approximate selection of spring constant value:
A typical race car weighs 1200 kg and net spring constant of the suspension system is around 3600 N/m.
Scaling this up for the aircraft, we can select spring constant values up to 900,000 N/m. 

Approximate selection of damping ratio value:
Datasheets suggest damping ratio value of 0.3 for auto shock absorbers.

Initial conditions of a plane landing:
In case of standard landing, the vertical speed of aircraft is 1 m/s.

Initially, just before impact, the spring will be in relaxed state and thus initial displacement will be equal to undamped amplitude for the above case. Solving the differential equation with zero damping and initial condition `\dotu_0 = 1`, we get maximum displacement to be around 0.6 m.

Thus initial conditions can be selected as `u_0 = 0.6` and `\dotu_0 = -1`.
Negative initial velocity means the mass will be pushed in opposite direction of the initial displacement, at the time of the impact.

Results:
1) For `\xi = 0.3`

2) For `\xi = 0.6`

3) For `\xi = 0.9`

4) For `\xi = 0.999`

5) For `\xi = 1.2`

Magnetic field due to straight current carrying wire

Biot-Savart Law is the basic formula used to calculate magnetic field due to different kinds of current configurations. Consider an arbitrary current carrying wire. Take a small portion of the wire having length `\vec(dL)` and a point of interest where you need to calculate the magnetic field due to this small section of wire. Assume the radius vector to the point of interest be `\vecr`.

Then, according to Biot-Savart Law, the magnetic field at that point will be given as:

If you want to calculate the magnetic field due to complete wire, then integrate this expression over whole length.

Let's consider a very common case of finite wire as shown below and try to apply B-S Law to calculate the field at point P:

Consider a small piece of wire `\vec(dL)` at a distance `\vecr` from the point of interest. According to the geometry shown above, the small magnetic field created by this small section of wire will be:

Using the geometry of the problem, we can write:

Thus initial magnetic field expression become:

Integrating this expression, we get:

Length of the wire can be calculated as:

Now, consider a special case when the wire is of infinite length. In that case:

Right hand rule

One ring to rule them all, One ring to find them;

One ring to bring them all and in the darkness bind them.

If a ring can do that, why a hand can't?

Of course, it can!

In physics, different books are full of a lot of hand rules which I personally find rather confusing and difficult to remember. Some of them are:

1) Fleming's Left Hand Rule - If a current carrying wire is placed in an external magnetic field, the wire experiences a force perpendicular both to the field and to the direction of current. The direction of force can be given by Fleming's left hand as illustrated below:

Fleming's Left Hand Rule

2) Fleming's Right Hand Rule - If a wire is moving in an external magnetic field, there is a current induced in the wire, direction of which can be given by Fleming's right hand rule as illustrated below:

Fleming's Right Hand Rule

These are two rules specific to two cases. Although they can be applied to few other scenarios as well, it's always confusing which hand to use in which situation. To get rid of this, I'll be explaining a simple method "Right Hand Rule" where you don't need to remember such complicated convention. In this method, you need to keep only one thing in mind - "Always use a right hand! If you don't have a right hand, borrow from someone else but always use a right hand only!"

Let's start with vector cross products. Example `\vec\tau = \vecR X \vecF`

Let's use our right hand to determine the direction of Torque when direction of Force and Radius vector are known. 

Step 1) Note down the first element of the cross product. Here it's `\vecR`. Place the fingers of your right hand along this vector such that your little finger lies on the first vector.

Step 2) Now, try to curl the fingers towards the second vector i.e. `\vecF`. Keep in mind that you are not defying the laws of nature while doing so. Your fingers should always curl towards a closed palm and not in the reverse direction. If other vector is on the back side of your palm, flip your hand and place the fingers on `\vecR` in such a manner that your index finger lies on it. Now you should be able to close the fingers without breaking them.

Step 3) Note down the angle your fingers moved. Ideally it should be between 0 and 180 degrees if you followed the steps correctly. Even if you get something larger than 180 degrees, the expression will take care of it so you don't need to worry.

Step 4) Once you are done curling your fingers, your thumb is pointing in the direction of the torque and it's magnitude is given by `\tau = RF\sin\theta` where `\theta` is the angle you got in step 3. If you got an angle larger than 180 degrees, this expression will be negative and thus the direction of torque will be in direction opposite to your thumb. Simple as that!

Now, let's see if we can follow the same procedure and get the results of Fleming's left hand rule:

Motion will be in the direction of force and the force is given by: `\vecF = I\vecLX\vecB` which is essentially a vector cross product. 

Keep your hand in the direction of current (because that is the direction of `\vecL`) i.e. along the middle finger of the first figure. Now curl the fingers towards the magnetic field which points along the index finger of the first figure. You will find that your thumb is pointing in the same direction as given by Fleming's left hand rule.

For the Fleming's right hand rule, we need to find the direction of current. To do this, let's find out where a positive charge placed inside the conductor would go (because that's the direction of the current). The positive charge will move in the direction of magnetic force which is given by: `\vecF = q\vecv X \vecB`. Velocity of the positive charge `\vecv` points in the direction of motion i.e. thumb in the second figure. Magnetic field points in the direction of index finger and thus when you curl your fingers, your thumb will be pointing in the direction of middle finger i.e. the current. Try it yourself.

Same concept can be used to tell the direction of magnetic field produced by a straight wire as well as a curved wire. Magnetic field due to a small element of wire (which is essentially straight) is given by:

 `\vecB = (i\vec(dL)   X   \vecr)/r^3`

This expression along with the concepts learnt above can determine the direction of magnetic field. However, direction of field can be easily decided as per the following method:

1) For a straight current carrying conductor, place your thumb along the wire in the direction of current. Your fingers will curl around the wire and will give the direction of magnetic field.

2) For a curved current carrying conductor such as a loop, place your fingers along the current (because you simply can't place a thumb along a curved line). your thumb will point in the direction of magnetic field.

Try it out a few times and you will never get it wrong. This process is so simple that you will have to try hard to get it wrong.

Angular Displacement - Demystified

There has been a long debate on various physics forums and pages whether "Angular Displacement" is a scalar or a vector.
We all know that angular displacement represents the rotational effects of a body and thus we can say that it's an axial vector direction of which can be determined using right hand principle. However, the trouble comes when we try to perform vector addition on two angular displacements. Refer to the figure given below:

The given object (a book) has been rotated 90 degrees about horizontal and vertical axes. It's clear that final configuration is different depending upon whether it was rotated about horizontal axis or vertical axis in the first step. Thus addition of the two angular displacements is not commutative i.e. `A+B!=B+A`. However, two vectors always follow commutative law of addition. Thus in this case we conclude that angular momentum is not a vector quantity.

Now consider a similar situation but instead of 90 degrees, the book is rotated only 45 degrees in a similar fashion as above:

Here you can see that although the addition is not commutative, but still they are coming closer to it. Now, consider the third case with 10 degree rotation:

In this case, you can't tell the difference with your naked eyes whether the addition is commutative or not. The final position of both books is almost exactly same in both the case. This can be interpreted as follows:

No doubt, angular displacements are not vector quantities when they are large or strictly speaking, finite. However as they approach infinitesimally small values, they do show vector-behavior and follow commutative law of vector addition. Thus  we can conclude that angular displacement is a vector if infinitesimally small and not a vector if finitely large.

Elastic Collision in One Dimension

Consider two bodies moving initially along the line joining their centers as shown in the figure below:

Assume initial direction of motion to be positive. Also assume that `u_1 > u_2` (so that collision may take place). Conservation of momentum gives:

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`     `...eq^n (i)`

`=> m_1(u_1-v_1)=m_2(v_2-u_2)`     `...eq^n (ii)`

In case of perfectly elastic collision, total kinetic energy also remains conserved i.e. KE will be same before and after the collision:

`\frac1 2 m_1u_1^2+\frac1 2m_2u_2^2=\frac1 2 m_1v_1^2+\frac1 2 m_2v_2^2`     `...eq^n (iii)`

`=> m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)`     `...eq^n (iv)`

Dividing `eq^n(iv)` by `eq^n(ii)`, we get:

`u_1+v_1=u_2+v_2`     `...eq^n (v)`

`=> (u_1-u_2)=(v_1-v_2)`     `...eq^n (vi)`

Thus, in 1-D perfectly elastic collision, "velocity of approach" before collision is equal to the "velocity of recession" after collision.

Now, let's multiply `eq^n (vi)` by `m_2` and subtract it from `eq^n (ii)`:

`(m_1-m_2)u_1+2m_2u_2=(m_1+m_2)v_1`     `...eq^n (vii)`

`v_1 = (\frac(m_1-m_2) (m_1+m_2)) u_1 + (\frac(2m_2) (m_1+m_2)) u_2`     `...eq^n (viii)`

Similarly, multiplying `eq^n(vi)` by `m_1` and adding it to `eq^n(i)`:

`2m_1u_1+(m_2-m_1)u_2=(m_1+m_2)v_2`     `...eq^n (ix)`

`v_2 = (\frac(2m_1) (m_1+m_2)) u_1 + (\frac(m_2-m_1) (m_1+m_2)) u_2`     `...eq^n (x)`

This is the derivation of final speeds of both objects after collision. `eq^n(viii)` and `eq^n(x)` give the final speed of `m_1` and `m_2` respectively. Care should be taken before applying these equations that these are valid only in case of perfectly elastic collisions.