Student Problems: 1. Below is the zonal momentum equation...

Student Problems: 1. Below is the zonal momentum equation that we will use to describe the mid-latitude, synoptic scale atmospheric motion: $\frac{Du}{Dt} = -\frac{1}{\rho}\frac{\partial p}{\partial x} + 2\Omega v \sin\phi$ a. Identify each of the three terms in the equation. b. Using this equation, list all processes that affect 'local' time changes of zonal wind.

Transcript

00:01 Hello students, in this question a data table is given in this four systems are there that is magnet low mass, magnet high mass, spring and clay where the initial speed and final speed are given, impulse and the change in momentum is given and also the percentage difference is given in the table.

00:19 For the first part of the question it is asked to compare the values of impulse compared to the change in momentum and we have to determine what does this implies about the equation 6 from the theory.

00:42 We know that the impulse which is represented by delta it is represented by j and this is the product of force and time and this will be equal to rate of change of momentum that is change in momentum delta p because the force can be written by rate of change of momentum that is delta p by delta t multiplied by t so here t and t will get cancelled and as a result we get impulse is equal to change in momentum delta p.

01:28 So first let's check the values of impulse and the change in momentum here it is 0 .382 this is 0 .379 and for second case it is 0 .485 and change in momentum is 0 .496 and for spring it is 0 .406 this is 0 .407 and for clay system the impulse is 0 .204 where as the change in momentum is 0 .234.

01:58 So from the table from the table the values of impulse are more or less equal to the value of change in momentum.

02:13 So therefore we can say that the equation from theory that is impulse is equal to change in momentum change in momentum holds true holds true in these cases in these cases and for question 2 here we have to find the comparison between the kinetic energies before and after collision that is kinetic energy initial and this is kinetic energy final.

03:10 Here initial kinetic energy is 1 by 2 m v i squared whereas the final kinetic energy is 1 by 2 m v f squared.

03:20 So here mass doesn't change so this half m terms half m terms will get cancelled and we should get the result that is v i is equal to v f squared.

03:36 So let's check these cases here the initial speed for the case is 0 .36 and this is 0 .32.

03:48 So for the first case 0 .36 squared and this is 0 .32 squared.

03:59 When we compare the results we get this is point this is v i squared let's draw this table and this is v f squared this is 0 .1296 whereas 0 .32 squared this gives 0 .1024.

04:23 Then for the second case the initial and final speeds are 0 .25 and 0 .22 respectively.

04:32 So this is 0 .25 and this is 0 .22...

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