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Keshav K.

Mentor for 12th Grade Mathematics and Physics

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Physics (Newtonian Mechanics)

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Question:

A bead, under the influence of gravity, slides along a frictionless wire whose height is given by the function V (x). Find an expression for the bead’s horizontal acceleration. (It can depend on whatever quantities you need it to depend on.) You should find that the result is not the same as the x'' for a particle moving in one dimension in the potential mgV (x), in which case x'' = −gV(0) . But if you grab hold of the wire, is there any way you can move it so that the bead’s x'' is equal to the x'' = −gV(0) ,result due to the one-dimensional potential, mgV (x)?

Keshav K.

Answer:

Consider the normal force, N, acting on the bead at a given point. Let θ be the angle that the tangent to V (x) makes with the horizontal, The horizontal F = ma equation is −N sin θ = mx'' (1) The vertical F = ma equation is N cos θ − mg = my'' =⇒ N cos θ = mg + my'' (2) Dividing eq. (1) by eq. (2) gives − tan θ = x'' / g + y'' (3) But tan θ = V(0) Therefore, x¨ = −(g + ¨y)V(0)( (4) We see that this is not equal to −gV . In fact, there is in general no way to construct a curve with height y(x) which gives the same horizontal motion that a 1-D potential V (x) gives, for all initial conditions. We would need (g + y'')y= V, for all x. But at a given x, the quantities V and y are fixed, whereas y'' depends on the initial conditions. For example, if there is a bend in the wire, then ¨y will be large if y' is large. And y' depends (in general) on how far the bead has fallen. Eq. (4) holds the key to constructing a situation that does give the x'' = −gV(0) result for a 1-D potential V (x). All we have to do is get rid of the y'' term. So here’s what we do. We grab our y = V (x) wire and then move it up (and/or down) in precisely the manner that makes the bead stay at the same height with respect to the ground. (Actually, constant vertical speed would be good enough.) This will make the y'' term vanish, as desired. (Note that the vertical movement of the curve doesn’t change the slope, V(0) , at a given value of x.0 Remark: There is one case where x'' is (approximately) equal to −gV(0) , even when the wire remains stationary. In the case of small oscillations of the bead near a minimum of V (x), y'' is small compared to g. Hence, eq. (4) shows that x'' is approximately equal to −gV(0). Therefore, for small oscillations, it is reasonable to model a particle in a 1-D potential mgV (x) as a particle sliding in a valley whose height is given by y = V (x).

C Programming

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Question:

What will be the output of this C program: #include<stdio.h> int main(){ int a[5]; a[5] = 5; 4[a] = 4; printf("%d\n%d\n",5[a],a[4]); return 0; }

Keshav K.

Answer:

The definition of a is just a pointer to the location of the head of the array. the way the arithmetic of works is it adds the index specified within the [] to the main address. As such a[5] means the same as 5[a] which is 5 + a or a + 5. hence the code prints out 5 and 4 as normal.

Algebra

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Question:

Suppose there are two candidates, A and B in an election, and A gets "a" votes, and B gets "b" votes, with a <= b. If the votes are counted in a random order, what is the probability that at some point during the count, the tally will be even ? (0 - 0 doesn't count).

Keshav K.

Answer:

Let's take a specific example: 6 votes, 4 for B, 2 for A: There are 15 possible orders in which votes can be tallied (6 choose 2 possible 'slots' where the a votes come up) They can be classified as follows: b b b * * * = 3 ways, won't happen b b a b * * = 2 ways, won't happen otherwise it will, hence Pr = (15-5)/15 = 10/15 = 2/3 chance it will happen. Counting them the other way: a a * * * * - yes (1) a b * * * * - yes (4) b a * * * * - yes (4) b b a a * * - yes (1) Total 10. To generalize to any a, b where a <= b: I don't now how to figure this out, so I wrote a program to generate and count the various sequences. Here are some more data points: a b: Pr(equal at some point) 2 2: 6/6 2 3: 8/10 2 4: 10/15 2 5: 12/21 2 6: 14/28 2 7: 16/36 2 8: 18/45 3 3: 20/20 3 4: 30/35 3 5: 42/56 3 6: 56/84 3 7: 72/120 4 4: 70/70 4 5: 112/126 4 6: 168/210 4 7: 240/330 So for 2 a votes, we have Pr = [ (b+1) * 2 / 1! ] / [ (a+b) C 2 ] For 3 a votes, we have [ (b+2) (b+1) * 2 / 2! ] / [ (a+b) C 3 ] For 4 votes, we have [ (b+3) (b+2) (b+1) * 2 / 3! ] / (a+b) C 4 ] For 5 a votes, we have [ (b+4) (b+3) (b+2) (b+1) * 2 / 4! ] / [ (a+b) C 5 ] And in general, the numerator will be (b+a-1) (b+a-2) ... (b+1) * 2 / (a-1)! the denominator will be (a+b) C (a).

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