## Saturday, September 14, 2013

### Applied Algebra Solutions (3)

1)A beam of electrons moving with v = 1.0 x 107 m/s enters midway between two horizontal plates in a direction parallel to the plates which are 5 cm long and 2 cm apart, and have a potential difference V between them. Find V, if the beam is deflected so that it just grazes the bottom plate. (Take the electron charge to mass ratio: e/ me = 1.8 x 1011 C/kg).

Solution:

The key piece of information (and key to the solution!) is that the beam just grazes the bottom plate. Since the total y -span = 2 cm, this means the effective beam deflection is one half this, or dy = 1 cm = 0.01m. Meanwhile, the electric field is:

E = V/d = V/ (0.02m)

The downward acceleration, from Newton's 2nd law, is:

ay = eE/m e

The vertical displacement in terms of this acceleration is:

y = ½ {ay t2) = ½ {eE/ m e } t2

The time, t,  applicable is:

t = x/v = (0.005m)/ 107 m/s = 5 x 10-9 s

and we know:

y = ½ (V/d)(e/ m e )t2

Substituting:

0.01m = 0.5(V/0.02m) [1.8 x 1011 C/kg] (5 x 10-9 s)2

for which V can be found:   V = 89 V

2) A heated filament emits electrons which are accelerated to the anode by a p.d. of 500 V. Find the kinetic energy and velocity of the electron as it strikes the anode.

Solution:

We have V = 500 V

But the electric field E is defined: E = 500V /d

Where d is the distance between plates

The acceleration will be:   a = eE / me   =  e (500V /d )/  me

Where (e/ me  ) is the charge to mass ratio.

Then    vy = ay t = (eE/ me) d/vy  =   [e (500V /d )/  me]  d/ vy  = e (500V  )/  me]  vy

And:

(vy)2  =  e (500V  )/  me

Where the charge to mass ratio is:  e/ me   = 1.8 x 1011 C/kg

So: (vy)2  =     (1.8 x 1011 C/kg)   (500V  ) =  9  x 10 14   m2/ s2

And: vy  =  3  x 107    m/ s

The KE is:  ½  me  (vy)2    =  ½ (9.1 x 10-31 kg) ( 9  x 10 14   m2/ s2 )

=  4.0 x 10 -16 J

3) Given that x' = 1/a (x - vt) and t' = 1/a (t - vx/c2), derive similar equations for x and t in terms of x' and t'.  (Let: 1/a = (1 - v2/c2)½)

If we now substitute x' = a(x - vt) and equation for t’  into the right hand side of:

r’2 = x’2 + y’2 + z’2 = c2 t’2

what do we get?

Solution:

Given x' =  1/a (x - vt) and t' = 1/a (t - vx/c2),

Then: x' = x/a - vt/a and t' = t/a - vx/ac2

and: x' + vt/a = x/a and t' + vx/ ac2  = t/a

so:

x = a(x' + vt/a) and t' = a(t' + vx/ ac2 )

finally: x = a(x' + vt) and t = a(t' + vx/c2)

b) On substituting x' = a(x - vt) and equation for t’  into the right hand side of:

r’2 = x’2 + y’2 + z’2 = c2 t’2

We get:

x2 + y2 + z2 -  c2 t2  = a2(x - vt)2 + y2 + z2

4) Let a quadratic model for central meridian transit of a sunspot group – indicating days past the central meridian (where the closer the value is to 0, the greater the incidence of likely flaring) be:

F(x) =  0.1x2 + 8x + 12

Is this complex group more or less likely to see powerful flares than the one given in the example?

Soln.

If it is presumed to incept more powerful flares it should be fewer days from the central meridian.

[-b + {b2 - 4ac}1/2]/ 2a

Where a = 0.1, b = 8 and c = 12

x1 = -1.529  and x2=  - 78.471

Choosing the lesser of the two and comparing it the lesser value in the text we find the latter has x1  = -1.396 so it is more likely to produce  powerful flares since the group is more proximate to the central meridian.

5) An athlete standing close to the edge on the top of a 160 ft. high building throws a baseball vertically upward. The quadratic function:

s(t) = -16 t2 + 64t + 160

models the ball’s height above the ground, with s(t) in feet, t seconds after it’s thrown:

a)After how many seconds does the ball reach its maximum height? What is the maximum height?
b)How many seconds does it take until the ball finally hits the ground? (Round to the nearest tenth of a second)

Solution:  We plot a graph of s(t) vs. t with t on the horizontal axis. We find on doing so that we obtain a parabola with a maximum at t = 2 s, and s(t) = 264’

Hence, the answer to (a) is t = 2 seconds and the maximum ht. = 264 ft.

To answer (b) we examine the graph to see when the motion ends (e.g. where parabola terminates with s(t) = 0)  and find it to be at t = 5.7 secs.  This can also be found from the quadratic formula, setting s(t) =  0 and solving for t.

6) Another form of Newton’s law of gravitation which incorporates Kepler’s third law is:

P2 = 4(p)2 a3/[G(M + Mp)].

where P is the planet's period, a the distance (i.e. semi-major axis, we want to solve for) and G the gravitational constant, M the mass of the Sun, and Mp the planet's mass.

Write an equation for the planet's  mass, i.e. which might be used to find it if all the parameters are known, and the units.

Solution:

Cross multiply to get:

P2  [G(M + Mp)] =  4(p)2 a3
Then divide both sides by  P2 :

[G(M + Mp)] =  4(p)2 a3 /  P2

Divide through by G:

M + Mp =     4(p)2 a3 / G P2

Solve for  Mp  :

Mp =     4(p)2 a3 / G P2   - M

Partial Fraction Decomposition Problem:

Add the following algebraic expressions and arrive at such a single rational expression:

3/ (x – 4)   +  2/ (x + 2)

While this is not an application per se, it IS frequently one of the algebraic manipulation tasks needed in calculus.

Then the common denominator is :

(x – 4)   (x + 2)  =  x2  - 2x  -8

So we can write:

3 (x + 2) +  2 (x – 4)/  (x2  - 2x  -8)

Or:

[3x + 6 + 2x – 8]/  (x2  - 2x  -8)  =  5x -2/ (x2  - 2x  -8)