Here we continue with the answers to the previous segment of the advanced physics test:
http://brane-space.blogspot.com/2011/01/advanced-physics-questions-2.html
16) This solution can be treated classically since 6 x 10^6 m/s is much less than c, the speed of light. Thus:
dv/dt = a and upon integration:
v - v(o) = at
and dx/dt = v
Integrating a second time one obtains:
x - x(o) = v(o)t + ½ at^2
Now: a = -1.2 x 10^14 m/s/s
v(o) = 6 x 10^6 m/s and the final velocity is: v(f) = 0
So: v(f) - v(o) = at = 0 - 6 x 10^6 m/s = (-1.2 x 10^14 m/s/s) t
whence: t = [0 - 6 x 10^6 m/s ] / (-1.2 x 10^14 m/s/s)
t = 5 x 10^-8 sec
Now: x(f) - x(o) = v(o)t + ½ at^2
= (6 x 10^6 m/s)(5 x 10^-8 sec) - ½((-1.2 x 10^14 m/s/s)(5 x 10^-8 sec)^2
x(f) - x(o) = 0.15 m = 15 cm
17) By Newton's law of universal gravitation, the force between the two masses is:
GMm/r^2
From the integration of the above the potential energy can be obtained as: -(GMm/R) and the kinetic energy as KE = GMm/2R.
let d = 2R then the total KE available is:
GMm/d
Now: Momentum must also be conserved so (assuming conservation of linear momentum along the line of action):
mv(m) = M v(M)
But: KE = ½ [mv(m)^2 + M v(M)^2] = GMm/d
Using algebra:
v(M)^2[1 + M/m] = 2GM/d and
v(m)^2 + v(M)^2 + (m/M)v(m)^2 + (M/m)v(M)^2= 2G/d{M + m}
v(m)^2 + v(M)^2 = 2G/d (M + m)
Therefore: v_r = [2G/d (M + m)]^½
18) We define current as the rate at which charge flows, thus:
I = dQ/ dt so dQ = I dt, or Q = It
since I = 5 A (amps) = 5 coulombs/sec
then for 4 minutes (= 4 x 60 sec = 240 secs):
Q = 5 coulombs/sec (240 sec) = 1200 coulombs
19) The effective value of a sinusoidal emf is its root mean square value, defined:
v(eff)^2 = v(rms)^2 = INT (0 to r) [v(max)^2 sin wt dt] T = ½ v(max)^2
so v(eff) = [½ v(max)^2]^½ = 1/ [2]^½ (v(max)) or
v(eff) = [2]^½ / 2 (v(max))
20) Potential difference is the work needed to move a test charge (q) from point A to B, or:
V = W/q
In this case, W = 2 J and q = 20 C, so:
V = 2J/ 20 C = 0.1 Volt
21) For a parallel plate capacitor,
C = k k(o) A/ d
where k(o) = 1 for air, and k is the relative dielectric constant: or
k = k(mica) k(o) = 6 k(o) = 6
Then C(mica) = 6 (15 uuF) = 90 uuF
(A/d is the same in each case)
22) The charge on the electron, q = 1.602 x 10^-19 C
Again, from Problem #18, the current is charge moved per unit time:
I = dQ/ dt
In this case, since the angular frequency is the key information given, and it is related to time via:
w = 1/T
then:
I = qw = (1.602 x 10^-19 C) (6.6 x 10^15 /sec)
I = 1.05 x 10^-3 amp, or I = 0.00105 A
23) For a plane mirror the applicable law of geometrical optics is that the image appears to be as far behind the mirror as the object is in front. Therefore, one's eyes must be focused to a distance of 40 cm.
24) We apply the lens makers' formula here:
1/ f = (n - 1) [1/r1 - 1/r2]
where f is the focal length of the convex lens (to be found)
n = 1.65 (the refractive index) and r1 = 40 cm
then r2 = -40 cm (recall for convex lenses, r2 = -r1)
Then:
1/f = (1.65 - 1) [1/40 cm - (1/ -40cm)]
1/f = 0.65/ 20 cm
So: f = 20 cm/ 0.65 = 31 cm
25) In this nuclear reaction, the charge and nucleons must be conserved, meaning that the upper index total on the right side must be equal the upper index total on the left, plus the lower number on the right must be equal to the sum of lower numbers on the left. The only one which fits this is (D) 8 O 17.
26) Here, we use the relativistic form for the kinetic energy:
K= Mc^2 - M(o)c^2
We are given the rest mass is 106 MeV so M(o)c^2 = 106 MeV
We also have: K = 4 MeV, then:
Mc^2 = K + M(o)c^2 = 4 MeV + 106 MeV = 110 MeV
We also know:
M = M(o)/ [1 - (v/c)^2)]^½
But K = M(o)c^2 [1/ (1 - (v/c)^2)]^½) - 1]
Therefore:
(v/c) = (110 MeV/ 106 MeV) - 1
(v/c) = 0.28
or
v = 0.28c = 84,000 km/sec
27) The system depicted in the accompanying graphic is the one used for solution here.
The Lagrangian is the difference between kinetic and potential energies:
L = T - V
And from the system:
T = ½ m1v1^2 + ½m2v2^2 = ½ (m1 + m2)x' ^2
V = (m1 x - m2 x)g
So:
L = ½ (m1 + m2)x' ^2 -(m1 x - m2 x)g
= ½ (m1 + m2)x' ^2 + (m2 x - m1 x)g
28) Applying force balance:
m2 g - T = m2 a
and:
m1 g - T = -m1a
(where T is the tension in the string.)
so: Using algebra to solve the simultaneous equation:
m1 m2 g - m1 T = m1 m2 a
m1m2 g - m2T = -m1m2 a
----------------------------
2 m1m2g - (m1 + m2) T = 0
and T = (2 m1m2 g)/ (m1 + m2)
29) The efficiency of a Carnot cycle engine is:
e = 1 - Q(R) / Q(W)
where Q(R) denotes the heat returned to a reservoir at T2, and Q(W) is the heat withdrawn from a reservoir at T1.
Then:
Q(R)/ Q(W) = T2 / T1
and
e = 1 - Q(R) / Q(W) = 1 - (T2/T1)
30) From our quantum principle for the total angular momentum:
L = L1 + L2........[L1 - L2]
L = 5, 4, 3, 2, 1
http://brane-space.blogspot.com/2011/01/advanced-physics-questions-2.html
16) This solution can be treated classically since 6 x 10^6 m/s is much less than c, the speed of light. Thus:
dv/dt = a and upon integration:
v - v(o) = at
and dx/dt = v
Integrating a second time one obtains:
x - x(o) = v(o)t + ½ at^2
Now: a = -1.2 x 10^14 m/s/s
v(o) = 6 x 10^6 m/s and the final velocity is: v(f) = 0
So: v(f) - v(o) = at = 0 - 6 x 10^6 m/s = (-1.2 x 10^14 m/s/s) t
whence: t = [0 - 6 x 10^6 m/s ] / (-1.2 x 10^14 m/s/s)
t = 5 x 10^-8 sec
Now: x(f) - x(o) = v(o)t + ½ at^2
= (6 x 10^6 m/s)(5 x 10^-8 sec) - ½((-1.2 x 10^14 m/s/s)(5 x 10^-8 sec)^2
x(f) - x(o) = 0.15 m = 15 cm
17) By Newton's law of universal gravitation, the force between the two masses is:
GMm/r^2
From the integration of the above the potential energy can be obtained as: -(GMm/R) and the kinetic energy as KE = GMm/2R.
let d = 2R then the total KE available is:
GMm/d
Now: Momentum must also be conserved so (assuming conservation of linear momentum along the line of action):
mv(m) = M v(M)
But: KE = ½ [mv(m)^2 + M v(M)^2] = GMm/d
Using algebra:
v(M)^2[1 + M/m] = 2GM/d and
v(m)^2 + v(M)^2 + (m/M)v(m)^2 + (M/m)v(M)^2= 2G/d{M + m}
v(m)^2 + v(M)^2 = 2G/d (M + m)
Therefore: v_r = [2G/d (M + m)]^½
18) We define current as the rate at which charge flows, thus:
I = dQ/ dt so dQ = I dt, or Q = It
since I = 5 A (amps) = 5 coulombs/sec
then for 4 minutes (= 4 x 60 sec = 240 secs):
Q = 5 coulombs/sec (240 sec) = 1200 coulombs
19) The effective value of a sinusoidal emf is its root mean square value, defined:
v(eff)^2 = v(rms)^2 = INT (0 to r) [v(max)^2 sin wt dt] T = ½ v(max)^2
so v(eff) = [½ v(max)^2]^½ = 1/ [2]^½ (v(max)) or
v(eff) = [2]^½ / 2 (v(max))
20) Potential difference is the work needed to move a test charge (q) from point A to B, or:
V = W/q
In this case, W = 2 J and q = 20 C, so:
V = 2J/ 20 C = 0.1 Volt
21) For a parallel plate capacitor,
C = k k(o) A/ d
where k(o) = 1 for air, and k is the relative dielectric constant: or
k = k(mica) k(o) = 6 k(o) = 6
Then C(mica) = 6 (15 uuF) = 90 uuF
(A/d is the same in each case)
22) The charge on the electron, q = 1.602 x 10^-19 C
Again, from Problem #18, the current is charge moved per unit time:
I = dQ/ dt
In this case, since the angular frequency is the key information given, and it is related to time via:
w = 1/T
then:
I = qw = (1.602 x 10^-19 C) (6.6 x 10^15 /sec)
I = 1.05 x 10^-3 amp, or I = 0.00105 A
23) For a plane mirror the applicable law of geometrical optics is that the image appears to be as far behind the mirror as the object is in front. Therefore, one's eyes must be focused to a distance of 40 cm.
24) We apply the lens makers' formula here:
1/ f = (n - 1) [1/r1 - 1/r2]
where f is the focal length of the convex lens (to be found)
n = 1.65 (the refractive index) and r1 = 40 cm
then r2 = -40 cm (recall for convex lenses, r2 = -r1)
Then:
1/f = (1.65 - 1) [1/40 cm - (1/ -40cm)]
1/f = 0.65/ 20 cm
So: f = 20 cm/ 0.65 = 31 cm
25) In this nuclear reaction, the charge and nucleons must be conserved, meaning that the upper index total on the right side must be equal the upper index total on the left, plus the lower number on the right must be equal to the sum of lower numbers on the left. The only one which fits this is (D) 8 O 17.
26) Here, we use the relativistic form for the kinetic energy:
K= Mc^2 - M(o)c^2
We are given the rest mass is 106 MeV so M(o)c^2 = 106 MeV
We also have: K = 4 MeV, then:
Mc^2 = K + M(o)c^2 = 4 MeV + 106 MeV = 110 MeV
We also know:
M = M(o)/ [1 - (v/c)^2)]^½
But K = M(o)c^2 [1/ (1 - (v/c)^2)]^½) - 1]
Therefore:
(v/c) = (110 MeV/ 106 MeV) - 1
(v/c) = 0.28
or
v = 0.28c = 84,000 km/sec
27) The system depicted in the accompanying graphic is the one used for solution here.
The Lagrangian is the difference between kinetic and potential energies:
L = T - V
And from the system:
T = ½ m1v1^2 + ½m2v2^2 = ½ (m1 + m2)x' ^2
V = (m1 x - m2 x)g
So:
L = ½ (m1 + m2)x' ^2 -(m1 x - m2 x)g
= ½ (m1 + m2)x' ^2 + (m2 x - m1 x)g
28) Applying force balance:
m2 g - T = m2 a
and:
m1 g - T = -m1a
(where T is the tension in the string.)
so: Using algebra to solve the simultaneous equation:
m1 m2 g - m1 T = m1 m2 a
m1m2 g - m2T = -m1m2 a
----------------------------
2 m1m2g - (m1 + m2) T = 0
and T = (2 m1m2 g)/ (m1 + m2)
29) The efficiency of a Carnot cycle engine is:
e = 1 - Q(R) / Q(W)
where Q(R) denotes the heat returned to a reservoir at T2, and Q(W) is the heat withdrawn from a reservoir at T1.
Then:
Q(R)/ Q(W) = T2 / T1
and
e = 1 - Q(R) / Q(W) = 1 - (T2/T1)
30) From our quantum principle for the total angular momentum:
L = L1 + L2........[L1 - L2]
L = 5, 4, 3, 2, 1
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