1 - Oak Park Independent



Multiple Choice

|1 |a |d = vot + ½at2. Let a = 2, then d = t2. When t = 0, 1 and 2, then d |

| | |= 0, 1 and 4, ∴ the graph curves upward. |

|2 |c |d = area under the v vs. t graph. |

| | |Area = ½BH = ½(4 s)(20 m/s) = 40 m |

|3 |b |vy = vsinθ = (250 m/s)sin60o = 217 m |

|4 |c |vt = vo + at → -217 m/s = 217 m/s + (-10 m/s2)t ∴ t = 43 s |

|5 |a |vx = vcosθ = (250 m/s)cos60o = 125 m/s |

|6 |a |dx = vxt = (125 m/s)(43 s) = 5375 m |

|7 |b |Ftot = (Fx2 + Fy2)½ = ((20 N)2 + (20 N)2)½ = 28 N (northeast) |

|8 |a |More that forces go in the same direction, the greater their sum. |

|9 |c |The forces are equal because of Newton's third law. |

|10 |c |F = Ff = μFn → 40 N = (0.05)Fn ∴ Fn = 800 N |

|11 |a |Because the velocity is constant, the force equals Ff |

| | |F = Ff = μFn = (0.67)(60 N) = 40 N |

|12 |d |Fs = kx → 10 N = k(0.02 m) ∴ k = 500 N/m |

|13 |d |J = Ft = mΔv → F(0.010 s) = (0.15 kg)(-20 m/s) |

| | |∴ F = -300 N |

|14 |b |F – Ff = ma → 50 N – Ff = (4.0 kg)(10 m/s2) = 40 N |

| | |∴ Ff = 10 N |

|15 |a |Us = ½kx2 = ½(80 N/m)(0.3 m)2 = 3.6 J |

|16 |b |The cart is slowing down, the force of friction, Ff, is greater than |

| | |the applied force, F, and is directed left. |

|17 |c |K = ΔUg = mgΔh ∴ K ∝ d |

|18 |a |E = mc2. The graph of E vs. m would be a straight line with slope c2|

| | |(y = mx + b). |

|19 |d |Inelastic collision: mAvA + mBvB = (mA + mB)v' |

| | |mv + 0 = (m + M)v' ∴ v' = mv/(m + M) |

|20 |a |P = W/t = Ug/t = mgh/t |

| | |P = (30,000 N)(800 m)/(480 s) = 50,000 W (5.0 x 104 W) |

|21 |b |K = Ug – Wf = mgh – Wf = (40 N)(8.0 m) – 50 J = 270 J |

|22 |c |W = F||d = (20 N)(cos25o)(4 m) = 73 J |

|23 |b |Total energy is constant where Ko = UH∴ half of the initial K is used|

| | |to reach ½H. |

|24 |b |Going up from 2.25 m on the x-axis until intersection, and then going|

| | |across to the y-axis you get 55 J. |

|25 |c |Ug = mgh → 25 J = m(10 m/s2)(1.0 m) ∴ m = 2.5 kg |

|26 |a |The acceleration gravity, which is constant and downward. |

|27 |a |vt = vo + at → 0 m/s = vo + (-10 m/s2)(3 s) ∴ vo = 30 m/s |

|28 |a |When masses are equal, all the velocity is transferred to the |

| | |stationary ball ∴ ptot = Mv and Ktot = ½Mv2. |

|29 |b |J = Ft = mΔv = area = ½(2 N)(1 s) + (2 N)(2 s) + ½(2 N)(1 s) |

| | |J = 6 N•s = (2 kg)Δv ∴ Δv = 3 m/s |

|30 |a |The direction of motion (tangential) is at right angles to the radial|

| | |force ∴ no work is done (W = F||d). |

|31 |c |The upper string supports all of the weight. |

| | |Fg = mg = (4 kg)(10 m/s2) = 40 N |

|32 |d |The lower string pulls down with force equal to Fg. |

| | |Fg = mg = (2 kg)(10 m/s2) = 20 N downward |

|33 |c |px = (1.5 kg)(2.0 m/s) = 3. py = (4.0 kg)(1.0 m/s) = 4 |

| | |ptotal = (px2 + py2)½ = (32 + 42)½ = 5 kg•m/s |

|34 |a |Ktotal = K1 + K2 |

| | |Ktotal = ½(1.5 kg)(2.0 m/s2)2 + ½(4.0 kg)(1.0 m/s2) = 5 J |

|35 |d |pboy = pgirl → mbvb = mgvg ∴ vb = 2mv/m = 2v |

| | |W = Kg + Kb = ½(2m)v2 + ½(m)(2v)2 = 3 mv2 |

|36 |b |Fg is the same for X and Y: FX = FY → mXaX = mYaY |

| | |aY = mXaX/mY = (mX)(2.2 m/s2)/2mX = 1.1 m/s2 |

|37 |d |Since the force is constant, then acceleration is constant. v2 = vo2 |

| | |+ 2ad = 0 + 2ad ∴ v2 ∝ d |

|38 |c |Momentum is conserved in the x and y directions, there must be some |

| | |+y velocity to cancel –y velocity |

|39 |d |P = W/t = K/t = mv2/2t |

| | |P = (900 kg)(20 m/s)2/(2)(5 s) = 36000 W |

|40 |b |In order for the block to stay put, then Ff = Fg. |

| | |μFn = mg (where Fn = ma) → μma = mg ∴ a = g/μ |

|41 |b |Velocity is along the perimeter of the circle and acceleration is |

| | |toward the center of the circle. |

|42 |d |Fg = mg → 1560 N = (60 kg)a ∴ a = 26 m/s2 |

|43 |c |Fg = GMm/r2: When R is doubled, gravity is (½)2 = ¼ as much. |

|44 |b |Fc = mv2/r = (5.0 kg)(2.0 m/s)2/0.70 m = 29 N |

|45 |d |Amplitude ∝ loudness. (Pitch ∝ frequency, but we don't have a |

| | |perception of speed or wavelength.) |

|46 |d |Δf/f = v/vw → Δf/1000 Hz = (30 m/s)/(330 m/s) ∴ Δf = 90 Hz |

| | |f' increases by 90 Hz because source is approaching. |

|47 |b |v = fλ (picking any point on the graph such as λ = 2.0 m |

| | |and f = 2.5 s-1): v = (2.5 s-1)(2.0 m) = 5 m/s |

|48 |b |F = ma ∴ the heavier box requires more force. Their orbital position|

| | |doesn't matter with respect to station. |

|49 |b |Tp = Ts → 2π(L/g)½ = 2π(m/k)½ → L/g = m/k ∴ L = mg/k |

|50 |c |Mass number (top) and charge number (bottom) are conserved. 22688Ra →|

| | |22286Rn + 42He |

|51 |d |Weight = Fg = GMm/r2 ∴ Fg ∝ M/r2. Without knowing the specific |

| | |values, you don't know overall effect. |

|52 |c |The time it takes the sound to go from shore to ship is half the |

| | |total time. d = vt = (340 m/s)(3 s) = 1020 m |

|53 |a |Photons travel at the speed of light, c, in a vacuum. |

|54 |a |Amplitude is distance between midpoint and extreme = 4 cm. |

|55 |c |v = λ/T = 10 cm/0.2 s = 50 cm/s |

|56 |c |τ = r⊥F, but since the masses are on opposite sides of the fulcrum, |

| | |then (0.60 m)(20 N) – (0.4 m)(10 N) = 8 m•N |

|57 |c |Tension ∝ speed (vw = (Ft/(m/L))½), speed ∝ frequency (v = fλ), but |

| | |only Length affects wavelength (λn = 2L/n). |

|58 |d |Binding energy separates a nuclide into protons and neutrons, and |

| | |visa versa = mlossc2. |

|59 |d |T = 2π(m/k)½ ∴ T2 = (4π2/k)m. In order to double T, the mass is |

| | |multiplied by 22: 4(2 kg) = 8 kg |

|60 |a |Fc = Fg → mac = GMm/r2 ∴ a = GM/R2 |

|61 |c |Resonance frequencies are harmonics where nodal points are ½λ apart =|

| | |0.05 m ∴ f = v/λ = 340 m/s/0.1 m |

|62 |a |The bell can't be heard because sound requires a medium. Light can |

| | |travel through a vacuum. |

|63 |c |c = fλ → 3 x 108 m/s = f(6 x 10-7 m) ∴ f = 5 x 1014 s-1 |

|64 |b |E = hf = (6.63 x 10-34 J•s)(5 x 1014 s-1) = 3.3 x 10-19 J |

|65 |c |Our eyes perceive energy (frequency) of light as color. |

|66 |d |Short wavelength = high energy (E = hc/λ). low to high: |

| | |red-orange-yellow-green-blue-violet. |

|67 |d |The different n for air and water bends light due to differences in |

| | |light speed. (nH2O > nair ∴ vH2O < vair). |

|68 |c |nisinθi = nRsinθR (angles are measured from normal) |

| | |(1.0)sin55 = (1.5)sinθR |

|69 |b |v = c/n = (3 x 108 m/s)/1.5 = 2 x 108 m/s |

|70 |b |The refracted ray is slower in glass compared to air ∴ it bends |

| | |toward normal. Frequency is unchanged.-. |

|71 |a |Angle of reflection = angle of incidence. Phase shift occurs when |

| | |reflecting surface has a higher index.. |

|72 |b |The index of refraction, n ∝ f. Red light has a lower frequency ∴ it|

| | |has a greater speed: v = c/n. |

|73 |c |Partially covered ∴ bright bands aren't as bright (less combined |

| | |light) and the dark bands are not as dark. |

|74 |d |Red (lower frequency) has less energy per photon. Same power, ∴ more|

| | |red photons per second. |

|75 |a |Reflection is always the same. Refraction is larger because nR < ni |

| | |(nisinθi = nRsinθR). |

|76 |a |1/do + 1/di = 1/f |

| | |1/0.30 m + 1/di = 1/0.40 m ∴ di = -1.2 m (to left) |

|77 |c |-di is virtual and upright. |

| | |di > do, the image is larger (M = hi/ho = -di/do). |

|78 |a |Work function is an energy quantity: E = hf |

| | |∴ φ = hf and f = φ/h |

|79 |b |E = hf ∴ E ∝ f. If E is doubled, then f = 2fo. |

|80 |c |θi = 60o, sinθC = nR/ni |

| | |sin60o = ½√3 = 1/ni ∴ ni = 2/√3 = 2/3√3 |

Free Response

1. a.

|Cart is at rest when graph crosses the x-axis |

|∴ 4 s and 18 s. |

b.

||Speed| increases when line is moving away from x-axis ∴ 4 s to 9 s and |

|18 s to 20 s. |

c.

|Acceleration is positive when the slope is positive |

|∴ 9 s to 12 s and 17 s to 20 s. |

d.

|x – xo = vot + ½at2 |

|x – 2 m = (0.8 m/s)(9 s) + ½(-0.2 m/s2)(9 s)2 ∴ x = 1.1 m |

e.

| a |

(2)

|x = voxt = (0.8 m/s)(0.28 s) = 0.22 m |

(3)

|K' = K + U = ½mv2 + mgh |

|K' = ½(0.50 kg)(0.8 m/s)2 + (0.50 kg)(10 m/s2)(0.40 m) |

|K' = 2.2 J |

2. a.

|a = Δv/t ∴ t = Δv/a = (5.0 m/s – 2.0 m/s)/(1.5 m/s2) = 2 s |

|vav = d/t ∴ d = (vav)t = (3.5 m/s)(2 s) = 7 m |

|vav = d/t ∴ t = d/vav = (15 m – 7 m)/5.0 m/s = 1.6 s |

|ttotal = 2 s + 1.6 s = 3.6 s |

b. (1)

|mAvA + mBvB = mAvA' + mBvB' |

|(250 kg)(5 m/s) + 0 = (250 kg)vA' + (200 kg)(4.8 m/s) |

|vA' = 1.2 m/s |

(2)

|To the right because vA' > 0 (same direction as vA) |

c.

|KA = ½mAvA2 = ½(250 kg)(5 m/s)2 = 3125 J |

|KA' = ½mAvA'2 = ½(250 kg)(1.2 m/s)2 = 180 J |

|KB' = ½mBvB'2 = ½(200 kg)(4.8 m/s)2 = 2304 J |

|3125 J ≠ (2304 + 180 J) ∴ kinetic energy is lost and the collision is |

|not perfectly elastic. |

3. a.

Fn Ft

8 kg 4 kg

Fs ≤ Ft ≤

Fg Fg

b.

|Ft = m4g = (4 kg)(10 m/s2) = 40 N |

c.

|Ft = Fs = kΔx ∴ k = Ft/Δx = 40 N/0.05 m = 800 N/m |

d.

|dy = vyot + ½at2 |

|0.70 m = 0 + ½(10 m/s2)t2 ∴ t = 0.37 s |

e.

|T = 2π(m/k)½ = 2π(8.0 kg/800 N/m)½ = 0.63 s |

|f = 1/T = 1/0.63 s = 1.6 s-1 |

f.

|K = Us → ½mv2 = ½kx2 |

|∴ v = A(k/m)½ = (0.05 m)(800 N/m/8.0 kg)½ = 0.50 m/s |

4. a.

Ft

Fw

Fg

b.

|tan 30 = Fh/Fg |

|Fh = mgtan30 = (1.8 kg)(10 m/s2)tan30 = 10 N |

c.

|Δh = 2.3 – 2.3cos30 = 0.3 m |

|Ug = K → mgh = ½mv2 |

|∴ v = (2gh)½ = [(2)(10 m/s2)(0.3 m)]½ = 2.5 m/s |

5. a.

Fn Ft

Ff

Fg

b.

|Ff = μFn = μm1gcosθ ∴ μ = f/m1gcosθ |

c.

|0 = Mg – m2gsinθ + 2f – m1gsinθ + f |

|∴ M = (m2 + m1)sinθ – 3f/g |

d.

|Fnet = m1gsinθ – f = m1a ∴ a = gsinθ – f/m1 |

6. a.

|vt = vo + at → 25 m/s = 0 + a(10 s) ∴ a = 2.5 m/s2 |

|F = ma = (800 kg)(2.5 m/s2) = 2,000 N |

b.

|F = Fs + Ff = kx + μFn |

|2,000 N = (5,000 N/m)(0.08 m) + μ(800 kg)(10 m/s2) |

|∴ μ = 0.20 |

c.

|Fnet = Fs – Ff = kx – μFn |

|Fnet = (5,000 N/m)(0.08 m) – (0.20)(800 kg)(10 m/s2) |

|Fnet = 400 N – 1600 N = < 0 ∴ No (Ff can't move crate) |

7. a.

candle lens screen

do di

b.

|Trial # |do (m) |1/do (m-1) |di (m) |1/di (m-1) |

|1 |0.40 |2.5 |1.10 |0.91 |

|2 |0.50 |2.0 |0.75 |1.3 |

|3 |0.60 |1.7 |0.60 |1.7 |

|4 |0.80 |1.2 |0.50 |2.0 |

|5 |1.20 |0.83 |0.38 |2.6 |

c. di

|3.5 | | | | | | | |

| | | | | | | | |

do

(1)

|The y-intercept equals 1/f (1/di = -1/do + 1/f). |

(2)

|y-intercept = 3.3 ∴ f = 1/3.3 = 0.30 m |

d.

F F

8. a.

|E = mc2 = 2(9.11 x 10-31 kg)(3 x 108 m/s)2 = 1.64 x 10-13 J |

b.

|½(1.64 x 10-13 J) x 1 eV/(1.60 x 10-19 J) = 5.1 x 105 eV |

c.

|E = 1240 eV•nm/λnm |

|λnm = 1240 eV•nm/5.1 x 105 eV = 2.4 x 10-3 nm |

d.

|p = h/λ = 6.63 x 10-34/2.4 x 10-12 = 2.7 x 10-22 kg•m/s |

e.

|Zero (p = 0 before the event ∴ after p = 0) |

9. a.

|Fc = mv2/r = m2g ∴ v =( m2gr/m1)½ |

|v = 2πr/T ∴ T = 2πr/v = 2πr(m1/m2gr)½ = 2π(m1r/m2g)½ |

b.

|T2 = 4π2m1r/m2g = (4π2m1r/g)(1/m2) |

|The quantities that should be graphed are T2 and 1/m2. |

c.

|1/m2 (kg-1) |50 |25 |17 |12.5 |

|m2 (kg) |0.020 |0.040 |0.060 |0.080 |

|T (s) |1.40 |1.05 |0.80 |0.75 |

|T2 (s2) |1.96 |1.10 |0.64 |0.56 |

d.

T2 (s2)

|2.0 | | | | | | |

| | | | | | | |

1/m2 (kg-1)

e.

|slope = (1.96 – 0.56 s2)/(50 – 12.5 kg-1) = 0.037 kg•s2 |

|slope = 4π2m1r/g ∴ g = 4π2m1r/slope |

|g = 4π2(0.012 kg)(0.80 m)/0.037 kg•s2 = 10 m/s2 |

10. a.

|L1 + L2 = L1' + L2' |

|rβ1m1v1 + 0 = 0 + rβ2m2v2' |

|d(1/3)M1v1 = d(1)M2v2' ∴ v2' = M1v1/3M2 |

b.

|Kr1 + Kr2 = Kr1' + Kr2' |

|½β1m1v12 + 0 = 0 + ½β2m2v2'2 |

|1/3M1v12 = (1)M2(M1v1/3M2)2 ∴ M1/M2 = 3 |

c.

|L1 + L2 = L1' + L2' |

|dM1v1/3 + 0 = 0 + xM1v2' ∴ v2' = dv1/3x |

|K1 + K2 = K1' + K2' |

|½β1M1v12 + 0 = 0 + ½β2M1v22 |

|1/3(M1v12) = (1)M1(dv1/3x)2 ∴ x = d/√3 |

11. a.

|E = 1240 eV•nm/λnm |

|λ = 1240 eV•nm/1.02 x 106 eV = 1.22 x 10-3 nm |

|1.22 x 10-3 nm x 1 x 10-9 m/nm = 1.22 x 10-12 m |

b.

|pγ = h/λ = (6.63 x 10-34 J•s)/(1.22 x 10-12 m) |

|pγ = 5.43 x 10-22 kg•m/s |

c.

|pγ = pAl = mv |

|5.43 x 10-22 kg•m/s = (4.48 x 10-26 kg)v |

|∴ v = 1.21 x 104 m/s |

d.

|K = ½mv2 |

|K = ½(4.48 x 10-26 kg)(1.21 x 104 m/s)2 = 3.28 x 10-18 J |

12. a.

[pic]

b.

|Virtual because the image forms on the same side of the lens as the |

|object. |

c.

|1/do + 1/di = 1/f |

|1/6 + 1/di = 1/10 ∴ di = -15 cm |

d.

|1/do + 1/di = 1/f |

|1/9 + 1/di = 1/10 ∴ di = -90 cm |

|The first situation: M = -di/do = -(-15)/6 = 2.5 |

|The second situation: M = -di/do = -(-90)/9 = 10 |

|The greater magnification, second situation, produced a larger image. |

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