AP Physics



AP Physics Name ________________________

Fall Semester Practice Free Response Period _____

Free Response: Answer the questions on a separate sheet of paper.

1. A physics student kicks a ball across the football field. It travels 60 m in 4.0 seconds.

a. What is the x-component of the initial velocity?

b. What is the y-component of the initial velocity?

c. What is the magnitude and direction of the ball’s initial velocity?

d. Sketch the following graphs: dx vs. t, vx vs. t, ax vs. t, dy vs. t, vy vs. t, and ay vs. t.

2. A 10-kg box is pulled with a horizontal force of 50 N along a horizontal surface. The force of friction is 10 N.

a. Label all the forces acting on the box.

b. Determine the acceleration of the box.

c. Determine how long it takes the box to travel 8 m.

d. Determine the velocity of the box after moving 8 m.

e. Determine the coefficient of kinetic friction μk.

3. A 0.002-kg bullet traveling at 1000 m/s hits a 2-kg block of wood and becomes embedded in it. Determine

a. The block's velocity after it is struck by the bullet.

b. The distance the block slides along a surface with a coefficient of friction, μ = 0.25.

c. The height of the block when it swings upward on the end of a 2-m string.

4. A 2.5-kg bob is suspended from a light spring (k = 25 N/m). The bob is pulled upward 4 cm and then released. The system is set into simple harmonic motion.

[pic]

a. What is the period of vibration?

b. What is the frequency of vibration?

c. What is the amplitude in meters?

d. What is the maximum acceleration?

e. What is the maximum velocity?

f. What is the maximum kinetic energy?

g. What is the maximum potential energy?

h. Complete the following chart.

|t |0.5 s |1 s |1.5 s |2 s |

|x | | | | |

|v | | | | |

|a | | | | |

|F | | | | |

|U | | | | |

|K | | | | |

5. A 800 kg crate is attached to the bed of a 5000 kg truck by a spring (k = 5,000 N/m).

[pic]

The truck accelerates to 25 m/s in 10 s along a straight, level highway,

a. What is the force on the crate during acceleration?

b. What is the coefficient of friction between the crate and truck bed if the spring stretches 0.08 m during the acceleration?

The truck reaches its cruising speed.

c. Does the crate move forward?

6. A 70 kg woman and her 35 kg son are standing at rest on an ice rink. They push against each other for a time of 0.60 s, causing them to glide apart. The speed of the woman immediately after they separate is 0.55 m/s. Assume that during the push, friction is negligible compared with the forces the people exert on each other.

[pic]

a. Calculate the average force exerted on the son by the mother during the push.

b. What is the son's speed after they separate?

The friction force, although small, is not negligible and the mother comes to rest after moving a distance of 7.0 m.

c. What is the coefficient of friction?

d. If their coefficients of friction are the same, how far does the son move after the push?

7. An experiment is performed using the apparatus. A small disk of mass m1 on a frictionless table is attached to one end of a string. The string passes through a hole in the table and an attached narrow, vertical plastic tube. An object of mass m2 is hung at the other end of the string. A student holding the tube makes the disk rotate in a circle of constant radius r, while another student measures the period T.

[pic]

a. Derive the equation T = 2π(m1r/m2g)½ that relates T and m2. (hint: Fg-m2 = Fc-m1)

Quantities related to m2 and T can be graphed so that the slope of the line can be used to compute an experimental value for g.

b. What quantities should be graphed to yield a straight line with a slope that could be used to determine g? (hint: rearrange the equation from part (a) into an equation for a straight line (y = mx) so that the slope of the best fit line can be used to calculate g.)

The procedure is repeated, and the period T is determined for four values of m2, where m1 = 0.012 kg and r = 0.80 m.

c. Use the blank rows in the data chart to calculate the quantities from part (b)

| | | | | |

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

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

| | | | | |

d. Plot the quantities calculated above, label the axes, and draw the best-fit line to the data.

e. Use your graph to calculate the experimental value of g.

8. A rod (β = 1/3) of mass M1 and length d rotates about point P and strikes ball of mass M2 elastically with velocity v1 (figure 1) After the collision the rod is stationary and the ball moves to the right. Answer the questions in terms of M1, M2, v1 and d.

Figure 1 Figure 2

P P

d x

d ϒ

M1

v1 ϒ v1

M1 M2 M1

a. Using the conservation of angular momentum, determine the velocity v2 of the ball after the elastic collision.

b. Using the conservation of rotational kinetic energy, determine the numerical value of the ratio M1/M2?

A new ball with the same mass, M1, as the rod is placed a distance x from the pivot, P, as shown in figure 2.

c. For what value of x will the rod stop moving after hitting the ball elastically? (Hint: repeat the calculations from part (a) and part (b) with the new conditions.)

9. Following a nuclear reaction, a nucleus of aluminum is at rest in an excited state represented by 2713Al*, as shown below left. The excited nucleus returns to the ground state 2713Al by emitting a gamma ray photon of energy 1.02 MeV, as shown below right. The aluminum nucleus in the ground state has a mass of 4.48 x 10-26 kg. [pic]

a. Calculate the wavelength of the emitted photon (in m).

b. Calculate the momentum of the photon (in kg•m/s).

c. Calculate the speed of the recoiling nucleus (in m/s).

d. Calculate the kinetic energy of the recoiling nucleus (in J).

10. A thin converging lens L of focal length 10.0 cm is used as a simple magnifier to examine an object O that is placed 6.0 cm from the lens.

a. Draw a ray diagram showing at least two incident rays and the position and size of the image formed.

[pic]

b. Justify whether the image is real or virtual.

c. Calculate the distance of the image from the center of the lens.

d. The object is now moved 3.0 cm to the right, as shown below. How does the height of the new image compare with that of the previous image? Justify your answer.

[pic]

Answers

|1 |a |vxo = dx/t = 60 m/4.0 s = 15 m/s |

| |b |vy = vyo + at → -vyo = vyo + (-10 m/s2)(4.0 s) ∴ vy = -20 m/s |

| |c |v = (vxo2 + vyo2)½ = (152 + 202)½ = 25 m/s |

| | |tanθ = (vyo/vxo) = (20/15) ∴ θ = 53o |

|2 |b |Fnet = ma → (50 N – 10 N) = (10 kg)a ∴ a = 4 m/s2 |

| |c |d = vot + ½at2 → 5 m = 0 + ½(4 m/s2)t2 ∴ t = 2 s |

| |d |vt = vo + at = 0 + (4 m/s2)(2 s) = 8 m/s |

| |e |Ff = μFn = μFg → 10 N = μ(100 N) ∴ μ = 0.1 |

|3 |a |mbulletvbullet + mblockvblock = (mbullet + mblock)v’ |

| | |(0.002 kg)(1000 m/s) + 0 = (0.002 kg + 2 kg)v’ ∴ v’ = 1 m/s |

| |b |K = W = Ffrd → ½mv2 = μmgd |

| | |½(1 m/s)2 = (0.25)(10 m/s2)d ∴ d = 0.2 m |

| |c |K = U → ½mv2 = mgh → ½(1 m/s)2 = (10 m/s2)h ∴ h = 0.05 m |

|4 |a |2 s |

| |b |f = 1/T = 1/2 s = 0.5 s-1 |

| |c |4 cm x 1 m/100 cm = 0.04 m |

| |d |aA = A(k/m) = (0.04 m)(25 N/m)/(2.5 kg) = 0.4 m/s2 |

| |e |vo = A(k/m)½ = (0.04 m)[(25 N/m)/(0.25 kg)]½ = 0.126 m/s |

| |f |Ko = ½mv2 = ½(2.5 kg)(0.126 m/s)2 = 0.02 J |

| |g |UA = ½kA2 = ½(25 N/m)(0.04 m)2 = 0.02 J |

| |h |t |0.5 s |1 s |1.5 s |2 s |

| | |x |0 m |-0.04 m |0 |0.04 m |

| | |v |-0.126 m/s |0 |0.126 m/s |0 |

| | |a |0 |0.4 m/s2 |0 |-0.4 m/s2 |

| | |F |0 |1 N |0 |-1 N |

| | |U |0 |0.02 J |0 |0.02 J |

| | |K |0.02 J |0 |0.02 J |0 |

|5 |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 |Fforward = Fs – Ff = kx – μFn |

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

| | |Fforward = 400 N – 1600 N = -1200 N∴ the crate will not move forward|

| | |because friction can't move the crate. |

|6 |a |FΔt = mΔv ∴ F = mwvw/Δt = (70 kg)(0.55 m/s)/0.60 s = 64 N |

| |b |mwvw = msvs → (70 kg)(0.55 m/s) = (35 kg)vs ∴ vs = 1.10 m/s |

| |c |K = Wf = Ffd → ½mv2 = μmgd |

| | |½(0.55 m/s)2 = μ(10 m/s2)(7 m) ∴ μ = 0.0022 |

| |d |K = Wf = Ffd → ½mv2 = μmgd |

| | |½(1.10 m/s)2 = (0.0022)(10 m/s2)d ∴ d = 28 m |

|7 |a |Fc = mv2/r = m2g ∴ v =( m2gr/m1)½ |

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

| |b |P2 = 4π2m1r/m2g = (4π2m1r/g)(1/m2) |

| | |The quantities that should be graphed are P2 and 1/m2. |

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

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

| |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 |

|8 |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 ® M1v12/3 = M2M12v12/9M22 ∴ 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 → M1v12/3 = M1d2v12/9x2 ∴ x = d/√3 |

|9 |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 = ½(4.48 x 10-26 kg)(1.21 x 104 m/s)2 = 3.28 x 10-18 J |

|10 |a |Parallel ray turns toward left focus, but image forms on the right |

| | |side, when the line is extended to the right. |

| | |Straight ray goes to the center, but intersects focus ray on the |

| | |right side, when the line is extended to the right. |

| | |Image forms at -15 cm. |

| |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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