Physics 110

Fall 2009

Exam #2

October 16, 2009

Name______________

In keeping with the Union College policy on academic honesty, it is assumed that you will neither accept nor provide unauthorized assistance in the completion of this work.

Part Multiple Choice / 10

Problem #1 / 18 Problem #2 / 18 Problem #3 / 36 Problem #4 /18

Total / 100

Part I: Free Response Problems Please show all work in order to receive partial credit. If your solutions are illegible

no credit will be given. Please use the back of the page if necessary, but number the problem you are working on. Each subpart of a problem is worth 9 points.

1. The Earth revolves about the Sun in a nearly circular orbit, one time every 365 days

at a mean orbital radius of 1.5x1011m.

a. Starting from Newton’s Law of Gravity, derive an expression for the orbital period of the Earth as a function of the Earth’s radius from the Sun.

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FG =G MEMS

r2= MEac = ME

vE2

r= ME

2πrT

2

r→ T 2 =

4π 2

GMS

r3

, which is called Kepler’s third law for planetary motion.

b. From the result in part a, what is the mass of the Sun?

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T 2 =4π 2

GMS

r3 → MS =

4π 2

GT 2

r3 =

4π 2

6.67 ×10−11 Nm 2

kg 231.7 ×106s( )2

1.5 ×1011m( )

3= 2 ×1030kg

2. A ballistic pendulum can be used to measure the speed of a projectile, such as a bullet. The ballistic pendulum, shown on the right, consists of a stationary 2.5kg block suspended by a wire of negligible mass. A 10mg bullet is fired into the block, and the block (with the bullet in it) swings to a maximum height of H = 0.65m above the initial position.

a. Write the equation that governs conservation of momentum for this situation.

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Δp = 0→ mbvv = mb + mp( )V

b. Using energy ideas after the collision and your result from part a, what is the initial velocity of the bullet?

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ΔKE + ΔUg = 12 mb + mp( )v f2 − 1

2 mb + mp( )V 2( ) + mb + mp( )gy f − mb + mp( )gyi( ) = 0

V = 2gH = 2 × 9.8 ms2× 0.65m = 3.56 m

s

∴vb =mb + mp

mb

V =

0.01kg + 2.5kg0.01kg

3.65 m

s = 916.2 ms Note: the bullet’s mass should be 10g not 10mg for a

reasonable speed. If you used 10mg then you get a factor of 1000 larger for the speed.

y = 4.9348x + 0.2961 

0 

1 

2 

3 

4 

5 

0  0.2  0.4  0.6  0.8  1 

3. A mass hung from a vertical spring

a. Suppose that you attach a massless spring to the ceiling and that you hang a 500g (= mh) mass from the spring. When the mass comes into equilibrium it is found that the mass stretches the spring by 0.61m. Starting from a free body diagram, what is the spring constant?

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ky −mg = 0→ k =mgy

=0.5kg × 9.8 m

s2

0.61m= 8.03 N

m

b. Suppose that the spring is pulled down a further 0.1m and released from rest. What are the magnitudes of the maximum velocity and maximum acceleration and the period of the resulting motion?

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vmax =ωA =kmA =

8 Nm

0.5kg× 0.1m = 0.4 m

s

amax =ω 2A =kmA =

8 Nm

0.5kg× 0.1m =1.6 m

s2

T = 2π mk

= 2π 0.5kg8 Nm

=1.57s

c. Suppose, as in real life, that the spring is not in fact massless, but has mass ms = 200g. We are going to try to determine how much of the spring’s mass contributes to the period of oscillation. Starting from the definition for the period of a mass-spring system, define an effective mass term, which is the sum of two contributions. One is due to the mass that is hanging, mh, and the other is due to the mass of the spring, αms, where α is a fraction of the springs mass and

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0 ≤α ≤1. Derive an expression for the square of the period of the mass-spring system versus the hanging mass, mh. (Hint, the result should be a linear function of mh.)

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T = 2π mtotal

k= 2π mh +αms

k→ T 2 =

4π 2

kmh +

4π 2

kαms

d. Below is data taken on the period for an oscillating mass on the end of a massive

spring. From the equation of the line given, what is the value of α, using your result from part c?

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4π 2

kαms = 0.2961s2 →α =

0.2961s2 × k4π 2ms

∴α =0.2961s2 × 8.0 N

m

4π 2 × 0.2kg= 0.3

T2 (s2)

mh (kg)

4. Suppose that you are given the arrangement of blocks as shown in the diagram below. Block #1 has mass m1 = 2kg and block #2 has mass m2 = 4kg. The blocks are connected by a massless rope and pass over a massless frictionless pulley. The surface that the blocks ride along is frictionless, block #1 is connected to a wall by a spring with stiffness k = 36N/m, and the track is inclined at an angle of θ = 30o.

a. Suppose that block #2 is released from rest, when the spring is initially

unstretched. Using energy ideas, what is the maximum extension of the spring?

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ΔKE1 + ΔKE2 + ΔUg,1 + ΔUg,2 + ΔUs = m2gy f −m2gyi( ) + 12 kxmax

2 − 12 kxi

2( ) = 0

∴ 12 kxmax

2 = m2gyi = m2gxmax sinθ → xmax =2m2gsinθ

k=2 × 4kg × 9.8 m

s2× sin30

36 Nm

=1.1m

b. Suppose again, that the spring is at its equilibrium position. If block #2 is released again from rest, what is the speed of block #2 when the spring has been stretched by an amount xf = 0.5m?

€

ΔKE1 + ΔKE2 + ΔUg,1 + ΔUg,2 + ΔUs = 12m1v1 f

2 − 12m1v1i

2( ) + 12m2v2 f

2 − 12m2v2i

2( ) + m2gy f −m2gyi( ) + 12 kxmax

2 − 12 kxi

2( ) = 0

∴ 12 m1 + m2( )v f2 = m2gyi − 1

2 kx f2 → v f =

2gm2yi − kx f2

m1 + m2

=2 × 4kg × 9.8 m

s2× 0.5m × sin30 − 36 N

m 0.5m( )2

6kg=1.3 m

s

m1 m2

k

θ

Part II: Multiple-Choice Circle the best answer to each question. Any other marks will not be given credit. Each multiple-choice question is worth 2 points for a total of 10 points.

1. A dart is loaded into a spring-loaded toy dart gun by pushing the spring in by a

distance d. For the next loading, the spring is compressed a distance 2d. How much work is required to load the second dart compared to that required to load the first?

a. It takes four times as much work. b. It takes two times as much work. c. It takes the same amount of work. d. It takes half as much work.

2. A person hoists a bucket of water from a well

and holds the rope, keeping the bucket at rest as in the left photo. Call this situation A. A short time later the person ties the rope to the bucket so that the rope holds the bucket in place as in the right photo. Call this situation B.

a. The tension in situation B is greater than in situation A.

b. The tension in situation B is equal to that in situation A. c. The tension in situation B is less than that in situation A. d. Three is no way that the individual tensions in the ropes can be determined. 3. For an object undergoing uniform circular motion

a. the velocity is always parallel to the centripetal acceleration. b. the displacement of the particle is always perpendicular to the velocity. c. the net acceleration is always vertically down. d. the velocity is not constant and thus the particle must accelerate.

4. Suppose that a bowling ball and a baseball are thrown off of a high building with the

same magnitude of the velocity. Let the bowling ball be thrown horizontally while the baseball is thrown upward at an angle θ with respect to the horizontal. Ignoring air resistance, the balls

a. have the same magnitude of the velocity at the bottom. b. vbaseball > vbowling ball at the bottom. c. vbowling ball > vbaseball at the bottom. d. cannot tell from the information given.

5. To accelerate a car from rest to a speed v requires the work W1. The work needed to accelerate the car from v to 2v is W2. Which of the following is correct?

a. W2 = ½ W1 b. W2 = W1 c. W2 = 2W1 d. W2 = 3W1

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ω = 2πf =2πT

TS = 2π mk

TP = 2π lg

v = ±kmA 1− x 2

A2

12

x t( ) = Asin 2πtT( )

v t( ) = A kmcos 2πt

T( )

a t( ) = −A kmsin 2πt

T( )

v = fλ =FTµ

fn = nf1 = n v2L

I = 2π 2 f 2ρvA2

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g = 9.8m s2 G = 6.67×10−11 Nm 2

kg2

NA = 6.02×1023 atomsmole kB = 1.38×10−23 J Kσ = 5.67×10−8 W m 2K 4 vsound = 343m s

Useful formulas: Motion in the r = x, y or z-directions Uniform Circular Motion Geometry /Algebra

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rf = r0 + v0rt + 12 art

2 ar =v 2

r

v fr = v0r + art Fr = mar = m v 2

r

v fr2 = v0r

2 + 2arΔr v =2πrT

FG =G m1m2

r2

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Circles Triangles SpheresC = 2πr A = 1

2 bh A = 4πr2

A = πr2 V = 43 πr

3

Quadratic equation : ax 2 + bx + c = 0,

whose solutions are given by : x =−b ± b2 − 4ac

2a

Vectors Useful Constants

Linear Momentum/Forces Work/Energy Heat

Rotational Motion Fluids Simple Harmonic Motion/Waves Sound