Quiz 1This is a photograph of an apple and feather free fall in a evacuated chamber. The

apple and feather are released from the top. Suppose the camera opened flash every 1 s.

(1) Suppose A1 is the distance from t = 0s to t = 1 s, A2 is the distance from t = 1s to t =2s, A3 is the distance from t = 2s to t = 3s ……

what is the ratio of A1: A2: A3…….(2) If camera opened flash every 0.06s, is the ratio of the distance between every

0.06 s the same as above or change? Why?

Answer: The ratio is 1:3:5 :7……(2n -1), here n = 1, 2, 3,

• Show it in general case, suppose the time interval is Δt

The distance from 0 –one Δt, A1 = ½*g*(Δt)2

(notice here I consider the distance, so I neglect (-) sign)

The distance from Δt---2 Δt, A2 = ½*g*(2Δt)2 -½*g*(Δt)2

= ½*g*(Δt)2 (22 -1)

Using the same way the distance from 2Δt---3 Δt,

A3 = ½*g*(Δt)2 (32 - 22)

A4 = ½*g*(Δt)2 (42 - 32) …….

An = ½*g*(Δt)2 (n2 – (n-1)2)

= ½*g*(Δt)2 (n2 – n2 +2n-1)

= ½*g*(Δt)2 (2n-1)

Therefore the ratio should be 1:3:5…….(2n-1), the ratio is independent of Δt

quiz2(1) A simple method of measuring the static coefficients of friction: Suppose a block is placed on a rough surface inclined relative to the horizontal as show in the Fig. The incline angle is increased until the block stars to move. Measuring the critical angle θc at which this slipping just occurs, you can obtain μsProof: μs=tan θc

(2) The 10.2 Kg block is held in place by massless rope passing over two massless, frictionless pulleys. Find the tensions T1, T2, T3, T4, T5 and magnitude of force F.

Solution:(1) If the block is not moving, Newton second law applied to the block. For this balanced situation

When the incline angle is increased until the blockIs on the verge of slipping, the force of static Friction has reached its maximum valueThe angle in this situation is the critical angle θc, in that situation, fsmax = μs N= μs mgcosθc = mg sinθc Solve the above equations, μs N= μs mgcosθc = mg sinθc - μs=tan θc

cos 0yF N mg

sin 0xF mg fs

(2)

Quiz 3 You have a new job designing rides for an amusement park. In one wide, the rider’s chair is attached by a 9.0-m-long chain to the top of a tall rotation tower. The tower spins the chair at rate of 1.0rev every 4.0 s. You’ve assumed that the maximum possible combined weight of the chair and rider is 150 Kg. You’ve found a great price for chain at the local discount store, but your supervisor wonders if the chain is strong enough. You contact the manufacturer and learn that the chain is rated to withstand a tension of 3000N. Will this chain be strong enough for the ride? Why?

Solution:

Newton’s second law along the r-axis isT sinθ = Fr = mω2rNotice here, r is the radius of the circle, it is not the

length of chain. r=L sinθ, also tension T is along the chain, it is not along the r-axis.

From above equation: T sinθ = mω2r = mω2Lsinθ→ T = mω2L = 3330N

Thus, the 3000 N chain is not strong enough for the ride.

Quiz4: Figure shows a contest in which a sphere, a cylinder, and circular hoop all of mass M and radius R( notice particle is not taking the race). are rolling down from rest at the same instant of time. ( please read text book P 366-367 first)

(1) What is the speed of center of mass for each of the object when they reach the bottom

(2) What is acceleration of the center of mass for each?(3) Which will win the downhill

race? why

If we choose the bottom of the ramp as the zero point of potential energy, from energy conservation

2 21/ 2 1/ 2 , c is constant

and

cm cm cm

cm

I MV Mgh I cMR

V R

Here C is a constant that depends on the object’s geometry. C = 1 for hoop, =1/2 for cylinder= 2/5 for sphere From above equation

for hoop, for cylinder, for sphere2 4 10

, 1 3 7

gh gh ghcm cm

cV V gh

2 sin2 / sin ,

1

gVcm a S S h a

c

for the hoop,

for the cylinder,

for the sphere

sin / 2

2 sin / 3

5 sin / 7

cm

cm

cm

a g

a g

a g

Continue:

1 2 (1 )/

sincm cm cm cm

h cV a t t V a

g

Since they have the same h and θ, the greater of c, the longer of time.Hoop’s c =1> cylinder’s c = ½, > sphere’s c = 2/5, that results inSphere taking shortest time, cylinder second, hoop is the last

2/ sin 1/ 2

1 2 (1 )2 /(sin )

sin

cm

cm

S h a t

h ct h a

g

Quiz 5: Two equal masses are attached to identical ideal springs next to one another. One mass is pulled so its spring stretches 20 cm and the other pulled so its spring stretches only 10 cm. The masses are released simultaneously. Which mass reaches the equilibrium point first? Explain your reasoning

Quiz 5: Answer: They reach the equilibrium point at the same time

2m

Tk

1

( ) cos( ) 0 cos( ) 0

cos (0)2

2/ 2, , / 4

2

x t A t t

t

t t t TT

If they have the same m and k, they will have the same period. The maximum displacement does not effect the period.

The time reaches the equilibrium position should be T/4, since two spring-mass systems have the same period, two masses should reach the equilibrium position at same time.

Note: at equilibrium position x(t) = 0, the function for this oscillation is

Quiz 6:(1)In Isothermal process, the work on the ideal gas as the volume changes from Vi to Vf is:

Derive it(2) A mole of gas at temperature 20oC compressing at constant temperature until the gas pressure double. What is the work done on the gas?What is the thermal energy change in this process? How much heat released?(1) start from the definition:

ln( )f

ViW nRT

V

(ln ln ) ln

vf vf

vi vi

ii f

f

nRTW pdV dV

VV

nRT V V nRTV

2. A mole of gas at temperature 20oC compressing at constant temperature until the gas pressure double. What is the work done on the gas?What is the thermal energy change in this process? How much heat released?

2. Answer:Notice when temperature is a constant, gas pressure double, the volume should be half, Vi/Vf = 2. From above equation:W = nRT ln(2) = 1* 8.31 * 293* ln(2) = 1687 J ( Work should be positiveThat means external force does the work on the gas)In isothermal process, ΔT = 0 so the thermal energy change ΔE = 0 too.ΔE = W + Q, so Q = -1687 J, negative sign means heat is released from gas