9702_m16_qp_52 Question 1

 

Defining the problem 

• k is the independent variable and h is the dependent variable 

• keep the mass of the object constant 

Methods of data collection

• Set-up the equipment as shown in the diagram.

• Fix a rod using a boss at the near top of a retort stand.

• Tie the elastic cord on the rod, The other end to be tied at the hook of the hanging mass.

• Fix a ruler with its 0 mark at the same level as the top fixed end of cord using clips or clamps. 

• Fix an electromagnet so that the end of the nail is at the same level as the fixed end of the cord and the 0 mark of ruler.This is needed to ensure that the starting point is the same throughout the experiment.

• Measure the mass m using a weighing scale. 

• To determine the spring constant k of each cord, 

• hang a few weights, one at a time, 

• determine the corresponding extensions by subtracting the original length L from the length after a weight is hanged, and

• plot a graph of weight against extension and determine the gradient

• The gradient is the k of a cord

• For each cord with its spring constant k already determined,

• tie the mass m previously selected, then measure the length L between the two ends

• release the mass m from the upper fixed end [the ruler’s 0 mark] of the cord  

• measure h or maximum distance fallen by the object using the ruler

Method of analysis 

•  Plot a graph of (h - L)2 / h against 1/ k 

•  To determine g

• calculate the gradient of the graph

• use this gradient to determine g

g = gradient / 2m  

•  The relationship is valid if the graph is 

• a straight line 

• passing through the origin.

Additional detail

• When releasing or dropping the mass, use safety goggles. Place a sand bag directly below the dropping so that it catches the mass m thereby protecting the investigator and the floor.

• Use cords that 

• are elastic but, with the assumed hanging mass m, will not hit the floor or the surface of the table where  the experiment is being done. Do trials first.

• obey Hooke’s Law, that is, with the hanging mass m chosen, it will not reach its elastic limit.

• The retort stand is usually vertical. In case unsure, check using a set square.

• For each cord, repeat the experiment several times, then determine the average h.

9702_s19_qp_41 Question 5

 

Answer:

• Electric field strength is the force per unit positive charge.

Answer:

• Electric field is a vector quantity and that at any distance x, there are two fields - one due to A and one due to B.

• These 2 fields are combined to determine the electric field strength E at point P.

• Based on the graph, 

• E is zero at x = 10 cm.

• E is positive at x < 10 cm.

• E is negative at x > 10 cm.

• These are only possible if charges A and B have the same sign.

Answer:

• If there is only one charge creating the field, then E and x follow the inverse-square law.

• In the situation in this question, the field E is due to a combination of two fields.

Answer:

• Electric field strength E is given by

E = kQ / x2 .

• Let EA and EB be the field due to charge A and charge B, respectively.

• We know from the graph that at x = 10 cm, EA and EB.

EA = EB

kQA / x2 = kQB / (15 - x)2

QA / 102 = QB / 52

QA / QB = 4.0

9702_s20_qp_42 Question 4

Answer:

• Angular frequency ω = 2π / T. 

• From the diagram, we can see that the time it takes for one complete oscillation or the period T is 2.2 s.

ω = 2π / T

ω = 2(3.14) / 2.2

ω = 2.9 rad s-1

Answer:

• The equation given for acceleration is a = - (g / R) x.

• The defining equation of simple harmonic motion is a = - ω2x.

• We therefore have

g / R = ω2

R = g / ω2

where g is 9.81 m s-2 and ω is the value calculated in (a)

R = 1.2 m

Answer:

• We know from Conservation of Energy that the speed of the ball at that point is maximum.

• At that maximum speed vO, the displacement is also maximum. This is called amplitude xO.

• We can use the formula,

vO = ωxO

vO = 2.9 x 3.0 x 10-2

vO = 0.087 m s-1

Answer:

• An example is the following

• The orange curve

• starts from (3.0 cm, 0 s)

• has the same period as the original graph

• displacements [including amplitudes] are lower since energy is lost / damping is happening

 

9702_w18_qp_42 Question 3

 

Answer:

• Specific latent heat of fusion is the energy per unit mass to cause change of state between solid and liquid at constant temperature.

Answer:

• The following are possible observations:

• Rate of increase in mass of beaker and water is constant

• Level of water rises at a constant rate

• Volume of water in beaker increases at a constant rate

• Constant time between drops

Answer:

• The question is about calculating power.

power P = ΔE / t

and since ΔE = mLf, then P = mLf / t

P = (185.0 - 121.5) (332) / (5.00) (60)

P = 70.3 W

Answer:

• The rate at which energy is transferred to the system is calculated above. 

• From the data collected, this energy could have come from the electrical power supplied.

P = IV

P = 12.8 x 4.60

P = 58.9 W

• The power supplied is less than the rate of energy transferred to ice. This means that another source of energy is present i.e. surrounding. The rate at which energy is gained from the surroundings, therefore is

70.3 W - 58.9 W

11.4 W

9702_w18_qp_42 Question 2

 

Answer:

• An ideal gas is a gas that obeys the equation pV/T = constant where p is pressure, V is volume, and T is temperature of the gas.

Answer:

• An equation linking the macroscopic properties p and V of an ideal gas to its microscopic properties is pV = ⅓ Nm<c2>, where

• p and V are pressure and volume, respectively

• Nm is the number of atoms times the mass of each atom, thus, the mass of the gas

• <c2> is the mean-square speed

• Substituting the given values, the <c2> is

• <c2> = 3pV/Nm

• <c2> = 3 (2.12 x 107) (1.84 x 10-2) / (3.20)

• The root-mean-square speed is the square root of <c2>, so the answer is 605 m s-1.

Answer:

• The amount n is calculated from the rearranged Ideal Gas equation.

n = pV/RT

• The equation gives the n when all quantities are expressed in SI units. This means that the temperature T must be converted to kelvin before plugging in values.

n = (2.12 x 107) (1.84 x 10-2) / (8.31 x 295)

• Therefore, n = 159 mol.

Answer:

• The 2 versions of Ideal Gas Law i.e. pV = nRT and pV = NkT, when combined gives

nR = Nk

• The mass M of a gas is equal to the number of atoms N in the gas multiplied by mass m.

M = Nm

so, N = M / m

• Combining the 2 equations above, we should get

nR = M k/ m

m = Mk / nR

• We also know that the Boltzmann constant k is the ratio of the gas constant R and Avogagadro’s number NA.

k = R / NA

k / R = 1 / NA

So our working equation becomes

mass of one atom m = M / (nNA)

where M is the mass of the gas and NA is the Avogadro’s constant.

• The mass m, therefore is, 

m = 3.20 / (159 x 6.02 x 1023) 

m = 3.34 x 10-26 kg