2015 a Level Physics H2 Paper 2 Answers

2015 A level Physics H2 Paper 2 Suggested answers 1a.

Energy stored E = ½ Fx = ½ (450 x 10-3 x 9.81) x (70 – 40) x 10-2 = 0.662 J

bi.

Between point of release and l = 60 cm (equilibrium position as obtained from graph): GPE converted to elastic pe and ke. Between l = 60 cm and lowest point: GPE and ke converted to elastic pe.

ii.

GPE lost = ke and elastic pe gained mgh = ½mv2 + E

E  2 gh   m v= 

0.662   2 9.81  30.0  10  2   300  10 3  

= = 1.21 m s-1 iii.

Using F = kx,

F k= x 450  10 3  9.81 2 =  70  40   10 = 14.72 N m-1 GPE lost = elastic pe gained mgh’ = ½kh’2

2mg Distance fallen, h’ = k 2  300 3  9.81 14.72 = = 0.400 m 2a.

If V  I, then V/I should be constant. Choose any 2 points on the line to check. (Remember, gradient  R) When V = 12.0 V, I = 2.50 A. V/I = 12.0/2.50 = 4.80 V A-1 When V = 6.0 V, I = 1.25 A V/I = 6.0/2.50 = 4.80 V A-1 1

bi.

Since V/I is a constant, I is proportional to V. RX = 4.80  (from ai)

ii1.

12 IAB = 4.0  5.0 = 1.33 A

ii2.

12 IX = 4.8  2.7 = 1.60 A

ii3

VAC = 0.75VAB = 0.75 IABRAB = 0.75 x 1.33 x 4.0 = 3.99 V VAD = IXRX = 1.60 x 4.80 = 7.68 V VCD = 7.68 – 3.99 = 3.69 V

3a.

The energy of each photon must be larger than the work function of the metal, which is the energy needed to liberate an electron from the surface of the metal. The excess energy becomes the ke of the liberated electron.

bi.

The minimum pd will cause all the ke of the fastest electrons to be converted to electrical pe.

ii.

This is because the collector is already capturing 100% of all the electrons which are emitted from the emitter.

c.

hf = eVs + 

eVs   h f= 1.60  10 19  2.2  1.8 6.63  10 34 = = 9.65 x 1014 Hz

d.

(Draw a graph with the same stopping potential, but twice the saturated current)

4a.

 2    r ac =  T 

2

2

2  5  =  1.53  10  

2

 4.22  10 8



= 0.712 m s-2 bi.

Centripetal force on moon is provided by gravitational force on moon. mac = mg  ac = g

GM Jupiter ii.

ac =

 ac  Amalthea     ac  Io  rIo rAmalthea



r2 2



rIo



rAmalthea 

 ac  Amalthea  ac  Io 3.87 0.712

= = 2.33 5ai.

ii.

FA = qE = 2 x 1.6 x 10-19 x 1.5 x 105 = 4.8 x 10-14 N

q2 2 FBA = 4 0 r

 2  1.6  10  4  8.85  10   4.0  10  19 2

= = 5.75 x 10-5 N b.

12

12 2

The forces that A and B exert on each other are considered to be internal forces, and cancel out. The forces that the electric field exerts on the charges are equal and opposite, and also cancel out, since the charges are equal and opposite in polarity.

c.

6a. b.

Torque = FA Lsin60o = 4.8 x 10-14 x 4.0 x 10-12 x sin60o = 1.66 x 10-25 N m 128 54

D C = C0 e-t

3

ln2  - 1/2 t = ln(C/C0) ln2  1/ 2   ln(C/C 0 ) t ln2  = ln(28/175 ) x (6000 - 2000) = 1.51 x 103 s 7ai.

The theory suggests that y1 = kMgl r d1-3b-1, and y2 = kMgl r d2-3b-1, which implies (y1/y2) = (d1/d2)-3. So, For the same l value, compare (y1/y2) with (d1/d2)3. If the 2 ratios are the same, then the theory is true. (d1/d2)3 = (5.00 x 10-3/6.00 x 10-3)-3 = 1.73 l/m 0.900 0.800 0.700 0.600 0.500 0.400

y1/y2 1.74 1.73 1.74 1.74 1.75 1.75

The experimental values are quite close to the theoretical value. So the data supports the suggestion. ii.

-3.170, -0.693

iv.

Gradient =[-1.16-(-3.96)]/[-0.04-(-0.94)] = 3.1

v.

Since y = kMgl r d-3b-1, then lny = rlnl + ln(kMgd-3b-1). Since plotting lny vs lnl gives a straight line, the equation is valid.

vi.

r = gradient =3

b.

From graph, y intercept = -1.04 = ln(kMgd-3b-1).

d 3 be 1.04 Mg k=

5.00  10 

3 3

= = 2.7 x 10-10  3 x 10-10

 3.00  10  2  e 1.04 0.500  9.81

4

This is because the equation is of the form a = - 2x, which shows that a and x are directly proportional to each other, and are in opposite directions.

ci. ii.

 2     T 

2



b kMd 3 l 3

T  2

kMd 3 l 3 b



3  10 10  0.500  5.00  10 3  2 3.00  10  2



3

= 0.58 s

0.600 3 Light bulb

8. Lid (with holes for wire, thermometer and stirrer)

A

Thermometer (held by retort stand)

V

Stirrer

Insulating cladding 





Beaker

Water

Put an empty beaker on a digital weighing machine. Press the tare button and set the reading to zero. Pour water at room temperature into the beaker. Note the reading on the weighing machine, which is the mass m of the water. Place the beaker in the insulating cladding. Solder connecting wires to the terminals of the bulb. Immerse the glass envelope of the bulb completely in water, taking care not to allow the water to short circuit the terminals. Thread the wires through the holes of the lid. Connect the bulb to the circuit shown. Place the thermometer and stirrer through the holes of the lid. Cover the beaker with the lid. Note the starting temperature of the water, i. Switch on the switch, and at the same time start a stopwatch. Gently stir the water occasionally. When the temperature reading rises by 10 oC, stop the stopwatch. Record the time as t. Also record the readings of the ammeter and voltmeter, I and V respectively, which are the current and pd across the bulb.

5

 

mc  is calculated by  = 1 - IVt , where c = specific heat capacity of water, and  = 10 oC. By adjusting the variable resistor, vary the pd across the bulb to get several sets of readings of  and V. Each time the starting water temperature should be the same. Since  = aVb, ln = blnV + lna Plot ln vs lnV. If a straight line is obtained, the equation is true, with b = gradient and lna = y intercept.

 

        

Control variables The same light bulb should be used throughout the whole experiment, as different bulb might have different efficiencies. The starting water temperature should be the same, to maintain the same temperature difference between the water and the surrounding, as it affects the rate of gain of temperature. Accuracy The bulb should not touch the side or bottom of the beaker, otherwise heat will be directly transferred from the bulb to the beaker, affecting the temperature of the water. The water should be gently stirred to ensure that the water temperature is uniform. The glass bulb should be fully immersed in water so as to maximize the transfer of thermal energy to the water. However, to get a large enough rise in temperature, the amount of water used should not be too much. The bulb should be immersed in the water before switching it on. Otherwise, if the hot bulb is suddenly immersed in water, the glass might break due to sudden contraction. The beaker should be insulated to reduce heat lost to surrounding. The thermometer should not touch the bulb, which is hotter than the water. Safety Care must be taken to prevent water from short circuiting the battery. To prevent the bulb from being short circuited, the wires can be taped against the lid after the position of the bulb has been adjusted to the desired position.

6