• Published on
05-Apr-2018

• View
218

0

Embed Size (px)

Transcript

• 7/31/2019 VCE Physics Answers 2012

1/7

Area of Study 1 - Motion in one and two dimensionsQuestion 1 a)The formula for kinetic energy should be used. Thus, the velocity is 3m/s. (You can check thisand see that * 1.2 * 9 is 5.4Question 1 b)Conservation of energy applies, and work converts elastic potential energy to kinetic energy. Sowork done is 5.4 J, as listed.Question 1c)The formula for elastic potential energy can be used here as conservation of energy applies. So

k = 1687.5 N/m.Question 1d)Impulse is equal to the change in momentum. This is mv, with v as previously calculated. Thus,it is 3.6 Ns.

Question 2The explanation should note that momentum is a vector. After the collision, the 1.2kg blockmoves backwards, and therefore has negative momentum (if the original direction of motion ispositive). Therefore, the 2.4kg block must move forward with a greater momentum than that ofthe initial momentum, to compensate for the negative momentum that the 1.2kg block has in thefinal state.

Question 3B is the correct answer. The two forces on the right have vertical components that can becancelled, and the horizontal contribution of these forces can be cancelled by the arrow on theleft.

Question 4 a)Cable A supports both spheres, so the magnitude is 30N. The direction is up.Question 4b)The reaction force is the force of the sphere on the earth (due to the spheres gravitational field),and this is directed upwards, or to the centre of mass of the sphere.

Question 5a)

As no acceleration occurs, the tension force is equal to the frictional forces, so 800 N.

Question 5b)Using Fnet = ma, F = ma + friction, which is equal to 300N + 400N = 700NQuestion 5c)

Smarter VCE Lectures

http://www.connecteducation.com.au/http://www.connecteducation.com.au/
• 7/31/2019 VCE Physics Answers 2012

2/7

Use kinematics equations.

v^2 = u^2+ 2ad= 16 + 20 m^2/s^2v = 6 m/sQuestion 5d)T1 pulls everything along, so rope 1 will be the first to break.2400N = ma + friction2400N = ma + 800N1600N = maa = 1600N / 1200kg = 1.33 m/s^2

Question 6a)Using kinematics equations(vsin60)^2 = 2ad = 20 * 15 = 300Thus v = 1/sin60 * (300)^0.5= 20m/sQuestion 6b)Time of the flight is equal to twice the time taken to reach the topd = 0.5at^2t = (2d/g)^0.5 = 3^0.5 = 1.732 sThe total time for the flight is twice this, so is 3.46 s

Question 7 a)The string will break when the centripetal force is at the breaking force.4N = mv^2/rv = (4N * 1.8m/0.2kg)^0.5v = 6m/sQuestion 7b)The ball will fly down the page, which is tangential to the motionQuestion 7c)The centripetal force is the sum of the gravitational force and the tension force, and at the top,mg is directed in the same direction as the centripetal force, whereas at the bottom, the weight

force is directed in the opposite direction, so the tension needs to be greater to "cancel out" theweight force.Question 8a)Using GM/r^2 = 4pi^2r/T^2 and rearranging for r^3 gives r^3 = GMT^2/4pi^2r^3 = 6.446 * 10^18 m^3r = 1.86 * 10^6 mSubtracting the radius of the moon gives 1.2 * 10^5 m.

Smarter VCE Lectures

http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/
• 7/31/2019 VCE Physics Answers 2012

3/7

Question 8b)

Weightlessness occurs when there is no net gravitational field, and apparent weightlessnessoccurs when there is no normal force. When the astronauts are in orbit, their a is the sameas g, and therefore they experience no normal force. Thus this is an example of apparentweightlessness, as gravitational fields do exist at their point in space.

Area of study 2 - Electronics and photonicsQuestion 1)The resistance of the resistors in parallel is 100 ohms. Therefore, the resistance of the set up is

250 ohms.

Question 2a)2V are dropped across D1. Therefore 10V is dropped across R1, and using V = IR, I = 10/100 =0.1 A = 100mAQuestion 2b)3V are dropped across D2, and therefore 5V is dropped across R2.

Question 2c)The current has contributions from each of the loops in parallel. In the first loop, 2V is droppedacross D1, so V across R1 is 6V. Thus the current from this circuit is 0.06A. In the second loop,

5V is dropped across R2, so I = 5/150 = 0.033A. Therefore, the total current is 0.093A, which is93mA.

Question 2d)The power in D1 = VI = 2 V* 0.1A = 0.2W.The power in D2 = VI= 3 V* 0.1A = 0.3W.The efficiency of D1 = 0.15/0.2 = 0.75. The efficiency of D2 = 0.2/0.3 = 0.67. Thus the efficiencyof D1 is greater.

Question 3a)V = IR(total)V/I = 3000 ohm + R

10/2,5 * 10^3 ohm = 3000 ohm + R4000 ohm = 3000 ohm + RR = 1000 ohmReading off the graph, this gives a light level of 10 lux.Question 3b)

A sample circuit would include a voltage divider, with voltage out across the LDR. Using thevoltage divider formula, the resistance of a resistor R to be placed in series, can be found

Smarter VCE Lectures

http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/
• 7/31/2019 VCE Physics Answers 2012

4/7

6V = 10V * 3000/(3000 + R)3000/(3000+R) = 0.63000 = 1800 + 0.6R1200/0.6 = RR = 2000 ohm

Question 4a)The voltage gain is 10/0.04 = 250Question 4b)The graph will be inverted and multiplied by 250. Points on the new graph with voltages smallerthan -10V and greater than 10V should not exist; clipping occurs at these voltages.

Question 5)Modulation device - manipulates the electrical input signal to produce a light output (convertsbetween forms of energy), potentially adding a carrier waveLight beam - transmits the information between physical locationsDemodulation device - converts the light signal to an electrical signal, and performs operationsto make this signal the same as the input signal (such an operation could be subtracting thecarrier wave)

Detailed study 1 - Einsteins special relativityQuestion 1)

A. This can be verified with formulae that are outside of the scope of the VCE course.Question 2)D. Relative motion (can imagine that particles are moving with the sound), so player runningtowards measures speed of 340m/s, and other player measures 330m/s.Question 3)C. You can go through the numbers here, but it is apparent that A and B are incorrect (as c isconstant in a vacuum), and D does not make sense as if we can see the quasar, the light hashad time to reach earth.

Question 4)B. No acceleration can be observed. Whilst D does suggest that no centripetal accelerationoccurs, it does not account for linear acceleration. D would be correct if it stated that it wastravelling at a constant speed in a straight line.

Question 5)C. The results would have differed if ether had been detected.

Smarter VCE Lectures

http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/http://www.connecteducation.com.au/
• 7/31/2019 VCE Physics Answers 2012

5/7

Question 6)

Total energy = rest energy + kinetic energyrest energy =mc^2 = 9 * 10^16 * 1.67 * 10^-27J = 1.5 * 10^-10 JTherefore A is correct (the contribution from the kinetic energy is negligible)

Question 7)C. Using calculations and substitutions, the ratio is the square root of 1/.16 = 1/0.4 = 2.5

Question 8)A. Proper time is shorter than other measurements, or equal to them.

Question 9)D. Length contraction is apparent, but only in the direction of motion.

Question 10)The velocity at which she passes is 0.866c. The length she observes is 10m, so t = 10m/(0.866* 3 * 10^8) = 3.85 * 10^-8s, which is C.

Question 11)Apparent mass increases as velocity increases, if mass is taken as gamma multiplied by therest mass. So A.Question 12)The work done is the difference in energies, which is gamma2mc^2 - gamma1mc^2gamma1 = 1.05

gamma2 = 1.15so the work done = 0.1 * rest mass energy = 5.98*10^-10J, so the answer is B.

Detailed Study 2 - Materials and their use in structures

Question 1)The pole is in compression, and each pole exerts a force of 70.71N down. So the correctanswer is B.

Question 2)A. The stiffest alloy is the one with the largest gradient. This is alloy A. Modulus is stress/strain

= 300 * 10^6/0.004 = 7.5 * 10^10PaQuestion 3)C. F = stress * area = 2 * 10^-3 * 250 * 10^6 N = 5.0 * 10^5NQuestion 4)D. From the graph, alloy D.