No Brain Too Small PHYSICS

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1 Level Physics: Mechanics Equilibrium Answers The Mess that is NCEA Assessment Schedules. Level Physics: AS 97 replaced AS In 9055, from 003 to 0, there was an Evidence column with the correct answer and Achieved, Merit and Excellence columns explaining the required level of performance to get that grade. Each part of the question (row in the Assessment Schedule) contributed a single grade in either Criteria (Explain stuff) or Criteria (Solve stuff). From 003 to 008, the NCEA shaded columns that were not relevant to that question. In 97, from 0 onwards, the answers/required level of performance are now within the Achieved, Merit and Excellence columns. Each part of a question contributes to the overall Grade Score Marking of the question and there are no longer separate criteria. There is no shading anymore. At least their equation editor has stopped displaying random characters over the units. And in 03, with 97, we still have no Evidence column with the correct answer and Achieved, Merit and Excellence columns explaining the required level of performance to get that part even though the other two Level Physics external examinations do!!. And now in 04-07, we have the Evidence column back Question Evidence Achievement Merit Excellence 06(3) τ = F d = = 470 N m = 500 N m Correct answer with correct units. Force at support A must be downwards. So the torque by the Force at A is in an opposite direction (anticlockwise) to the total clockwise torque created by Sarah and the weight force of the board. Correct direction (with an attempt to give some correct explanation). Correct answer with correct reasoning. 05(4) (d) Taking B as pivot (or moments about B), F A.5 ( ) ( ) F A F A N.5 F A 00 N Correct substitution for one direction of torque about specified end. AND correct torques. One mistake. E.g. final answer for FB (3 N) instead of FA All correct.

2 04() Correct labelled diagram correct mathematical solution: (Either m = 0 kg W = 98 N as long as unit is correct.) Correct labelled diagram AND correct mathematical solution. c ac (m 9.8) 0.45 (30 9.8) 0.5 m 0 kg W 98 N

3 Question Achievement Merit Excellence 03() All four vectors correctly drawn and labelled. Fs must be larger. All four vectors correct without labels. three correct vectors labelled. ( ) F F.5 = 058 (450 +F.5 =080) 94 +Fs = Taking moments / calculating torques wrt to Dad s end: ( ) + ( ) (FS.5) = 0 where FS is the support force at the pivot when Dad s end is on the ground =.5 FS hence FS = = 76 N F F F (F.5 = 630) Total FUP = total FDOWN FS + FGROUND = (30 9.8) + (60 9.8) + (7 9.8) = = 588 N F N (F = 40 N) Hence FGROUND = = 4 N

4 (c) TWO of: The only unbalanced force acting on the ball is the force of gravity. Gravity acts downwards. This unbalanced force causes the ball to decelerate or accelerate downwards. Velocity at the top is zero. The only unbalanced force acting on the ball is gravity, which acts downwards. This causes the ball to decelerate or accelerate downwards. Hence the ball slows down to a stop when it reached maximum height. 0() c ac ( ) (0.5 0) F 3 83 = F 3 F = = 394 N F= 394 N Direction = up 390 ( sig figs) AND Same as the least accurate data.

5 Question Evidence Achievement Merit Excellence 0() At equilibrium, clockwise torques = anticlockwise torques Taking A as pivot, ( ) + ( ) = FB 5 Correct answer to sum of downward torques = Nm.(or 07 Nm) (Must use weight force for calculation not just mass). *Accept Nm if using g = 0ms Correct answer to support force at either A or B. Accept N for FB and 4376 N for FA(g = 0ms ). Correct answer to support force at both A and B. FB = 96 68/5 FB = = 3900 N FA = ( ) FA = 488 N FA = 4300 N s.f. for both FA and FB The lowest number of sf in the question is 5.0 which is s.f. The final answer cannot be any more accurate than the least accurate piece of data in the question. Correct rounding to sf. Reason. Correct rounding and reasoning.

6 00(5) (c) The total force is zero and the total torque is zero. One correct statement. Both statements correct. (d) Correct equation and substitution (line ). Even though g cancels out, it must be included in the equation unless it is clearly stated that it cancels. Correct answer for W = N If used g = 0 m s, then W = 06.6 N Got one error in substitution but got rest of method correct. Correct answer. Accept either.3 kg or 0.6 kg m = 0.6 kg (if used with g = 0 N kg ) 009(3) Sum of translational forces is zero. Sum of torques is zero. One correct statement. **Accept total force up = total force down Anticlockwise torque = clockwise torque Both statements correct (or words to that effect). (i) Fw = mg = = 3.9 N (ii) Fw = mg = =.96 N Both correct including unit. *NA if they have used g = 0 ms - (c) F Correct substitution. Correct answer. **Consequential marking if they have used g=0 for this one as well in addition to.

7 009() (i) (j) sin 37 F v 500 cos 53 F v 500 Fv = 500 sin 37 Fv = 500 cos 53 Fv = 90.7 N τc = τac (Ft 3) + (343.5) = Ft = 700 3Ft = 85.5 Ft = 78.5 N Correct working. Determines a correct torque. Correct working and answer. 007() 0 N 0 N 0 N Arrows are of the correct size in relation to each other. Arrows are not to scale but have force values indicating their size. Arrows are of the correct size in relation to each other and appropriately labelled. Other suitable labels include: Weight, Force of gravity, support.

8 3 FA = ( ) + (.5 0) + (.5 600) 3 FA = FA = 80 3 FA = 760 N 3 FB = ( ) + (.5 0) + (.0 750) States that clockwise and anticlockwise torques are equal. Calculate ANY correct torque. Correct formula and substitution but incorrectly determines ONE distance forgets to include the torque due to the beam. Calc Στac about B Calc Στc about A Correct answer 3 FB = FB = FB = 80 N ΣF = 0 therefore FA = 760 N 006() (f) = = 5 Nm Calculates torque due to Steve ( = 5 Nm). Correct working and answer. = 5 Nm F 3.5 = 5 F = 64 N Fd = Fd = F 3.5 F = 64 N

9 004(3) (i) All four arrows shown, all pointing in the correct direction. Achievement plus the weight of the beam is shown in the centre, Nadia s weight is in the correct position and support forces are shown at the supports. (ii) (ii) 550 N 900 N Both answers correct (iii) Correct calculation of torque of either weight force about B. Correct calculation of torques of both weight forces about B. Correct answer (iv) Since the beam is in equilibrium, the sum of the clockwise torques is equal to the sum of the anticlockwise torques about any chosen point. Balanced torques the beam is in equilibrium. Both statements correct