Assessment Schedule 2007 Physics: Demonstrate understanding of mechanics (90255)
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1 NCEA Level Physics (9055) 007 page of 5 Assessment Schedule 007 Physics: Demonstrate understanding of mechanics (9055) Evidence Statement Q Evidence Achievement Achievement with Merit Achievement with Excellence ONE A projectile has only the force of gravity acting on. An aircraft is powered as it has its own motor and hence is acted on by more than one force. The path of a projectile is a parabola that of an aircraft need not be so. Describe the motion of the aircraft or the projectile IN TERMS OF: force(s) (gravity, thrust, engine force, drag, air resistance, lift, etc) Parabolic path shape. Compares motion of aircraft to motion of projectile with reference to gravity and other force(s) Parabolic path. (b) By Principal of Conservation of Energy Total E KV at bottom = E P at top + E K horizontal ½ mv Total = mgh + ½ mv H v Total = (gh + v H) v Total = 5 m s By Principal of Conservation of Energy VERTICALLY E KV at bottom = E P at top v = gh = = ms VERTICAL velocity when it reaches the ground 35 ms - v f = v i + ad v f = v f = m s vert. down VERTICAL velocity when it V R reaches the ground d = v i t + ½ at hence t = 600/0 = s v f = v i + at v f = v f = m s vert. down THEN Horizontal velocity = 35 m s Final velocity on reaching the ground = = 4.99 m s 09.5 ms - Calculated, with arrows and labels. Calculated and labels. Calculates the final speed. Calculated and labels. Calculates the final speed. Vector diagram, with arrows and labels, but misunderstanding of air/ground speed on horizontal. and all correct labels.
2 NCEA Level Physics (9055) 007 page of 5 Any identified angle calculated Achieved plus Bearing correctly shown on diagram or clear from calculation. Doesn t need to be given to 3 figures. ' & 40 # θ = sin $! = 3.58 Bearing = 90 4 = 66 % 00 " ' & 40 # φ = cos $! = 66.4 Bearing = 66 % 00 " (d) a = (5 80) / 8 = a = 6.9 m s a = 6.9 m s in opposite direction to velocity. Calculates acceleration using (initial final) Correct working BUT the stated direction is inconsistent with the sign. Correct working (using change = final initial), and final answer is a valid physics statement with respect to sign and direction if stated. (d) sf sf (any correctly rounded answer) TWO 0 N 0 N 0 N Other suitable labels include: Weight, Force of gravity, support. 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. (b) 3 F A = ( ) + (.5 0) + (.5 600) 3 F A = F A = 80 3 F A = 760 N 3 F B = ( ) + (.5 0) + (.0 750) 3 F B = F B = F B = 80 N ΣF = 0 therefore F A = 760 N 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 Στ ac about B Calc Στ c about A Correct answer
3 NCEA Level Physics (9055) 007 page 3 of 5 THREE 7.0 m Suitcase Any arrow that shows a tangential direction. (Accept if arrow drawn tangentially at the end of the radius arrow). (b) This is because the object is continually changing direction even though the speed remains the same. A change in direction amounts to a change in velocity as velocity is a vector. The rate of change of velocity is acceleration. States direction is continually changing. States that a centripetal force is acting on the suitcase. Links changing direction to the vector nature of velocity. Links changing direction to centripetal force. MERIT plus Links changing velocity to acceleration. Links centripetal force to centripetal acceleration. mv F = hence v = r Fr m 5.5! 7.0 v = = d = πr = π 7.0 = d T = v T = = s Correct velocity. Correct circumference. Any two correct processes. Correct working and answer (Do not accept rounding for excellence if it causes a change in the significant figure answer. (d) F = 5cos 40 = 9.5 N W = Fd W = W = 5.3 J Calculated work done without using the horizontal component of the force (0 Nm). Calculate horiz. force component only. Calculates work using an incorrect force component. Correct answer. (e) Momentum is conserved. (33 3.6) + (35.0) = (33.4) + 35v = v v = 3.3 m s. E k of trolley B = ½ mv = E k =7.6 J States that momentum is conserved (maths or words). Determines Total p before Total p after Δp A Δp B Correct answer for velocity. Merit plus correct calculation of kinetic energy of trolley B. (Do not accept rounding for excellence if it causes a change in the significant figure answer. (f) Momentum is conserved assuming the absence of external forces such as friction or people pushing. Isolated system No external forces. No friction / push. Links an isolated system / absence of external forces to the friction / push. Identify both external forces.
4 NCEA Level Physics (9055) 007 page 4 of 5 (g) Elastic collisions Momentum is conserved, and Kinetic energy is conserved. Inelastic collisions Momentum is conserved, but Kinetic energy decreases as it is converted to heat/sound/elastic. Momentum is conserved in all collisions. E K is conserved in Elastic collisions. E K is not conserved in inelastic collisions. Momentum is conserved in all collisions Both kinetic energy statements. Clear definitions of elastic and inelastic in terms of momentum and kinetic energy and identifies that E K is transformed to heat/sound/elastic in inelastic. (DO NOT ACCEPT ALL kinetic energy for E). FOUR x = 7 5 = cm = 0. m F = = 4.0 N k = = 9 N m Calculates k using either cm (0.9048) or g (9 048). Uses correct units but takes length as extension (5.56). Correct answer. N m (or kg s ) DO NOT ACCEPT lower case n for newtons. Correct unit. (b) m= = F = = 7.00 N k = 9.0 N m x = F / k = 7.00 / 9.0 = m E p = ½ kx = =.9 J Calculates new extension. Correct working and answer. A spring with double the spring constant for the same weight force would mean half the extension as F = kx. As E P k, E P doubles when k doubles for the same extension. As E P x it decreases by four when x is halved for the same spring constant. Overall a spring with double the spring constant for the same weight force would mean half the extension and hence half the E p. Has identified that extension and/or energy stored has decreased. Has qualitatively described and explained the effect of a higher spring constant on extension energy stored for the same weight force. Has quantitatively described explained the effect on BOTH the extension energy stored with a greater spring constant for the same weight force.
5 NCEA Level Physics (9055) 007 page 5 of 5 Judgement Statement Achievement Achievement with Merit Achievement with Excellence Criterion One 4 A 3 A + M 3 A + M + E Criterion Two 4 A 3 A + 3 M 3 A + 3 M + E Marking Codes: Working must be shown for Excellence and Merit answers. Consequential marking When a candidate is required to use the answer to a previous question that they answered incorrectly, the candidate will not be penalised again, provided the current solution is completely correct. Rounding error Inappropriate rounding of an early step in a calculation results in a significant change in the final answer. Transcription error Candidate demonstrates complete understanding of solution but incorrectly transcribes a value during one step of the processing. Solution error Minor computational errors will not be penalised. A wrong answer will be accepted as correct provided there is sufficient evidence that the mistake is not due to lack of understanding. Sufficient evidence implies that the last written step before the answer is given, has no unexpanded brackets or terms and does not require rearranging.
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