Level Physics: Mechanics Motion Answers The Mess that is NCEA Assessment Schedules. Level Physics: AS 97 replaced AS 9055. 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. There is no spoon. 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 07() F = mg = 55.0 x 9.8 = 539 = 540 N (SF). Single arrow pointing downwards. (b) Vv is calculated correctly. Incorrect Vv, but consequently correct time. Correct Vv Correct time. 06() vf = vi + ad.5 = a 0.50 a =.3 m s
06() vv = 0 sin 40 =.856 =.9 m s Correct working. (b) vf = vi + at 0 =.9 + 9.8 t t =.3 s (or.3 if unrounded vv used) Correct equation and correct substitution. time of flight = t =.3 =.64 s (or.6 if unrounded) vh = 0 cos 40 = 5.3 = 5.3 m s Correct total time. Correct horizontal velocity. dh = vh time of flight = 5.3.64 = 40.4 m (or 40. if unrounded) Horizontal velocity remains constant, as there are no external forces in the horizontal direction, air resistance is negligible. Going up, vertical velocity decreases/ball decelerates as the weight force/gravity acts downwards/in an opposite direction to the motion Coming downwards, the vertical velocity increases/ball accelerates as the weight force/gravity is acting downwards/in the same direction as the motion. ONE correct statement with correct reason. Correct description of both velocities. TWO correct statements with correct reasons. Comprehensive explanation.
05() Fg pointing vertically down for C and G. Same size for both positions. Horizontal arrow vh of the same size for both positions, B and H. Arrow vv pointing vertically up at D and pointing vertically down at F, same length. Two out of three correct. All three correct including relative size of arrows. (b) Vh = 6 cos 4 =.89 m s (= m s - ) Vv = 6 sin 4 = 0.7 m s (= m s - ) Both correct. The vertical velocity of the ball is zero at the top, but the horizontal velocity will be.9 m s. There are no horizontal forces. Gravity acts vertically down. Vertical velocity is zero. Horizontal velocity is constant. One correct velocity with correct explanation. Complete answer giving reasons why horizontal velocity is constant (no horizontal force) but vertical velocity decreases to zero (gravity). Time taken to reach maximum height: vf = vi + at 0 = 0.7 9.8t t =.09 s Total time =.09 =.8 s Range = d = vt d =.89.8 = 6 m Correct time of.09 s Correct total time or one error in calculation. Two correct steps. 04() v f = v i + ad.0 =.0 + a 7 a = 3.33 m s - Correct acceleration ONE error in calculation. Correct acceleration and force. F = ma F = 00 3.33 F = 3666 N F = 3700 N
Question Achievement Merit Excellence 03() vhorizontal = 6.5 cos60 = 3.5 m s v vertical = 5.63 m s **Watch out for values being swapped around. vhorizontal = 6.5 cos60 = 3.5 m s v vertical = 5.63 m s t = d v 3.0 3.5 = 0.93 s Time to reach max height = 0.57s, so max height =.6m, so will go only.85 m across and so will not go through hoop. d = v i t + at d = 5.63 0.93-0.5 9.8 0.93 d =.0m This is less than.35 m hence ball will not go through hoop. Vertical velocity at.35 m height v =.9 ms- Time taken for vf to reach.9ms - v f = v i + at ±.9 = 5.63-9.8t 5.63±.9 t = = 0.808 s or 0.34 s 9.8 Horizontal distance travelled in 0.808s or 0.34 s or. m This is less than 3.00 m, so ball will not go through the hoop. 03() v f = v i + at 0 = v i -.5 4. v i = 0.5ms - Added (44. +.05)to get incorrect answer of 66.5 m *circle + sign Assumed v i =0 and worked out answer. d = v i t + at d = 0.5 4. -.5 4. d = 44..05 d =.05 m d = v f - v i t d = 0.5+ 0 4. d =.05 m
0() v v = 5sin70 v v = 4.095 v f = v i + at 0 = 4.095-9.8t t = 4.095 9.8 t =.4 s Horizontal velocity is 5. m s to the right. Her vertical velocity is zero. Horizontal velocity is 5. m s to the right. Her vertical velocity is zero. One reason to support either 5. m s or 0 m s. Horizontal velocity is 5. m s to the right. Her vertical velocity is zero. There is no horizontal force acting on her, so her horizontal velocity is constant. She is constantly being accelerated at 9.8m s downwards. hence her vertical velocity at the top is 0 m s.
Question Evidence Achievement Merit Excellence 0() v f = v i + at Þ v f = 0 +. 4 Þ v f = 6.8 ms - (must show in some way that v i = 0) 0(3) Horizontal component = 4 cos36 = 9.4 m s Vertical component = 4 sin 36 = 4. m s Time taken to travel 35 m = 35/9.4 =.80 s Calculates horizontal and vertical components. Calculates horizontal component of velocity and time to travel 35 m. Calculates time taken to travel 35 m. Calculates initial vertical velocity correctly. Calculates height of ball at 35 m range. States the ball goes over Ernie s head. d = (4..80) (/ 9.8.80 ) d = 9.5 m The ball will go over Ernie s head. 00() As it rises, kinetic energy changes to gravitational potential energy. As it falls, gravitational potential energy is converted to kinetic energy. As it rolls, kinetic energy is converted to heat. Correct except for one error. Description all correct. (b) Downwards 9.8 m s Both correct. Must have size direction. Accept 0 m s.
v v h Calculates speed using t =.4 s. v = 3.5ms - Used an incorrect value for time. Correct vertical component used v =3.5 to calculate v (v = 39.5 m s ) Correct working and answer. Consider vertical motion If there is a lift force, the total force downwards is smaller. So the downward acceleration is smaller. So the discus is in the air for a longer time. So the range is greater. The lift force allows discus to fly higher and hence it is in the air for longer. Since horizontal velocity is constant, then a greater distance is travelled in greater time. Hence range is greater. Greater range. Greater range because the discus is in the air for a longer time. Full explanation. (accept either reason reason ) a combination of and so long as they have 4 bullet points.
009() On its way up, the velocity keeps decreasing at a constant rate of 9.8 m s until it reaches zero. Once its velocity is zero it goes no higher, but starts falling down again with increasing velocity at the rate of 9.8 m s until it hits the ground. The acceleration is constant throughout its motion and acts in a downward direction at 9.8 m s. Speed at start is equal to speed at stop, assuming that these positions are the same vertical height. Describes decreasing velocity going up Describes increasing velocity on its way down Describes constant acceleration downwards. Describes constant acceleration downward Speed at start equal to speed at stop assuming the vertical height is the same. Either decreasing velocity on its way up Increasing velocity on its way down. Describes constant acceleration downward Decreasing velocity on its way up Increasing velocity on its way down Speed at start equal to speed at stop assuming the vertical height is the same. Calculates final vertical velocity. **NA if they have used 7.8 ms as initial velocity vi. Achievement plus calculates resultant velocity without direction. Correct answer including direction of resultant. Resultant velocity = 5.4 + 7.8 v = 9.40 m s Direction = tan 5.4 7.8 = 34
008() valid alternative. d = 0 to s.f. sig. fig. (b) Kinetic Heat (+ Sound) EK = ½ mv EK = ½ 65 8.0 EK = 080 J 008() (m) The ball lands in the same place relative to Rua. The ball has an initial horizontal velocity. There are no horizontal forces acting on the ball in flight so it keeps moving at the same horizontal speed as the trolley and lands where it left. Lands in same place relative to Rua. Lands in same place because it keeps moving horizontally at the same/constant speed as Rua. Lands in same place because there are no horizontal forces acting on Rua or the ball Lands in same place because it keeps moving horizontally at the same/constant speed as Rua as there are no horizontal forces acting on Rua or the ball. (n) vf = vi + at 9.8 = 9.8 9.8t Correctly determines the time to maximum height. Correct working and answer. t =.0 s
007() a = (5 80) / 8 = 6.875 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. s.f. s.f. (any correctly rounded answer) 007(3) F = 5cos 40 = 9.5 N W = Fd W = 9.5 0.80 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. 006() a = v t a = 0.90 m s sf (regardless of answer to a) Correct significant figures.
(b)
Calculates the force only. (F = 08 N) Correctly calculates the work done (W = 5 J) Correct working and answer. Calculate average velocity only. (vave =.5 ms ) Shows awareness that Power can be calculated by considering the EK gained. Combines power and Ek formula, but unable to solve. Calculates the force the average velocity, but is unable to combine to calculate Power. Correctly calculates the kinetic energy (EK = 5J) W watts Js Nms Correct unit. (regardless of answer to c)
Friction (and drag) opposes the motion. Therefore, Steve must do extra work to overcome the work done by friction. Recognises the effect of fiction on Steve s motion / Friction does work against the boat. Achieved plus Links applied forces to work done by Steve. Merit plus Links the rate of work done to power. Hence as Power is the rate at which work is done the Power that he produces must be greater than calculated in part c. Friction causes some energy to be wasted as heat. Some energy wasted as heat (and sound). Achieved plus Links EIN = EK + EHEAT Merit plus Links the rate of change of energy to power. Hence not all the energy Steve puts in is transformed into kinetic energy. Therefore as Power is the rate at which energy is changed the Power he produces will be greater than the Power output calculated in part c. (e) Net force = zero 006(3) VH = 6.4 cos 0 Correct working. (b) VV = 6.4 sin 0 Correct working.
vf = vi + at 0 =. 9.8 t t = 0. s total time = 0.44 s Calculates time to highest point. Calculates total time Correctly uses d = vt but with time to max height. Correct working and answer. dh = vh t = 6.0 0.44 d h =.69 m Accept calculation using g =0ms - so dh =.64 m 9.8 ms, downwards (accept 0 down). Do not accept 9.8 down. (e) As friction is negligible there are no other forces acting horizontally. A zero net force means Marama will not experience any acceleration. Horizontal net force is zero. The only force acting on Marama is in the vertical plane Just quotes Newton s st or nd law but doesn t apply. Clearly applies Newton s st or nd law to the idea of a zero net force horizontally on Marama. 005(3) The forces are equal in size because the balloon is rising at a steady speed, thus there is no unbalanced force acting on it. Forces are equal. Achievement plus constant speed means no unbalanced force.
Correct answer. No Brain Too Small PHYSICS (b) Correct calculation of time to ground. Correct answer 004() Correct answer (b) Correct answer.
Correct answer Unit: joules joules (accept J or Nm) (e) Calculation of t =.67 s Calculation of W = Pt = 6000 x.67 = 6000 J Correct answer for force. Derivation of F = Pv Correct answer for force. 004() Parabola Parabola (parabolic) (b) One force only, vertically downwards, labelled gravity or weight force. Arrow must be below javelin & near centre. Arrow not necessarily touching
Correct working. Calculation of vvert = 9.3 m s Calculation of time of flight = 3.86 s Or Complete calculation without doubling t. (Note: accept use of either rounded or unrounded data.)