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Common problems from 2010 S3 Physics EOY

Common problems from 2010 S3 Physics EOY. P1 : 8, 10, 12, 14 P2: 3b, 10c(ii), 10d, 12c( i ), 12c(ii). P1 Q8. Read as: Rate of change of displacement is velocity :. displacement. Velocity. Acceleration. P1 Q10. Draw horizontal forces acting on Y Draw horizontal forces acting on Z

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Common problems from 2010 S3 Physics EOY

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  1. Common problems from 2010S3 Physics EOY P1 : 8, 10, 12, 14 P2: 3b, 10c(ii), 10d, 12c(i), 12c(ii)

  2. P1 Q8 • Read as: Rate of change of displacement is velocity: displacement Velocity Acceleration

  3. P1 Q10 • Draw horizontal forces acting on Y • Draw horizontal forces acting on Z • Identify the direction (right or left) of motion and thus acceleration! • Therefore, direction of net force = direction of acceleration • Apply Newton’s 2nd Law!

  4. P1 Q10 continued… • On Y • On Z 12 N F (exerted by Z on Y) Same type of force; Exerted by one on the other Action – Reaction PAIR! SAME MAGNITUDE! OPPOSITE DIRECTION F (exerted by Y on Z)

  5. P1 Q10 continued… Direction of acceleration • On Y • On Z 12 N Net force = 12 - F F (exerted by Z on Y) Net force = F F (exerted by Y on Z)

  6. P1 Q10 continued… Direction of acceleration Acceleration = 3.0 m/s2 • On Y • On Z Net force = 12 – F By Newton’s 2nd law, Net force = ma 12 – F = ma 12 – F = (1.0) (3.0) F = 9.0 N Net force = 12 - F Net force = F Net force = F By Newton’s 2nd law, Net force = ma F = ma F = (3.0) (3.0) F = 9.0 N

  7. P1 Q12 • Draw diagram • Identify pivot • Draw weight

  8. P1 Q12 continued… A: C.G. can be above the pivot C: clockwise moments = 0; anticlockwise moments = 0. Therefore, statement is not very good. D: WEIGHT IS ACTING ON BOOK! x x Pivot Weight

  9. P1 Q14

  10. P1 Q14 continued… • As ball falls, GPE → KE Loss in GPE = Gain in KE Initial GPE – Final GPE = ½ mv2 mgh – Final GPE = ½ mv2 Final GPE = mgh – ½ mv2 Thus, final GPE = (0.50)(10)(2.0) – ½ (0.50)(4.0)2 =6.0 J

  11. P2 Q3b • Draw all the forces acting on the object (Free Body Diagram). Identify those angles! T2 T1 Weight

  12. P2 Q3b continued… Use sum of angles in triangle!

  13. P2 Q3b continued… 2. (optional) From free body diagram, draw all the forces acting from one point. 10o 10o T2 T2 20o 20o T1 T1 Weight Weight

  14. P2 Q3b continued… 3. Shift the forces one by one until it forms a closed triangle, starting from the weight! 10o 10o T2 T2 20o 20o T1 T1 10o Weight Weight

  15. P2 Q3b continued… 4. Tidy up the triangle! It should look like this. T2 10o Weight 20o T1

  16. P2 Q3b continued… • When you construct: (i) draw the Weight, (ii) measure 10o, draw straight line (for T2) (iii) measure 20o, draw straight line (for T1) (iv) where the two lines intersect, erase off the excess and add in the arrows for the forces! T2 10o Weight 20o T1

  17. P2 Q 10cii To find angle of tilt.

  18. P2 Q 10cii continued… • This is the angle of tilt, use trigonometry! 40 cm 11 cm

  19. P2 Q10d

  20. P2 Q10d continued… • Identify the pivot. (How will the block rotate up if I pulled it up as seen in the diagram?) • Find CW moments, • Find ACW moments • Equate using principle of moments! NOTE: Remember moments = force x DISTANCE!

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