Working out resultant forces in a straight line

Answers

1. Resultant force = 25 − 10 = 15 N to the right.
2. Resultant force = 40 − 18 = 22 N forwards.
3. Resultant force = 75 − 20 = 55 N forwards, so the trolley speeds up.
4. Resultant force = 160 − 95 − 25 = 40 N forwards.
5. Resultant force = 2400 − 2400 = 0 N, so the car moves at a steady speed.
6. Resultant force = 180 + 140 − 210 = 110 N forwards.
7. Resultant force = 850 − 420 − 130 = 300 N forwards, so the boat accelerates forwards.
8. Resultant force = 120 + 90 − 70 − 15 = 125 N forwards.
9. Resultant force = 6500 − 1900 − 2100 − 800 = 1700 N forwards, so the lorry speeds up.
10. Resultant force = 450 + 320 − 250 − 380 = 140 N to the right.

Finding the resultant force of 2 perpendicular forces – by pythagoras or scale drawing

Answers

1. Resultant force = √(30² + 40²) = 50 N.
2. Resultant force = √(12² + 16²) = 20 N.
3. Resultant force = √(60² + 80²) = 100 N.
4. Resultant force = √(90² + 120²) = 150 N.
5. Resultant force = √(150² + 200²) = 250 N.
6. Resultant force = √(180² + 240²) = 300 N, acting diagonally forwards and sideways.
7. Resultant force = √(800² + 600²) = 1000 N.
8. Resultant force = √(1200² + 900²) = 1500 N, acting diagonally.
9. Resultant force = √(2400² + 700²) = 2500 N; the current makes the ship move diagonally, not exactly forwards.
10. Resultant force = √(3600² + 1500²) = 3900 N; the drone must adjust because the crosswind pushes it sideways.

 

Speed = Distance /Time Questions

Answers to Speed = Distance/Time Questions

1. Speed = 100 ÷ 12.5 = 8.0 m/s
2. Speed = 2400 ÷ 360 = 6.7 m/s
3. Speed = 1800 ÷ 240 = 7.5 m/s
4. Distance = 18 × 150 = 2700 m
5. Time = 12000 ÷ 10 = 1200 s (20 min)
6. Distance = 4.5 × 40 = 180 m
7. Time = 72 ÷ 3.0 = 24 s
8. Speed = 9000 ÷ 720 = 12.5 m/s
9. Speed = 1800 ÷ 24 = 75 m/s

10a Time = 54000 ÷ 60 = 900 s

10b Time = 900 ÷ 60 = 15 min

Challenge: Speed = (4.2 × 10⁷) ÷ (3.5 × 10⁴) = 1.2 × 10³ m/s = 1200 m/s

17 Questions using v² = u² + 2as

Answers to v² = u² + 2as Questions

1. v² = 2² + (2 × 1.5 × 16) = 52, so v = 7.2 m/s.
2. v² = 4² + (2 × 2 × 24) = 112, so v = 10.6 m/s.
3. v² = 8² + (2 × 1.5 × 48) = 208, so v = 14.4 m/s.
4. v² = 10² + (2 × 3 × 75) = 550, so v = 23.5 m/s.
5. v² = 12² + (2 × 5 × 80) = 944, so v = 30.7 m/s.
6. v² = 0² + (2 × 1 × 800) = 1600, so v = 40 m/s.
7. a = (70² – 30²) ÷ (2 × 500) = 4 m/s².
8. a = (32² – 12²) ÷ (2 × 220) = 2 m/s².
9. a = (11² – 5²) ÷ (2 × 24) = 2 m/s².
10. a = (0² – 20²) ÷ (2 × 40) = -5 m/s², so deceleration = 5 m/s².
11. a = (0² – 14²) ÷ (2 × 28) = -3.5 m/s², so deceleration = 3.5 m/s².
12. a = (0² – 35²) ÷ (2 × 490) = -1.25 m/s², so deceleration = 1.25 m/s².
13. s = (0² – 25²) ÷ (2 × -6.25) = 50 m.
14. s = (0² – 80²) ÷ (2 × -2) = 1600 m.
15. a = (150² – 0²) ÷ (2 × 750) = 15 m/s².
16. u² = 42² – (2 × 3.5 × 210) = 294, so u = 17.1 m/s.

17a. a = (0² – 90²) ÷ (2 × 1800) = -2.25 m/s², so deceleration = 2.25 m/s².

17b. s = (0² – 90²) ÷ (2 × -1.5) = 2700 m.

17c. Reducing stopping distance is important because the train stops sooner, reducing the chance of a collision.

10 Ramped Questions on Calculating acceleration

Answers to acceleraiton Questions

1. Acceleration = (5 – 1) ÷ 4 = 1 m/s².
2. Acceleration = (10 – 0) ÷ 8 = 1.25 m/s².
3. Acceleration = (12 – 4) ÷ 5 = 1.6 m/s².
4. Acceleration = (44 – 20) ÷ 3 = 8 m/s².
5. Acceleration = (32 – 8) ÷ 6 = 4 m/s².
6. Final velocity = 0 + (2.5 × 12) = 30 m/s.
7. Final velocity = 5 + (1.5 × 20) = 35 m/s.
8. 18 km/h = 5 m/s and 72 km/h = 20 m/s, so acceleration = (20 – 5) ÷ 10 = 1.5 m/s².
9. Time = (90 – 30) ÷ 4 = 15 s.
10. a) Acceleration = (360 – 0) ÷ 30 = 12 m/s².

12 Questions on F=ma (In these questions resultant force F also has to be calculated)

Answers to F=ma resultant force questions

1. Resultant force = 80 – 30 = 50 N, so acceleration = 50 ÷ 25 = 2 m/s².
2. Resultant force = 180 – 60 = 120 N, so acceleration = 120 ÷ 75 = 1.6 m/s².
3. Resultant force = 250 – 100 = 150 N, so acceleration = 150 ÷ 50 = 3 m/s².
4. Resultant force = 900 – 300 = 600 N, so acceleration = 600 ÷ 400 = 1.5 m/s².
5. Resultant force = 5000 – 2000 = 3000 N, so acceleration = 3000 ÷ 1250 = 2.4 m/s².
6. Resultant force = 1500 – 4500 = -3000 N, so deceleration = 3000 ÷ 1500 = 2 m/s².
7. Resultant force = 100000 – 40000 = 60000 N, so acceleration = 60000 ÷ 80000 = 0.75 m/s².

8a. Resultant force = 20000 – 140000 = -120000 N.

8b. Deceleration = 120000 ÷ 60000 = 2 m/s².

9a. Resultant force = 12000 – 2000 = 10000 N.

9b. Acceleration = 10000 ÷ 1000 = 10 m/s².

9c. Final velocity = 8 + (10 × 5) = 58 m/s.

10a. Resultant force = 750000 – 250000 = 500000 N.

10b. Acceleration = 500000 ÷ 25000 = 20 m/s².

10c. Final velocity = 0 + (20 × 20) = 400 m/s.

11a. Resultant force = -(18000 + 2000) = -20000 N.

11b. Deceleration = 20000 ÷ 800 = 25 m/s².

11c. Final velocity = 72 – (25 × 2) = 22 m/s.

12a. Resultant force = 50000 – 250000 = -200000 N.

12b. Deceleration = 200000 ÷ 200000 = 1 m/s².

12c. Final velocity = 30 – (1 × 10) = 20 m/s.

12d. No, because after 10 s the train is still travelling at 20 m/s.

Questions on W =mg

Answers to W = mg

1. W = mg = 2 × 10 = 20 N
2. W = mg = 0.75 × 10 = 7.5 N
3. W = mg = 12 × 10 = 120 N
4. W = mg = 68 × 10 = 680 N
5. W = mg = 950 × 10 = 9500 N
6. m = W ÷ g = 240 ÷ 10 = 24 kg
7. m = W ÷ g = 180 ÷ 10 = 18 kg
8. m = W ÷ g = 12000 ÷ 10 = 1200 kg
9. W = mg = 75 × 1.6 = 120 N
10. m = W ÷ g = 3420 ÷ 3.8 = 900 kg

Hooke’s Law (F = kx) Ramped Calculation Questions

Answers to Force, spring constant and extension questions

1. k = F ÷ x = 8.0 ÷ 0.040 = 200 N/m
2. x = F ÷ k = 15 ÷ 250 = 0.060 m = 6.0 cm
3. F = kx = 400 × 0.030 = 12 N
4. F = kx = 8000 × 0.015 = 120 N
5. Extension = 18 − 12 = 6 cm = 0.06 m, F = kx = 300 × 0.06 = 18 N
6. Extension = 28 − 20 = 8 cm = 0.08 m, k = F ÷ x = 12 ÷ 0.08 = 150 N/m
7. Extension = 2.25 − 2.00 = 0.25 m, F = kx = 120 × 0.25 = 30 N
8. Compression = 25 − 19 = 6 cm = 0.06 m, k = F ÷ x = 1800 ÷ 0.06 = 30 000 N/m

9a. x = F ÷ k = 350 ÷ 5000 = 0.070 m = 7.0 cm

9b. Final length = 15 cm + 7 cm = 22 cm

10a. Extension = 52 − 40 = 12 cm = 0.12 m

10b. F = kx = 7500 × 0.12 = 900 N

10c. m = F ÷ g = 900 ÷ 10 = 90 kg

Easy Moments Questions

Answers to Easy Moments Questions

1. Moment = 25 × 0.80 = 20 Nm.
2. Moment = 300 × 1.2 = 360 Nm.
3. Moment = 45 × 0.30 = 13.5 Nm.
4. Moment = 180 × 0.17 = 30.6 Nm.
5. Moment = 60 × 1.4 = 84 Nm.
6. Moment = 90 × 0.70 = 63 Nm.
7. Force = 96 ÷ 1.2 = 80 N.
8. Force = 18 ÷ 0.25 = 72 N.
9. Distance = 750 ÷ 300 = 2.5 m.
10. Distance = 84 ÷ 140 = 0.60 m.

Balancing Moments Questions

Answers to Balancing Moments Questions

1. Anticlockwise moment = 100 × 2.0 = 200 Nm, so clockwise moment must be 200 Nm; distance = 200 ÷ 200 = 1.0 m.
2. Anticlockwise moment = 300 × 1.0 = 300 Nm, so clockwise moment must be 300 Nm; distance = 300 ÷ 150 = 2.0 m.
3. Anticlockwise moment = 60 × 1.5 = 90 Nm, so clockwise moment must be 90 Nm; distance = 90 ÷ 90 = 1.0 m.
4. Anticlockwise moment = 400 × 0.75 = 300 Nm, so clockwise moment must be 300 Nm; distance = 300 ÷ 250 = 1.2 m.
5. Anticlockwise moment = 120 × 0.80 = 96 Nm, so clockwise moment must be 96 Nm; force = 96 ÷ 1.6 = 60 N.
6. Anticlockwise moment = 180 × 1.5 = 270 Nm, so clockwise moment must be 270 Nm; force = 270 ÷ 2.0 = 135 N.
7. Total clockwise moment = (50 × 0.40) + (30 × 0.60) = 38 Nm.
8. Anticlockwise moment = (100 × 0.50) + (60 × 1.0) = 110 Nm, so clockwise moment must be 110 Nm; distance = 110 ÷ 110 = 1.0 m.
9. Anticlockwise moment = (300 × 1.0) + (200 × 1.5) = 600 Nm, so clockwise moment must be 600 Nm; distance = 600 ÷ 600 = 1.0 m.
10. Anticlockwise moment = (40 × 0.75) + (60 × 1.2) = 102 Nm, so clockwise moment must be 102 Nm; distance = 102 ÷ 90 = 1.13 m.

Questions on Gears for AQA GCSE Physics 

Answers to Gears Questions

1. The larger gear turns more slowly than the smaller input gear.
2. The smaller gear turns more quickly than the larger input gear.
3. Gear B turns anticlockwise.
4. Gear B turns more slowly because it has more teeth.
5. Gear A turns 3 times for Gear B to turn once.
6. Gear B turns once.
7. The larger gear turns more slowly than the small gear.
8. The turning force at the output increases.
9. A lower gear gives a larger turning force, so it is easier to pedal uphill.
10. Gears reduce the speed but increase the turning force.
11. The pivot point is the centre of the gear.
12. The force acts at a distance from the centre, so it creates a moment.
13. The moment increases when the force acts further from the centre.
14. Moment = 50 × 0.10 = 5 Nm.
15. Moment = 80 × 0.15 = 12 Nm.
16. Gear B produces the bigger moment because it has the larger radius.
17. Moment = 60 × 0.08 = 4.8 Nm.
18. Radius = 10 ÷ 100 = 0.10 m.
19. Input moment = F × 0.04 and output moment = F × 0.12, so the output moment is 3 times bigger.
20. The larger gear has a bigger radius, so it turns more slowly but produces a larger moment.

Real Life Pressure = Force/ Area Questions

Answers

1. P = 500 ÷ 0.05 = 10 000 Pa
2. P = 400 ÷ 0.02 = 20 000 Pa
3. P = 250 ÷ 0.10 = 2 500 Pa
4. P = 20 000 ÷ 2.5 = 8 000 Pa
5. P = 600 ÷ 0.0004 = 1 500 000 Pa
6. P = 12 ÷ 0.0000015 = 8 000 000 Pa
7. F = 80 000 × 0.015 = 1 200 N
8. A = 40 000 ÷ 200 000 = 0.20 m²
9. A = 120 000 ÷ 3 000 000 = 0.040 m²
10. A = 50 000 000 ÷ 250 000 = 200 m²

Momentum Calculations using mv

Answers to Momentum Questions

1. p = mv = 0.45 × 10 = 4.5 kg m/s
2. p = mv = 80 × 5 = 400 kg m/s
3. p = mv = 25 × 2.5 = 62.5 kg m/s
4. p = mv = 200 × 12 = 2400 kg m/s
5. p = mv = 300 × 4 = 1200 kg m/s
6. p = mv = 1800 × 15 = 27 000 kg m/s
7. p = mv = 25 000 × 20 = 500 000 kg m/s
8. v = p ÷ m = 18 000 ÷ 900 = 20 m/s
9. m = p ÷ v = 50 000 ÷ 25 = 2000 kg
10. v = p ÷ m = 180 000 ÷ 12 000 = 15 m/s

17 Conservation of Momentum Ramped Questions

Answers to Conservation of Momentum Questions

1. Total mass = 3 kg, total momentum = 6 kg m/s, so v = 6 ÷ 3 = 2 m/s
2. Momentum transfers to second ball, so v = 8 m/s
3. Total mass = 80 kg, total momentum = 240 kg m/s, so v = 240 ÷ 80 = 3 m/s
4. Total mass = 1500 kg, total momentum = 10 000 kg m/s, so v = 10 000 ÷ 1500 = 6.7 m/s
5. Total mass = 120 kg, total momentum = 350 kg m/s, so v = 350 ÷ 120 = 2.9 m/s
6. Total mass = 75.15 kg, total momentum = 4.5 kg m/s, so v = 4.5 ÷ 75.15 = 0.060 m/s
7. Total mass = 2000 kg, total momentum = 18 000 kg m/s, so v = 18 000 ÷ 2000 = 9 m/s
8. Total mass = 3000 kg, total momentum = 24 000 kg m/s, so v = 24 000 ÷ 3000 = 8 m/s
9. Bag momentum = 5 × 8 = 40 kg m/s backwards, so canoeist momentum = 40 kg m/s forwards and v = 40 ÷ 75 = 0.53 m/s forwards
10. Tool momentum = 2 × 15 = 30 kg m/s forwards, so astronaut momentum = 30 kg m/s backwards and v = 30 ÷ 50 = 0.6 m/s backwards
11. Total momentum = 30 000 + 5000 = 35 000 kg m/s, total mass = 2500 kg, so v = 14 m/s
12. Total momentum = 21 600 − 8000 = 13 600 kg m/s east, total mass = 2000 kg, so v = 6.8 m/s east
13. Change in momentum = 0.2 × (-15 − 25) = -8 kg m/s, so the size of the change is 8 kg m/s
14. Total momentum = 18 000 − 8000 = 10 000 kg m/s east, total mass = 2500 kg, so v = 4 m/s east
15. Total momentum = 1000 − 400 = 600 kg m/s, total mass = 450 kg, so v = 1.3 m/s in the first dodgem’s direction
16. Total mass = 5000 kg, total momentum = 18 000 kg m/s, so v = 18 000 ÷ 5000 = 3.6 m/s
17. Total momentum = 24 000 − 15 000 = 9000 kg m/s, total mass = 2700 kg, so v = 3.3 m/s in the electric car’s direction

10 Ramped Questions on Force, time and Change in Momentum

Answers to Force, change in momentum and time

1. Δp = mv = 0.40 × 15 = 6.0 kg m/s, F = Δp ÷ t = 6.0 ÷ 0.20 = 30 N
2. Δp = mΔv = 0.060 × 20 = 1.2 kg m/s, F = 1.2 ÷ 0.040 = 30 N
3. Δp = mΔv = 1200 × 20 = 24 000 kg m/s, F = 24 000 ÷ 5.0 = 4800 N
4. Δp = mΔv = 90 × 8.0 = 720 kg m/s, F = 720 ÷ 2.0 = 360 N
5. Δp = mΔv = 80 × 7.0 = 560 kg m/s, F = 560 ÷ 0.50 = 1120 N
6. Δp = mΔv = 75 × 12 = 900 kg m/s, F = 900 ÷ 0.15 = 6000 N
7. Δp = 900 kg m/s, F = 900 ÷ 0.60 = 1500 N, so the airbag reduces injury by increasing stopping time and reducing force

8a. Δp = mΔv = 1500 × 25 = 37 500 kg m/s

8b. F = Δp ÷ t = 37 500 ÷ 0.80 = 46 875 N

9a. Δp = mΔv = 200 000 × 20 = 4 000 000 kg m/s

9b. F = Δp ÷ t = 4 000 000 ÷ 40 = 100 000 N

10a. p = mv = 250 000 × 18 = 4 500 000 kg m/s

10b. Δp = 4 500 000 kg m/s

10c. F = Δp ÷ t = 4 500 000 ÷ 25 = 180 000 N

10d. If the stopping time doubled, the braking force would halve.