Work, energy, power answers, Summaries of Physics

Download Work, energy, power answers and more Summaries Physics in PDF only on Docsity!

OCR Maths M

Topic Questions from Papers

Energy, Work and Power

Answers

PhysicsAndMathsTutor.com

6 (i) ½.700.20^2 or ½.700.15^2 B1 either K.E.

700x9.8x400sin5° B1 correct P.E.

½.700.15^2 +700.9.8.400sin5°= M

½.700.20^2 + W.D.

for 4 terms with W.D.

W.D. = 178,000 J A1 4 or 178 kJ

(ii) D=200 + 700.9.8sin5° M

D = 798 N A1 may be implied

P = Dx15 = 12,000 = 12kW A1 3 AG (11,968W)

(iii) D’ = 11,968÷20 = 598 M

D’-700.9.8sin5°-200 = 700a M

a = 0.285 ms -2^ (±) A1 3 allow 0.283 (from 12kW) 10

Alternative for false assumption of constant acceleration

(i) D-700 x 9.8sin5° = 700a and

152 = 20^2 + 2a. 400

M1 (D = 445, a = -0.21875)

W.D. = 400xD = 178,000 A1 2 marks (out of 4)

maximum

7 (i) 50x9.8x2 = Rx3.75 +80x9.8x0.25 M1 moments about D.

“ A1 SR/no g/ R = 21.

(M1A1A0)

R = 209 N A1 3

(ii) 130 = 50 x 2 + 80 x 4.

A

moments about BC or

FE…..

x M

x =80x0.25 + 50x2.

x = 3.385 A1^ x = 1.

130 y = 50 x 0.125 + 80 x 0.25 M

A

moments about EC

y = 0.202 A

tanθ = 0.615/0.202 M

θ = 71.8° to the horizontal A1^8 71.6° to 72.0°^11

8 (i)^ x = 49cosθ^. t^ B

y=49sinθ.t - ½.9.8.t 2 B

y=xtanθ - 4.9x^2 /49^2 .cos^2 θ M

aef (eliminating t)

y=xtanθ – x^2 (1+tan^2 θ)/490 A1 4 AG

(ii) 30 = 70tanθ – 10(1+tan^2 θ) M

tanθ=(70±√3300)÷ 20 M (6.37/0.628)

81.1° A1^ θ^1 or^ θ 2

32.1° A1^4 “

(iii) x^2 (1+tan^2 θ)/490 = xtanθ M set y = 0

x = 490tanθ/(1+tan^2 θ) A

x = 75.0 A

(Q6, June 2005)

4729 Mark Scheme January 2006

2 (i)^ 0 = 50 sin25° t – 4.9t 2 M1^ or 0=50sin25° - 9.8t &2t :2x2. A t = 4.31 s A1 3 (ii) (^) d = 50cos25° x 4.31 M1 (^) or u 2 sin(2x25°)/g 195 m (^) A1 2 50cos25° x their t 5

3 (i)a 100 J B1 1 b 7500 Nm B1 1 (ii) 400cosα x 25 =7500 + 100 M for = a + b (^) A

sc N II gets M1A1only.This M for total M (a=0.08)&A1for α α = 40.5 A1 3 or 0.707 rads (^) 5

4 (i) horiz comps in opp direct B1 at E & F Right at E + Left at F B1 2 (ii) 1.6x9.8x30 = 20X or M1 or 10X + 1.6gx30 = 30X M(A) 0.5x30g + 0.7x30g + 0.2x60g = 20X

A1 or 10X + (... = 470.4) = 30X M mark ok without g but 3 parts X = 23.5 N A1 3

(iii) 1.6 y = M1 must be moments with vert dists

20x0.2 + 20x0.2 +40x0.5 A1^ or 1.6^ y^ =20x0.2x2+40x0.7(22.5)

y = 17.5 cm A1 3 8

5 (i) 6m = 3mx + 2my M1 - 3mx ok if clear on diagram 6 = 3x + 2y A1 m must have been cancelled M 57 B or ½. 3m.2 2 = ½.3mx 2 + ½.2my 2

e = 1 = (y-x)/

A1 6 = 3x 2 /2 + y 2 aef x = 0.4 or 2/5 A y = 2.4 or 12/5 A1 6

sc A1A0 if x = 2, y = 0 not rejected

(ii) 4.8m or 24m/5 (^) B1 2m x their y or 3m(2-their x) same as original dir. of A B1 2 use their diagram(or dir. of B) (iii) e = (2.8 – 1.0)/2.4 M 0.75 watch out for ± fiddles (^) A1 2 (1.8/their y) with 0  e : 1 10

(^1) tanθ = ⅓ (θ = 18.4° at B) B1 (^) 71.6° at C 3 x T sinθ = 20 x 1.5 must M1 M(A) (d=3/√10) have two distances and no g (^) A T = 31.6 N A1 (^4) 4

(Q3, Jan 2006)

4729 Mark Scheme January 2006

6 (i) x = 7t B y = - 4.9t 2 or -½gt^2 M1 some attempt at vertical motion A y = - x 2 /10 AG (no fiddles) A1 4

sc y=xtanθ ─gx 2 /(2V^2 cos^2 θ) with θ=0 M1 then A1 (max = 2) (ii) -20 = -x 2 /10 M1 or t=√(20/4.9) & x=7t 14.1 m A1 2 sc B1 for 14.1 after wrong work (iii) ½mv 2 = ½m7 2 + mgx20 n.b. v 2 =u 2 M1 OR v (^) h = 7 (B1) +2as gets M0 A1 v (^) v = ±19.8 (B1) 14√2,2√98 etc v = 21 ms-1^ A1 v = 21 (B1) dy/dx = -2x/10 & tanθ M A

OR tanθ = 19.8/7 or cosθ=7/21 or sinθ=19.8/ 70.5° to horizontal A1^6 or 19.5° to vertical^12

7 (i) F = 300/12 M R = 25 A1 2 (ii) P=17.5x12 (R 2 =17.5 & F 2 = 17.5) M P = 210 W A1 2

n.b. B1 only for 210 W without working (iii) 500 = Fx12 M F = 41.67 or 500/12 aef A 41.67 – 25 – 75x9.8sin1°= 75a M A 0.0512 ms-2^ A1 5 or 0. (iv) (^) PE = 75x9.8x200sin10° (25530) B WD = 200x120 (24000) B

OR 75x9.8sin10° -120= 75a (M1 + A1) ½.75v 2 = M1 a = 0.102 (A1) ½.75.13 2 +75x9.8x200sin10°-200.120 A1 v 2 =169+2x0.102x200 (M1) 14.5 ms-1^ A1 5 v = 14.5 14

8 (i) (^) R cos30° = 0.1 x 9.8 M1 resolving vertically

A R = 1.13 N A1 3 (ii) (^) r = 0.8cos30° = 0.693 or 2√3/5 B1 may be implied Rcos60° = 0.1 x 0.693 ω^2 M1^ or 0.1v^2 /r^ &^ ω^ = v/r A ω = 2.86 A1 4

(iii) T = 1.96 N B1 1 (iv) (^) Rcos30° = Tcos60° + 0.1x9.8 M A R = 2.26 N A

(Q7, Jan 2006)

1 mgh = 35 x 9.8 x 4 M A mgh/t = 1372/10 M1 watch out for extras 137 W A1 4 or 0.137 kW (^) 4

2 v^2 = 2gh^ M1^ kinematics or energy 2

u=√4g or √39.2 or 6.26 A1 speed of impact (±) v=√2.8g or √27.44 (5.24) A1 speed of rebound (±) I =C 0.3(6.26 + 5.24) M1 must be sum of mags. of vels.

(Q1, June 2006)

PhysicsAndMathsTutor.com

(ii) –25 = 3 x – x^2 /10 (must be -25) M1 or method for total time (5.26) solving quadratic M1 or 7 x total time 36.8 m A1 3 8

5(i) ½. 70 .4^2 M 560 J A1 2 (ii) 70 x 9.8 x 6^ M

4120 A1 2 4116

(iii) 60d B 8000 = 560 + 4120 + 60d M1 4 terms A1 their KE and PE 55.4 m A1 4 8 (Q5, June 2007)

10 (i) 4729 Mark Scheme January 2008

4729 Mechanics 2

1 (i) 12 x cos55° M

6.88 m s -1^ A1 2

(ii) 12 x cos55° x 0.65 M

(±) 4.47 m s-1^ A1^^2 0.65 x their (i) 4

2 F = 0.2 mg cos30° M1 =

A1 = (1.6974m) (49√3/50m)

0.2mgcos30° x d B1^ a=0.2gcos30°+gsin30°

mg x d x sin30° B1^ a= (±) 6.

d=½x25/(0.2x9.8cos30°+9.8xsin30°) M1^ 0 = 5^2 – 2x6.60d

1.89 m A1 6 6

3 direction of R perp. to wall B

R at 70° to rod B1 10 ° to horiz.

0.8 x 25cos60° = 1.6 x R sin70° M1^ moments about A

0.8 x 25 cos60° A

1.6 x R sin70° A

R = 6.65 N A1 6 6

4 (i) 45 000/v = kv M

k = 50 A1 2 AG

(ii) 45 000/20 – 50x20 = 1200a M

A

a = 1.04 m s -2^ A1 3

(iii) P/15 = 50x15 + 1200x9.8sin10° M

A

41 900 W A1 3 8

5 (i) 2mu – 3kmu = –mu + kmv M

v = ….. M1 attempting to make v the subject

v = 3u(1 – k)/k A

3u/k – 3u

( 0 < ) k < 1 A1 4 not ≤ 1

(ii) I = mu – – 2mu M1 or km(3u/k - 3u + 3u)

3mu A1 2 + only

(iii) v = ± 3u B

e = (u/2 + 3u)/4u M

e = 7/8 or 0.875 A1 3 9

(Q2, Jan 2008)

4729 Mechanics 2

1 (i) 12 x cos55° M

6.88 m s -1^ A1 2

(ii) 12 x cos55° x 0.65 M

(±) 4.47 m s-1^ A1^^2 0.65 x their (i) 4

2 F = 0.2 mg cos30° M1 =

A1 = (1.6974m) (49√3/50m)

0.2mgcos30° x d B1^ a=0.2gcos30°+gsin30°

mg x d x sin30° B1^ a= (±) 6.

d=½x25/(0.2x9.8cos30°+9.8xsin30°) M1^ 0 = 5

2 – 2x6.60d

1.89 m A1 6 6

3 direction of R perp. to wall B

R at 70° to rod B1^10 ° to horiz.

0.8 x 25cos60° = 1.6 x R sin70° M1^ moments about A

0.8 x 25 cos60° A

1.6 x R sin70° A

R = 6.65 N A1 6 6

4 (i) 45 000/v = kv M

k = 50 A1 2 AG

(ii) 45 000/20 – 50x20 = 1200a M

A

a = 1.04 m s -2^ A1 3

(iii) P/15 = 50x15 + 1200x9.8sin10° M

A

41 900 W A1 3 8

5 (i) 2mu – 3kmu = –mu + kmv M

v = ….. M1 attempting to make v the subject

v = 3u(1 – k)/k A

3u/k – 3u

( 0 < ) k < 1 A1 4 not ≤ 1

(ii) I = mu – – 2mu^ M1^ or km(3u/k - 3u + 3u)

3mu A1 2 + only

(iii) v = ± 3u B

e = (u/2 + 3u)/4u M

e = 7/8 or 0.875 A1 3 9

(Q4, Jan 2008)

12 (i)

4729 Mark Scheme June 2008

4729 Mechanics 2

(^1) 200cos35ϒ B 200cos35ϒ x d = 5000 M d = 30.5 m A1 3 3

2 0.03R =^ ½x0.009(250^2 – 150^2 )^ M1^1502 = 250^2 + 2a x 0. 0.03R B1 a = ±2x10^6 /3 or ±666,667 (A1) either K.E. B1 F = 0.009a (M1) R = 6000 N (^) A1 4 unit error s 4

3 (i) D = 12000/20 B 12000/20=k x 20 + 600 x 9.8 x 0.1 M k = 0.6 A1 3 AG (ii) 16000/v = 0.6v + 600 x 9.8 x 0.1 M 0.6 v^2 + 588v – 16000 = 0 M1 attempt to solve quad. (3 terms) v = 26.5 m s-1^ A1 3 (iii) 16000/32 – 0.6 x 32 = 600a M A a = 0.801 m s -2^ A1 3 0.80 or 0.8 9

2 2

(Q1, June 2008)

4729 Mark Scheme June 2008

4729 Mechanics 2

(^1) 200cos35ϒ B 200cos35ϒ x d = 5000 M d = 30.5 m A1 3 3

2 0.03R = ½x0.009(250^2 – 150^2 ) M1 1502 = 250^2 + 2a x 0. 0.03R B1 a = ±2x10^6 /3 or ±666,667 (A1) either K.E. B1 F = 0.009a (M1) R = 6000 N (^) A1 4 unit error s 4

3 (i) D = 12000/20 B 12000/20=k x 20 + 600 x 9.8 x 0.1 M k = 0.6 A1 3 AG (ii) 16000/v = 0.6v + 600 x 9.8 x 0.1 M 0.6 v^2 + 588v – 16000 = 0 M1 attempt to solve quad. (3 terms) v = 26.5 m s-1^ A1 3 (iii) 16000/32 – 0.6 x 32 = 600a M

(Q2, June 2008)

PhysicsAndMathsTutor.com

A1 4

3 (i) D = 12000/20 B 12000/20=k x 20 + 600 x 9.8 x 0.1 M k = 0.6 A1 3 AG (ii) 16000/v = 0.6v + 600 x 9.8 x 0.1 M 0.6 v^2 + 588v – 16000 = 0 M1 attempt to solve quad. (3 terms) v = 26.5 m s-1^ A1 3 (iii) 16000/32 – 0.6 x 32 = 600a M A a = 0.801 m s -2^ A1 3 0.80 or 0.8 9

4 (i) 0 = 35sinθ x t – 4.9t^2 M1 R=u^2 sin2θ/g only ok if proved t = 35sinθ/4.9 50sinθ/7 A1 or 70sinθ/g aef R = 35cosθ x t aef B their t R = 35^2 sinθ.cosθ/4.9 M eliminate t R = 125sin2θ A1 5 AG (ii) 110 = 125sin2θ M θ = 30.8ϒ or 59.2ϒ A1+ t = 3.66 s or 6.13 s A1+1 5 10

5 (i) 3/8 x 3 (1.125) B1 c.o.m. hemisphere 0.53d = 5x0.02 + (10 + 3/8x3) x 0.5 M1^ 0.53e=3x5/8x0.5+8x0.02+13x. A1 0.53f=3x3/8x0.5-5x0.02-10x0. d = 10.7 A1 4 AG ( e = 2.316 f = 0.684 ) (ii) Attempt to calc a pair relevant to P,G M1 distance / angle OP=0.9 (pair) , p= 73.3ϒ q=16.7ϒ r=76.9ϒ (77.2ϒ) , s=13.1ϒ(12.8ϒ) AC=0.86 , BC=0.67 , AD=10.4 BD=10.

A1 not a complimentary pair

r > p , s < q , p + s < 90 , 0.67 < 0.86 , 10.2 < 10.

M1 make relevant comparison 0.7 < 0.9 (OG < OP) 10.7 < 10. it is in equilibrium A1 (^4 )

(Q3, June 2008)

15 (i)

T sin30° x 12 = 8 x 2 x 9.8^ M1^ ok if g omitted

A1 ft ft their x

T = 26.1 A1 4 4

3 (i) 140 x X = 40 x 70 M X = 20 N A at F 20 N to the right B1 inspect diagram at G 20 N to the left B1 4 SR B1 for correct directions only (ii) đ = (2x40sinΠ/2)÷ 3 Π/2 M1^ must be radians A đ = 17.0 A1 16.98 160/3Π ( 8/15Π m )

70 y = 100x60 + 217 x 10 M

A1 ft ft 200 + their đ or 2 + their đ (m)

y = 117 A1 6 116.7 10

4 (i) (^) P /10 – 800x9.8sin12° – 100 k = 800x0.25 M1 P/10 = D 1 ok A1 D 1 ok P /20 – 400 k = 800x0.75 M1 P/20 = D 2 ok A1 D 1 = 2D 2 needed for this A solving above M k = 0.900 A1 AG 0. P = 19 200 A1 7 or 19.2 kW (maybe in part (ii)) (ii) 0.9 v^2 = 28 800/ v M1 ok if 19200/ v solving above M1 * ( v^3 = 32 000) v = 31.7 m s-1^ A1 3 10

5 (i) 0.8 S B1 vert comp of S 0.6 T B1 vert comp of T S cos α = T cos β + 0.2 x 9.8 M

0.8 S = 0.6 T + 1.96 aef A1 4 AG 4 S = 3 T + 9. (ii) 0.6 S B 0.8 T B 0.2 x 0.24 x 8^2 B1 3.072 384/ S sin α + T sin β = 0.2 x 0.24 x 8 2 M1 must be mrω^2 6 S + 8 T = 30.72 A1 aef eliminate S or T M S = 3.4 N A1 3. T = 1.3 N A1 8 1.282 12

(Q4, Jan 2009)

4729 Mark Scheme June 2009

4729 Mechanics 2

1 (i) (^) ½×75×12 2 or ½×75×3^2 (either KE) 75×9.8×40 (PE) R ×180 ( change in energy = 24337) ½×75×12 2 =½×75×3 2 +75×9.8×40– R × R = 135 N

B

B

B

M

A1 5

M1 12^2 = 3 2 + 2 a × 180 A1 a = 0.375 (3/8) M1 75 × 9.8 × sin θ – R = 75 a A1 R = 135 (max 4 for no energy) 5

2 (i) (^) R = F = P / v = 44 000/ v = 1400 v = 31.4 m s-

M

A1 2

(ii) (^) 44 000/ v = 1400 + 1100 × 9.8 × 0.

v = 22.7 m s-

M

A

A1 3

must have g

(iii) (^) 22 000/10 + 1100×9.8×0.05 – 1400 = 1100 a a = 1.22 m s-

M

A

A1 3

3 (i) (^) cos θ = 5/13 or sin θ = 12/13 or θ = 67.4q

0.5× F sin θ = 70×1.4 + 50 × 2. F = 516 N

B

M

A

A1 4

any one of these moments about A (ok without 70) 0.5sin θ = 0. SR 1 for 303 (omission of beam) (ii) (^) F sin θ = 120 + Y (resolving vertically)

Y = 356 their F × 12/13 – 120 X = F cos θ (resolving horizontally) X = 198 their F × 5/ Force = √(356 2 + 198^2 ) 407 or 408 N

M

A

M

A

M

A1 6

M1/A1 for moments (B) Y ×2.8+1.4×70=2.3×516 ×12/ (C) 0.5× Y = 0.9×70 + 2.3× (D) 1.2 X = 1.4×70 +2.8×

4 (i) (^) T = 0.4 × 0.6 × 2 2 T = 0.96 N

M

A1 2

(Q1, June 2009)

4729 Mark Scheme June 2009

4729 Mechanics 2

1 (i) (^) ½×75×12 2 or ½×75×3^2 (either KE) 75×9.8×40 (PE) R ×180 ( change in energy = 24337) ½×75×12 2 =½×75×3 2 +75×9.8×40– R × R = 135 N

B

B

B

M

A1 5

M1 12^2 = 3 2 + 2 a × 180 A1 a = 0.375 (3/8) M1 75 × 9.8 × sin θ – R = 75 a A1 R = 135 (max 4 for no energy) 5

2 (i) (^) R = F = P / v = 44 000/ v = 1400 v = 31.4 m s-

M

A1 2

(ii) (^) 44 000/ v = 1400 + 1100 × 9.8 × 0.

v = 22.7 m s-

M

A

A1 3

must have g

(iii) (^) 22 000/10 + 1100×9.8×0.05 – 1400 = 1100 a a = 1.22 m s-

M

A

A1 3

3 (i) (^) cos θ = 5/13 or sin θ = 12/13 or θ = 67.4q

0.5× F sin θ = 70×1.4 + 50 × 2. F = 516 N

B

M

A

A1 4

any one of these moments about A (ok without 70) 0.5sin θ = 0. SR 1 for 303 (omission of beam) (ii) (^) F sin θ = 120 + Y (resolving vertically)

Y = 356 their F × 12/13 – 120 X = F cos θ (resolving horizontally) X = 198 their F × 5/ Force = √(356 2 + 198^2 ) 407 or 408 N

M

A

M

A

M

A1 6

M1/A1 for moments (B) Y ×2.8+1.4×70=2.3×516 ×12/ (C) 0.5× Y = 0.9×70 + 2.3× (D) 1.2 X = 1.4×70 +2.8×

(Q2, June 2009)

18 (i)

4729 Mark Scheme January 2010

4729 Mechanics 2

1 75×9.8×40^ B1^ Average Speed = 40√ 120 (75×9.8×40)√ 120 M1 (75×9.8)×(Average speed) 245 W A1 [3] 3

2 (i) v 2 = 2×9.8×3 or 2×9.8×1.8 M1 Kinematics or energy

v 1 = 6 g or 58. 8 or 30

or 7.

A1 Speed of impact (±)

v 2 = 3. 6 g or 35. 28 or 2

or 5.

A1 Speed of rebound (±)

(Q1, Jan 2010)

PhysicsAndMathsTutor.com

AG = 0.5 m A [5]

This may only be seen in (ii), allow M1A1 in this case. (ii) v = 0.5x v = 1.5 ms -

M A [2]

Allow use of candidate’s 0.2, 0.4, 0.3, 0.

2 (i) (k25 3/2) x 25 = 15000 k = 4.8 AG

M A A [3]

Tractive force x speed = power

(ii) R = 4.8 x 16 3/ T – 4.8x163/2^ + 700gx1/15 = 700x0. P = 59.9 x 16 P = 958 W

B M A M A [5]

N2L, 4 terms to find tractive force (T) Allow cv(R), R not 600; (T = 59.866..) 16xTractive force

3

(Q2, Jan 2011)

4729 Mark Scheme January 2011

3 (i) T (^) Acos30 + TBcos60 = 0.4g 2Tcos30 + Tcos60 = 0.4g T (^) B = 1.76 N T (^) A = 3.51 N

M A A A [4]

Resolves vertically, 3 terms T = 1.756. Watch for MR of Tcos30 + 2Tcos60 = 0.4g Accept 3. (ii) r = 0.5sin30 (= 0.25) 3.51sin30 + 1.76sin60 = 0.4ω^2 0.5sin ω = 5.72 rad s -

B M A1ft A [4]

N2L radial, 3 terms cv(1.76, 3.51, 0.25) Accept 5.

4 (i) WD = 100cos20 x 30

WD = 2820 J

M A [2]

Product of 3 relevant elements. Angle could be 5, 25 or complements 2819.1... (ii) PE = 25g x 30sin PE = 641

M A [2]

Product of weight and vertical height. Allow without g

(iii)

OR

2819.1 = 640.

• 30x70 + 25v^2 / v = 2.51 ms - 25a = 100cos20 -70 -25gsin a = 0. v^2 = 2 x 30 x ‘a’ v = 2.

M A1ft A A [4] M A depM A [4]

4 term energy equation ft(cv 2820 and cv 641) cao 4 term equation Allow 0.1 here Or equivalent complete method cao

4

(Q4, Jan 2011)

4729 Mark Scheme June 2011 1 i

PE = 70x3g KE change =70x(2.1 2 – 1.4 2 )/ PE change + KE change 2143.75 J

B B M A [4]

2058

Must include evaluation Accept 2140. Allow all values to be negative. ii

OR

20(90 + T) = 2143. T = 17.1875 N

70g.0.15 – 90 – T = 70.(-0.06125) T = 17.1875 N

M A1ft A [3] M A A [3]

Work done = Energy change used ft(cv(2143.75)) accept 17. Use of v^2 =u 2 + 2as to find a AND use of N2 law(4 terms) accept 17.

2 i

21000/ 0 = 21000/25 – 25k – 1250gsin k = 16.

B M A A [4]

Use of force = power/speed 3 terms cv(21000/25)

ii 21000/v= 16.5v v = 35.7 ms -

M A1ft A [3]

ft on cv(k)

3 i - (8cos30/3)(8 2 sin60/2)

• (4)(8 2 ) = (8 2 +8 2 sin60/2)(x (^) G ) x (^) G = 2.09 cm

M A A A A [5]

Table of moments idea, may include g and/or density. -2.309 x 27.

ii tanθ = (2.09/4) θ = 27.6°

M A1ft [2]

ft cv(x (^) G )

1

(Q1, June 2011)

25 i

4729 Mark Scheme June 2011 1 i

PE = 70x3g KE change =70x(2.1 2 – 1.4 2 )/ PE change + KE change 2143.75 J

B B M A [4]

2058

Must include evaluation Accept 2140. Allow all values to be negative. ii

OR

20(90 + T) = 2143. T = 17.1875 N

70g.0.15 – 90 – T = 70.(-0.06125) T = 17.1875 N

M A1ft A [3] M A A [3]

Work done = Energy change used ft(cv(2143.75)) accept 17. Use of v^2 =u 2 + 2as to find a AND use of N2 law(4 terms) accept 17.

2 i

21000/ 0 = 21000/25 – 25k – 1250gsin k = 16.

B M A A [4]

Use of force = power/speed 3 terms cv(21000/25)

ii 21000/v= 16.5v v = 35.7 ms -

M A1ft A [3]

ft on cv(k)

3 i - (8cos30/3)(8 2 sin60/2)

• (4)(8 2 ) = (8 2 +8 2 sin60/2)(x (^) G ) x (^) G = 2.09 cm

M A A A A [5]

Table of moments idea, may include g and/or density. -2.309 x 27.

ii tanθ = (2.09/4) θ = 27.6°

M A1ft [2]

ft cv(x (^) G )

1

(Q2, June 2011)

26 i

PhysicsAndMathsTutor.com

R = 0 B1 May be implied. T 1 sin T = 0.2 g B T 1 cos T = 0.2 × v^2 / r or 0.2 × r Z 2 M1 Attempt at resolving. v = 4.2 ms^1 A [5] 5 (i) 25000/10 B 1500 g sin5 B1 1281. M1 Attempt at N2L with 4 terms. 2500 – 750 – 1500 g sin5 = 1500 a A1 cv(1500 g sin5); cv(2500) not 25000. a = 0.313 A1 Allow 0. [5] 5 (ii) WD against resistance = 750 d B1 750 h /sin WD by engine = 25000×28 ( = 700000) B Change in PE = 1500 g × d sin5 B1 1500 g × h Change in KE = ±½ ×1500 × (20^2 – 10^2 ) B1 Use of correct formula for KE. M1 Use conservation of energy, at least 3 used including WD by engine. 25000×28 = ½ ×1500 ×(20^2 – 10^2 ) + 750 d + 1500 g × d sin

A d = 234 A [7]

6

(Q5, Jan 2012)

4729 Mark Scheme June 2012 Question Answer Marks Guidance 1 (i) Speed = 1.2 ms^1 B1 May be seen anywhere, even in (ii) ; allow -1. Impulse = 0.8 × ± (4  1.2) M1^ Difference between momenta, allow 0.8 × ± (4  1.2) ±4.16 Ns A [3] 1 (ii) KE lost = ½ × 0.8 × (4^2 –(±1.2) 2 ) M 5.82(4) J A1 Allow -5.82(4) [2] 2 (i) Driving Force = 20000/20 (= 1000) B M1 Attempt at N2L with 3 terms. Signs may not be correct at this stage. 20000/20 – 800 = 1600 a A1 Using their 20000/20, but not 20000 a = 0.125 ms^2 A1^ Allow 81 [4] 2 (ii) 20000/ v B M1 3 terms with attempt at resolving weight; g can be omitted at this stage; if F = ….. then F = 0 somewhere to award M DF – 800 – 1600 g sin4 = 0 A1 aef v = 10.6 ms^1 A [4] 3 (i) M1 Attempt at moments about A, g can be omitted at this stage T cos30 × 1.5sin 30 = 15 g × 2 A T = 453 A [3] 3 (ii) X = T (^) c sin30 (=226) B1ft M

Using their value T or taking moments about P Attempt to resolve vertically or taking appropriate moments Y + T (^) c cos30 = 15 g A1ft Using their value T ; expect Y = -245 or better Either or both of these equations can be replaced with moments about an appropriate point eg P , Q , B , c of m of beam. R = √(226 2 + 245^2 ) or tan T = 245/226 M1^ Any relevant angle R = 334 A1 Allow 333 T = 47.3 below horizontal (to the left) A1^ Allow 47.2, 42.7 to the downward vertical [6] SC: If 392 in (i) leading to Y= ± 245 only in (ii) max M1A

5

(Q2, June 2012)

4729 Mark Scheme January 2013

5

Question Answer Marks Guidance 1 (i) 18cos15 x 6 104 J

M A A [3]

Force component x distance

1 (ii) 18cos15 x 6/5 or ans(i)/ 20.9 W

M A [2]

Force component x distance/ Allow 20.

2 (i) DF = 15000/ DF – k x 151/2^ = 1500 x 0. k = 103

B M A A [4]

N2L, 3 terms and attempt at DF. Numerical DF Allow 80 √^15 / (^3)

2 (ii) P/30 = k 30 1/ P = 17000W

M A A [3]

Using cv(k) Allow 17(.0)kW, 16900W, 16.9kW, 12000√2W 3 (i) a = gsin

1+ u = 0.4(2+2gsin30) u = 3.72 ms -

B M A A [4]

Using NEL with uA from cv(a), u (^) A ≠ 0 cwo

3 (ii) Use v^2 = u^2 – 2(gsin30) s s = 1.41 m

M A [2]

Using v = 0, cv(a) from (i) or correct a SC If a not found in (i), allow a=g for M1A0. 3 (iii) Use of conservation of momentum 0.5 x 2gsin30 - 2 m = m - 0.5 x 3. m = 2.

M A1ft A [3]

Using cv(a) ft cv(u) from (i) Aef(raction) eg 2 19 / 75 or 169 / 75

(Q1, Jan 2013)

4729 Mark Scheme January 2013

5

Question Answer Marks Guidance 1 (i) 18cos15 x 6 104 J

M A A [3]

Force component x distance

1 (ii) 18cos15 x 6/5 or ans(i)/ 20.9 W

M A [2]

Force component x distance/ Allow 20.

2 (i) DF = 15000/

DF – k x 151/2^ = 1500 x 0. k = 103

B M A A [4]

N2L, 3 terms and attempt at DF. Numerical DF Allow 80 √^15 / (^3)

2 (ii) P/30 = k 30 1/ P = 17000W

M A A [3]

Using cv(k) Allow 17(.0)kW, 16900W, 16.9kW, 12000√2W 3 (i) a = gsin 1+ u = 0.4(2+2gsin30) u = 3.72 ms -

B M A A [4]

Using NEL with uA from cv(a), u (^) A ≠ 0 cwo

3 (ii) Use v^2 = u^2 – 2(gsin30) s s = 1.41 m

M A [2]

Using v = 0, cv(a) from (i) or correct a SC If a not found in (i), allow a=g for M1A0. 3 (iii) Use of conservation of momentum 0.5 x 2gsin30 - 2 m = m - 0.5 x 3. m = 2.

M A1ft A [3]

Using cv(a) ft cv(u) from (i) Aef(raction) eg 2 19 / 75 or 169 / 75

(Q2, Jan 2013)

PhysicsAndMathsTutor.com