OCR Maths M2. Topic Questions from Papers. Projectiles. Answers

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1 OCR Maths M Topic Questions from Papers Projectiles Answers PhysicsAndMathsTutor.com

2 1 v = x9.8x10 energy:½mv =½mu + mgh v = 14 ½v = ½ x10 speed = ( ) (must be 6 ) v = =3 speed = 15. ms -1 tanθ = 14/6 cos -1 (6/15.) etc θ=66.8 (below) horiz. 6 or 3. to the vertical 6 (Q, June 005) 8 (i) x = 49cosθ. t y=49sinθ.t - ½.9.8.t y=xtanθ - 4.9x /49.cos θ M aef (eliminating t) 1 y=xtanθ x (1+tan θ)/490 4 AG (ii) 30 = 70tanθ 10(1+tan θ) M 1 tanθ=(70± 3300) 0 M 1 (6.37/0.68) 81.1 θ1 or θ (iii) x (1+tan θ)/490 = xtanθ M 1 set y = 0 x = 490tanθ/(1+tan θ) x = 75.0 x = 1 (0.6) d = 146 m 5 13 (iii) Alternatively (1 st marks) t=49sinθ/4.9 and (9.88/5.31) x=49cosθ.t 1 tanθ x = = 490sinθcosθ ⅓ (θ = 18.4 at B) sinθ at C M 1 s=ut+½at and x=49cosθ.t or R = u sinθ/g (precise) 3 x T sinθ = 0 x 1.5 must M(A) (d=3/ 10) (Q8, June 005) have two distances and no g T = 31.6 N = 50 sin5 t 4.9t 3 (i) or 0=50sin5-9.8t &t :x.16 t = 4.31 s 3 (ii) d = 50cos5 x 4.31 or u sin(x5 )/g 195 m 50cos5 x their t 5 (Q, Jan 006)

3 64 (i) x = 7t y = - 4.9t or -½gt y = - x /10 AG (no fiddles) 4 some attempt at vertical motion sc y=xtanθ gx /(V cos θ) with θ=0 then (max = ) PhysicsAndMathsTutor.com (ii) -0 = -x /10 or t= (0/4.9) & x=7t 14.1 m sc for 14.1 after wrong work (iii) ½mv = ½m7 + mgx0 n.b. v =u OR v h = 7 () +as gets M0 v v = ±19.8 () 14, 98 etc v = 1 ms -1 v = 1 () dy/dx = -x/10 & tanθ OR tanθ = 19.8/7 or cosθ=7/1 or sinθ=19.8/ to horizontal 6 or 19.5 to vertical 1 (Q6, Jan 006) 75 (i) vsin50 initial vertical component 0=v sin 50 -x9.8x13 (must be 13) or mx9.8x13 = ½m(vsin50 ) v = 0.8 ms -1 3 sin/cos mix ok for above (ii) 45 = vcos50.t see alternative below t = 3.36 their v (3.13 for other methods include other t s v=.4) s = vsin50 x t - ½ x 9.8 x t ignore ht adjustments can be their v and their t s = 1.6 to.0 inclusive ( 1.68) can be implied from next ht above ground = 0.30 m 6 (iii) v v = vsin50-9.8xt or v v =g(15-their ans to ii) v v = their v,t(-13.5 above for v v for.4) speed= (v v +(vcos50 ) ) or ½mv mgx1.68 = speed = 1.6 ms -1 their v 4 ½mx0.8 (4 marks) / s,v / and v v (19.7 for v =.4) solve/ (ii) y = xtanθ gx /v cos θ Alternative 1 st 5 marks y=45tan50 - substitute v and 50 and x= /.v cos 50 can be their v calculate y y = 1.6 to.0 inclusive should be (Q7, June 006)

4 86 (i) v v = 4sin30 (=1) 0 = 1 x9.8xh h =.5 3 (ii) v h = 4cos30 (=36.4) v v = ±v h x tan10 v v = ±6.41 or 1 3 tan10 or 4cos30.tan = 4sin30 9.8t ** must be 6.41(also see or x ) t =.80 ** y=4sin30 x.8 4.9x.8 ** y = 0.4 ** their t x = 4cos30 x.80 x = 10 their t (x + y ) d = or 6.41 = 1 + x -9.8s ** vert dist first then time s = 0.4 ** 0.4 = 1t + ½. -9.8t ** t =.80 ** or.5 s and 6.41 =x9.8s ** dist from top (s =.096) y = 0.4 **.5 &.1 = ½.9.8t ** separate times (.143, 0.654) t =.80 ** alternatively (ii) y = x/ 3 x /70 aef y=xtan30 9.8x /.4.cos 30 dy/dx = 1/ 3 x/135 for differentiating aef dy/dx = tan10 must be tan10 1/ 3 x/135 = tan10 solve for x x = 10 on their dy/dx y = x/ 3 x /70 y = 0.4 their x (x + y ) d = 104 (11) (Q8, Jan 007) 7 0 = 1sin7 t 4.9t any correct. or R = u sinθ/g (B) t = method for total time correct formula only R = 1cos7 x t 1 x sin54 / 9.8 sub in values (Q, June 007) 48 (i) x = 7t y = 1t 4.9t or g/ y = 1.x/7 4.9 x /49 y = 3x x /10 5 AG (ii) 5 = 3x x /10 (must be -5) or method for total time (5.6) solving quadratic or 7 x total time 36.8 m 3 8 (Q4, June 007)

5 91 (i) 1 x cos m s -1 (ii) (Q1, Jan 008) 10 7 (i) 0 = (175sinθ) x9.8x650 θ = (ii) Attempt at t 1, t, t top or t total 650 = 175sin55.t - 4.9t etc 5.61, 3.65, 14.63, 9.6 t t 1 or (t top t 1 ) or t total t 1 time difference = (iii) v h = 175cos55 (100.4) or KE ½mv v v = 175sin x 5.61 () PE mx9.8x650 speed = ( ) v = (175 x9.8x650) 134 m s (Q7, Jan 008) 11 4 (i) 0 = 35sinθ x t 4.9t R=u sinθ/g only ok if proved t = 35sinθ/4.9 50sinθ/7 or 70sinθ/g aef R = 35cosθ x t aef their t R = 35 sinθ.cosθ/4.9 eliminate t R = 15sinθ 5 (ii) 110 = 15sinθ θ = 30.8 or t = 3.66 s or 6.13 s AG (Q4, June 008) 1 (ii) (ii) v h = or () ½mx v v = x 9.8 x 4 () ½mxv v v = 8.85 (14 10/5) () mx9.8x4 speed = ( ) v = ( + x9.8x4) 9.08 m s -1 tan -1 (8.85/) or cos -1 (/9.08) 77.3 to horizontal to vertical 13 (Q7, June 008)

6 113 (0 sin θ) = x 9.8 x 17 or B for max ht = v sin θ/g PhysicsAndMathsTutor.com sin θ = (x9.8x17) 0 subst. values in above θ = (Q1, Jan 009) 14 6 (i) x = vcosθ t y = vsinθ t ½ x 9.8 t or g substitute t = x/vcosθ y = xtanθ 4.9x /v cos θ 4 AG (ii) Sub y = -h, x = h, v = 14, θ =30 signs must be correct h = h/ 3 h /30 aef solving above h = (iii) v v = (14sin30 ) x9.8x(-47.3) 14cos30 t=47.3 ft & v v =14sin30-9.8t (double negative needed) ft their ft t = 3.90 (or dy/dx=1/ 3 x/15 etc ft) v v = ±31. v v = ±31. (tanα = 1/ /15) tan -1 (31./14cos30 ) tan -1 (31./14cos30 ) α= 68.8 below horiz/1. to d vert /.. (iv) ½mx14 + mx9.8x47.3 = ½mv ft ( ) v = (Q6, Jan 009) 15 7(i) (i) 9 = 17 cos5 t t = (or 9/17cos5 ) d = 17sin ½ 9.8x (d = ht lost (5.87) h =.13 (ii) v h = 17cos5 (15.4) v v = 17sin or v v = (17sin5 ) v v = 1.9 tanθ = 1.9/15.4 θ = 40.0 below horizontal (iii) speed = ( ) ½mv = ½m v = 16.8 m s y=xtanθ-4.9x /v cos θ / y=9tan(-5 ) /17 cos 5 y = / dy/dx = tanθ 9.8x/v cos θ dy/dx = tan -1 (-.838) or 50.0 to vertical (0.1) NB 0.3 instead of 0.7 gives 11.0 (M0) 14 (Q7, June 009)

7 16 6 (i) 30 = V 1 sin θ ½m V 1 = ½m 50 + m V 1 sin θ 1 = 5800 AEF V 1 cos θ 1 = 40 V 1 = 86.0 AG θ 1 = 6.3 AG (ii) 0 = 5800 t p 4.9t p 30 = V 1 sin θ 1 9.8t t p = 15.5 t = = t q t q = 10.8 (iii) R = R = 61 (60, 6) V cos θ 10.8 = 61 V cos θ = = V sin θ V sin θ = 53.1 or 53.0 (5.9,53.1) Method to find a value of V or θ θ = to 4.9 V = 78. m s -1 or 78.1 m s -1 [8] or cos = 3/5 or sin = 4/5 or tan = 4/3 = angle to vertical (Q6, Jan 010) 117 v = x 9.8 x 10 Using v = u +as with u = 0 v = 14 m s -1 speed = ( ) Method to find speed using their v 15.7 or 7 5 m s -1 tan -1 (14/7) or tan -1 (7/14) Method to find angle using their v 63.4 to the horizontal to vertical 6 (Q1, June 010)

8 718 (i) R = 0. x 9.8 x cos30 (= 1.70) F = 0.1 x 9.8 x cos30 (= 0.849) FT FT on their R, but not R =0.g Use of conservation of energy ½ x 0. x 11 - ½ x 0. v = 0. x 9.8 x 5sin x v = 5.44 m s -1 6 AG Or F + 0.gsin30 = 0.a Use of NL, 3 terms last 4 a = 9.1 marks v = 11 + x a x 5 Complete method to find v of (i) v = 5.44 m s -1 (ii) t = 5cos30 /5.44cos30 time to lateral position over C t = s u = 5.44sin30 (=.7) s =.7 x x s = -1.6 (or better) Ht dropped Ht drop to C = 5sin30 =.5 m Ball does not hit the roof 7 13 Or y = xtanθ gx sec θ/v first substitute values 5 V = 5.44 θ = 30 x = 5cos30 all 3 correct marks y =.5 9.8x5x3/4x4/3 / (x5.44 ) of (ii) y = -1.6 (or better) OR (ii) u = 5.44sin30 (=.7) -.5 = 5.44sin30t 4.9t aef t = 1.04 time to position level with AC x = 5.44cos30 x 1.04 = 4.9 (or better) Horizontal distance from B to C = 5cos30 = 4.3(or better) Ball does not hit the roof 7 OR (ii) y = xtanθ gx sec θ/v substitute values -.5 = 0.577x 0.1x aef Attempt to solve quadratic for x x = 4.9 (or better) Horizontal distance from B to C = 5cos30 = 4.3 (or better) Ball does not hit the roof 7 OR (ii) u = 5.44sin30 = = 5.44sin30t 4.9t aef t = 1.0 (or better) time to position level with AC T = 5cos30 /5.44cos30 T = 0.9 (or better) time to lateral position over C Ball does not hit the roof 7 (Q7, June 010)

9 19 6 (i) 0 = (14sin30) gh h =.5 m (ii) 0.4x15 = 0.4(14cos30) + I I = 1.15 (iii) v = (14sin30) + 15 v = 16.6 ms -1 tanθ = 14sin30/15 OR tan =15/14sin30 θ = 5(.0) OR = 65(.0) (iv) t = 14sin30/g (= 1/1.4 = ) T = 1.43 s R = 14cos30/ /1.4 R = 19.4 m [] [3] h = (14sin30)x1/1.4 g(1/1.4) / or use (u sin )/g Impulse = change in momentum Not 14 or 0 for horizontal speed before impulse aef Not (14sin30) + (14cos30) Allow 74 Correct trig to find an appropriate angle; not 14cos 30 for 15 Rise or fall time (not to be given in (i)) Accept 10/7 (14 sin(x30) sin(x5))/g. 14 resolved, 15 not (Q6, Jan 011) 5 i 0 i ii iii OR x = (7cos30)t y = (7sin30)t gt / y = xtan30 - gx /(x7 cos 30) x /15 x/ = 0 or 9.8t 7t + 1. = 0 x = 1.73 m or 3 m.6(0) m or 3 3/ m v = (7sin30) x9.8x0.6 v = 0.7 ms -1 tanθ = 0.7/(7cos30) θ = 6.59 to horizontal or 83.4 to vertical Attempt to differentiate equation of trajectory tan30 - gx/(7 cos 30) Substitute x = 3 and equate to tan θ = 6.59 to horizontal or 83.4 to vertical Attempt to eliminate t y = x/ 3 x /15 or y = 0.577x 0.133x aef Create a 3 term Q.E. in x or t with y = 0.6 Solve 3 term Q.E. for x or t Using v = u -gs with u a component of 7; can find t first from their x in (i), and then use v = u + at. Use component of 7 Allow 1/ 3 4x/15 or y = x (Q5, June 011) 1 v x = 40cos35 Expect 3.8, need not be evaluated. 1 v y = 40sin Expect 6.46, need not be evaluated. v = or tan = 6.46/3.8 Use of Pythagoras or relevant trig on cv(v x ) and cv(v y ) v = 33.4 ms 1 = 11.1 below horizontal AEF; allow 11. (i) Seen anywhere and in any form. (Q1, Jan 01)

10 7 (i) For P 4.9t = 60 Signs must be consistent. t = 3.5(0) aef For Q 0 = 5sin t ½ 9.8 t = 43.3 PQ = (5cos 15) t c = 11. [6] 7 (ii) 5cos (t) = 15(t) and solving for Equating horizontal components of velocity (or displacement) and solving for = 53.1 For Q s y1 = 5sin t ½ 9.8 t For P s y = ±½ 9.8 t Using s y1 + s y = 60 * Solving for t dep* t = 3 v = 5sin Other methods include finding time to max height for Q. v = 9.4 therefore falling. [9] OR (ii) 5cos (t) = 15(t) and solving for Equating horizontal components of velocity (or displacement) and solving for = 53.1 gx For Q y xtan (5) cos gx For P y (60 ) (15) Equate y and solve for x Use x = ucos t to find t t = 3 v = 5sin v = 9.4 therefore falling. * dep* [9] Must include 60. PhysicsAndMathsTutor.com Other methods include finding time to max height for Q. (Q7, Jan 01) 3 4 (i) ½ 9.8 t = 0. Using SUVAT to find t, consistent signs for g and 0. t = 0.(0) aef s = 14.4 t c Using their value of t s =.91 m OR Use equation of trajectory for correct equation of the trajectory seen anywhere but award in part (ii) unless different method seen; consistent signs for g and 0. 4 (ii) -0. = xtan0 gx sec 0/(x14.4 ) Solve quadratic for x x =.91 Using s = ut + ½ at with s = ±0. and a = ±g Usin15 t ½ 9.8 t = -0. Ucos15 t = 6 Eliminate t Dep* Eliminate U Attempt to solve to find U Dep* Attempt to solve to find t(=0.607) U = 10. ms 1 [6] OR y = xtanθ gx sec θ/u * * Substitute values for y, x, Dep* -0. = 6tan15 g.6 sec 15/U Attempt to solve for U Dep*M U = 10. ms 1 [6] (Q4, June 01) 4 7 (i) x = ucos t y = usin t ½gt Eliminate t Get y = xtan gx sec /u [AG] 7 (ii) Substitute x =, y = -.1 and u = 14 Use sec = 1 + tan Tidy to 1.1tan tan + 10 = 0 [AG] Solve QE for tan = (iii) t = /14cos t =.1s [] www May start again of course www allow in radians (0.738) May work vertically, but must solve for t to get (Q7, Jan 013)

11 5 7 (i) ucos = 14cos 0 U x = Horizontal component of initial velocity, could use U x Complete method to find vertical component of initial velocity, could use U y 14sin 0 = usin 1.4g U y = u = (1.4g 14sin 0) + (14cos 0) Method to find u u = 15.9 AG cwo tan = (1.4g 14sin 0)/14cos 0 Method to find or a relevant angle = 34. SC for tan0= (usin 1.4g)/ucos OR 14 = (usin 1.4g) + (ucos ) for both. [7] 7 (ii) Method to find Level of P above A ½ m( ) = mgy y =.9 m [3] OR (ii) Use constant acc formulae, a complete method needed. (14sin0) = (15.9sin ) gs or s = 15.9sin 1.4 ½g 1.4 ft ft their from (i). no value used then A0. s =.9 m [3] 7 (iii).9 = vsin0.t 9.8t / ft ft their.9.9tan0 = vcos0.t ft ft their.9 Eliminate t to obtain equation in v only Eliminate v to obtain equation in t only and solve for t Solve for v Substitute t to find v v = 1.37 OR (iii) M Using equation of trajectory method..9 = (.9tan0) tan0 g(.9tan0) /v cos 0 ft Solve for v v = 1.37 OR (iii).9/cos 0 = ½gcos0 t ft 0 = vt ½gsin0 t Eliminate t t = Solve for v v = (iv) e = ft ft their v from (iii), must be v/14. [1] (Q7, June 013)

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PhysicsAndMathsTutor.com 4729 Mark Scheme June 2005 1 (i) use of h/4 com vert above lowest pt of contact can be implied r = 5 x tan24 r = 2.2 4 2.226 (ii) No & valid reason (eg 24 26.6 ) 1 Yes if their