Nội dung text Physics ÔÇó Final Step-C Solutions.pdf
Vidyamandir Classes APP | Kinematics [Motion in 1-D & 2-D] 1 Solution | Physics Solutions to Advanced Problem Package | Physics KINEMATICS [MOTION IN 1 & 2 DIMENSIONS] 1.(D) dv v a bx dx 2 2 v 2ax bx Now v = 0 at x = 0 and x = l (l : the distance between the stations) l = 2a b also max a v b 2.(A) The particle is moving with constant acceleration therefore velocity time graph of the particle will be straight line. From t = 0 s to t = 1 s slope of given displacement-time graph is negative and decreasing. From t = 1 s to t = 2 s slope is positive and is decreasing. At time t = 0 and t = 2 s slope of displacement time graph is zero therefore velocity at that moment will also be zero. 3.(C) The graph, given in the question shows that a s dv Ks dt (where K is some constant) dv v Ks ds v s 0 0 vdv K sds 2 2 v Ks v Ks v K's K K ' v s 4.(C) 2u T a 2 (usin ) T g cos In both the cases , u and a is same, so the time 1 2 2u sin T T T g cos (ii) v = u + at
Vidyamandir Classes APP | Kinematics [Motion in 1-D & 2-D] 2 Solution | Physics x y u u cosα g sin θ t u u sin α g cos θ t For maximum height to the incline Uy = 0 usinα g cosθt u sinα t gcosθ ; 1 2 h ut at 2 h = 2 usinα 1 u sin α usinα gcosθ gcosθ 2 gcosθ 2 2 u sin α h 2g cosθ As y u is same for both the cases and ay is same so 2 2 1 2 u sin α h h 2gcosθ (iii) 1 2 ( cos ) ( sin ) 2 R u T g T ; As T is same for both 2 u u g t cos sin ; 1 u u g t cos sin 2 1 usinα u ucosα g sin θ gcosθ usin α u ucosα g sin θ gcosθ ; Clearly 1 2 u u 5.(B) 6.(A) The relative velocity V makes an angle with AB, where u cos V The distance travelled during the period A arrives at nearest distance = d cos Required time = 2 d cos du V V (A) 7.(B) P1: A(5, 3) P2 : (B) (7, 3) 1 2 ˆ ˆ ˆ ˆ v 2i 3j v xi yj At t = 2 seconds they collide. It means that their S is same. 5 2 2 7 x 2, 3 y 2 3 3 2 x = 1, y = 3 8.(B) Displacements of B and C in horizontal direction is same. VC = VB cos60o C B v 1 v 2 ......(1) Displacement of A and B in vertical direction is same to A B v t v sin 60o t A B v 3 v 2 ....(2) From (2) and (1)
Vidyamandir Classes APP | Kinematics [Motion in 1-D & 2-D] 3 Solution | Physics v : v : v 3 : 2:1 A B C 9.(B) dv a v vdv adx dx 2 2 f i v v v Area under a –x curve. 10.(B) Let speed of man be ‘v’ m/sec. Then, v(time of flight of projectile) = 9 m 2usin θ v 9 g ; v = 5 m/sec 11.(C) (1) As the bird starts it’s motion at the same time as the ball, therefore, Vx = u. (2) For all values of ‘x’ except h = ‘Hmax”, the ball will touch the bird twice. h = Hmax = 2 y u 2g Range = v v x y 2h 2 2u g g 12.(A) Time period = time of projectile = 2 1 10 1 2 2 10 u sin / sec g 13.(A) Average velocity = displacement , cos x AB x AB AB S u u t t 2 sin roots are and 2 2 OA OB h g u t t t t 2 2 0 AB gt u sin . t h t diff. in roots 2 2 2 2 2 2 sin 2 sin 2 sin 2 AB u u u gh t g g 2 2 u sin T / g . As acc. is uniform = g av. Acc. in any interval of time is also g . Direction of ins. Velocities at A & B are different. 14.(C) 2 3 5 2 x u a S t t ; 2 4 5 5 y u S t t 3 5 u t 2 2 a t 4 5 u t 2 5t 7 5 2 5 at u t 14 10 5 u at t 3 5 x u V at ; 4 10 5 y u V t ; 3 4 10 5 5 u u at t 10 5 add. 14 10 5 u t at u t at ; 20 3 t u 10t a t 20 t 3 5 4 26 10 / 3 3 a m s 2 26 3 a m / s 15.(D) Let V : velocity of buggy Velocity highest point of rear wheel = 2V 2 wr. t buggey 2 2 2 y g S b a t y x
Vidyamandir Classes APP | Kinematics [Motion in 1-D & 2-D] 4 Solution | Physics 2 2 S C b a Vt x 2 4 g C b a C b a b a V 4 g C b a C b a V b a 16.(ABCD) Change in velocity = final velocity – initial velocity ˆ ˆ ˆ ˆ u cosθi u cosθi u sin θj u sin θj (A) is correct Average velocity = (total displacement)/(time taken) = Ri / Time of flight ˆ = ˆ u cos i (B) is correct. Change in velocity = final velocity – initial velocity = ˆ ˆ ˆ ˆ ucos i usin j u cos i usin j = ˆ 2u sin j (C) is also correct. Rate of change of momentum = force Constant gravitational force is acting on the projectile. (D) is also correct. 17.(AC) At t = 2 sec, projectile reverses its motion. Maximum displacement in initial velocity = 1 2 10 2 5 2 2 = 10 m. Distance travelled = Displacement from t = 0 to t = 2 added to magnitude of displacement from t = 2 to t = 3 = 12.5m 18.(BC) a = dv V ds Or, vdv ads area under a curve Or, 2 2 f i v v 1 10 2 10 4 For S 10m 2 2 f v 10m / s i v 0 B is correct. Max. velocity is attained at S = 30. 2 2 V V max i 1 30 6 2 2 max v 13.4 19.(BD) For A : A,y river width V time = 10 1 m / s 120 12 And, A,x, river A, river 1 V 0 V m / s 12 V V V A,x,earth r A,x,river = Vr A,x,earth r 30 30 120 V V r 1 V m / s 4 B is correct. For B : B,y,earth 10 1 V m / s 120 12 B,x,earth B,x,river r 25 5 V V V m / s 120 24