Title 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    