Title 1-kinematics-motion-in-1--2-dimensions-ba0b30f8-b2fb-449d-864b-55867afc49e4.pdf ✅

KINEMATICS [MOTION IN 1 & 2 DIMENSIONS] 1. (D) v dv dx = a − bx ⇒ v 2 = 2ax − bx 2 Now v = 0 at x = 0 and x = l(l: the distance between the stations ⇒ l = 2a b also vmax = a √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 dt = Ks (where K is some constant) ⇒ v dv ds = Ks ⇒ ∫0 v vdv = K∫0 s sds ⇒ v 2 = Ks2 ⇒ v = √Ks ⇒ v = K ′ s (√K = K ′ ) ⇒ v ∝ s 4. (C) T = 2u⊥ a⊥ T = 2 × (usin α) gcos θ In both the cases, u⊥ and a⊥ is same, so the time T1 = T2 = 2usin α gcos θ = T (ii) v = u + at ux = ucos α + (−gsin θ)t uy = usin α + (−gcos θ)t For maximum height ⊥ to the incline Uy = 0 ∴ usin α = gcos θt t = usin α gcos θ ; h = ut − 1 2 at2 h = usin α × ( usin α gcos θ ) − 1 2 × (gcos θ) ( usin α gcos θ ) 2 h = u 2 sin2 α 2 gcos θ As uy is same for both the cases and ay is same so h1 = h2 = u 2 sin2 α 2gcos θ R = (ucos α)T + 1 2 (−gsin θ)T 2 ; As T is same fo r bothu2 = ucos α + gsin θt ; u1 = ucos α − gsin θt u2 = ucos α + gsin θ × usin α gcos θ u1 = ucos α − gsin θ × usin α gcos θ Clearly u1 ≠ u2 As T is same for both
5. (B) 6. (A) The relative velocity V makes an angle θ with AB, where cos θ = u V The distance travelled during the period A arrives at nearest distance = dcos θ ∴ Required time = dcos θ V = du V2 7. (B) P1: A(5,3) P2: (B)(7,3) v1 = 2iˆ + 3ˆj v2 = xiˆ + yjˆ s1 = 5iˆ + 3ˆj + (2iˆ + 3ˆj)t ∣ s2 = 7iˆ + 3ˆj + (xiˆ + yjˆ)t ⟹ s2 = (5 + 2t)iˆ + (3 + 3t)ˆj ⟹ s2 = (7 + xt)iˆ + (3 + yt)ˆj 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(cos 60∘ ) vC vB = 1 2 Displacement of A and B in vertical direction is same to vA × t = vB × sin 60∘ × t vA vB = √3 2 ... . (2) From (2) and (1) ∴ vA: vB: vC = √3: 2: 1 9. (B) a = v dv dx ⇒ ∫ vdv = ∫ adx ⇒ [ v 2 2 ] vi vf = Area under a − x curve. 10. (B) Let speed of man be ' v ' m/sec. Then, v × ( time of flight of projectile ) = 9 m v × 2usin θ g = 9; ∴ 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 = uy 2 2 g Range = 2 (vx )(vy) g = 2u√ 2 h g 12. (A) Time period = time of projectile = 2usin θ g = 1(10)1/√2 10 = √2sec 13. (A) Average velocity = displacement tAB = Sx,AB tAB = ux = ucos θ h 2 = usin θt − g 2 t 2 ⇒ roots are tOAand tOB ⇒ gt 2 − 2usin θ ⋅ t + h = 0 ⇒ tAB = diff. in roots ⇒ tAB = 2√u 2sin2 θ − gh g = 2√u 2sin2 θ − u 2sin2 θ 2 g = √2usin θ g = T/√2 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) Sx = − 3u 5 t + a 2 t 2 ; Sy = 4u 5 t − 5t 2 ⇒ −3u 5 t + a 2 t ⧸2 = 4u 5 t − 5t 2 ⇒ at 2 + 5t = 7u 5 ⇒ at + 10t = 14u 5 Vx = −3u 5 + at ; − Vy = 4u 5 − 10t; −3u 5 + at = −4u 5 + 10t 10t − at = u 5 10t + at = 14u 5 ] add. ; 20t = 3u ⇒ 10t = at ′ = 20t 3 × ⧸ ⇒ a = 10 − 4 3 = 26 3 m/s ⇒ a = 26 3 m/s 2X
15. (D) Let V : velocity of buggy ⇒ Velocity highest point of rear wheel = 2 V ⇒ wr. t buggey Sy = −(2b − 2a) = −g 2 t 2 Sx = √C 2 − (b − a) 2 = Vt ⇒ 4(b − a) = g[C − (b − a)][C + (b − a)] V2 ⇒ V = √g√(C − b + a)(C + b − a) √4(b − a) 16. (ABCD) Change in velocity = final velocity - initial velocity = ucos θiˆ − (ucos θiˆ + usin θˆj) = −usin θˆj ∴ (A) is correct Average velocity = ( total displacement )/( time taken ) = (Riˆ/ Time of flight ) = ucos θiˆ is correct. Change in velocity = final velocity - initial velocity = (ucos θiˆ − usin θˆj) − (ucos θiˆ + usin θˆj) = −2usin θˆj (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 = 2sec, projectile reverses its motion. Maximum displacement in initial velocity = (10 × 2 + ( 1 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 = 12.5 m 18. (BC) a = V dv ds Or, ∫ vdv = ∫ ads = area under a − curve Or, vf 2−vi 2 2 = ( 1 2 × 10 × 2 + 10 × 4) [ For S = 10 m] ⇒ vf = 10 m/s [∵ vi = 0] ∴ B is correct. Max. velocity is attained at S = 30. ∴ Vmax 2 − Vi 2 2 = 1 2 × 30 × 6 ⇒ vmax = 13.4 19. (BD) For A: VA,y = river width time = 10 120 = 1 12 m/s And, VA,x, river = 0 ∴ VA, river = 1 12 m/s VA,x, earth = Vr + VA,x, river = Vr ∴ 120 = 30 VA,x, earth = 30 Vr ⇒ Vr = 1 4 m/s ∴ B is correct. For B : VB,y, earth = 10 120 = 1 12 m/s VB,x, earth = (VB,x, river + Vr) = 25 120 = 5 24 m/s 20. (BD) For first one Minute : h1 = 0 + 1 2 × 10 × (60) 2 = 18,000 m = 18 km V1 = 0 + 10 × 60 = 600 m/s After first one minute : Rocket motion is under gravity. V 2 = V1 2 − 2gh2 ⇒ h2 = v1 2 2 g = (600) 2 2 × 10 = 18000 m( V = 0) = 18 km ∴ Max. ht. reached, Hmax = h1 + h2 = 18 + 18 = 36 km Using h⃗ = u⃗ × t + 1 2 ⃗g t 2 ⇒ 18000 = −600 × t2 + 1 2 (+10) × t2 2 ⇒ t2 2 − 120t2 − 3600 = 0 ⇒ t2 = (60 + 60√2). ∴ Total time required, T = t1 + t2 = 60 + (60 + 60√2) = (120 + 60√2)r 21. (ACD) v A,Board = v A, earth − v⃗ Board, earth = 2v − v = v v⃗ B, Board=−2v−v=−3v ∴ T = L v⃗ A,B = L v − (−3v) = L 4v dB,Board = VB,Board × T = 3v × L 4v = 3 L 4 and dA, Board = VA, Board × T = v × L 4v = L 4
22. (ABC) Δv⃗ = v⃗ f − v⃗ i = (60cos 60∘ iˆ + 60sin 60∘ ˆj) − (60iˆ) = (−30ˆi + 30√3ˆj) ∴ |Δv | = √(30) 2 + (30√3) 2 = 60 m/s a = a r + a⃗ t a = √ar 2 + at 2 = ar [at = dv dt = 0] = v 2 R = (60) 2 0.3 × 1000 = 12 m/s 2 |a arg| = |v⃗ f − v⃗ i | Δt = 60 ( π 3 × 300) /60 = 11.5 m/s 2 23. (ABCD) v⃗ A = 4iˆ + 4kˆ v⃗ B = 3ˆj + 4kˆ ∴ v⃗ A,B = (4iˆ + 4kˆ ) − (3ˆj + 4kˆ ) = 4iˆ − 3ˆj ∴ |v⃗ A,B| = 5 m/s (A is correct) As the initial velocity and acceleration of both the particles in vertical direction are equal, so they would hit the ground at same time, B is correct. Time of projectile = 2×4 10 = 4 5 sec Distance covered by A = 4 × 4 5 = 16 5 m Distance covered by B = 3 × 4 5 m = 12 5 m. Separation = √( 16 5 ) 2 + ( 12 5 ) 2 = √ 162+122 5 = 20 5 = 4 m C is correct. r A,B = r A − r B = v A × t − v B × t = v A,B × t = (4iˆ − 3jˆ) × t 24. (ACD) HA = 2 3 RB ⇒ u 2 sin2 α 2g = 2 3 u 2 sin2 p g ⇒ 3sin2 α = usin β ⇒ 3(1 − cos2 α) = 8sin2 β sin2 α = 1 ⇒ βmax = 1 2 sin−1 3 4 RAmax = u 2 sin 90∘ g = u 2 g 25. (ABCD) T = Rθ 2v0 + 2Rcos (θ/2) v0 DT dθ = 0 ⇒ R 2v0 − Rsin (θ/2) v0 = 0 ⇒ sin θ 2 = 1 2 ⇒ θ = 60∘ ⇒ T = R v0 [ π 6 + √3] Distance = R × π 3 + 2R × √3 2 26. (ABC) x 2 a 2 + y 2 b 2 = 1; So the path is an ellipse Vx = −apsin pt, Vy = bpcos pt ax = −ap 2 cos pt, ay = −bp 2 sin pt So, V⃗ ⋅ a = 0 as V⃗ ⊥ a So, = a 2p 3 sin pt ⋅ cos pt − b 2p 3 sin pt ⋅ cos pt = a 2p 3 sin pt ⋅ cos pt − b 2p 3 sin pt ⋅ cos pt As, a ≠ b So, sin pt. cos pt = 0 ⇒ sin p2t = 0 ⇒ p2t = π, 2π ⇒ t = π 2p − − − − The motion is similar to motion of each around sun. So force always towards focus and hence acceleration. At t = 0 particle is at (a, o) At t = π 2p particle is at (o, b) So distance travelled along X-axis is a not the actual distance, which is the length of the part of the ellipse between (a, o) to (o, b) you can try out for distance by following method ds = √dx 2 + dy 2 ⇒ ds dx = √1 + ( dy dx) 2 ⇒ ∫0 s ds = ∫a 0√1 + ( dy dx) 2 ⋅ dx 27. (ABD) If u is the initial speed and θ the angle of projection. Then vy = usin θ − gt i.e., vy − t graph is a straight line with negative slope and positive intercept. x = (vcos θ)t i.e., x − t graph is a straight line passing through the origin. y = (usin θ)t − 1 2 gt 2 i.e., y − t graph is parabola i.e., vx − t graph is a straight line parallel to t-axis. 28. (ABC) 1 2 = u1t + 1 2 a1t 2 and − 1 2 = −u1t + 1 2 (−a2 )t 2