Tiêu đề 3-motion-in-two-dimensions-.pdf ✅

Motion in Two Dimensions 1. (A) Let V : final velocity, a; tangential acceleration ⇒ V 2 = 2at(πR) ⇒ ar = V 2 R = 2atπ Angle between velocity vector acceleration vector = Angle between tangential acceleration and total acceleration u = tan−1 ar at = tan−1 (2π) 2. (C) Particle may have its velocity continuously changing a⃗ = dv⃗⃗ dt 3. (AB) H1 > H2 V1 2 sin2 θ1 2g > V0 2 sin2 θ2 2g ⇒ θ1 > θ2 ⇒ T1 > T2 4. (BCD) r⃗ = R(cos ωtiˆ + sin ωtjˆ) → position vector V⃗⃗ = dr⃗ dt = −Rω sin ωtiˆ + Rω cos ωtjˆ → Velocity vector a⃗ = dv⃗ dt = d 2 r⃗ dt 2 − Rω 2 cos ωtiˆ − Rω 2 sin ωtjˆ → acc. vector = −Rω 2 (cos ωtiˆ + sin ωtjˆ) a⃗ = −ω 2 r⃗ → Variable as r⃗ varies with time. Magnitude |a⃗| = ω 2 |r⃗| = ω 2R |v⃗| = ωR = Speed const d⃗a dt⃗ = −ω 2 dr⃗ dt = −ω 2v⃗ | da⃗ dt| = ω 2 |v⃗| = ω 2ωR = ω 3R( ∴ Speed = Const = ωR d|v⃗| dt = Rate of change of speed = 0 5. (A) Final velocity vector makes angle 90 − 30 = 60∘ with horizontal, when it becomes (perpendicular to initial velocity vector) Let it be V ⇒ 20cos 30∘ = Vcos 60∘ = Vcos 60∘ ⇒ V20√3 m/s ⇒ Vy = −20√3 sin 60∘ = − 30 m s , uy = 20 sin 30∘ = 10 m s (v = u + at) ⇒ − 30 = 10 − 10t ⇒ t = 4sec horizontal displacement = uxt = 20cos 30∘ × 4 = 40√3 m 6. (A) Let initial point A is on −y-axis and at t = T/4, it is at +x-axis. ⇒ At t = T/4, a ̅ = −V 2 R iˆ ⇒ At t = 0, V ̅ = Viˆ; at t = T/4, V ̅ = Vjˆ ⇒ ΔV ̅ = Vjˆ − Viˆ Average velocity from t = 0 to t = T/4 is equal to Riˆ−(−Rjˆ) T/4 = 4R T (iˆ + jˆ) ⇒ θ1 = angle between vectors −iˆ and iˆ + jˆ = 135∘ and θ2 = angle between vector −iˆ and jˆ − iˆ = 45∘ 7. (C) tan θ = dy dx = 2x a {= 2 at (a, a)} where θ is angle made by tangent at (a, a) with x axis i.e. angle made by normal with y-axis ⇒ Ncos θ = mg and Nsin θ = mω 2a ⇒ tan θ = ω 2a g ⇒ ω = √ gtan θ a = √ 2g a
8. (C) Energy is conserved but K.E of the particles is not zero at P(vx ≠ 0) 9. (C) As horizontal component of vertical remains same ⇒ VVcos φ = ucos θ ⇒ v = ucos θsec φ 10. (ABC) HP < HA Acceleration is along radius vector 11. (BD) Horizontal component w.r.t. ground is along the tangent and its net initial velocity makes some angle θ(≠ 90) with horizontal ⇒ path is parabolic. 12. (ABD) u 2 sin 2θ1 g = u 2 sin 2θ2 g ⇒ sin 2θ1 = sin (180 − 2θ2 ) ⇒ θ1 + θ2 = 90∘ t1 = 2usin θ1 g and t2 = 2usin θ2 g ⇒ t1 sin θ1 = t2 sin θ2 = 2u g t1/t2 = sin θ1 sin θ2 = sin θ1 cos θ1 = tan θ1 = cot θ2 13. (A) t = 2sec Vsin 45∘ + usin θ − g.t Vcos 45∘ = ucos θ t = 3sec usin θ − g. 3 = 0 Vsin 45∘ = 10 = Ucos θ usin θ = 30 ucos θ = 10 (From 1, 2 and 3) Vsin 45∘ = 10 = Ucos θ usin θ = 30 ucos θ = 10 ⇒ tan θ = 3 and u = 10√10tan θ = 3 and u = 10√10 14. (d) ⇒ dx dt = V1 , dy dt = V2 velocity of mid point = 1 2 dx dt iˆ + 1 2 dy dt jˆ ⇒ magnitude = √V1 2 + V2 2 2 15. (B) ⇒ w.r.t. elevator, ux = 4cos 30∘ = 2√3 m/s uy = 4sin 30∘ = 2 m/s ax = 0 m/s 2 and ay = (−10) − (+2) = −12 m/s 2 Sy = 0 ⇒ uyt + 1 2 ayt 2 = 0 ⇒ t = −2uy ay = 1/3 ≃ 0.33sec 16. (D) Let u : velocity of projection and θ : angle of projection ⇒ R = u 2 sin 2θ g and H = u 2 sin2 θ 2g If horizontal acceleration = g/4 is imparted : Let T : time of Flight Sy = 0 ⇒ 0 = usin θT − 1 2 gT 2 ⇒ T = 2usin θ g New range : Sx = ucos θT + 1 2 g 4 T 2 = 2u 2 sin θcos θ g + 1 2 g 4 ⋅ 4u 2 sin2 θ g2 = R + H From initial point of projection to maximum height gain, Vy = 0
⇒ Vy 2 = uy 2 + 2aysy ⇒ 0 = u 2 sin2 θ − 2gsy ⇒ sy new Height gain = u 2 sin2 θ 2g = H 17. (B) ar ∝ t 2 ⇒ ar = ct 2 [ c is constant] V 2 R = ct 2 az = d|V| dt = √cR = Constant of time 18. (b) 30 = ucos θt 40 = usin θt − 10t 2 2 ⇒ 40 = 30tan θ − 5(30) 2 u 2cos2 θ ⇒ 4 = 3tan θ − 450 u 2 (1 + tan2 θ) ⇒ 450tan2 θ − 3u 2 tan θ +450 + 4u 2 = 0 tan θ is real ⇒ D ≥ 0 ⇒ 9u 4 −1800(450 +4u 2 ) ≥ 0 ⇒ u 4 − 800u 2 − 90,000 ≥ 0 ⇒ u 4 −900u 2 +100u 2 − 90,000 ≥ 0 ⇒ u 2 (u 2 − 900) +100(u 2 − 900) ≥ 0 ⇒ u 2 = −100 or 900 ⇒ u = 30m/s 19. (A) Let angle made by velocity vector with horizontal is θ when speed is √125 m/s ⇒ 10√2cos 45∘ = 5√5cos θ ⇒ cos θ = 2/√5 ⇒ sin θ = 1/√5 ⇒ vertical component of velocity = ±√125sin θ = ±5 m/s ⇒ along vertical ±5 = 10√2sin 45∘ − gt ⇒ t1 = 0.5sec,t2 = 15sec ⇒ time interval = 1sec. 20. (B) Along AB, relative velocity of A w.r.t. B = u√3cos 30∘ + ucos 60∘ = 2u ⇒ time = relative horizontal displacement relative horizontal velocity = x 2u 21. (D) For total time of flight (T), y = 0 ⇒ kt(1 − αt) = 0 ⇒ t = 0 or 1/α At maximum height, dy dt = 0 ⇒ K(1 − 2αt) = 0 ⇒ t = 1 2 α ⇒ maximum height = K 2α (1 − α ⋅ 1 2α ) = K 4α 22. (B) As, initial velocity v = (i + 2j)m/s For projectile : Motion along vertical horizontal direction, x = t Motion along vertical direction, y = ut − 1 2 gt 2 Where, u is a vertical component of velocity y = 2t − 1 2 10t 2 From equation (i) and (ii), we get : y = 2x − 5x 2 23. (D) u 2 sin 2θ g − 6 = √3u 2 2g u 2 sin 2θ g + 9 = u 2 g .... Subtract . 15 = u 2 g (1 − √3 2 ) u 2 = 15g 1 − √3/2 u 2 sin 2θ g = 15 1 − √3/2 − 9 sin 2θ = 4 + 3√3 10 ⇒ θ = 1 2 sin−1 ( 4 + 3√3 10 ) 24. (B) Let ωAO = dθ/dt ⇒ ωAB = d(θ/2) dt = 1 2 dθ dt = 1 2 ωAO ⇒ ωAO ωAB = 2 25. (d) ⇒ a⊥ = 10cos 60∘ = 5 m/s ⇒ R = V 2 a⊥ = 10×10 5 = 20
26. (A) Two vector A⃗ and B⃗⃗ are orthogonal to each other, if their scalar product is zero i.e. A⃗ ⋅ B⃗⃗ = 0 Here, A⃗ = cos ωtiˆ + sin ωtjˆ and B⃗⃗ = cos ωt 2 iˆ + sin ωt 2 jˆ ∴ A⃗ ⋅ B⃗⃗ = (cos ωtiˆ + sin ωtjˆ) ⋅ (cos ωt 2 iˆ + sin ωt 2 jˆ) = cos ωtcos ωt 2 + sin ωtsin ωt 2 (∵ iˆ ⋅ iˆ = jˆ ⋅ jˆ = 1 and iˆ ⋅ jˆ = jˆ ⋅ iˆ = 0) = cos (ωt − ωt 2 ) (∵ cos (A − B) = cos Acos B + sin Asin B) But A⃗ ⋅ B⃗⃗ = 0 (as A⃗ and B⃗⃗ are orthogonal to each other) ∴ cos (ωt − ωt 2 ) = 0 cos (ωt − ωt 2 ) = cos π 2 or ωt − ωt 2 = π 2 ωt 2 = π 2 or t = π ω 27. (A) Here R⃗⃗ = 4sin (2πt)iˆ + 4cos (2πt)jˆ The velocity of the particle is v = dR⃗⃗ dt = d dt [4sin (2πt)iˆ + 4cos (2πt)jˆ = 8πcos (2πt)iˆ − 8πsin (2πt)jˆ] Its magnitude is |v| = √(8πcos (2πt)) 2 + (−8πsin (2πt)) 2 = √64π 2cos2 (2πt) + 64π 2sin2 (2πt) = √64π 2[cos2 (2πt) + sin2 (2πt)] = √64π 2 (assin2 θ + cos2 = 1) = 8πm/s 28. (A) The equation of trajectory is y = xtan θ − gx 2 2u2cos2 θ Where θ is the angle of projection and u is the velocity with which projectile is projected. For equal trajectories and As per question, 9.8 5 2 = g ′ 3 2 g u 2 = constant Where g ′ is acceleration due to gravity on the planet. g ′ = 9.8 × 9 2.5 = 3.5 ms−2 29. (D) At time t = 0, the position vector of the particle is r⃗1 = 2iˆ + 3jˆ At time t = 5s the position vector of the particle is r⃗2 = 13iˆ + 14jˆ Displacement from r⃗1 to r⃗2 is Δr⃗ = r⃗2 − r⃗1 = (13iˆ + 14jˆ) − (2iˆ + 3jˆ) = 11iˆ + 11jˆ ∴ Average velocity, vav = Δr⃗ Δ′ 11iˆ+11jˆ 5−0 = 11 5 (iˆ + jˆ) 30. (A) At point BX component of velocity remains unchanged while Y component reverses it direction. ∴ The velocity of the projectile at point B is 2iˆ − 3jˆ m/s 31. (B) Here, u⃗⃗ = 2iˆ + 3jˆ, a = 0.3iˆ + 0.2jˆ,t = 10 s As v⃗ = u⃗⃗ + a⃗t ∴ v = (2iˆ + 3jˆ) + (0.3iˆ + 0.2jˆ)(10) = 2iˆ + 2jˆ + 2jˆ = 2iˆ + 5jˆ |v⃗| = √(5) 2 + (5) 2 = 5√2 units 32. (D) Here, Radius, R = 5 cm = 5 × 10−2 m Time period, T = 0.2πS Centripetal acceleration ac = ω 2R = ( 2π T ) 2 = R = ( 2π 0.2π ) 2 (5 × 10−2 ) = 5 m/s 2 As particle moves with constant speed, therefore its tangential acceleration is zero. So, at = 0 The acceleration of the particle is a = √ac 2 + at 2 = ac = 5 m/s 2 . It acts towards the centre of the circle. 33. (A) Here, u = 20 m/s, g = 10 m/s 2 For maximum range, angle of projection is θ = 45∘ ∴ Rmax = u 2 sin 90∘ g = u 2 g {∵ R = u 2 sin 2θ g } = (20 m/s) 2 (10 m/s 2) = 40 m 34. (D) Velocity towards east direction v⃗1 = 30iˆ m/s Velocity towards north direction, v⃗2 = 40jˆ m/s Change in velocity Δv⃗ = v⃗2 −