Title 2-dynamics-of-a-particle-8318b4e3-70a2-476b-83e8-9a1a4ad145ea.pdf ✅

DYNAMICS OF A PARTICLE 1. (A) Displacement of M with respect to ground = displacement of block with respect to M ∴ (A) is correct. Acceleration of m with respect to ground a⃗G − (a − acos α)iˆ − asin αjˆ |a⃗⃗G | = 2asin α 2 2. (C) For the equilibrium of block μ(150cos 45∘ + 50cos 45∘ ) = 150sin 45∘ − 50sin 45∘ ⇒ μ = 0.5 3. (B) F.B.D. of man and plank are For plank be at rest, applying Newton's second law to plank along the incline Mgsin α = f And applying Newton's second law to man along the incline. Mgsin α + f = ma a = gsin α (1 + M m ) down the incline 4. (C) Till both Blocks remain in equilibrium : F − T − fr1 = 0 T − fr1 − fr2 = 0 5. (B) x 2 + y 2 = l 2 ∴ 2x dx dt + 2y ⋅ dy dt = 0 ⇒ dy dt = −x y , dx dt −vy = −cot 60∘ × √3 ⇒ vy = 1 m/s ∴ v = √vx 2 + vy 2 = 2 m/s 6. (C) F.B.D. of A 2 T − 10 g = 10aA mA = 10 kg ⋅ mB = 5 kg ∴ let acc of A, B and C be aA, aB, aC F.B.D. of B T − 5 g = 5aB From (1) and (2) 10aA = 10aB aA = aB
aA: aB = 1: 1 7. (B) 8. (B) Draw force diagram of M and se that net force on M in both the cases is zero. 9. (B) Let B : foot of perpendicular drawn from A on the ground. C : foot of perpendicular drawn from B to OX AC = 5sin 30∘ = 2.5 m AB = ACsin 30∘ = 1.25 m ⇒ sin θ = AB OA = 1 4 ⇒ acc. of the block = gsin θ = 2.5 m/s 2 (s = ut + 1/2at 2 ) ⇒ 1 2 (2.5)t 2 ⇒ t = 2sec 10. (B) Along horizontal MV = m(Vrcos θ − V) ⇒ Vr = 10 √3 m/s = 5.77 m/s 11. (BD) 12. (ACD) Maximum fr force between m2 and m1 = m2 gμ = 10 × 0.2 ⇒ 2 N F − fr − T = m2a2 If in equilibrium F = fr + T T < 2 N [ If F < 4 N] For m1 in this case :
T < Frmax ∴ fr and T will be equal ∴ It is not necessary if F > 4 T = 4 N. (B) is not correct. If F > 4, max Fr = 2 N ∴ and system will accelerate ∴ system will not be in equilibrium (C) is correct If F = 6 N : Fr will be at max value. Fr = 2 N F − T − Fr = a2 4 − T = a2 Since blocks are connected by string there acceleration will be equal fr ∴ fr m1 ⟶ T T − fr = a T − 2 = a 4 − T = a 2a = 2, a = 1 and, T = 3 N 13. (AC) Due to symmetric Structure wedge will be in equilibrium and therefore acceleration of wedge is = 0 For B : ∴ mg − T = ma (1) For A: m m → T T = ma ... . . (2) From (1) and (2) ; ∴ a = g 2 14. (ABCD) If Fr is not present system will move towards right ∴ fr will act on P towards left Max fr = 40 × 0.6 = 24 N System will be in equilibrium If mqg = fr + mRg 40 = f + 20 f = 20 N Since for is within limiting value ∴ f = 20 N system is in equilibrium Q is in equilibrium R is in equilibrium ∴ TA = 20 N ∴ TB = 40 N R is in equilibrium fr = 20 N Contact force = √402 + 202 = √2000 = 20√5 15. (ABC) System will be in equilibrium until A is in equilibrium. Max Fr force on A = μs × 0.5 g = 2 N F = Kt = t (k = 1) If T = f2 B is in equilibrium f2max0.2 × 0.5 g = 1 N
∴ For T ≤ 3 : system as a whole is in equilibrium ∴ For t upto 3sec. system is in equilibrium and is at rest. ∴ Options A, B and C are correct. For t > 3sec : F on A = 1 N Fr force max while motion of A ⇒ μk × N = 1 T − 1 = t − T − 1 = a 2 T − 1 = a 2 t − 2 = a = dv dt or ∫3 10(t − 2)dt = ∫0 v dv [ t 2 2 − 2t] 3 10 = (V − 0) = 31.5 m/s ∴ D is incorrect 16. (BCD) To maintain constant velocity, Fnet = 0 ⇒ P = fr always 17. (BC) For 10 kg block : kx = 10 × 12 = 120Nt. For 20 kg block : ⇒ 200 − kx = 20 × a a = 200 − 120 20 = 4 m/s 2 kx 18. (AD) Suppose blocks A and B move together. Applying NLM on C, A + B, and D 60 − T = 6a T − 18 − T ′ = 9a T − 10 = 1a Solving a = 2 m/s 2 To check slipping between A and B, we have to find friction force in this case. If it is less than limiting static friction, then there will be no slipping between A and B. Applying NLM on A. T − f = 6(2) AsT = 48 N f = 30 N and fs = 42 N hence A and B move together. And T ′ = 12 N. 19. (ABC) F.B.D of block B w.r.t. wedge For block A Ncos 45∘ = 1.7a for block B 0.6 gsin 45∘ + 0.6acos 45∘ = 0.6 b N + 0.6acos 45∘ = 0.6 gcos 45∘ ... By solving (i), (ii), and (iii) a = 3 g 20 and b = 23 g 20√2 Now vertical component of acceleration of B = bcos 45∘ = 23 g 40 And horizontal component of acceleration of B = bsin 45∘ − a = 17 g 40 20. (BD) by string constrain vA + u − vB = 0 Or vB + u − vA Differentiating both side ; aB = 0 + aA 21. (AC) (1) Balancing forces perpendicular to incline N = mgcos 37∘ + masin 37∘ N1 = 4 5 mg + 3 5 ma