Title 4-dynamics-of-a-particle-.pdf ✅

16. (ABCD) So as F increase ⇒ f increase ⇒ R increase When friction is limiting R = mg√μ 2 + 1 ⇒ R ≤ mg√μ 2 + 1 and tan θ ≤ μmg mg 17. (B) fmax = Tmax 4μmg = mg + mvmax 2 l ; vmax = A√ g l ⇒ μ = 1 4 (1 + A 2 l 2 ) 18. (AC) Just after thread is burnt : am1 = 0; am2 = m1g−m2g m2 = ( m1 m2 − 1) g As tension in BC is zero and tension in after string remaining ' mg '. 19. (b) Fcos θ = μM N + Fcos θ = mg μ = tan θ 20. (b) Ny = masin 53∘ = 40N 21. (D) 22. (C) a = mgsin θ 2m = gsin θ 2 m T T = ma = mgsin θ 2 23. (AB) If they slide together N ≠ 0 mgsin θ + N − f1 = ma mgsin θ − N − f2 = ma (i) If μ1 > μ2 ⇒ f1 > f2 (ii) If μ1 < μ2 ⇒ f1 < f2 So always in contact (Block 1 slower) So different acceleration (Block 1 faster) 24. (B) 2T = 2t ⇒ T = t T − g = 1a1 T − 2g = 2a2 t − g = a1 t − 2g = 2a2 Since a ∝ t ⇒ V ∝ t 2 so parabolic. A will leave the ground first. 25. (D) μmgcos θ = mgsin θ = 2(mgsin θ − μmgcos θ) tan θ = 3μ 26. (A) x + y + √d 2 + x 2 = constant dx dt + dy dt + 2x 2√d 2 + x 2 dx dt = 0 −v + u − vcos θ = 0 ⇒ u v = 1 + cos θ
27. (BD) 28. (BC) A will never move up if mgsin θ2 ≤ mgsin θ1 + μmgcos θ1 29. (C) Tmax = 40 sin 30∘ = 80 N Tmax − mg = mamax ⇒ amax = 6 m/s 2 30. (C) Let v be tangential speed of heavier stone. Then, centripetal force experienced by lighter stone is (FC )lighter = m(nv) 2 r and that of heavier stone is (Fc )heavier = 2mv 2 (r/2) But (FC )lithter = (FC )heavier (given) ∴ m(nv) 2 r = 2mv 2 (r/2) n 2 ( mv 2 r ) = 4 ( mv 2 r ) n 2 = 4 or n = 2 31. (D) Option not matching with answer F − 14 Here, MA = 4 kg, MB = 2 kg, MC = 1 kg, F = 14 N Net mass : M = MA + MB + MC = 4 + 2 + 1 = 7 kg Let a be the acceleration of the system. Using Newton's second law of motion, F = Ma 14 = 7a ∴ a = 2m−2 Let F ′ be the force applied on block A by block B i.e. the contact force between A and B. Free body diagram for block A Again using Newton's second law of motion, F − F ′ = 4a 14 = F ′ = 4 × 2 ⇒ 14 − 8 = F ′ ∴ F ′ = 6N 32. (A) 33. (C) Force of friction of mass m1 = μm2g; Force of friction of mass m3 = μm2g Let a be common acceleration of the system : ∴ a = m1g−μm2g−μm2g m1+m2+m3 Here, m1 = m2 = m3 = 3 = g(1−2μ) 3 Hence, the downward acceleration of mass m1 is g(1−2μ) 3 34. (C) Change in momentum = Area under F − t graph in that interval = Area of △ ABC − Area of rectangle CDEF + Area of rectangle FGHI = 1 2 × 2 × 6 − 3 × 2 + 4 × 3 = 12Ns 35. (A) Let F be the upthrust of the air. As the balloon is descending down with an acceleration a. ∴ mg − F = ma Let mass m0 be removed from the balloon so that it starts moving up with an acceleration a. Then,