Title 2-kinematics-of-a-particle-.pdf ✅

Kinematics of a Particle 1. (AC) Just after release particle accelerates till equilibrium position and beyond that its speed decreases until it reaches other extreme 2. (A) Let total time taken : t ⇒ tAB = 5/2 3 = 5 6 sec,tBC = tCD = t − tAB 2 = t − 5/6 2 BD = BC + CD ⇒ s/2 = ( t − 5/6 2 ) 4.5 + ( t − 5/6 2 ) 7.5 ⇒ s/t = 4 m/s 3. (ABCD) Ball moving in direction of train Vb = 10 + 1 = 11 m/s Ball moving opposite to direction of train Vb = 10 − 1 = 9 m/s Time duration between collisions = 10 1 = 10 s 4. (A) Slope is continuously decreasing 5. (C) ut − 1 2 gt 2 = u(t − T) − 1 2 g(t − T) 2 = H 6. (B) ucos θ = v ⇒ θ = cos−1 v u 7. (A) Let C : highest point, A to B, (v/2) 2 = v 2 + 2(−10)(+3) B to C: 0 = ( v 2 ) 2 + 2 − (−10)s ⇒ s = 1m 8. (C) VS ̅ = VSR ̅̅̅̅ + VR ̅̅̅ VR > VSR Swimmer can not cross the river along shortest path i.e., along the width of the river. Possible shortest path length will be greater than width of the river. To travel along this path Vs must be ⊥ r to VSR ∴ VSR = 4 km/hr S = 120 × 5 3 θ = 37∘ = 200 m 9. (C) Time taken to reach the maximum height : t1 = u g If t2 is the time taken to hit ground i.e., − H = ut2 − 1 2 gt 2
But t2 = nt1 So, −H = u nu g − 1 2 g n 2u 2 g2 −H = nu 2 g − 1 2 n 2u 2 g 2gH = nu 2 (n − 2) [Given] 10. (BC) Component of acceleration parallel to velocity is rate of charge of speed. (Perpendicular component of acceleration is responsible for changing the direction. 11. (B)−da dt = ka ⇒ −∫a0 a da a = k∫0 t dt ⇒ a = a0e −kt a = a0/2att = t0 ⇒ ln2 = kt0 ⇒ k = ln2/t0 ⇒ a = a0e −(ln 2)t/t0 ie a = a = a02 −t/t0 ⇒ ∫0 v dv = a0∫0 ∞ e (ln2)t/t0dt ⇒ v = a0 | e −(ln2)t/t0 −(lln 2)/t0 | 0 ∞ = a0t0 ln2 12. (ABC) dv ̅ dt = a ̅ ⇒ | dv dt| = |a ̅|, |v ̅| = speed ⇒ d|v ̅| dt = rate of change of speed i.e. magnitude of acc. If a ̅ ⊥ v ̅ then |a ̅| ≠ 0 but d dt |v ̅| = 0 13. (AB)v dv dx = 4 − 2x ⇒ ∫0 v vdv = ∫0 x (4 − 2x)dx ⇒ v 2 2 = 4x − x 2 ⇒ v = ±√8x − 2x 2 = 0 ⇒ x = 4m, v is max = dv dx = 0 ⇒ x = 2m ⇒ max velocity = √16 − 8 = 2√2 m/s 14. (C) Let v: maximum speed ⇒ l = 1 2 vt1, 2l = v(t2 − t1 ), 3l = 1 2 u(t3 − t2 ) ⇒ t1 = 2l/v,t2 = 4l v ,t3 = 10l/v ⇒ av. velocity = l + 2l + 3l 10l/v = 3v 5 = av. velocity max. velocity = 3 5 15. (C) a = v dv dx = { ( 8 10 x) 8 10 ... ... ... ... ... ... ... ... 16 (1 − x 20) ( −8 10) ... ... ... .10 ≤ x < 20 i.e. { 16x 25 ... ... . . x ≤ 10 16 25 (x − 20) ... ... .10 < x ≤ 20 16. (D) Let they cross each other after time t at (x, 30) ⇒ For A: x = 3t & For
B: { x = 1 2 asin θt 2 ⇒ x = 0.2sin θt 2 30 = 1 2 acos θt 2 ⇒ 30 = 0.2cos θt 2 ⇒ t = 300sec = 5 min. 17. (C) a = ks ⇒ v dV ds = ks ⇒ ∫0 v Vdv = k∫0 s sds ⇒ v 2 = ks 2 i.e. v = √ks 18. (A) a 10 + t 11 = 1 ⇒ v is max ⇒ dv/dt = 0 i.e. a = 0 ⇒ t = 11sec. dv dt = 10(1 − t/11) ⇒ vm = ∫0 1110(1 − t/11)dt = |10t − 5t 2 11 | 0 11 = 110 − 55 = 55 m/s 19. (B) x = 9t − t 2 ... .... ( Let east is positive) and initial position is x = 0. v = 0 ⇒ t = 4.5sec ⇒ distance traveled is 5sec. = |x(4.5)− x(4)| + |x(5)−x(4.5) | ... ... where x(t) distances position at time t. = |20.25 − 20| + |20 − 20.25| = 0.25 + 0.25 = 0.5 20. (C) For upward motion, net down ward acc = W+F m ; m is mass. Let h : height gain ⇒ 0 = V 2 − 2 ( W + F m ) h ⇒ h = mv 2 2(W + F) For downward motion, net downward acc = W−F m . Let V ′ ; final velocity and displacement = −h ⇒ V ′2 = 0 2 − 2 {− W − F m } { −mv 2 2(W + F) } ⇒ V ′ = V√ W − F W + F 21. (C) As the force is exponentially decreasing, so is acceleration, i.e., rate of increase of velocity will decrease with time. Thus, the graph of velocity will be an increasing curl with decreasing slope with time. ∴ a = F M = F0 m e −bt = dv dt ⇒ ∫0 v dv = ∫0 t F0 m e −btdt ⇒ v = F0 m [( 1 −b ) e −bt] 0 t = F0 mb [e −bt]t 0 = F0 mb (e 0 − e −bt) = F0 mb (1 − e −bt) ⇒ vmax = F0 mb 22. (B) − h1 = − 1 2 gt1 2 , −h2 = − 1 2 gt2 2 t1 = 5sec and t2 = 3sec ⇒ h1 − h2 = g/2(5 2 − 3 2 ) = 80m 23. (AC) 2x dx dt = 2t ⇒ xv = t differential wrt t ⇒ v 2 + xa = 1 ⇒ a = 1−v 2 x ⇒ a = 1−(t/x) 2 x = x 2−t 2 x 3 24. (D) According to question, velocity of unit mass varies as v(x) = βx −2n dv dx = −2nβx −2n−1 Acceleration of the particle is given by a = dv dt = dv dx × dx dt = dv dx × v Using equation (i) and (ii), we get : a = (−2nβx 2n−1 ) × (βx 2n) = −nβ 2x −4n−1 25. (D) Distance covered by the stone in first 5 seconds (i.e. t = 5s ) is h1 = 1 2 g(5) 2 = 25 2 g Distance travelled by the stone in next 5 second (i.e. t = 10 s ) is h1 + h2 = 1 2 g(10) 2 = 100 2 g Distance travelled by the stone in next 5 seconds (i.e. t = 15 s ) is : h1 + h2 + h3 = 1 2 g(15) 2 = 225 2 g Subtract (i) from (ii), we get :
(h1 + h2 ) − h1 = 100 2 g − 25 2 g = 75 2 g h2 = 75 2 g = 3h1 Subtract (ii) from (iii), we get : (h1 + h2 + h3 ) − (h2 + h1 ) = 225 2 g − 100 2 g h3 = 125 2 g = 5h1 F rom (i), (iv) and (v), we get : h1 = h2 3 = h3 5 26. (D) Given : x = 8 + 12t − t 3 Velocity, v = dx dt = 12 − 3t 2 When v = 0,12 − 3t 2 = 0 or t = 2s a = dv dt = −6t ⬚| t=2s = −12 ms−2 Retardation = 12 ms−2 Retardation = 12 ms−2 27. (A) Distance, x = (t + 5) −1 Velocity, v = dx dt = d dt (t + 5) −1 = −(t + 5) 2 Acceleration, a = dv dt = d dt [−(t + 5) −2 ] = 2(t + 5) −3 From equation (ii), we get : v 3/2 = −(t + 5) −3 Substituting this in equation (iii) we get : Acceleration, a = −2v 3/2 Or a ∝ (velocity) 3/2 From equation (i) we get : x 3 = (t + 5) −3 Substituting this in equation (iii), we get : Acceleration, a = 2x 3 Hence option (A) is correct 28. (A) Let the two balls meet after ts at distance x from the platform. For the first ball u = 0,t = 18 s, g = 10 m/s 2 Using h = ut + 1 2 gt 2 ∴ x = 1 2 × 10 × 182 For the second ball u = v,t = 12 s, g = 10 m/s 2 Using h = ut + 1 2 gt 2 ∴ x = v × 12 + 1 2 × 10 × 127 From equations (i) and (ii), we get : 1 2 × 10 × 182 = 12v + 1 2 × 10 × (12) 2 or12v = 1 2 × 10 × [(18 + 12)(18 − 12)] 12v = 1 2 × 10 × 30 × 6 or v = 1 × 10 × 30 × 6 2 × 12 = 75 m/s 29. (B) Given u = 0. Distance travellend in 10s, S1 = 1 2 a ⋅ 102 = 50a and distance travelled in 20 s, S2 = 1 2 a ⋅ 202 = 200a ∴ S2 = 4S1 30. (D) Let vx be the velocity of the scooter, the distance between the scooter and the bus = 1000 m, The velocity of the bus = 10 ms−1 Time taken to overtake = 100 s Relative velocity of the scooter with respect to the bus = (vs − 10) ∴ 1000 (vs − 10) = 100s ⇒ vs = 20 ms−1 31. (C) Given : At time t = 0, velocity, v = 0. Acceleration f = f0 (1 − t T ) At f = 0,0 = f0 (1 − t T ) Since f0 is a constant, ∴ ∫0 v2 dv = ∫t=0 t=T fdt = ∫0 T f0 (1 − t T ) dt ∴ vx = [f0t − f0t 2 2T ] n T = f0T − f0T 2 2T = 1 2 f0T 32. (D) Average speed = totaldistance travelled total time taken = s+s t1+t2 = 2s s vn + s vd = 2vuvd vd+vu 33. (A) Given : x = 9t 2 − t 3 Speed v = dx dt = d dt (9t 2 − t 3 ) = 18t − 3t 2 For maximum speed, dx dt = 0 ⇒ 18 − 6t = 0 ∴ t = 3 s ∴ xmax = 81m − 27m = 54m (From x = 9t 2 − t 2 ) 34. (C) Distance travelled in one rotation (lap) = 2πr ∴ Average speed = distance time = 2πr t = 2×3.14×100 62.8 = 10 ms−1 Net displacement in one lap = 0. Average velocity = net displacement time = 0 t = 0 35. (D) x = ae −αt + be βt dx dt = −aαe −αt + bβe βt v = −aαe −αt + bβe βt For certain value of t velocity will increases. 36. (B) Let total height = H Time of ascent = T