PHYSICS-09-07- 11th (PQRS) SOLUTION

XI(PQRS) PHYSICS SOLUTIONS REVIEW TEST-2 Note: Take g = 10 m/s2 wherever required. Q.1 Use T if statement is true and F if it is false. [1 + 1] (a) Aball is thrown vertically upward in air. If the air resistance cannot be neglected (assume it to be directly proportional to velocity), then the acceleration of the ball at the highest point is g. (b) Trajectory of a particle thrown upwards as seen by a person moving horizontally on ground will be a straight line. [Sol:(a) (T) At highest point, velocity is zero. So, retardation due to air resistance is zero and only 'g' is acting. (b) (F) Vrelative = Vball Vman Hence, trajectory will be parabolic relative to man. Q.2 A particle is moving along x-axis. Initially it is located 5 m left of origin and it is moving away from the origin and slowing down. In this coordinate system, the signs of the initial position, initial velocity and acceleration, are [2] [Sol:  ˆ x0 = 5i m V0 = V0 iˆ m / s (along –ve x) a0 = a0 iˆ m / s2 (opposite to V i.e. along +ve x) Ans.] Q.3 A stone is thrown vertically upward with an initial velocity v0. Find the distance travelled in time 3v0 2g . [3] [Sol: t = 3V0 2g V0 V0 = g + 2g V0 5V2 [Ans. 0 ] 8g Time to attain maximum height = g V 2  h = 0 2g At B, u = 0, 1  V 2 V h1 = 0 + g  0  2  2g  [Distance travelled in 0  h1 V 2 = 0 8g  Total distance travelled = h + h1 V 2 2 = 0 + 0 2g 8g 5V 2 = 0 Ans. ] 8g Q.4 Two particles A & B are projected from the same point in different directions in such a manner that vertical components of their initial velocities are same. Which one of them will have more time of flight and which one of them will have more range? [Sol: Vertical component equal  VAY = VBY [2 + 2] x [Ans. Both will have same time of flight; B]  T = 2VAY g 2VBY = g so time of flight is same for both Ans. RA = VAx T RB = VBxT VBx > VAx  RB > RA Ans. ] Q.5 A particle moves in xy plane with a velocity given by  = (8t  2)ˆi + 2ˆj . If it passes through the point (14, 4) at t = 2 sec., then give equation of the path [4] [Ans. x = y2 – y + 2] dx [Sol: Vx = dt dy = 8t – 2 Vy = dt = 2 x t y t   dx =  (8t  2)dt   dy =  2dt 14 2 4 2 x t  x =(4t 2  2t) 14 2 y t  y = 2t 4 2  x – 14 = (4t2 – 2t) –12  y – 4 = 2t – 4  x = 4t2 – 2t + 2  y = 2t  t = y/2  y 2  x = 4  2   y  – 2  2  + 2      x = y2 – y + 2 equation of Trajectory Ans. ] Q.6 To a man running upwards on the hill, the rain appears to fall vertically downwards with 4 m/s. The velocity vector of the man w.r.t. earth is (2i + 3j) m/s. If the man starts running down the hill with the same speed, then determine the relative speed of the rain w.r.t. man. [4] [Sol: Case I : Man going up the incline [Ans. m/s ] V = (2iˆ + 3 ˆj) Vrm = 4 ˆj Vrm = Vr Vm  Vr = Vrm + Vm  Vr = 2iˆ  ˆj Case II: Man running down the plane with the same speed Vm' = Vm = (2iˆ + 3 ˆj) Vr ' = Vr = 2iˆ  ˆj Vrm' = Vr ' Vm' = 4iˆ + 2 ˆj  relative speed of rain w.r.t. man  = Vrm' = = m/s Ans. ] Q.7 Two particles A & B start from rest from x = 0 m & x = –1 m respectively at t = 0. If the velocity variation of A with time is VA = 4t3m/s & that of B is VB = 4t m/s. Find when and where A & B meet? [4] [Sol: dxA = 4t3 dt dxB dt = 4t x A t xB t  dxA =  4t3dt  dx =  4t dt B 0 0 1 0 xA = t4 xB = 2t2 – 1 For meeting , xA = xB  t4 = 2t2 – 1  t4 – 2t2 – 1 = 0  t = 1 sec, x = 1mt. ] Q.8 A projectile is thrown from a point A as shown. Find the co-ordinates of the point where it strikes the inclined plane considering point O as origin. [4] [Sol: (10 + h) = 12 t ; h = 9t – Solving both 1 gt2 2 h = 209  3 10 h  1.145  (1.145, 1.145) ] Q.9 A diwali rocket is launched vertically with its fuse ignited at time t = 0, as shown above. The fuels provides constant acceleration for 2 sec. till rocket attains V = 40 ms–1. Afterwards rocket continues to move freely under gravity. [2 + 3] [Sol: (a) Draw labelled a-t graph from launching till it reaches ground. (b) Draw labelled v-t graph Let, a = acceleration  40 = 0 + 2a  a = 20 m/s2 1 h1 = 2 × 20 × 4 = 40 m At C, v = 40m/s, at B, v = 0 40  0 = 40 – gt2  t2 = g = 4s 1 h2 = 40 × 4 – 2 × 10 × 16 = 80 m t = time taken to reach ground again h = 1 gt2  120 = 2 1 × 10 × t2  t = 2 = 2 S we note following velocity at A' upon return, 2 = 0 + 2gh = 2 × 10 × 120  vA' = 20 (downward) we note: ] Q.10 A projectile is shot at an angle  with respect to the ground from the top of an inclined plane (see figure). You know that the projectile lands at the base of the inclined plane. What was the original speed v0 in terms of h, x0 and ? [5] [Sol: y : –h = V sin t – 1 gt2 0 2 x0 x : x0 = V0cos t  t = 0 cos  –h = (tan) x0  V0 = gx2 – 0 0 Ans.] Q.11 Achild in a boat needs to cross the river. The speed of the current in the river is k (k < 1) times the speed of the boat in still water. If a child crosses the river in such a way as to minimize the lateral displacement (drift), it takes time T to cross. Find the minimum time required to cross the river in terms of T & k. [6] [Sol: vR = kv ˆi | vb | = v Suppose width of river = 'd' vb / R = –(vsin) ˆi + (vcos) ˆj vR = kv ˆi  vb / g = vb / R + vR = (kv – vsin) ˆi + (vcos) ˆj If it cross the river to minimize the drift then for zero drift kv – v sin = 0 or sin = k ...(1) In this case time taken T = d v cos  T = v d 1 k2 ...(2) In general time required to cross the river d t = vcos for min t, cos = 1  t = d/v Putting values from equation (2) t = T ] Q.12 A particle moves with uniform speed of  m/s along a path comprising of two semicircles from Ato B and then from B to C as shown in figure. (a) Find magnitude of average velocity for the entire journey (b) Find magnitude of displacement of the particle from A, 3 sec 2 after starting [3 + 3] [Sol: t1 = time taken to move from A to B t2 = time taken to move from B to C  t1 =  R1  = R1 = 2S t2 =  R2  = R2 = 1.5 S  t = t1 + t2 = 3.5 S AC = = = 5m AC VAV = t 3 (b) = 2  5 = 3.5 3 10 m/s = 7 m/s Ans. R = = 2  2 = 3   = 3 2 4   =  –  =  4  OQ = OPcos = 2cos   OQ = 4  QP = OP sin = 2sin 4  QP =  AQ = AO + OQ = 2 + 2  Magnitude of displacement AP = =  AP = = 2 Ans.] Q.13 The graph shows the motion of a person over 9 seconds. Indicate on the graph one region in which the person had [6] (a) positive velocity (b) zero velocity (c) negative velocity (d) positive acceleration. (e) zero acceleration (f) negative acceleration. (Label the points v > 0, v = 0, v < 0, a > 0, a = 0, and a < 0.) [Sol: (a) V > 0 : A ; (b) V = 0: B, E, G (c) V < 0  C, D, F (d) a > 0 : D (e) a = 0  A, B, E, G; (f) a < 0  C, F ] da Q.14 The time derivative of the acceleration is called "jerk" , i.e. j = dt . Two cars initially at rest start a race. Car A accelerates at constant rate a, while car B moves with constant jerk j and zero initial acceleration. Part way through the race, at t = 1 s, the cars are at same position. Who was ahead at t = 0.5 s? [6] [Sol: car A : xA = 1 at2 2 a t da car B : dt B = j   da 0 vB = j  dt 0 t  aB = jt   dvB 0 = j tdt 0 2 B dx t t2dt  vB =   0 B = j  0 jt 3  xB = 6 1 j At t = 1 S, xA = xB  2 a = 6  j = 3a  xB = 1 at3 2  xB = txA At t = 0.5, xB = 0.5 xA  xA > xB  Car A is ahead at t = 0.5 s Ans.] Q.15 A platform is moving upwards with a constant acceleration of 2m/sec2. At time t = 0, a boy standing on the platform throws a ball upwards with a relative speed of 8m/sec. At this instant platform was at the height of 4m from the ground and was moving with a speed of 2m/sec. Take g = 10m/sec2. Find (a) When and where does the ball strikes the platform? (b) Maximum height attained by the ball from the ground. [4 + 3] 4 [Ans.76/9 m 3 sec, 9 m] [Sol:(a) Acceleration of ball w.r.t. platform, aBP = (10 + 2) = 12m/s2 (downwards) Let, t = Time taken to strike back the platform 2uBP t = aBP 2  8 = 12  t = 4 S 3 1 h = u t + a t2 = 2 × 4 + 1 16 × 2 × = 8 16 + P 2 P  h = 40 m 9 3 2 9 3 9  4 + 40  76  height of platform from ground = 4 + h =   4  m = m 9  76 i.e. the ball hits back the platform after 3 S of throwing at an height of 9 (b) uB = uBP + uP = 10m/s (w.r.t. ground) from ground. 02 = u 2 –2gHm  Hm u2 = B = 2g 100 2 10  maximum height attained by ball from ground = (4 + 5) m = 9m ] Q.16 Ronaldinho who is at rest 25 meters away. The ball is hit at an angle of 370 to the horizontal and with a speed of 20 meters/sec. Immediately sensing that the ball is going to over–shoot him, Ronaldo begins to run, with constant acceleration, horizontally in the same direction the ball is travelling. Amazingly, he traps (stops the ball with foot) the ball. (You may assume that the ball is kicked from and trapped at the same height) [2 + 2 + 3] (a) How far did the receiver have to run? (b) What was the acceleration of Ronaldo? (c) At the time of the trapping, which horizontal component of velocity is larger, the ball's or the receiver's? By how much? [Sol. A  Ronaldinho ; B  Ronaldo ux = 20cos370 = 16 m/s uy = 20sin370 = 12 m/s 2uy T = g = 2 12 10 = 2.4 S Range, AB2 = ux T = 16 × 2.4 = 38.4 m (a) Reciever has to run a distance S = B1B2 = 13.4 m Ans. (b) S = 0 + 1 aT2  a = 2 2S T 2 = 2 13.4 2.4  2.4 = 4.7 m/s2 Ans. (c) VBall =ux = 16m/s V = 0 + aT 2 = 4.7 × 2.4 = 11.3 m/s  VBall > VB by 4.7 m/s ]