PHYSICS-27-08- 11th (J-Batch)

= u sin 60° – 10 t A the moment of collision t = 2 and vy = 0  0 = u 3 – 10 × 2  u = 2 (b) x = (u cos ) t 40 1 = 3 × 2 × 2 = 40/ m (c) Vertical displacement of ball I. y1 = uyt – 1 gt2 2 = 20×2 – 1 ×10 × (2)2 = 20 m (upward) 2 Vertical displacement of ball II y2 = – 1 gt2 = – 2 1 2 × 10 × (4) = – 20 m ( downward) height of tower h = (y1) + (y2) = 20 + 20 = 40 m Ans. ] Q.15kin A police officer sits on a parked motorcycle. Acar travelling at a constant speed of 0 = 40.0 m/s passes by at t =0. At that same instant (t =0), the officer accelerates the motorcycle at a constant rate, and at time t1 = 20.0s overtakes the speeder. (a) (i) Find the acceleration of the motorcycle. (ii) Find the speed of the motorcycle at the instant it overtakes the car. (b) Sketch of the position of the car and that of the motorcycle as functions of time is shown, t2 indicate the time at which car and motorcycle are travelling at the same speed. Find  and t2. [2+2+2+2] [Sol. (a) (i) v0 1 t1 = 2 a t 2  a = 2v0 t1 2  40 = 20 = 4 m/s2 (ii) vm = a t1 = 4 × 20 = 80 m/s (b) a t2 = v0 v0  t2 =  40 = 4 = 10 c tan  = v0   = tan–1 (40) ] [Sol. Considering horizontal as X and vertical as Y and point of projection as origin ux = 20 uy = 0 ax = 0 ay = – g = – 10 vx = ux = 20 vy = – 10 t at t = 2 At the moment of collision with incline a angle with horizontal  = – vy tan  = x  – tan  = 10 (2) 20  tan  = 1   = 45° Displacement along horizontal x = uxt = 20 × 2 = 40 m Displacement along vertical y = 0 × 2 – 1 ×10(2)2 2 = – 20 m Net displacement (OP) = = = 20 5m Alternative method ux' = u cos  = 20 cos  ax' = – g sin  = – 10 sin  vx' = ux' + ax' t = 20 cos  – 10 sin  t At the moment of collision vx' = 0  20 cos  – 10 sin  (2) = 0  tan  = 1   = 45°] Q.12kin A train is moving forward at a velocity of 2.0 m/s. At the instant the train begins to accelerate at 0.80 m/s2, a passenger drops a coin which takes 0.50 s to fall to the floor. (a) Calculate the height in the train from where it is dropped. (b) Where does it lands relative to a spot on the floor directly under the coin at release. (c) The dropped coin is viewed by an observer standing next to the tracks. By how much horizontal distance relative to this observer, the coin moves before landing. [2+2+2] 5 [Ans. (a) 4 mts. (b) 0.10 m toward the rear of the train (c) forward 1.0m] 1  1 2 1 1 5 [Sol. (a) h = 2 g 2  = 10  = 2 4 4 mts.   (b) An observed is train frame for coin ux = 0 uy = 0 ax = –0.8 ay = –10 5 v = 0 at t = 2 5 1 at t = 2 , x = – 4 distance travelled = 6.5 mts. ] Q.9 Sketch the position x and the acceleration a as a function of time for a ball whose velocity v as a function of time is shown. Use the same scale for the time axis of all graphs. The ball is located at x = 2 m when t = 0 s. [2+4] [Sol. x = 2 + area of curve v–t t 0 1 2 3 4 5 6 x 2 4 6 7 6 4 2 = 14 ˆi + 2ˆj  Fmt ˆ ˆ a = = (14i + 2 j) m/s ] m Q.4 A lift is moving with uniform downward acceleration of 2m/s2. A ball is dropped inside the lift from a height 2 metre from the floor of lift. Find the time after which ball will strike the floor. [3] [Sol. geff = g – a = 8 Mt/s2 SB, L = 2 metre downward  is positive uB, L = 0 2 = 1 8 t2 2  t = sec ] Q.5 A “moving sidewalk” in an airport terminal moves at a speed of 1.0 m/s relative to the ground and is 80m long. A person runs on the side walk from its one end to the other at a speed of 4.0 m/s relative to the moving sidewalk in the same direction as the sidewalk. (a) How much time, t, does it take to exit the sidewalk? (b) Suppose the person performs a round trip on the sidewalk by moving from one end to the other and then returning back to the starting point, such that his speed always remains 4 m/s relative to the sidewalk.What is the total time for such a round-trip? [3] [Ans: (a)16sec (b) 42.67 sec] [Sol: (a) Vp, g = Vp, sidewalk + Vsidewalk , g = 4iˆ +1iˆ 80 = 5iˆ m/s t = 5 s = 16 s (b) While moving in opposite direction Vp, g = 3iˆ m/s – 80 t =  3 = 26.67 Total time (T) = 16 + 26.67 = 42.67 ] Q.6kin A passenger who just missed the train stands on the platform, sadly watching the last two boggies of the train. The second last boggy takes time 3 sec. to pass by the passenger, and the last one takes time 2 sec. to pass by. How late is the passenger for the departure of the train? Assume that the train accelerates at constant rate. [3] [Sol. Let acceleration be a. Man is late by T and length of each boggy is L. Velocity of train when man enters = 0 + aT = aT Considering time taken to cross IInd last boggy L = (aT)3 + 1 a(3)2 (1) 2 Considering time taken to cross last boggy REVIEW TEST-2 Class : XI (J-Batch) Time : 100 min Max. Marks : 75 General Remarks: INSTRUCTIONS 1. The question paper contain 15 questions and 12 pages. All questions are compulsory. Please ensure that the Question Paper you have received contains all the QUESTIONS and Pages. If you found some mistake like missing questions or pages then contact immediately to the Invigilator. 2. Each question should be done only in the space provided for it, otherwise the solution will not be checked. 3. Use of Calculator, Log table and Mobile is not permitted. 4. Legibility and clarity in answering the question will be appreciated. 5. Put a cross ( × ) on the rough work done by you. Name Father's Name Class : Batch : B.C. Roll No. Invigilator's Full Name For Office Use ……………………………. Total Marks Obtained………………… Q.No 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Marks