PHYSICS-24-09-11th (J-Batch)

XI(J) PHYSICS REVIEW TEST-3 Take g = 10 m/s2 where ever required in this paper. Q.1 The mass in the figure can slide on a frictionless surface. The mass is pulled out by a distance x. The spring constants are k1 and k2 respectively. Find the force pulling back on the mass and force on the wall. [Sol. Springs are in series  keq = k1k2 k1 + k2 [Ans. – k1k2 x k1 + k2 ˆi , [3] k1k2 x k1 + k2 ˆi ]  F = –k x ˆ = – k1k2xˆi block eq i k1 + k2 k1k2xˆi Fwall = k1 + k2 Q.2 The system of two weights with masses m1 and m2 are connected with weightless spring as shown. The system is resting on the support S. The support S is quickly removed. Find the accelerations of each of the weights right after the support S is removed. [4] (m1 + m2 )g [Sol. Initially m1g = kx [Ans. a1 = 0, a2 = ] 2 When support is removed, spring force does not change. New FBD For m1 : m1g – kx = m1a1  a1 = 0 For m : m g + kx = m a  a = (m1 + m2 )g ] 2 2 2 2 2 m2 Q.3 A particle of mass 5 kg is pulled along a rough horizontal surface by a string which is inclined at 60° to the horizontal. If the acceleration of the particle is g/3 m/s2 and the coefficient of friction between the particle and the plane is 2/3, find the tension in the string. [6] [Sol. [Ans. ] T sin 60° + N = mg T cos 60° – N = mg/3  + N = 50 T 2 50 2 – 3 N = 3 3T T 3  4 + 2 = 75  T = ] Q.4 Consider the pulley system in the diagram below. The unknown force F being applied is just sufficient to hold the system in equilibrium. The block has mass M, while the pulleys and ropes have negligibly small masses. Draw the free body diagram of M. What is the tension T in the upper cable (i.e. the cable connecting the top pulley to the ceiling) in terms of M and the acceleration due to gravity g only. [2+4] [Sol. In equilibrium T1 = 2F (1) T2 = 2T1 = 4F (2) T = 2T2 = 8F (3) For M T2 + T1 + T = Mg 4F + 2F + F = Mg 7 F = Mg Mg 4Mg [Ans. T = 7 ] F = 7 8Mg T = 7 putting in (3) ] Q.5 In the situation given, all surfaces are frictionless, pulley is ideal and string is light. If F = Mg/2, find the acceleration of both the blocks in vector form. [3+3] [Ans. a = g ˆj + g ˆi , a = g ˆi ] 1 2 4 2 4 [Sol. Sol. Mg–T = May (1) N = Max (2) N1 = Mg + F (3) F–N = Max (4) Solving (1), (2), (3), (4) g g ax = 4 ; ay = 2 a = g ˆi + g ˆj 1 a2 = 4 2 g ˆi ] 4 Q.6 In figure, aA & vB are unknown but initial velocity of A & constant acceleration of B are known. Find the time in which block A comes down by a distance of 2 m. [6] [Sol. TSB – 2TSA = 0  vB = 2vA aB = 2aA  aA = 2 m/s2 2 = 1×t + 1 × 2 × t2 2 t2 + t –2 =0 t2 + 2t – t – 2 = 0 (t–1) (t+2) = 0  t = 1 , –2  t = 1 sec. ] Q.7 A man is coming down an incline of angle 30°. When he walks with speed 2 3 m/s he has to keep his umbrella vertical to protect himself from rain. The actual speed of rain is 5 m/s. At what angle with vertical should he keep his umbrella when he is at rest so that he does not get drenched? [6] [Sol.  vrm  = xˆj =   ˆ ˆ ˆ ˆ [Ans. 37°] vm = 2 3[cos 30 i + sin 30 j] = 3i + j   ˆ ˆ vr = 3i + (x  3) j 5 =  16 = (x   4 + 3 = x 3)2  ˆ ˆ vr = 3i + 4 j tan  = 3/4   = 37° ] Q.8 Two ports, A and B on a North-South line are separated by a river of width D. The river flows east with speed v. A boat cross the river from port A to port B. The speed of the boat relative to the water is v & that of river is v/2 towards east. State all your answer in terms of v and D. (a) What is the direction of the rowing of the boat so that it crosses directly on a line from A to B? How long does the trip take? (b) Suppose the boat wants to cross the river from A to the other side in the shortest possible time. In what direction should he row the boat? How long does the trip take? [3+3] [Sol.(a) To cross directly [Ans. (a) t = , (b) t = D/v] v sin  = v/2  sin  = 1/2   = 30° vy = v cos  = t = (b) To have shortest time, he should row perpendicular to river vy = max = v t = D/v ] Q.9 A block of mass 1 kg is kept on the tilted floor of a lift moving down with 3 m/s2. If the block is released from rest as shown, what will be the time taken by block to reach the bottom. What is the normal reaction on the block during the motion? [4+4] [Sol. 7 cos37° = N  N = 7 × 4 5 = 5.6 N 1 × arel = 7 sin37°  arel = 7 × 3 5 = 4.2 2.1 = 3 = 1 × 2.1 × t2 2  t = 1 sec. ] Q.10 Two particles A and B of mass 5 kg and 9 kg respectively rest on the smooth faces of a fixed wedge as shown in the figure. They are connected by a light inextensible string passing over a smooth pulley at C and are released from rest from the position shown in the diagram. In the subsequent motion B hits the ground and does not rebound. Find (a) the speed of the particles when B hits the ground. (b) the acceleration of A after B hits the ground. (c) the distance of A from C when A first comes to rest. [4+1+3] [Sol.(a) 9g sin 45 – T = 9a T – 5g sin 45 = 5a [Ans. (a)  35  , (b) – g / , (c) 2/7 m] 2 g = 14 a a = v2 = 2 × g 2 7 × 0.4  v = (b) String becomes slack & acceleration of A = –g sin 45 (c) 02 = v2 – 2 × × s  4 2g = s 35  s = 4/35 m (where s is the distance travelled by A up the incline after B hits the ground) Thus, distance b/w A and C now = 0.6 – 0.4 – 4/35 = 3/35 m ] Q.11 A block of mass 3 kg is released from the top of a smooth inclined plane of inclination 37° and length 2m. At the bottom of the track, the block collides with a stopper so that its velocity is reversed in direction and halved in magnitude. (a) Find the time taken by the particle to slide down before collision and time taken to slide up after the collision. (b) Find the magnitude of the average velocity of the block till it stops for the first time after the first collision. [4+4] 6 [Ans. (a) t1 = sec , t2 = sec, (b) 2 m/s ] [Sol. a1 = 6 m/s2  v2 = 2 × 2 × 6  v = 2 m/s v' = 2 6 = 2 m/s  t1 = v = 2 6 = 1 6 sec  a2 = – 6 m/s2 0 = – 6 t2  t2 = sec. 1 1 1 1 S2 = × 6 – 2 × 6 × 6  2 m net S = 1.5 m vav = 1.5 = 6 m/s ] 2 Q.12 A particle P is projected from a point A. Particle strikes the ground at B. (a) Show that the time taken for P to reach B from A is 6 . (b) Find the horizontal component of the velocity of P. (c) Find also the tangent of the acute angle between horizontal and velocity of P at time 3 after P leaves A. [4+2+2] –1  2 + 7  [Sol. 8h = [Ans. Horizontal velocity component of P = 2 y 2g ,  = tan   ]  4   vy = 4 –6h = vyt – 1 gt2 2  –6h = 4 t – 1 gt2 2  gt2 – 8 t – 12 h = 0 8  t = vx × t = 8h vx = gh  64gh + 48gh 2g = (4 + 28) vy = 4 gh – g × 3 = gh –1  2 + 7  tan  = vy/vx   = tan   ]  4 

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