PHYSICS-13-08- 11th (PQRS) SOLUTION

XI(PQRS) PHYSICS REVIEW TEST-3 Take g = 10 m/s2 where ever required in this paper. Q.1 Answer following short questions (Show all important steps involved): (i) If the extension of the spring in figure 1 is x1 and in figure 2 is x2, the systems being in constant acceleration, then find the ratio x1/x2. Pulley and strings are ideal. [2] (ii) Consider the two configurations shown in equilibrium. Find ratio of TA/TB. (Ignore the mass of the rope and the pulley) [2] (iii) Four identical 2 kg blocks are in equilibrium as shown. All surfaces are rough. Aforce of 10 N is applied to block C parallel to incline as shown. Rank the NORMAL on the BOTTOM surface of each block? [2] (iv) Find the acceleration of the blocks. [2] (i) (ii) (v) Four identical balls of mass m are arranged as shown. The mass of the block in figure (3) is 4 m and having acceleration a, the angle of the incline is =30° in figure (2) and all the surfaces are frictionless. Rank the TENSIONS in the labelled ropes (from lower to higher)? Consider wedge and table to be fixed. [2] (1) (2) (3) [Sol: (i) x1 = 32 50 K ; x2 = K ; (ii) TA = Mg ...(1) 2TB = Mg ...(2)  TA TB = 2 Ans. (iii) N1 = 20, N2 = 40, N3 = 20 + 10sin N4 = 20cos  N4 < N1 < N3 < N2 Ans. 4F (iv) (i) a = M  ; (ii) a = 4F  Mg  M (v) TB = mg ...(1) TA = 2mg ...(2) TC = mgsin  TC = mg/2 ...(3) Ans. ] Q.2 Answer following short questions: (i) Two blocks of masses M1 and M2 are connected to each other through a light spring as shown in the figure. If we push mass M1 with a force F and cause acceleration a1 in it, what will be the acceleration of M2? [3] (Assume friction less surface everywhere). (ii) A particle of mass 10 kg is acted upon by a force F along the line of motion which varies as shown in the figure. The initial velocity of the particle is 10 ms–1. Find the maximum velocity attained by the particle before it comes to instantaneous rest. [3] (iii) For the given system of blocks, massless and frictionless pulley, and ideal strings. (a) Find the minimum coefficient of friction (min) for which the system stays in equilibrium. (b) If  = min/2 find the acceleration of the system, tensions in string A and B. [1+2] (iv) If Block B comes to rest instantaneously after striking the ground, find the time taken by the block A to strike the pulley. (Assume friction less surface everywhere). [3] (v) Two blocks A & B are kept on a partically rough surface. Two forces F1 & F2 act as shown on blocks A & B respectively. Find the acceleration of the system & friction force on block A.(with direction) [2+1] (vi) An aeroplane A is flying horizontally due east at a speed of 400 km/hr. Passengers in A, observe another aeroplane B moving perpendicular to direction of motion at A. Aeroplane B is actually moving in a direction 30° north of east in the same horizontal plane as shown in the figure. Determine the velocity of B. [3] [Sol.(i) F – kx = M1a1 kx = M2a2 kx = F – M1a1 ...(1) F – M1a1 = M2a2  Ans. (ii) change in velocity is area under a-t graph. a is +ve till t = 10s. area till 10s ½ × 10 × 2 = 10 Vmax – 10 = 10 (iii) (a) T2 = minMg T2 = T1 + g T1 = g  T = 2g  2g = min Mg (b)  = min = 1 2 M fs = Mg = g T2 – g = Ma (1) T1 + g – T2 = a g – T1 = a 2g – T2 = 2a  T2 = 2g – 2a (2) Eq. (1) and (2) implies (iv) (v) V = = m/s 0.5 = ½ × g/2 × t2  t = s T = +  VB 3 ˆ = VB ˆ (vi) VB = i + j 2 2 VA = VAiˆ   VB 3  ˆ VB ˆ V = VB / A =  B / A  2 VA  i + j  2  VB = =  ] Q.3nl Krrish is saving a child in a building caught by fire. He and child are trapped and he ties a rope and drops it out of window. Unfortunately, the tensile strength of the rope is 900 N and Krrish & child's combine mass is 100 kilograms. He therefore decides to let the rope slide through his hands, pulling just hard enough on it so it doesn't break. What will be his speed at the bottom of the rope as he hits the ground? Assume the rope is 12 meters long and that Krrish's vertical speed is zero as he leaves the window at the top of the rope. [3] 1000  900 [Sol. a = 100 = 1 m/s2 v2 = 2 × 1 × 12 v = m/s Ans. ] Q.4nl A small cubical block is placed on a triangular block M so that they touch each other along a smooth inclined contact plane as shown. The inclined surface makes an angle  with the horizontal. Ahorizontal force F is to be applied on the block m so that the two bodies move without slipping against each other. Assuming the floor to be smooth also, determine the (a) normal force with which m and M press against each other and (b) the magnitude of external force F. Express your answers in terms of m, M,  and g. [3+3] [Sol. F = (M + m) a (1) mg sin  + ma cos  = F cos  (2) m cos F cos  F = mg sin  + m + M (m + M) cos   mcos  F =   (m + M)  = mg sin   (m + M) mgsin  F = M cos  = M F mg (m + M) tan  M N sin  = M + m  N = ] Q.5nl The system shown in the figure is in equilibrium. Find the initial acceleration of A, B and C just after the spring-2 is cut. [2+2+2] [Sol. 3mg = KX1 (1) 2mg + KX1 = KX2  2mg + 3mg = KX2  5 mg = KX2 (2) KX3 = KX2 + mg  KX3 = 5mg + mg = 6 mg (3) when spring 2 is cut KX3 – mg = ma3  KX1 + 2mg = 2ma2  acceleration of 3 m will be ] Q.6nl A truck undergoes constant acceleration, a = 5 m/s2. On the back of truck a two-sided, frictionless ramp is fixed, with each side of the ramp at an angle  from the horizontal. Two masses, m1 and m2, sit on either side of the ramp as shown, connected by a massless string wound over a massless pulley. (a) Draw free body diagram (FBD) of m1 & m2 in ground frame. (b) Calculate the ratio of masses, m2/m1, for which the masses do not slide on the ramp when released. [3+3] [Sol. (a) (b) m2 g sin = m2a cos + T m2g sin  = T + m2 a cos  (1) m1g sin  + m1a cos  = T (2) From eg. (1) & (2) m2g sin  = m2 a cos  + m1 g sin  + m1 a cos  m2[g sin  – a cos ] = m1 [g sin  + a cos ] m2  m1 g sin  + a cos  = g sin   a cos  = ] Q.7 A time - dependent force, F = 10t, starts to act on a block of mass 5 kg at t = 0 as shown in the figure. (i) When does it lose contact with the ground? (ii) Draw the a – t graph for 0-10 seconds. (iii) Find the speed of the block at t = 10 sec. [2 + 3 + 1] [Sol: (i) (ii) (iii) ] Q.8nl A block with mass m is pushed along a horizontal floor by a force P that makes an angle  with the horizontal as shown. The coefficients of kinetic and static friction between the block and the floor are k and s, respectively. (a) Find the maximum force that can be applied without moving the block. (b) The applied force P is larger than the maximum force calculated in (a). As a consequence, the block will start to move. Find the acceleration of the block. [Sol. (a) Max. force that can be applied P cos  = sN = s (mg + P sin ) [3+3] P [ Cos  – s sin ] = s mg ; (b) Now block starts moving so friction acting on the block is kinetic P cos  – k (P sin  + mg) = ma ; ] Q.9nl The system shown in the figure is connected by flexible inextensible cord.The coefficient of friction between block C & the rigid surface is 0.3. The system starts from rest when height of block A above ground is d. Pulley and strings are ideal. (a) Find velocity of C just before A is going to touch ground in terms of d and the other given values. (b) Find acceleration of B after A touches the ground and comes to rest instantaneously. (c) Find the initial distance 'd' between Aand the ground, so that the system comes to rest when body B just touches A. [3+2+2] [Sol. fmax = 0.3 × 5 mg = 1.5 mg common acceleration a = 3mg 1.5mg 8m 1.5mg = 8m = 15 m/s2 8 After vertically a distance 'd' v2 = 0 + 2 × 15 × d ( v2 = u2 + 2as) 8 v2 = 15 d (1) 4 After this 2 m comes to rest and system retards Retardation = 1.5mg  mg 5 6m = 6 m/s2 Now After travelling 1.8 m, masses come to rest  O = v2 – 2 × 5 × 1.8  v2 = 3 6 Putting the value of v2 from eq. (1) 15 4 4 d = 3  d = 5 m Ans. ] Q.10327kin/sss An aircraft is flying horizontally with a constant velocity = 200 m/s, at a height = 1 km above the ground. At the moment shown, a bomb is released from the aircraft and the cannon-gun below fires a shell with initial speed 200 m/s, at some angle . (a) For what value of '' will the projectile shell destroy the bomb in mid-air? (b) If the value of  is 53°, find the minimum distance between the bomb and the shell as they fly past each other. Take sin 53° = 4/5. [3+4] [Sol.(i) Suppose shell destroy the bomb at time 't' then for horizontal motion t(200 + 200cos) =  t(1 + cos) = 5 For vertical motion × 1000 ...(1) 1 gt2 + (200sin)t – 1 gt2 = 1000 2 2  (sin)t = 5 ...(2) from (1) and (2) sin  1+ cos  = (ii) On solving,  = 60° Ans. vA = 200 ˆi vB = –200cos53 ˆi + 200sin53 ˆj = –200 × 3 ˆi + 200 × 4 ˆ 5 5 = –120 ˆi + 160 ˆj vA / B = vA – vB = (200 + 120) ˆi – 160 ˆj tan =  = 1 2 AB = 2 km BP = minimum distance =AB sin(30° – )  BP = 2[sin30°cos – cos30°sin] = 2  1  2 – 3  1  Ans. ]    