PHYSICS-13-08- 11th (PQRS)

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 arranged as shown. All surfaces are rough. A force 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) Four identical balls of mass m are arranged as shown. The mass of the block in the last diagram is 4 m and having acceleration a, the angle of the incline is =30° and all the surfaces are frictionless. Rank the TENSIONS in the labeled ropes? Consider wedge and table to be fixed. [2] (v) A child is belted into a Ferris Wheel seat that rotates counterclockwise at constant angular speed in vertical plane at an amusement park. At the location shown, which direction best represents the total force exerted on the child by the seat and belt? [2] [Sol.(i) 8g–T1 = 8a1 (A) T1–2g = 2a1 (B) Now, a1 = 80  20 8 + 2 60 = 10 = 6 a2 = 0 kx1 – 20 = 2 × 6 kx2 = 50  x1 = 32 50 k x2 = k Thus x1 32 x2 = 50 16 = 25 Ans. (ii) TA = Mg ...(1) 2TB = Mg ...(2)  TA TB = 2 Ans. (iii) (iv) NA = 20, NB = 40, NC = 20 + 10sin ND = 20cos  ND < NA < NC < NB Ans. TB = mg ...(1) TA = 2mg ...(2) mg – TD = ma   T = 4m × g g D 5 TC = mgsin TD = 4ma   a = 5 TD = 4mg/5 ...(5)  TC = mg/2 ...(3) TC < TD < TB < TA Ans. (v) Total force exerted on child by seat and belt is resultant of N & F so 'D' best represents this force ] 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] (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] [Ans. 20 ms–1] (iii) An object is moving in a circle at constant speed v. The magnitude of the rate of change of momentum of the object is proportional to vn. Find value of n. [3] (iv) A car is moving in a circular path of radius 50 m, on a flat rough horizontal ground. The mass of the car is 1000 kg. At a certain moment, when the speed of the car is 5 m/s, the driver is increasing speed at the rate of 1 m/s2. Find the value of static friction on tyres (total) at this moment, in Newtons. [Ans. 500 ] [3] (v) Coefficient of friction between 5 kg and 10 kg block is 0.5. If friction between them is 20 N. What is the value of force being applied on 5 kg. The floor is frictionless [3] (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 (ii) F = 2t  a2 = F  M1a1 M2 Ans. a = F/m = 2t/10 Max. velocity will be attained at t = 10 sec because after that force start acting in opposite direction v 10 t 10   dv =  5 dt  v – 10 = 10 0 0 v = 20 m/s Ans. (iii) Given  dp = cvn dt mv2 = cvn r On comparing n = 2 Ans. (iv) Fnet = m = m = m 5 2 = 500 N Ans. (v) 20 N is static friction so there will be no relative motion between blocks and acceleration of both will be same  F–20 = 5 × 2 F = 10 + 20 = 30 N Ans. (vi) VA = 400 ˆi ....(i) VB = VB cos 30 ˆi + VB sin 30 ˆj V = VB B 2 3 ˆi + VB ˆj 2 ....(ii) VBA = VB  VA  VBA = ˆi +  j – 400 ˆi      VB V =  – 400 i + VB ˆj BA  2  VB  800 = 0, VB = V = 400 ˆi + 400 ˆj B 3 Ans.] 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 v2 = 2 × 1 × 12 = 1 m/s2 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. A horizontal 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. (a) Considering motion of triangular block M F = (M + m)a M F  M F  N sin  = M + m  N =  M + m  cosec  Ans.   (b) F = (M + m) a (1) mg sin  + ma cos  = F cos  (2) From eq. (1) & (2)  Fcos  = mg sin  + (m + M) cos   mcos  m cos F m + M F  (m + M)  = mg sin   F = mg (m + M) tan  Ans. ] M 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 = KX3 (1) 2mg + KX3 = KX2  2mg + 3mg = KX2  5 mg = KX2 (2) KX1 = 6 mg (3) when spring 2 is cut spring force in other two strings remain unchanged. KX1 – mg = ma3  a3 = 5g  5g KX3 + 2mg = 2ma2  a2 = 2  acceleration of 3 m will be zero. ] Q.6nl Atruck 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 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. from figure-1 from figure-2 T cos  – N1 sin  = m1a ....(1) N2sin  – T cos  = m2a ....(2) N1 cos  = m1g ....(3) N2 cos  = m2g ....(4) from (1) and (2) N2sin  – N1 sin  = (m1 + m2)a putting N1 and N2 from (3) and (4) m2g sin  – cos m1g sin  = (m cos 1 + m2)a m2g sin  – m1g sin  = m1a cos + m2a cos m2(g sin  – a cos) = m1(a cos + g sin ) m2 m1 = g sin  + a cos  m2 g sin   a cos  ;  is 53° so m = 11 5 Ans. ] Q.7 A bead of mass m = 300 gm moves in gravity free region along a smooth fixed ring of radius R = 2m. The bead is attached to a spring having natural length R and spring constant k = 10 N/m. The other end of spring is 6R connected to a fixed point O on the ring. AB = 5 . Line OB is diameter of ring. Find (a) Speed of bead at A if normal reaction on bead due to ring at A is zero (b) The rate of change in speed at this instant. [Sol. Elongation in spring = – R = 8R  R 5 [3+3] 3R = 5  3R  12kR Radial component of spring force = k   5  cos = 25 As normal reaction is zero 12kR 25 mv2 = R  v = = 8 m/s  3R  Tangantial component of force = k    sin 5  dv 9kR m dt = 5 Rate of change in speed dv 9kR =  120 m/s2 ] dt 5m 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. [3+3] [Sol. (a) Max. force that can be applied is P then P cos  = sN = s (mg + P sin ) P [ Cos  – s sin ] = s mg ; P = s mg cos  s sin Ans. (b) Now block starts moving so friction acting on the block is kinetic P cos  – k (P sin  + mg) = ma ; a = P cos k (Psin  + mg) ] m 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. (c) Find the initial distance 'd' between A and 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 = 1.5mg = 15 m/s2 8m After travelling 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 8m 8 Retardation = 1.5mg  mg = 5 m/s2 Ans. m Now After travelling 1.8 m, masses come to rest  O = v2 – 2 × 5 × 1.8 Putting the value of v2 from eq. (1) 15 4 d = 18 18 4 d = 15 24 = 5 = 4.8 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 . For what value of '' will the projectile shell destroy the bomb in mid-air? 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) = × 1000  t(1 + cos) = 5 3 ...(1) For vertical motion 1 gt2 + (200sin)t – 2 1 gt2 = 1000 2  (sin)t = 5 ...(2) from (1) and (2) sin  1+ cos  = On solving,  = 60° Ans. (ii) 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 = – 160 = 1 320 2 AB = 2 km BP = minimum distance = AB sin(30° – )  BP = 2[sin30°cos – cos30°sin] = 2  1  2 – 3  1  = 2  3 Ans. ]    