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

SOLUTIONS (PHYSICS) RTP-3 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 due to springs on the mass and the wall. [Sol: K1x1 = K2x2 x1 + x2 = x K1x [1.5+1.5] x2 = K1 + K2  K1x1 = K2x2 = K1K2 x 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 immediately after the support S is removed. [2+2] [Sol: Spring force do not change immediately Initial force in spring = m1 g a = 0 1 a = (m1 + m2 ) g ] m2 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. (i) Draw the free body diagram (ii) Find the tension in the string. [2+4] [Sol: N + = 50  N = 50 – T  2 N = 5g 2 3 3 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: 8Mg T = 7 Mg F = 7 ] 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] [Sol: a = g a = g 1 2 a = g iˆ , a 4 = g iˆ + g ˆj ] A 4 B 4 2 Q.6 In figure, aA & vB are unknown but velocity of A& acceleration of B are known at an instant. Find aA & vB at the same instant. [3+3] [Sol: 2aB = aA 2vB = vA 1  v = m/s, a = 8 m/s2 ] B 2 A Q.7 Ablock of mass 1 kg is kept on a rough floor (  = 0.7 ) and is at rest initially. A force F starts to act on it in horizontal direction as shown. Given the variation of the force F with time, calculate the velocity of the block just after t = 2 sec. [6] [Sol: A block of mass 1 kg is kept on a rough floor (m = 0.7) and is at rest initially. A force F starts to act on it in horizontal direction as shown. Given the variation of the force F with time, calculate the velocity of the block just after t = 2s. Area = 3 – 2 = 1m/s ] 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/2. A boat cross the river from port A to port B. The speed of the boat relative to the water is v State all your answer in terms of vand D. (a) What is the direction of the rowing of the boat so that it crosses directly on a line from Ato B? How long does the trip take? (b) Suppose the boat wants to cross the river from Ato 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: Done in class 1 (i) cos = 2  = 600 i.e he should head at 1200 with the flow D tcrossing = V sin 600 = D (ii) tmin = V ] Q.9 A block of mass 1 kg is kept on the inclined floor of a lift moving down with 3m/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? [5+3] [Sol: geff = 7m/s2 V = (mgeff ) (cos370) = 28 5 Newton 1 2.1 = (g ) (sin) t2 2 eff t = 1 sec. ] Q.10 Two blocks of masses 2 Kg and 5 Kg are kept on a smooth horizontal surface. The co-efficient of friction between the blocks is 0.1. A massless & inextensible string is tied on the upper block as shown. (a) Calculate the minimum value of a horizontal force P = Pmin such that the blocks start slipping over each other. (b) Obtain the tension in the string if P = Pmin [5+3] 3T [Sol: N + 5 = 20 ..................(1) 4T = N (2) 5 100 80 Solving (1) & (2), T = 43 N, N = 43 N  N = P  P = 80 Newtons T = 100 N ] 43 43 Q.11 A block of mass 3 kg is released at t = 0, 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 from t = 0 untill it stops for the first time after the first collision. [4+4] [Sol: (i) 2 = 1 (g) (sin 370) t 2 2 d td = sec. V = g(sin370) td  3  V = 10  5    = 2 m/s    after collision V = m./s V ' ta = g sin 370 ta = S = sec. (V ')2 2g sin 370 1 = 2 mt Vavg = = m/s ] 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 9 . (b) Find the horizontal component of the velocity of P. (c) Find the tangent (tan ) of the acute angle between the horizontal and velocity of P at time 3 after the [Sol: projection. [4+2+2] u2 1 = 8h (1) 2g u2t = 8h (2) – 9h = u (t ) 1 g(t)2 ................(3) 2 1 2 gt2 – 2ut – gh = 0 gt2 – 2(4) ( t = t = 9 gh )t – gh = 0 2g (ii) u2 = = (iii)  = 8 gh iˆ + (4  3 gh ) ˆj, tan  = 9 9 8 24-09-06 SOLUTIONS (PHYSICS) RTP-3 ACME Q1. K1x1 = K2x2 x1 + x2 = x K1x x2 = K1 + K2  K1x1 = K2x2 = Q.2 Spring force do not change immediately Initial force in spring = m1 g a = 0 1 a = (m1 + m2 ) g m2 m2 Q.3 N + = 50  N = 50 – T  2 N = 5g 2 3 3 T = Q.4 8Mg T = 7 Mg F = 7 Q.5 a = g a = g 1 2 4 Q.6 2aB = aA 2vB = vA 1  v = m/s, a = 8 m/s2 B 2 A Q.7 Area = 3 – 2 = 1m/s 1 Q.8 (i) cos = 2  = 600 i.e he should head at 1200 with the flow D tcrossing = V sin 600 = D (ii) tmin = V Q.9 geff = 7m/s2 N = (mgeff ) (cos370) = 1 2.1 = (g ) (sin) t2 2 eff t = 1 sec. 3T Q.10 N + 5 = 20 ..................(1) 4T = N (2) 5 100 Solving (1) & (2), T = 43  N = P  Q.11 (i) 2 = 1 (g) (sin 370) t 2 80 N, N = 43 N ] 2 ; V = g(sin370) td  3   V = 10  5  d = 2 m/s    after collision V = m./s V ' ta = g sin 370 (V ')2 S = 2g sin 370 1 = 2 mt Vavg = = u2 Q.12 (i) 1 = 8h 2g u2t = 8h ................(1) ................(2) – 9h = u (t ) 1 g(t)2 ................(3) 2 1 2 gt2 – 2ut – gh = 0 gt2 – 2(4) ( t = gh )t – gh = 0 2g (ii) u2 = = (iii)  = 8 gh iˆ + (4 – 3 gh ) ˆj, 9

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