5.MECHANICS-SOLUTIONS-PART TEST-01

SOLUTIONS TO PART TEST-01 TOPIC : MECHANICS (PHYSICS) 1. As the particle of mass m looses contact with the sphere, no normal reaction is exerted by sphere on the particle. The only force acting on the particle is its weight. Hence its acceleration is g( acceleration due to gravity) 2. F.B.D. for minimum speed (w.r.t. automobile) : = N – mg cos  –  mv2 mv2 R sin  = 0. fx' = R cos  + N – mg sin  = 0  mv2 R cos  + (mg cos  + mv2 R sin ) – mg sin  = 0  v2 = (Rgcos   Rgsin) (cos   sin) for  = 45º and  = 1 : v Rg  Rg = 1 1 = 0 F.B.D for maximum speed (w.r.t. automobile) x' = mv2 R cos  – mg sin  – (mg cos  + mv2 R sin ) = 0 for  = 45º and  = 1 v =  (infinite) m v 2 v 2 3. V = ; B < mg ; R > B B R g  R > 20 m 4. Reading of the weighing machine = 2T + weight of the machine. As weight of the machine is constant 2m1m2 T T T = m1  m2 So reading is maximum for the case m1m2 is maximum as m1 + m2 in all cases is same. 5. Work done in changing speed from 0 to V is - 1 W = 2 mV2 work done in changing the speed from V to 2V is W = 1 2 2 1 m (2V)2 – 2 1 mV2 = 2 3 mV2  W1 1 W2 = 3 6. Let v be the speed of B at lowermost position, the speed of A at lowermost position is 2v. From conservation of energy 1 1 2 m (2v)2 + 2 mv2 = mg (2𝑙) + mg𝑙. Solving we get v = 7. The work done by force from time t = 0 to t = t sec. is given by shaded area in graph below. Hence as t increases, this area increases.  Work done by force keeps on increasing. 8. v = 50 ˆj vf = – 50 ˆi 50 km/h = vj –50 i →  →  → = –50 ˆi – 50 ˆj 50 km/h = vi v v vf vi v = 50 towards south west –50 j 9. x = T × V x = × 2gh x = 2h 10. Relative to the person in the train, acceleration of the stone is ‘g’ downward, a (acceleration of train) backwards. 1 1 X a g According to him : x = at2 , Y = 2 gt2 = 2 Y  Y = a x  straight line. 11. Velocity of approach of P and O is dx – dt = v cos 60° = 5 m/s It can be seen that velocity of approach is always constant.  P reaches O after = 100 = 20 sec. 5 12. Figure shows one of the legs of the mosquito landing upon the water surface. Therefore, T. 2 a × 8 = W = weight of the mosquito. 13. W = TA 2 × 10–4 = (60 – 30) × 10–4 T × 2 2 T = 30  2 1 = 30 = 3.3 × 10–2 N/m 14. Due to surface tension, the surface area tries to minimize itself. 15. In equilibrium, torques of forces mg and Mg about an axis passing through O balance each other. mg. L cos30° = Mg 2 L cos60°  M  2 m 16. Let v1 and v2 be the velocity of efflux from square and circular hole respectively. S1 and S2 be cross-section areas of square and circular holes. v1 = and v2 = The volume of water coming out of square and circular hole per second is Q1 = v1S1 = L2 ; Q2 = v2S2 = R2  Q1 = Q2  R = . L 17. By momentum conservation, the momentum of other fragment will be –p Total KE = p2 2m1 (p)2 + 2m2 18. The COM will fall at distance R = u2 sin 2 g from the initial point, irrespective of explosion. Now one R 3R particle is at 2 , the other should be at 2 . 19. Let the velocity of boat be 'V' opposite to the man. By momentum conservation, –40V + 80 (6 – V) = 0  V = 4 m/s  velocity of the man = 6 – 4 = 2m/s 20. P =   F. V = T V cos 21. The direction of L is perpendicular to the line joining the bob to point C. Since this line keeps changing its orientation in space, direction of L keeps changing however as  is constant, magnitude of L remain constant. Aliter : The torque about point is perpendicular to the angular momentum vector about point C. Hence it can only change the direction of L, and not its magnitude. 22. The only force acting on the body is the viscous force 0 x Here, m vdv = –6rv = – rv mdv   rdx  x = mv . dx v 0 r 23. As collision is elastic, freely suspended mass moves acquiring velocity of colliding mass after the collision. Also acquired velocity must be equal to to complete the circular motion. Hence , mgL + 1 mu2 = 2 1 m(5g𝑙) 2 u = 24. For the ring to move in a circle at constant speed the net force on it should be zero. Here spring force will provide the necessary centripetal force.  kx = mx2   = = = 10 rad/sec. Ans. 25. W = F.ds = (3t ˆi  5 ˆj) . (4t dt ˆi ) 2 12 t3 2 = 12 t2 dt = 0 0 = 32 J 3 26. VA sin 60º = VB  VA  VB 1  sin 27. Balancing the force : T.4a cos 120º + 𝑙  a2g = a2 h g T.2a = a2  g (𝑙 – h) (𝑙 – h) = 2T ag 28. For a rigid body in uniform pure rolling, there is no friction, hence irrespective of the nature of surface on which it rolls (as long as the surface is rigid), it will continue in its state of pure rolling with uniform velocity. 29. Conservation of angular momnentum  o  (t)R2 m R2    = m R2  mo  t = 2  2  2 o 2 30. mg sin Potential energy of spring = 1 Kx2 , 2 Workdone by gravity = mg sin x. By energy conservations, mg x sin  = 1 Kx2 2 2mg sin  = K . SOLUTIONS TO PART TEST-02 TOPIC : MECHANICS (PHYSICS) 1. In the frame (inertial w.r.t earth) of free end of spring, the initial velocity of block is 3 m/s to left and the spring unstretched 1 mv2 = 2 1 kA2 2 or A = v = 3 = 6cm. 2. Let h be the height of water surface, finally a2h = a . a . a ; h = 2 2 a 4  a  a   3a 5a  C.M. gets lowered by a –   8  = a – 8 = 8  Work done by gravity = mg 5a 8 3. Acceleration of two mass system is a = FBD of block A F leftward 2m N cos 60° – F = ma = mF solving N = 3 F 2m 4. Relative to lift initial velocity and acceleration of coin are 0 m/s and 1 m/s2 upwards  8 = 1 2 (1) t2 or t = 4 second m 5. Tension at point P in the rod is T = 𝑙 Extension of width dx length of rod is = x g Tdx Yr 2 = mg Yr 2𝑙 xdx 𝑙 mg mg𝑙 Total extension in rod =  Yr 2𝑙 xdx = 2r 2 Y 6. p1 4T = pa + 2 = pa + 2T pa 4T p2 = p1 + 1 p2 = pa + 2T + 4T p2 – pa = 6T 4T = 6T r = 4/6 cm = 0.0067 m r mg 7. The initial extension in spring is x0 = k Just after collision of B with A the speed of combined mass is v . 2 For the spring to just attain natural length the combined mass must rise up by x0 = comes to rest. mg(sec fig.) and k Applying conservation of energy between initial and final states 1  v 2 2m  2  + 1  mg 2 k  k   mg  = 2mg  k  2   2     Solving we get v = Alternative solution by SHM v 2 = ; v = = ] 8. Pressure at 'A' from both side must balance. Figure is self–explanatory.  h g = h g  2 R sin (45º + ) = R [cos – sin]  [cos + sin] = [cos – sin]    tan =    9. As shown in the figure , at A at = g . Here, a is maximum because at other place, a = gcos  . a and a will be equal at some point. a will have extreme value at lowest points and there at = 0. 10. x = u (t – 2) + a(t – 2)2 Velocity of particle at time t dx dt = u + 2a (t – 2) dx Velocity at t = 0 dt = u – 4a Acceleration of particle d2x dt2 = 2a xt = 2 = 0 So correct ans is (C) and (D). 11. The pressure at any point can never have different values. Hence (A) & (D) are not possible. (Calculate the pressures at points A & D from both their left and right) In case of insufficient length of capillary tube the shape of meniscus is as below : 12. arel = g – g = 0  v = constant acm = g = constant  v  constant 13. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 14. Only in uniform circular motion force must be directed perpendicular to the velocity all the time. However in non-uniform circular motion, force will always not be perpendicular to velocity. Hence statement-1 is false. 15. Centre of mass and centre of gravity of a uniformly distributed mass shall coincide if acceleration due to gravity at all point of mass distribution is same. Hence statement-2 is false. 16. As the oil is poured till it covers the object completely, pressure in water at all points keeps on increasing. As a result upward force on object exerted by water increases and the object moves up for the given duration. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. 17. The time in which the planet rotates about its axis is not given for either planet. 18. For geostationary satellite, time period = 1 planet day (by def.) Let T = 1 planet day T0 = 1 planet year 42 3 42 3  m  42 3 3 Now T2 = Gm rG = Gm r  M  = GM r  T0  T = T0   19. The energy of any geostationary satellite is the sum of kinetic energy of satellite, interaction energy of satellite and its own planet and interaction energy of satellite and star. Both planets have same mass and same length of day. Geostationary satellite - planet system will have same interaction energy in either planet. Also kinetic energy of both satellites will be same. But the satellite-sun system will account for the energy difference. U = – GMm0 + U i 2r GMm 0 satellite – planet Uf = – 2 (4r) + Usatellite – planet Emin = Uf – Ui = 3 GMm 0 8 r du 20. If the particle is released at the origin, it will try to go in the direction of force. Here dx is positive and hence force is negative, as a result it will move towards – ve x-axis. 21. When the particle is released at x = 2 +  ; it will reach the point of least possible potential energy (– 15 J) where it will have maximum kinetic energy.  1 m v2 = 25  v = 5 m/s 2 max max 22. At x = – 5 m , U is minimum  Stable At x = + 10 m , U is maximum  Unstable 23. P.E. = – GMm  K.E. = r 1 mV2 2 Total energy =  GMm r + 1 mV2 2 T.E. = 0 if  GMm  1 mV 2  0  v = r 2 For v < , T.E. is – ve for v > , T.E. is + ve If V is i.e. equal to orbital velocity, path is circular. If T.E. is negative, path is elliptical. If T.E. is zero, path is parabolic. If T.E. is positive, path is hyperbolic. 24. Use equation of continuity and concept "pressure is greater at lower and broader section". SOLUTIONS TO PART TEST-03 TOPIC : MECHANICS (PHYSICS) 1. When the acceleration of bob is horizontal, net vertical force on the bob will be zero. T cos  – mg = 0 The tangential force at that instant is = mg sin  = mg = mg T 2. For the driver to observe the rain move vertically upward after the elastic collision, rain shoud come at an angle 30º with the horizontal (as clear from figure). Let , velocity of rain w.r.t. car be Vm/c → → → VR / G = VR / C + VC / G → → → VR / Gx = VR / Cx  + VC / Gx  → But VR / Gx = 0 , since rain fall vertically down. → →  VC / Gx  = – VR / Cx  = –(V cos 30º) →  | VC / Gx  | = V cos 30º (i) → Now, VR / Gy → = VR / Cy  –10 = –V sin 30º + 0 [Since ; VR/G(y) = – 10 m/s ; VC/G(y) = 0]  V sin 30º = 10  V = 20 m/s. Substituting V = 20 m/s in equation (i) → VC / Gx  = V cos 30º = 20 × → 3 = 10 2 m/s.  VC / G = 10 ˆi m/s. 3. Vefflux = time of fall t = x = Vefflux t = 2 the roots of x are (0,4) and the maximum of x is at h = 2. The permitted value of h is 0 to 1 clearly h = 1 will give the maximum value of x is this interval. Aliter : If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum range x. Farther the hole is from this midpoint, lower the range. Here the nearest point possible to this midpoint is the base of the container. Hence h = 1m. 4. For the given situation, liquid of density 2  should be behind that of . From right limb : P = P +  gh PB = PA +  a 𝑙 = P atm + gh +  a 𝑙 PC = PB + (2) a 𝑙 = P atm +  gh + 3  a 𝑙 (1) But from left limb : P = P + (2 ) gh (2) From (1) and (2) : 3 3a Patm +  gh + 2  a 𝑙 = P atm + 2  gh  h = 2g 𝑙 Ans. v mv3 v pdx m dv  v dx mv2 dV   10 3 5. Area under P–x graph =  =   =  dt  1 =  3 = 1 7  3 (v – 1) from graph ; area = 1 (2 + 4) × 10 = 30  v = 4 m/s 2 ALITER : from graph dv P = 0.2 x + 2 or mv dx or mv2 dv = (0.2 x + 2) dx v = 0.2 x + 2 v Now integrate both sides, mv 2dv 1 10 = (0.2 x  2) dx 1  v = 4 m/s. 6. (Pure rotation about instantaneous point of contact) Note : If line of action passes through point of contact, it only spins. 7. Resultant Acceleraton = = g 2 g.sin150º 2 tan  = 3g  g = .cos150o  = 30º 2 2 i.e. Resulting acceleration is in vertical direction. 8. For rigid body separation between two point remains same. v1 cos60° = v2 cos30° v1 = 2 3 v 2  v = v 2 1 2  v 2 sin30  v1 sin 60 v2  3v1 2 2 v2  3  3v2 2v 2 v 2 disc = d = d = 2d = 2d = d v 2 disc = d 9. For minimum velocity. at A ; 1 mV 2 = mgR  VA = 2 A Now ; 1 m V 2 + mgR' = 1 m V 2 2 B 2 E As , VB = For looping the loop ; VE =  1 m2gR + mgR' = 2 1 m 5gR'  2 R' = 2 R 3 2 And also , N – mg = E R' N – mg = N = 6mg m 5gR' R' 10. = 2t – 3 x = (2t – 3)2 dx Vel. v = dt = 2(2t – 3) × 2 = 4 (2t – 3) if v = 0 t = 3/2 acceleration dv a = dt = 4 × 2 = 8 so correct ans is (A) and (B) 11. The resultant force can be accelerating or decelerating, hence the momentum can increase or decrease. Hence (A) is wrong. Since Fnet = M acm  a  0 ; hence vcm must change Hence (B) In case of a circular motion of centre of mass about a point the distance of centre of mass will remain constant. Hence (C) Kinetic energy of some particles may increase and of some particles may decrease at the same time. 12. For the whole system, 2T – 1500 = 150 × 5  T = 1125 N For the person, T – 1000 + N = 100 × 5  N = 1500 – 1125 = 375 N 13. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 14. The equation of distance travelled in nth second is dimensionally correct because the interval of time 1 second has already been substituted into the equation and its dimension should be taken into account. Therefore statement-1 is false. 15. As T = 2  m = T 2K 42 0.2  0.2  1000 = 4 10 = 1 kg 16. Immediately after the collision, suppose velocities of the blocks are V1 and V2 as shown 1 v1 v2 2 vel. of approach = velocity of separation.  5 = V2 – V1 (1) Using principle of conservation of momentum for the collision 2 = 0.2 V1 + V2 or 10 = V1 + 5V2 (2) On solving V2 = 2.5 m/s; V1 = – 2.5 m/s Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s. The maximum velocity of B = 2.5 = A  A = V = 2.5 m = 2.5 cm 17. Time of flight = = 1 sec. d = (2.5 m/s) × 1 s = 2.5 m 18. In all four situation of column-I, angular momentum of the disc about a point on ground is conserved. Take angular momentum out of the paper as positive (A) Initial angular momentum about its point of contact on ground = 1 mR2 –mR(2R ) = negative. 2 o o Hence final state of the disc is as shown if figure B. Hence angular velocity shall first decrease and then increase in opposite sense. The velocity of centre shall decrease till the disc starts rolling without slipping. (B) The initial angular momentum about its point of contact on ground =0. Hence angular speed and velocity of centre simultaneously reduce to zero without a change in direction. (C) Because v0 > R0, velocity of centre of mass will decrease and angular velocity will increase without a change in direction till disc starts rolling without slipping. (D) Because v0 < R0, velocity of centre of mass will increase and angular velocity will decrease without a change in direction till disc starts rolling without slipping. 19. Velocity of  particle after time t  = (v cos  ˆi + v sin ˆj ) – (g ˆj ) t Velocity of  particle after time t  = (v cos 2 ˆ + v sin 2 ˆj ) – (g ˆj ) t V2 i To be parallel of V1 and V  v cos  v cos 2 v sin  gt = v sin 2  gt Solving above equation we can get result. 20. Method - 1 If the river is still, the man will be at a distance 3 metres from origin O after 1 second. The locus of all the point where man can reach at t = 1 second is a semicircle of radius 3 and centre at O (dotted semicircle shown in figure). The river flows to right with a speed 1 m/s. Hence there shall be additional shift in position by 1 m/s × 1 sec = 1m towards right. Hence the locus of all points giving possible position after one second will be the dotted semicircle shifted to right by 1 m as shown in figure. Hence locus all the points where the man can be at t = 1 sec. is a semicircle of radius 3 and centre at O (1 m, 0 m)  Equation of locus of all the points is (x – 1)2 + (y – 0)2 = 32 or (x – 1)2 + y2 = 9 Method - 2 Let the relative velocity of the man make angle ‘’ with the x-axis. Then at time ‘t’ : x = (3 cos + 1) t and y = 3 sin t  (x – t)2 + y2 = (3 cos)2 t2 + (3 sin)2 t2 (x – t)2 + y2 = 9t2 at t = 1 sec. the required equation is (x – 1)2 + y2 = 9. 21. (figure - 1) Let u and v be the speed of wedge A and block B at just after the block B gets off the wedge A. Applying conservation of momentum in horizontal direction, we get. mu = mv (1) Applying conservation of energy between initial and final state as shown in figure (1), we get mgh = 1 mu2 + 2 1 mv2 (2) 2 solving (1) and (2) we get v = .............(3) At the instant block B reaches maximum height h on the wedge C (figure 2), the speed of block B and wedge C are v. Applying conservation momentum in horizontal direction, we get mv = (m + m) v (4) Applying conservation of energy between initial and final state 1 mv2 = 2 1 (m + m) v2 + mgh (4) 2 22. Solving equations (3) (4) and (5) we get h = 1 = 100 rad/sec 2 = 0 h Ans. 4 (a1 = a2)  fR = 1 fR = 2 1 = 2  = 2 red/sec2 For A cylinder :  = 0 – t  = 100 – 2t ...(i) For B cylinder  = 0 – t 0 = 0  = t  = 2 t (ii) From (i) and (ii)  = 100  –  2 = 100  = 50 From (ii) euqation 50  = 2 t t = 25 sec