Newton's Laws of Motion Exercises

Newton's Laws of Motion  Marked Questions can be used as Revision Questions. PART - I : SUBJECTIVE QUESTIONS Section (A) : Type of forces, Newton’s third law, Free body diagram : A-1. Which type of forces does a proton exerts on a proton inside nucleus ? A-2. A block ‘A’ exerts a force on ‘B’ of magnitude 20 N. Calculate the magnitude of force exerted by ‘B’ on ‘A’. A-3. Two forces of same magnitude act on an isolated body in opposite directions to keep it at equilibrium position, is this true according to Newton’s third law ? A-4. Draw F.B.D. of the sphere of mass M placed between a vertical wall and ground as shown in figure (All surfaces are smooth). A-5. A block of mass 1 kg placed on ground is pulled by a string by applying 10 N force : (g = 10 m/s 2) (i) Draw F.B.D. of block. (ii) Give action–reaction pair involved in the above problem. A-6. Draw free body diagrams for masses m and M shown in figure. Identify all action-reaction pairs between two blocks. The pulley is frictionless and massless and all surfaces are smooth. A-7. Draw the FBD for the following individual parts of the systems : (Pulley are massless and friction less) (a) (b) (c) (e) (f) (g) (h) (i) Section (B) : Calculation of normal reaction (d) Newton's Laws of Motion B-1. A block of mass ‘m’ is placed on ground and an additional force F = mg is applied on the block as shown in figure. Calculate contact force between ground and block. B-2. Two blocks of masses m1 and m2 are placed on ground as shown in figure. Two forces of magnitude F act on m 1 and m2 in opposite directions. (i) Draw F.B.D. of masses m1 and m2. (ii) Calculate the contact force between m1 and m2. (iii) What will be the value of normal force between m1 and m2. (iv) Calculate force exerted by ground surface on mass m1 and m2 B-3. A sphere of mass ‘m’, radius ‘R’ placed between two vertical walls having separation ‘d’ which is slightly greater than ‘2R’ : (i) Calculate force exerted by walls on the sphere. (ii) Calculate force exerted by ground surface on the sphere. B-4. A cylinder of weight w is resting on a fixed V-groove as shown in figure. Smooth fixed groove 60 60 0 0 60 0 /////////////////////////// mg (a) Draw its free body diagram. (b) Calculate normal reactions between the cylinder and two inclined walls. B-5. The 50 kg homogeneous smooth sphere rests on the 30º incline A and bears against the smooth vertical wall B. Calculate the contact forces at A and B. B-6. A spherical ball of mass m = 5 kg rests between two planes which make angles of 30° and 45° respectively with the horizontal. The system is in equilibrium. Find the normal forces exerted on the ball by each of the planes. The planes are smooth. 45° 30° Section (C) : Calculation of Tension C-1. A one meter long massless string fixed with a wall is pulled horizontally by applying a force of magnitude 10 N. Calculate : (a) the tension at a point 0.5m away from wall. (b) the tension at a point 0.75 m away from wall. (c) force exerted by string on the rigid support. Newton's Laws of Motion C-2. A block of mass 1 kg is suspended by a string of mass 1 kg, length 1m as shown in figure. (g = 10 m/s2) Calculate: (a) the tension in string at its lowest point. (b) the tension in string at its mid–point. (c) force exerted by support on string. C-3. A block of mass 0.3 kg is suspended from the ceiling by a light string. A second block of mass 0.2 kg is suspended from the first block through another string. Find the tensions in the two strings. Take g = 10 m/s2. C-4. Two unequal masses m1 and m2 are connected by a string going over a /////////////////////////////////// clamped light smooth pulley as shown in figure m1 = 3 kg and m2 = 6 kg. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds. (b) Find the tension in the string. (c) Find the force exerted by the clamp on the pulley. (g = 10 m/s 2) m1 m2 C-5. A mass M is held in place by an applied force F and a pulley system as shown in figure. The pulleys are massless and frictionless. (a) Draw a free body diagram for each pulley (b) Find the tension in each section of rope T1, T2, T3, T4 and T5. (c) Find the magnitude of F. Section (D) : Constrained motion D-1. In the figure shown, blocks A and B move with velocities v 1 and v2 along horizontal direction. Find the v ratio of 1 . v2 Newton's Laws of Motion D-2. In the figure shown, the pulley is moving with velocity u. Calculate the velocity of the block attached with string. D-3. The velocity of end ‘A’ of rigid rod placed between two smooth vertical walls is ‘u’ along vertical direction. Find out the velocity of end ‘B’ of that rod, rod always remains in contact with the vertical wall and also find the velocity of centre of rod. Also find equation of path of centre of rod. y x D-4. If block A has a velocity of 0.6 m/s to the right, determine the velocity of block B. Light rod D-5. Find the acceleration of rod A and wedge B in the arrangement shown in figure if the mass of rod equal that of the wedge and the friction between all contact surfaces is negligible and rod A is free to move downwards only. Take angle of wedge as 37o. D-6. Velocities of blocks A, B and pulley p2 are shown in figure. Find velocity of pulley p1 and block C. 10 m/s p2 p1 C B 20 m/s D-7. Find the acceleration of A in term of b. A 10 m/s Newton's Laws of Motion Section (E) : Calculation of Force and Acceleration E-1. In the figure the tension in the string between 1 and 2 is 60 N. (a) Find the magnitude of the horizontal force F1 and F2 that must be applied to hold the system in the position shown. (b) What is the weight of the suspended block? E-2. A 3.0 kg mass is moving in a plane, with its x and y coordinates given by x = 5t 2 – 1 and y = 3t3 + 2, where x and y are in meters and t is in second. Find the magnitude of the net force acting on this mass at t = 2 sec. E-3. A constant force F = m1g / 2 is applied on the block of mass m2 as shown in figure. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m 2. E-4. A chain consisting of five links each with mass 100gm is lifted vertically with constant acceleration of 2m/s2 () as shown. Find : (g = 10 m/s2) : (a) the forces acting between adjacent links (b) the force F exerted on the top link by the agent lifting the chain (c) the net force on each link. E-5. A 2 kg toy car can move along an x axis. Graph shows force F x, acting on the car which begins at rest at time t = 0. The velocity of the car at t = 10 s is : Fx(N) 4 0 -2 t(s) 4 8 9 10 11 Newton's Laws of Motion E-6. Find out the acceleration of the block B in the following systems : (a) E-7. (b) Find out the mass of block B to keep the system at rest : (g = 10 m/s 2) (a) (b) Section (F) : Weighing machine, Spring related problems and Spring balance F-1. A man of mass 60 kg is standing on a weighing machine placed in a lift moving with velocity ‘v’ and acceleration ‘a’ as shown in figure. Calculate the reading of weighing machine in following situations : (g = 10 m/s2) (i) a = 0, v = 0 (ii) a = 0, v = 2 m/s (iii) a = 0, v = – 2m/s (iv) a = 2m/s2, v = 0 2 (v) a = – 2 m/s , v = 0 (vi) a = 2 m/s2, v = 2 m/s 2 (vii) a = 2 m/s , v = – 2 m/s (viii) a = – 2 m/s2, v = – 2 m/s F-2. What will be the reading of spring balance in the figure shown in following situations (g = 10 m/s2) (i) a = 0, v = 0 (ii) a = 0, v = 2 m/s (iii) a = 0, v = – 2m/s (iv) a = 2m/s2, v = 0 (v) a = – 2 m/s2, v = 0 (vi) a = 2 m/s2, v = 2 m/s (vii) a = 2 m/s2, v = – 2 m/s (viii) a = – 2 m/s2, v = – 2 m/s F-3. Three identical balls 1, 2, 3 are suspended on springs one below the other as shown in the figure. OA is a weightless thread. The balls are in equilibrium (a) If the thread is cut, the system starts falling. Find the acceleration of all the balls at the initial instant (b) Find the initial accelerations of all the balls if we cut the spring BC, which is supporting ball 3, instead of cutting the thread. Newton's Laws of Motion Section (G) : Newton’s law for a system G-1. Man ‘A’ of mass 60 kg pushes the other man ‘B’ of mass 75 kg due to which man ‘B’ starts moving with acceleration 3 m/s2. Calculate the acceleration of man ‘A’ at that instant. G-2. Find the acceleration of the 6 Kg block in the figure. All the surfaces and pulleys are smooth. Also the strings are inextensible and light. [Take g = 10 m/s2] G-3. A painter of mass M stands on a platform of mass m and pulls himself up by two ropes which hang over pulley as shown. He pulls each rope with the force F and moves upward with uniform acceleration ‘a’. Find ‘a’ G-4. Three monkeys A, B and C with masses of 10, 15 & 8 Kg respectively are climbing up & down the rope suspended from D. At the instant represented, A is descending the rope with an acceleration of 2 m/s 2 & C is pulling itself up with an acceleration of 1.5 m/s 2. Monkey B is climbing up with a constant speed of 0.8 m/s. Calculate the tension T in the rope at D. ( g = 10 m/s −2) D A B C Section (H) : Pseudo Force H-1. An object of mass 2 kg moving with constant velocity 10iˆ m/s is seen in a frame moving with constant velocity 10iˆ m/s. What will be the value of ‘pseudo force’ acting on object in this frame. H-2. In the adjoining figure, a wedge is fixed to an elevator moving upwards with an acceleration ‘a’. A block of mass ‘m’ is placed over the wedge. Find the acceleration of the block with respect to wedge. Neglect friction. m Newton's Laws of Motion PART - II : ONLY ONE OPTION CORRECT TYPE Section (A) : Type of forces, Newton’s third law, Free body diagram A-1. Which figure represents the correct F.B.D. of rod of mass m as shown in figure : (A) (B) (C) (D) None of these A-2. When a horse pulls a cart, the force needed to move the horse in forward direction is the force exerted by : (A) the cart on the horse (B) the ground on the horse (C) the ground on the cart (D) the horse on the ground Section (B) : Calculation of normal reaction B-1. Two blocks are in contact on a frictionless table. One has mass m and the other 2m. A force F is applied on 2m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio of force of contact between the two blocks will be : (A) same B-2. (B) 1 : 2 (C) 2 : 1 (D) 1 : 3 Two forces of 6N and 3N are acting on the two blocks of 2kg and 1kg kept on frictionless floor. What is the force exerted on 2kg block by 1kg block ? 6N 2kg 1kg 3N /////////////////////////////////////////////// (A) 1N B-3. (B) 2N (C) 4N (D) 5N A constant force F is applied in horizontal direction as shown. Contact force between M and m is N and between m and M’ is N’ then (A) N = N’ (B) N > N’ (C) N’ > N (D) cannot be determined Section (C) : Calculation of Tension C-1. A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force F. If the rope AB makes an angle  with the vertical in equilibrium, then the tension in the string AB is : (A) F sin  (B) F/sin  (C) F cos  ///////////////////////////// A  B M (D) F/cos  F Newton's Laws of Motion C-2. C-3. Two persons are holding a light rope tightly at its ends so that it is horizontal. A 15 kg weight is attached to the rope at the mid point which now no longer remains horizontal. The minimum tension required to completely straighten the rope is : 15 kg 2 (A) 15 kg (B) (C) 5 kg (D) nfinitely large (or not possible) n the system shown in the figure, the acceleration of the 1 kg mass and the tension in the string connecting between A and B is : B 3kg 3kg C A 1kg (A) C-4. g 8g g g downwards, (B) upwards, 4 7 4 7 (C) g 6 g downwards, g (D) upwards, g 7 7 2 A body of mass 8 kg is hanging from another body of mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 m s–2. The tension T1 and T2 will be respectively : (use g = 9.8m/s2) (A) 200 N, 80 N (B) 220 N, 90 N (C) 240 N, 96 N (D) 260 N, 96 N C-5. A particle of small mass m is joined to a very heavy body by a light string passing over a light pulley. Both bodies are free to move. The total downward force on the pulley due to string is nearly (A) mg (B) 2 mg (C) 4 mg (D) can not be determined C-6. A heavy block kept on a frictionless surface and being pulled by two ropes of equal mass m as shown in figure. At t = 0, the force on the left rope is withdrawn but the force on the right end continues to act. Let F1 and F2 be the magnitudes of the forces by the right rope and the left rope on the block respectively. (A) F1 = F2 = F for t < 0 (B) F1 = F2 = F + mg for t < 0 (C) F1 = F, F2 = F for t > 0 (D) F1 < F, F2 = F for t > 0 Newton's Laws of Motion Section (D) : Constrained motion D-1. A block is dragged on smooth plane with the help of a rope which moves with velocity v. The horizontal velocity of the block is : V  m (A) v D-2. v sin (C) v sin  (D) v cos Two masses are connected by a string which passes over a pulley accelerating upward at a rate A as shown. If a1 and a2 be the acceleration of bodies 1 and 2 respectively then : (A) A = a1 – a2 D-3. (B) //////////////////////////// ////////////////// (B) A = a1 + a2 (C) A = a1 − a 2 2 (D) A = a1 + a 2 2 In the arrangement shown in fig. the ends P and Q of an unstretchable string move downwards with uniform speed U. Pulleys A and B are fixed. Mass M moves upwards with a speed. (A) 2 U cos  (B) U cos  (C) 2 U/cos  (D) U/cos  D-4. A system is shown in the figure. A man standing on the block is pulling the rope. Velocity of the point of string in contact with the hand of the man is 2 m/s downwards. The velocity of the block will be: [Assume that the block does not rotate] (A) 3 m/s (B) 2 m/s (C) 1/2 m/s (D) 1 m/s Newton's Laws of Motion D-5. A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity 3 m/s towards right. The velocity of end B of rod when rod makes an angle of 60º with the ground is: (A) D-6. (B) 2 m/s (C) 2 3 m/s (D) 3 m/s System is shown in figure and wedge is moving towards left with speed 2 m/s. Then velocity of the block B will be: (A) D-7. 3 m/s 3 m/s (B) 1 m/s (C) 2 m/s (D) 0 m/s In the figure shown the velocity of lift is 2 m/s while string is winding on the motor shaft with velocity 2 m/s and block A is moving downwards with a velocity of 2 m/s, then find out the velocity of block B. (A) 2 m/s (B) 2 m/s (C) 4 m/s (D) 8 m/s D-8. Two beads A and B move along a fixed semicircular wire frame as shown in figure. The beads are connected by an inelastic string which always remains tight. At an instant the speed of A is u, BAC = 45° and BOC = 75°, where O is the centre of the semicircular arc. The speed of bead B at that instant is : (A) 2u (B) u (C) u 2 2 (D) 2 u 3 Section (E) : Calculation of Force and Accelertion E-1. A particle is moving with a constant speed along a straight line path. A force is not required to : (A) increase its speed (B) decrease its momentum (C) change the direction (D) keep it moving with uniform velocity E-2. An object will continue accelerating until : (A) resultant force on it begins to decrease (B) its velocity changes direction (C) the resultant force on it is zero (D) the resultant force is at right angles to its direction of motion Newton's Laws of Motion E-3. n which of the following cases the net force is not zero ? (A) A kite skillfully held stationary in the sky (B) A ball freely falling from a height (C) An aeroplane rising upwards at an angle of 45º with the horizontal with a constant speed (D) A cork lying on the surface of water E-4. Two masses M1 and M2 are attached to the ends of a light string which passes over a massless pulley attached to the top of a double inclined smooth plane of angles of inclination  and . If M2 > M1 and M (A) E-5. M2 g ( sin  ) M1 + M2 (B) M1g ( sin  ) M1 + M2 M2 1  >  then the acceleration of block M2 down the inclined will be :  fixed   M sin  – M1 sin   (C)  2 g M1 + M2   (D) zero The pulley arrangements shown in figure are identical, the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with a constant downward force F = 2mg, where g is acceleration due to gravity. The acceleration of mass in case I is : (A) zero (C) less than that in case II (B) more than that in case II (D) equal to that in case II E-6. A force produces an acceleration of 4 ms–2 in a body of mass m1 and the same force produces an acceleration of 6 ms–2 in another body of mass m2. If the same force is applied to (m1 + m2), then the acceleration will be: (A) 10 ms–2 (B) 2 ms–2 (C) 2.4 ms–2 (D) 5.4 ms–2 E-7. A body of mass M is acted upon by a force F and the acceleration produced is a. If three coplaner forces each equal to F and inclined to each other at 120º act on the same body and no other forces are acting. The acceleration produced will be : (A) 2 a (B) a/ 3 (C) 3a (D) zero E-8. A fireman wants to slide down a rope. The rope can bear a tension of With what minimum acceleration should the fireman slide down : g g g (A) (B) (C) 4 3 6 E-9. 3 4 th (D) of the weight of the man. g 2 A force F = 6iˆ − 8ˆj + 10kˆ newton produces acceleration 1 m/s2 in a body. The mass of the body is (in kg) : (A) 6iˆ − 8ˆj + 10kˆ kg (B) 10 2 kg (C) 100 kg (D) 10 kg Newton's Laws of Motion E-10. A body is moving with a speed of 1 m/s and a constant force F is needed to stop it in a distance x. If the speed of the body is 3 m/s the force needed to stop it in the same distance x will be (A) 1.5 F (B) 3F (C) 6 F (D) 9F E-11. Two blocks, each having mass M, rest on frictionless surfaces as shown in the figure. If the pulleys are light and frictionless, and M on the incline is allowed to move down, then the tension in the string will be : M  (A) E-12. 2 Mg sin  3 (B) 3 Mg sin  2 fixed (C) M Mgsin  2 (D) 2Mg sin  A force of magnitude F1 acts on a particle so as to accelerate it from rest to a velocity v. The force F 1 is then replaced by another force of magnitude F2 which decelerates it to rest. (A) F1 must be the equal to F2 (B) F1 may be equal to F2 (C) F1 must be unequal to F2 (D) None of these Section (F) : Weighing machine, Spring related problems and Spring balance F-1. n the given figure, what is the reading of the spring balance ? 1kg (A) 10 N F-2. 1kg (B) 20 N (C) 5 N (D) zero Two blocks of masses M1 and M2 are connected to each other through a light spring as shown in figure. If we push mass M1 with force F and cause acceleration a1 in right direction. What will be the magnitude of acceleration in M2? F M1 M2 /////////////////////////////////////////////////////////////// (A) F/M2 (B) F/(M1 + M2) (C) a1 (D) (F – M1a1)/M2 F-3. A massless spring balance is attached to 2 kg trolley and is used to pull the trolley along a flat surface as shown in the fig. The reading on the spring balance remains at 10 kg during the motion. The acceleration of the trolley is (Use g = 9.8 ms–2) 2kg • (A) 4.9 F-4. ms–2 (B) 9.8 ms–2 • (C) 49 ms–2 (D) 98 ms–2 The ratio of the weight of a man in a stationary lift & when it is moving downward with uniform acceleration 'a' is 3 : 2 . The value of 'a' is : (g = acceleration. due to gravity) (A) (3/2) g (B) g (C) (2/3) g (D) g/3 Newton's Laws of Motion F-5. A body of mass 32 kg is suspended by a spring balance from the roof of a vertically operating lift and going downward from rest. At the instants the lift has covered 20 m and 50 m, the spring balance showed 30 kg & 36 kg respectively. The velocity of the lift is: (A) decreasing at 20 m & increasing at 50 m (B) increasing at 20 m & decreasing at 50 m (C) continuously decreasing at a constant rate throughout the journey (D) continuously increasing at constant rate throughout the journey Section (G) : Newton’s law for a system G-1. Two masses of 10 kg and 20 kg respectively are connected by a massless spring as shown in figure. A force of 200 N acts on the 20 kg mass at the instant when the 10 kg mass has an acceleration of 12 ms–2 towards right, the acceleration of the 20 kg mass is : 10kg 20kg 200 N (A) 2 ms–2 G-2. (B) 4ms–2 (C) 10ms–2 (D) 20ms–2 In the arrangement shown in the figure all surfaces are frictionless, pulley and string are light. The masses of the block are m1 = 20 kg and m2 = 30 kg. The accelerations of masses m 1 and m2 will be if F = 180 N is applied according to figure. G-3. (A) am1 = 9m/ s2 , am2 = 0 (B) am1 = 9m/ s2 , am2 = 9 m/s2 (C) am1 = 0 , am2 = 9m / s2 (D) None of these Three masses of 1 kg, 6 kg and 3 kg are connected to each other with threads and are placed on table as shown in figure. What is the acceleration with which the system is moving ? Take g = 10m s –2. (A) Zero (B) 1 m s–2 (C) 2 m s–2 (D) 3 m s–2 Section (H) : Pseudo force H-1. Figure shows a wedge of mass 2kg resting on a frictionless floor. A block of mass 1 kg is kept on the wedge and the wedge is given an acceleration of 5 m/sec2 towards right. Then : (A) block will remain stationary w.r.t. wedge (B) the block will have an acceleration of 1 m/sec2 w.r.t. the wedge (C) normal reaction on the block is 11 N (D) net force acting on the wedge is 2 N Newton's Laws of Motion H-2. A triangular block of mass M rests on a smooth surface as shown in figure. A cubical block of mass m rests on the inclined surface. If all surfaces are frictionless, the force that must be applied to M so as to keep m stationary relative to M is : (A) Mg tan 30° H-3. (B) mg tan 30° (C) (M+m)g tan 30° (D) (M+m)g cos 30° A block of mass m resting on a wedge of angle  as shown in the figure. The wedge is given an acceleration a towards left. What is the minimum value of a due to external agent so that the mass m falls freely ? A m a  B (B) g cos  (A) g C (C) g cot  (D) g tan  PART - III : MATCH THE COLUMN 1. Column-I gives four different situations involving two blocks of mass m 1 and m2 placed in different ways on a smooth horizontal surface as shown. In each of the situations horizontal forces F 1 and F2 are applied on blocks of mass m1 and m2 respectively and also m2 F1 < m1 F2. Match the statements in column I with corresponding results in column-II Column  (A) Column  . Both the blocks  F1 F2  −    m1 m2  (p) m1m2 m1 + m2 (q) m1m2  F1 F  + 2   m1 + m2  m1 m2  (r) m1m2 m1 + m2 are connected by massless inelastic string. The magnitude of tension in the string is (B) . Both the blocks are connected by massless inelastic string. The magnitude of tension in the string is (C) . The magnitude  F2 F  − 1   m m 1   2 of normal reaction between the blocks is (D) . The magnitude of normal reaction between the blocks is  F +F  (s) m1m2  1 2   m1 + m2  Newton's Laws of Motion 2. In column  four different sitiuation are given and in column  tension in string (which is not connected with spring) & extention in spring at equilibrium is given Match the statements in column  with corresponding results in column-II Column- Column- (A) (p) mg (B) (q) 2mg (C) (r) mg k (D) (s) 2mg k Newton's Laws of Motion 3. In column- four different situation are given column- corresponding result are given. Match the statements in column I with corresponding results in column-II in the question below take the unit vector as follows : ^j ^i Column- Column- (A) The acceleration of B w.r.t ground if  = 60° & a = 2m/s2 (p) 4 ĵ (B) The acceleration of A w.r.t. ground if b = 1m/s2 (q) – 2 ĵ (C) The acceleration of C w.r.t. ground if a = b = 1m/s2 (r) −ˆi − 3 ˆj (D) The acceleration of wedge A if  = 45° & b = 2m/s2 (s) î – 4 ĵ  Marked Questions can be used as Revision Questions. PART - I : ONLY ONE OPTION CORRECT TYPE 1. Consider the shown arrangement. Assume all surfaces to be smooth. If ‘N’ represents magnitude of normal reaction between block and wedge then acceleration of ‘M’ along horizontal equals: N sin  along + ve-x-axis M N sin  (C) along – ve x-axis M (A) Ncos  along-ve x-axis M N sin  (D) along –ve x-axis m+M (B) Newton's Laws of Motion 2. A cylinder rests in a supporting carriage as shown. The side AB of carriage makes an angle 30 o with the horizontal and side BC is vertical. The carriage lies on a fixed horizontal surface and is being pulled towards left with an horizontal acceleration 'a'. The magnitude of normal reactions exerted by sides AB and BC of carriage on the cylinder be NAB and NBC respectively. Neglect friction everywhere. Then as the magnitude of acceleration 'a ' of the carriage is increased, pick up the correct statement: (A) NAB increases and NBC decreases. (C) NAB remains constant and NBC increases. 3. Two blocks ‘A’ and ‘B’ each of mass ‘m’ are placed on a smooth horizontal surface. Two horizontal forces F and 2F are applied on the two blocks ‘A’ and ‘B’ respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is : (A) F 4. (B) Both NAB and NBC increase. (D) NAB increases and NBC remains constant. (B) F/2 (C) F (D) 3F 3 A flexible chain of weight W hangs between two fixed points A and B at the same level. The inclination of the chain with the horizontal at the two points of support is . What is the tension of the chain at the endpoint : (A) 5. W cosec  2 W sec  2 (C) W cos  (D) W sin  3 Two masses m and M are attached with strings as shown. For the system to be in equilibrium we have (A) tan  = 1 + 6. (B) 2M m (B) tan  = 1 + 2m M (C) tan  = 1 + M 2m (D) tan  = 1 + m 2M Three blocks A, B and C are suspended as shown in the figure. Mass of each block A and C is m. If system is in equilibrium and mass of B is M, then : /////////////////////////////////// A (A) M = 2 m (B) M < 2 m B C (C) M > 2 m (D) M = m Newton's Laws of Motion 7. ///////////////////// In the arrangement shown in figure, pulleys are massless and frictionless and threads are light and inextensible. Block of mass m1 will remain at rest if 1 1 1 (A) = + m1 m2 m3 4 1 1 (B) = + m1 m2 m3 (C) m1 = m2 + m3 1 2 3 (D) = + : m3 m2 m1 A B m1 8. 9. m2 m3 A uniform rope of length L and mass M is placed on a smooth fixed wedge as shown. Both ends of rope are at same horizontal level. The rope is initially released from rest, then the magnitude of initial acceleration of rope is (A) Zero (B) M( cos – cos ) g (C) M( tan  – tan )g (D) None of these Block B moves to the right with a constant velocity v0. The velocity of block A relative to B is : v0 A B ///////////////////////////////////////// v (A) 0 , towards left 2 10. 3u 4 (C) 3v 0 , towards left 2 (D) 3v 0 , towards right 2 (B) 3u 2 (C) u 4 (D) none of these In the figure shown, find out the value of  at this instant [assume string to be tight] (A) tan−1 12. v0 , towards right 2 System is shown in the figure and man is pulling the rope from both sides with constant speed ' u'. Then the speed of the block will be (M moves vertical): (A) 11. (B) 3 4 (B) tan−1 4 3 (C) tan−1 3 8 (D) none of these A hinged construction consists of three rhombus with the ratio of sides (5 : 3 : 2). Vertex A3 moves in the horizontal direction with velocity V. Velocity of A2 will be : A1 (A) 2.5 V (B) 1.5V A2 A3 (C) (2/3)V V (D) 0.8V Newton's Laws of Motion 13. A balloon of gross weight w newton is falling vertically downward with a constant acceleration a( m2. System is held at rest by thread BC. Just after the thread BC is burnt : (A) acceleration of m2 will be upwards  m − m2  (B) magnitude of acceleration of both blocks will be equal to  1 g  m1 + m2  (C) acceleration of m1 will be equal to zero spring k m1 A (D) magnitude of acceleration of two blocks will be non-zero and unequal. B m2 //////////// C 2. A particle is resting on a smooth horizontal floor. At t = 0, a horizontal force starts acting on it. Magnitude of the force increases with time according to law F = .t, where  is a constant. For the figure shown which of the following statements is/are correct ? (A) Curve 1 shows acceleration against time (B) Curve 2 shows velocity against time (C) Curve 2 shows velocity against acceleration (D) none of these 3. A light string is wrapped round a cylindrical log of wood which is placed on a horizontal surface with it's axis vertical and it is pulled with a constant force F as shown in the figure.(Friction is absent everywhere) (A) tension T in the string increases with increase in  (B) tension T in the string decreases with increase in  (C) tension T > F if  > /3 (D) tension T > F if  > /4 4. Two blocks A and B of mass 10 kg and 40 kg are connected by an ideal string as shown in the figure. Neglect the masses of the pulleys and effect of friction. (g = 10 m/s2) B A Fixed 45° 5 (A) The acceleration of block A is ms–2 45° 5 (B) The acceleration of block B is 2 2 2 (C) The tension in the string is 125 N (D) The tension in the string is 2 5. Two blocks of masses 10 kg and 20 kg are connected by a light spring as shown. A force of 200 N acts on the 20 kg mass as shown. At a certain instant the acceleration of 10 kg mass is 12 ms–2 towards right direction. (A) At that instant the 20 kg mass has an acceleration of 12 ms–2. (B) At that instant the 20 kg mass has an acceleration of 4 ms–2. (C) The stretching force in the spring is 120 N. (D) None of these 150 2 N ms–2 Newton's Laws of Motion 6. In the figure shown all the surface are smooth. All the blocks A, B and C are movable X-axis is horizontal and y-axis vertical as shown. Just after the system is relased from the position as shown. (A) Acceleration of 'A' relative to ground is in negative y-direction (B) Acceleration of 'A' relative to B is in positive x-direction (C) The horizontal acceleration of 'B' relative to ground is in negative x-direction. (D) The acceleration of 'B' relative to ground directed along the inclined surface of 'C' is greater than g sin . y A B  x C  Horizontal Surface 7. A particle stays at rest as seen from a frame. We can conclude that (A) the frame is inertial. (B) resultant force on the particle is zero. (C) if the frame is inertial then the resultant force on the particle is zero. (D) if the frame is noninertial then there is a nonzero resultant force. 8. A trolley of mass 8 kg is standing on a frictionless surface 37° F inside which an object of mass 2 kg is suspended. A constant force F starts acting on the trolley as a result of which the string 2kg stood at an angle of 37° from the vertical (bob at rest relative to trolley) Then : (A) acceleration of the trolley is 40/3 m/sec2. (B) force applied is 60 N (C) force applied is 75 N (D) tension in the string is 25 N 8kg 9. A reference frame attached to the earth (A) is an inertial frame by definition (B) cannot be an inertial frame because the earth is revolving around the sun. (C) is an inertial frame because Newton’s laws are applicable in this frame. (D) cannot be an inertial frame because the earth is rotating about its axis. 10. A particle is found to be at rest when seen from a frame S 1 and moving with a constant velocity when seen from another frame S2. Markout the possible options. (A) Both the frames are inertial (B) Both the frames are noninertial. (C) S1 is inertial and S2 is noninertial. (D) S1 is noninertial and S2 is inertial. 11. Figure shows the displacement of a particle going along the X-axis as a function of time. Find the region where force acting on the particle is zero (A) AB 12. (B) BC (C) CD In the Figure, the pulley P moves to the right with a constant speed u. The downward speed of A is vA, and the speed of B to the right is vB. (A) vB = vA (B) vB = u + vA (C) vB + u = vA (D) the two blocks have accelerations of the same magnitude (D) DE Newton's Laws of Motion 13. In an imaginary atmosphere, the air exerts a small force F on any particle in the direction of the particle’s motion. A particle of mass m projected upward takes time t 1 in reaching the maximum height and t2 in the return journey to the original point. Then : (A) t1 < t2 (B) t1 > t2 (C) t1 = t2 (D) t1 = t2 g +F/m g −F/m PART - IV : COMPREHENSION Comprehension-1 Figure shows a weighing machine kept in a lift. Lift is moving upwards with acceleration of 5 m/s2. A block is kept on the weighing machine. Upper surface of block is attached with a spring balance. Reading shown by weighing machine and spring balance is 15 kg and 45 kg respectively. Answer the following questions. Asume that the weighing machine can measure weight by having negligible deformation due to block, while the spring balance requires larger expansion :(take g = 10 m/s2) M Weighing Machine 1. Mass of the object in kg and the normal force acting on the block due to weighing machine are : (A) 60 kg, 450 N (B) 40 kg, 150 N (C) 80 kg, 400 N (D) 10 kg, zero 2. If lift is stopped and equilibrium is reached. Reading of weighing machine and spring balance will be : (A) 40 kg, zero (B) 10 kg, 20 kg (C) 20 kg, 10 kg (D) zero, 40 kg 3. Find the acceleration of the lift such that the weighing machine shows true weight of block. (A) 45 m/s2 4 (B) 85 m/s2 4 (C) 22 m/s2 4 (D) 60 m/s2 4 Comprehension-2 In the figure shown all the surfaces are smooth. The blocks A, B and C have the same mass m. F is floor and W is the wall. 4. Find the magnitude of the contact force (in newton) between wall and block C.? mg sin2 2mgcos  mg sin2 2mgcos  (A) (B) (C) (D) 2 2 2 1 + sin  1 + cos  1 + sin  1 + cos2  5. If wall removed then find acceleration block B ? 2gsin  (A) (B) 2gsin (C) 1 + 3 sin2  6. gsin  1 + 3 sin  2 (D) 2gsin  1 + sin2  A horizontal force ‘F’ applied on block C such that block B only move in downword direction then minimum value of F. (A) mgsin (B) mgcos (C) mgtan (D) mgcot Newton's Laws of Motion Comprehension-3 An object of mass m is suspended in equilibrium using a string of length constant K (< 2 mg/ ) and unstreched length /2. 7. 8. 9. Find the tension in the string in newton ? k (A) mg − (B) mg–k 2 (C) 2 mg − k 2 and a spring having spring (D) 2mg – k Find the acceleration of block just after cut the string ? k k k (A) 2g − (B) g − (C) 2g − 2m 2m m (D) g − k m What happens if K > 2 mg/ ? (A) at equilibrium tension in string is negative (B) at equilibrium position change in length of spring is greater then 2 (C) at equilibrium tension in string is zero. (D) If we cut the sting, block will accelerate in upword direction.  Marked Questions can be used as Revision Questions. PART - I : ADVANCED) /PROBLEMS (PREVIOUS YEARS) 1. System shown in figure is in equilibrium and at rest. The spring and string are massless. Now the string is cut. The acceleration of mass 2m and m just after the string is cut will be : (A) g/2 upwards, g downwards (C) g upwards, 2g downwards 2. (B) g upwards, g/ 2 downwards (D) 2g upwards, g downwards Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance ‘a’ from the centre P (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x, is (A) F a 2m a2 − x 2 (B) F x 2m a2 − x 2 (C) F x 2m a F m a F a −x 2m x 2 (D) m P a 2 Newton's Laws of Motion 3. A piece of wire is bent in the shape of a parabola y = kx2 (y-axis vertical) with a bead of mass m on it. The bead can slide on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is –1] (A) a gk (B) a 2gk (C) 2a gk PART - II : MAIN / PROBLEMS (D) a 4gk (PREVIOUS YEARS) 1. A ball of mass 0.2 kg is thrown vertically upwards by applying a constant force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2m height further, find the magnitude of the force. Consider g = 10 m/s2. 1] (1) 20 N (2) 22 N (3) 4 N (4) 16 N 2. A block of mass m is connected to another block of mass M by a string (massless). The blocks are kept on a smooth horizontal plane. Initially the blocks are at rest. Then a constant force F starts acting on the block of mass M to pull it. Find the force on the block of mass m (1) 3. mF m (2) (M + m)F m mF (m + M) (4) MF (m + M) Two fixed frictionless inclined planes making an angle 30º and 60º with the vertical are shown in the figure. Two blocks A and B are placed on the two planes respectively. What is the relative vertical acceleration of A with respect to B? (1) 4.9 ms–2 in horizontal direction (3) Zero 4. (3) (2) 9.8 ms–2 in vertical direction (4) 4.9 ms–2 in vertical direction A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F(t) = F 0e–bt in the x direction. Its speed v(t) is depicted by which of the following curves ? (1) (2) (3) (4) Newton's Laws of Motion EXERCISE–1 PART - I Section (A) A-1. A-2. A-3. Gravitational, Electromagnetic, Nuclear. 20 N No (c) (d) T2 A-4. T1 (e) (f) (i) m1 m 2g m 1g Vertical wall does not exert force on sphere (N' = 0). A-5. T1 m2 (N = 0). (g) (ii) Gravitational between earth and block . A-6. (h) Normal reaction A-7. (a) (i) Section (B) : (b) B-1. 2 mg B-2. (i) (ii) N = F B-3. (i) zero (ii) mg (iii) F (iv) m1g, m2g. Newton's Laws of Motion B-4. 2 D-3. u tan  ˆ  2 u   i  , x + y2 =   u tan .  ( −ˆj) + 2 2   2 D-4. VB = 3 VA = 1.8 m/s in downward direction. D-5. aA = D-6. VP1 = 5 m/s downward (a) (b) equal magnitude w . 9g 12g , aB = 25 25 VC = 25 m/s upward B-5. NA = B-6. N45 = 1000 N30 = 3 N, NB = 50 2 3 −1 100 3 −1 500 N 3 D-7. Section (E) : = 96.59 N ; = 136.6 N Section (C) : C-1. (a) 10 N, (b) 10 N, (c) 10 N. C-2. (a) 10 N (b) 15 N (c) 20 N C-3. T1 = 5N, T2 = 2N C-4. (a) C-5. 2g = 6.7 m (b) 40 N 3 (c) 80 N E-1. (a) | F1 | = | F2 | = 30 2 N (b) W = 30 2 N E-2. |F| = (30)2 + (108)2 = 112.08 N E-3. m1g 2(m1 + m2 ) E-4. (a) 4.8N, 3.6N, 2.4N,1.2N (b) F = 6 N(c) 0.2 N E-5. 6.5 m/s E-6. (a) 2g/3, E-7. (a) mB = 10 kg (b) mB = 10 kg (b) 3g/4 Section (F) : Mg (b) T1 = T2 = T3 = , 2 (a) b î + b ĵ F-1. (i) 600 N, (iv) 720 N, (vii) 720 N, (ii) 600 N, (v) 480 N, (viii) 480 N (iii) 600 N, (vi) 720 N, F-2. (i) 100 N, (iv) 120 N, (vii) 120 N, (ii) 100 N, (v) 80 N, (viii) 80 N. (iii) 100 N, (vi) 120 N, F-3. (a) 3 g 0, 0, (b) 0, g  g  Section (G) : G-1. 15 m/s2, opposite direction. 4 G-2. a = 3 m/s2 G-4. 322 N 4F – g. M+m G-3. a= H-2. (g + a) sin  Section (H) : T5 = Mg and T4 = 3Mg 2 (c) F = Mg 2 H-1. F=0 PART - II Section (A) : Section (D) : D-1. cos 2 cos 1 A-1. D-2. 2u (C) A-2. (B) B-2. (C) Section (B) : B-1. (B) B-3. (B) Newton's Laws of Motion Section (C) : 1. (AC) C-1. (B) C-2. (D) C-3. (C) C-4. (C) C-5. (C) C-6. (A) 4. 7. 10. 13. Section (D) : D-1. (B) D-2. (C) D-3. (D) D-4. D-7. (B) (D) D-5. D-8. (B) (A) D-6. (C) Section (E) : 2. (ABC) 3. (AC) (ABD) 5. (CD) 8. (BC) (CD) 6. 9. (ABCD) (BD) (AB) (BD) (BD) 12. (BD) 11. PART - IV 1. 4. (B) (B) 2. 5. (D) (A) 3. 6. (A) (D) 7. (A) 8. (B) 9. (C) E-1. (D) E-2. (C) E-3. (B) E-4. E-7. (C) (D) E-5. E-8. (C) (C) E-6. E-9. (C) (B) EXERCISE–3 E-10. (D) E-11. (C) E-12. (B) PART - I 1. (A) 2. Section (F) : F-1. F-4. (A) (D) F-2. F-5. (D) (B) F-3. G-1. (B) G-2. (A) G-3. (C) H-2. (C) H-3. (C) 1. 4. Section (H) : H-1. (C) PART - III 1. 2. (A) – q ; (B) – r ; (C) – q ; (D) – r (A) – p,s ; (B) – q, r ; (C) – p,s ; (D) – q, r 3. (A) – r ; (B) – p ; (C) – s ; (D) – q EXERCISE–2 PART - I 1. 4. 7. 10. 13. 16. 19. 22. (C) (A) (B) (A) (C) (B) (A) (C) 2. 5. 8. 11. 14. 17. 20. (C) (A) (A) (A) (A) (B) (D) 3. 6. 9. 12. 15. 18. 21. (D) (B) (B) (D) (C) (C) (D) PART - II 1. 5 2. 6 3. 3 4. 4 5. 1 6. 17 7. 10. 60 6 8. 11. 32 1 9. 12. 23 2 13. 40 PART - III 3. (B) 3. (4) PART - II (C) Section (G) : (B) (2) (3) 2. (3)