# 4-Laws of motion-4-Done.doc

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### 4-Laws of motion-4-Done.doc

• 1. LAWS OF MOTION SYLLABUS Force and inertia : Newton's Laws of Motion, Conservation of linear momentum and its applications rocket propulsion, Laws of friction. FORCE A force is a push or pull acting on a body. It is a vector quantity i.e. it has both magnitude and direction. Unit of Force : Its unit is Newton in SI system and dyne in CGS system. Dimensions : MLT 2 (a) Contact Forces: Tension, Normal Reaction, Friction etc. Forces that act between bodies in contact. (b) Field forces (non-contact forces): Weight, electrostatic forces, etc. Forces that act between bodies separated by a distance without any actual contact. CONTACT FORCES (i) Tension (T) When a string, thread, wire or a spring is held taut, the ends of the string or thread (or wire) pull on whatever bodies are attached to them in the direction of the string. This force is known as Tension. A B O T T T T m R T (ii)Normal Reaction When two surfaces are in contact, then the surfaces exert forces on each other. The forces are opposite to each other in direction and are equal to each other. The direction is always at right angles to the surfaces in contact. N1 N1 N2 N2 (iii) Friction Friction is a force that acts between bodies in contact with each other, A B NEWTON’S FIRST LAW Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by an external force. INERTIA Inertia is the property of a body by virtue of which a body retains its state: either of rest or uniform motion along a straight line.
• 2. MOMENTUM It is equal to the product of the mass and the velocity of the body. Its direction is same as the velocity of the body. v m p    where p  = momentum of the body. m = mass of the body. v  = velocity of the body. Unit : Its unit is kg-m/s in SI system and gm-cm/s in CGS system. Dimensions : Its dimensions is MLT-1 NEWTON'S SECOND LAW OF MOTION Newton’s second law states that the rate of change of momentum of a body is directly proportional to the applied force. If a body of mass m is moving with a velocity v  , then its momentum is, v m p    . . . (i) According to Newton's 2nd law, if a net force F  acts on the body, dt p d F    . . . (ii) dt p d k F    . . .. (iii) Where k is the constant of proportionality. Unit force is now defined as that force which produces a unit rate of change of momentum in a body i.e. F  = 1 unit if dt dp = 1 unit. This makes k = 1. In this system of units, dt p d F    . . . (iv) If the mass of the body m is constant, then ) v m ( dt d F    = m dt v d  = ma  . . .(v) NEWTON’S THIRD LAW Every action has an equal and opposite reaction Example 1: A block of mass m = 10 kg is pulled by a force F = 100 N at an angle  = 30o with the horizontal along a smooth horizontal surface. What is the acceleration of the block? (g = 10 m/s2)  F m Solution: The forces that act on the body can be decomposed along x and y axis. As there is no acceleration along y-axis, net force acting along vertical or y axis should be zero i.e. F sin  F cos  N mg y x FY = N + Fsin – mg = 0 . . . (1) The body accelerates along x-axis. Therefore Fx = Fcos = ma . . . (2)
• 3. where the acceleration along x-axis is a.  F Cos  = ma  a = m cos F  = 2 o s / m 3 5 10 2 3 100 10 30 cos 100     The acceleration of the block is 2 s / m 3 5 Directed towards right. Since F sin < mg & the surface is rigid, the block remains in equilibrium along y-axis. Example 2: In the figure, the tension in the horizontal cord is 30N. Find the weight of the body B. B cord 1 cord 2 30 N P 45 Solution: (i) Isolate the point P (ii) The forces acting on it are : Unknown tension T2 in cord 2 Unknown tension T1 in cord 1 Known tension of 30N in the horizontal cord. (iii)T2 is resolved along x and y axes. (iv)Condition of equilibrium : Fx = 0  30 – T2sin45 = 0 (1) and Fy = 0  T2cos45 - T1 = 0 (2) For body B Since the body B is also in equilibrium, Hence T1 = W (3) (v) After solving these equations, we get W = 30N T2 T1 = W X Y 30 N 45 W T1 Example 3: A block of mass m1 and weight w1 = m1g moves on a level frictionless surface. It is connected by a light flexible cord passing over a small frictionless pulley to a second hanging block of mass m2 and weight w2 = m2g. What is the acceleration of the system and what is the tension in the string? m2 m1 w1=m1g w2=m2g T T N Solution: The diagram shows the forces acting on each block. The forces exerted on the blocks by the cord can be considered an action reaction pair, fx = T = m1ax fy = N – mg = m1ay = 0 Applying Newton’s second law to the hanging block, we obtain fy = m2g – T = m2ay T = m1a . . . (1) m2g – T = m2a . . . (2) Adding two equations eliminates T giving
• 4. m2g = (m1 + m2)a a = 2 1 2 m m m  g Substituting the value of a in equation (1) gives the tension as T = g m m m m 2 1 2 1  Example 4: Two blocks of masses m1 = 2 kg and m2 = 1 kg are in contact on a smooth horizontal surface as shown in the figure. A horizontal force F = 3N is applied on the block m1. Find the contact force between blocks. m1 m2 F Solution: Let contact force between the blocks be N. Since N is responsible for the acceleration of m2 hence it will be in the direction of acceleration of m2 and on m1it will be opposite to the acceleration Equations of motion F N = m1a . . . (1) N = m2a . . . (2) Where a = acceleration of the blocks. N1 F N m1g N2 N m2g From (1) and (2) , we get a = 2 1 m m F   N = 2 1 2 m m F m  = 1N. SPRING FORCE The extension or compression produced in a spring is directly proportional to the force applied on it. If x is the deformation in the spring then F  x or F = – k x where k is constant and is called the spring constant or force constant. The negative sign shows that the force and the extension or compression are opposite in their directions. FRICTIONAL FORCE Frictional force comes into play between two surfaces whenever there is relative motion or a tendency of relative motion between two surfaces in contact. Frictional force has the tendency to oppose relative motion between the surfaces in contact. Thus, friction can be classified as (a) Static Friction: which acts between surfaces in contact but not in relative motion, it opposes the tendency of relative motion (b) Kinetic Friction: which acts between surfaces in contact which are in relative motion, it opposes the relative motion between the surfaces. F f 45 fk f f (0, 0) F
• 5. LAWS OF STATIC FRICTION Static Friction, acting between the surfaces in contact, (not in relative motion) opposes the tendency of relative motion between the surfaces. It is independent of the area of contacting surfaces. Now, fs(max) N where f = fs(max) fs(max) = sN Here s = co-efficient of static friction. N = normal reaction on the block from the surface. 0  fs  N When F exceeds f block starts moving and frictional force decreases to a constant value fk. fk is called kinetic friction and it has unique value which is given by fk = k N Here k = co-efficient of kinetic friction. N = normal reaction. ANGLE OF FRICTION The angle made by the resultant reaction force with the vertical (normal reaction) is known as the angle of the friction. Now, in the triangle OAB,      tan AB OB cot OB AB or tan = N f  N f O A B ANGLE OF REPOSE Consider a body of mass m resting on an inclined plane of inclination . The forces acting on the body are shown – Ff being the force of friction. If friction is large enough, the body will not slide down. mg  f N y x Along x: mg sin  – f = 0 …(1) Along y: N –mg cos = 0 …(2) i.e. N = mg cos  and f = mg sin  Thus, f  SN gives, mg sin   S mg cos  or, tan   S, the coefficient of static friction between the two surfaces, in order that the body doesn’t slide down. When  is increased so that tan  > , then sliding begins, and the angle r = tan-1 , where sliding begins is known as the angle of repose. Example 5: A block weighing 2kg rests on a horizontal surface. The coefficient of static friction between the block and surface is 0.40 and kinetic friction is 0.20.
• 6. (a) How large is the friction force acting on the block ? (b) How large will the friction force be if a horizontal force of 5N is applied on the block? (c) What is the minimum force that will start the block in motion? Solution: (a) As the block rests on the horizontal surface and no other force parallel to the surface is on the block, the friction force is zero. (b) With the applied force parallel to the surfaces in contact 5N, opposing friction becomes equal and opposite. Further the limiting friction is sN = sMg = 8N  Force of friction is 5N. (c) The minimum force that can start motion is the limiting one. sN = smg = 8N Example 6: In the figure suppose that the block weighs 2kg, that the friction T can be increased to 8N before the block starts to slide, and that a force of 4N will keep the block moving at constant speed once it has been set in motion. Find the co- efficient of static and kinetic friction. N 2kg T Solution: There is no acceleration of the body along y-axis. Hence fy = N – W = N – 20N = 0  N =20N fX = T – fs = 8N – fs = 0  fs =8N Hence, we have fs = sN As the motion takes place at constant velocity and there is no motion along y-axis. fy = N – W = 0  N = mg =20N There is no acceleration along x-axis  fX = T – fk = 4N – fk = 0  fs =8N fk = kN Hence, 20 . 0 N 20 N 4 N fk   Example 7: Block A in the figure weighs 0.4 kg and block B weighs 0.5kg. The coefficient of sliding friction between all surfaces is 0.25 (a) find the force F necessary to drag block B to the left at constant speed. (b) find the tension in the string. A B F Solution: For A as there is no motion along horizontal or vertical. Let m1 be mass of A g m N 1 1  = 0  N1 = m1g and T – kN1 = 0  T – km1g = 0  T =km1g= 0.25  0.410 = 1N T km1g N1 m1g
• 7. For B as there is no motion along vertical. Let m2 be mass of B N2 – N1 = m2g  N2 = (m1+m2)g As there is no acceleration along x-axis F – f1 – f2 = 0  F = f1 + f2 = kN1 + kN2 = kq (2m1+m2) = 0.25 (0.8 + 0.5)10 = 3.25 N N1 f2 N m2g F N2 f1 FRAME OF REFRENCE It is a conveniently chosen co-ordinate system, which is used to describe the position and motion of a body. Inertial Frame of Reference Any frame of reference in which Newton’s first laws are valid is an inertial frame (IFR). Example 8: A block of mass m is placed on an inclined plane. With what acceleration a, towards right should the system move on a horizontal surface so that m does not slide on the surface of inclined plane? Assume all surfaces are smooth. Solution : the normal reactions and a pseudo force of magnitude ma towards left. Rcos = mg  a = gtan R sin = ma R mg ma R cos R sin ma mg Example 9: A pendulum of mass m is hanging from the ceiling of a car having an acceleration ao with respect to the road in the direction shown. Find the angle made by the string with the vertical. a0  Solution: Since bob of the pendulum is stationary relative to car Hence T sin = mao (pseudo force) (i) T cos = mg (ii) Dividing (i) by (ii), we get tan = g ao   = tan-1 g ao mao mg Tsin Tcos T  CENTRIPETAL FORCE & CENTRIFUGAL FORCE If a body is moving with a constant speed in a circle, as seen from an inertial frame it is continuously accelerated towards the centre of rotation. The magnitude of the centripetal acceleration for a body moving with a tangential velocity v is given by v2/r. If the angular velocity of the body is  then the centripetal acceleration is mr2. According to Newton’s law force causes acceleration and so, the net centripetal force
• 8. F = ma = r mv2 = m2 r a  = a  T + a  R  ma  = ma  T + ma  R Where T a  is tangential and, R a  is centripetal are radial acceleration and a  is total acceleration Example 10: Find the ratio of radius of curvature at the highest point of projectile to that just after its projection if the angle of projection is 300. Solution : If 0 v  is the initial velocity vp = v0 cos  Normal acceleration at O = g cos  Normal acceleration at P = g Hence if r0 and rp be radii of curvatures at O & P respectively. r0 =  cos g v2 0 , rp = g cos v 2 2 0  Hence the required ratio = 0 p r r cos3  = 8 3 3 . O  P v0  vP  (an)p x y Example 11: A 1200 kg automobile rounds a level curve of radius 200m, on an unbanked road with a velocity of 72 km/hr. What is the minimum coefficient of friction between the tyres and road in order that the automobile may not skid (g= 10m/s2)? Solution: In an unbanked road the centripetal force is provided by the frictional force.  ffriction = r mv2 But flimiting friction  ffriction or mg  ffriction or mg  r mv2  min = 2 . 0 200 10 20 20 gr v2     Rocket 1. Acceleration of rocket 0 M u t a M M t t d × d = d æ ö - × ç ÷ è ø d 2. Thrust on rocket M F u t d = d 3. Impulse of force av F t p I = ´ = D
• 9. 4. Recoil velocity of the gun b b g g m v v m = OBJECTIVE 1. A reference frame attached to the earth is: (A) an inertial frame sometimes (B) an inertial frame always (C) a non-inertial frame (D) may be inertial or non-inertial. Ans. (C) Solution: As the earth itself is rotating on its axis the frame attached to it should be a non-inertial one. 2. A vehicle is moving on a rough road in a straight line with uniform velocity. (A) No force is acting on vehicle (B) A force must act on the vehicle (C) An acceleration is being produced in the vehicle (D) An centriprital acceleration is being produced in the vehicle Ans. (A) Solution: As the vehicle is moving with uniform velocity, net force acting on it should be zero. 3: Two identical particles A and B, each of mass m, are interconnected by a spring of stiffness k. If the particle B experiences a force F & the elongation of the relative spring is x, the relative acceleration between the particles is equal to: (A) F/2m (B) F - kx m (C) F - 2kx m (D) kx m Ans. (C) Solution : Equation of motion for A : T = ma  a = m kx m T  for B : F – T = ma  a = m kx F  A B F m T T m  The relative acceleration ar = | a  a | = m kx 2 F  . 4. A 60 kg man stands on a spring balance in a lift. At same instant he finds that the reading on the scale has changed from 60kg to 50 kg for a while and then comes back to the original mark. What is his conclusion? (A) The lift was in constant motion upwards (B) The lift was in constant motion downwards (C) The lift while in motion downward suddenly stopped. (D) The lift while in motion upward suddenly stopped. Solution: When lift accelerates up W = W          g a 1
• 10. When lift decelerates up W = W          g a 1 Here weight decreases i.e. lift decelerates up. Initially lift was moving with constant velocity up and then it suddenly stopped i.e. decelerated, 5. A body is accelerated by applying a force of 30N. The change in the momentum of the body after 2sec is: (A) 7.5 kg-m/s (B) 30 kg-m/s (C) 120 kg-m/s (D) 60 kg-m/s Solution: P = Ft = 302 = 60 kg m/s Hence D 6. Two blocks of masses 2kg and 5kg are connected by a light string passing over a frictionless pulley. The tension in the cord connecting the masses will be (A) 20N (B) 30N (C) 28 N (D) 50 N Solution: T = g m m m m 2 2 1 2 1  8 . 9 2 5 2 5 2      = 28N Hence C 7. A vessel containing water moves with constant acceleration towards the right, along a horizontal straight line. Which of the following figures: a, b and c represents the surface of the liquid. None of the above (A) (B) (C) (D) Solution: When the vessel accelerates forward, the pseudo force acts on water backward and the resultant of the weight of water and the pseudo force, acts on the left downward. The surface of water is perpendicular to this resultant. Hence C 8: A bob hanging from the ceiling of the car acts as an accelerometer. Then the relation expressing horizontal acceleration a of the car and the angle  made by bob with the vertical is (A) a = gtan (B) a = gsin (C) a = gcot (D) None Hence A Solution: tan = Mg Ma  a = gtan   Ma a Mg
• 11. 9. A weight W is attached to two weightless strings AB and AC as shown in figure. The tension in each string will be: (A) W 4 (B) W 2 (C) 3 W 2 (D) W T T C B 120 W A Solution: For vertical equilibrium 2T cos60 = W T =  60 cos 2 W = W Hence D 10. A body of mass 2kg is acted upon by two forces each of magnitude 1N, making an angle 60 with each other. The net acceleration of the body in m/s2 is (A) 0.1 (B) 1.0 (C) 3 2 (D) 2 3 Solution: Net force F =    60 cos F F 2 F F 2 1 2 2 2 1 = N 3 2 1 1 2 1 1 2 2      a = 2 s / m 2 3 2 3 M F   Hence C 11. A scooter of mass 120kg is moving with a uniform velocity of 108km/hr. The force required to stop the scooter in 10s is (A) 180N (B) 208N (C) 360 N (D) 720 N Solution: F = t mv t mv 0 t P        =   10 18 5 108 kg 120        = 360N Ans. (C) 12. Two blocks A and B of masse 4kg and 12 kg are placed on a smooth horizontal plane a force F of 16N is applied on A, as shown, then the contact force between the blocks is (A) 4N (B) 8N (C) 12N (D) 16N F = 16N A B Solution: Acceleration
• 12. a = 12 4 16 M M F 2 1    = 1m/s2 P = 12a (for block B)  F = 16 – M1a = 16 – 4  1 = 12N Hence C 13. N bullets, each of mass m, are fired with a velocity v at the rate of n bullets/sec upon a wall. The bullets are stopped by the wall. The reaction offered by the wall to the bullets is (A) Nmv n (B) nNmv (C) Nv n m (D) nmv Solution: F = t Δ P Δ Moment transferred to the wall, P = mv, t = n 1  F = nmv A 1 mv  Hence D 14. A knife edge of mass M is dropped from a height ‘h’ on a wooden floor with its tip pointing of downward. If the blade penetrates a distance ‘s’ into the wood, the average resistance offered by the wood to the blade is (A) Mg (B) Mg h 1+ s       (C) Mg h 1- s       (D) Mg 2 h 1+ s       Solution: Velocity of knife edge when it hits the floor v = gh 2 Retardation, a = s gh s 2 gh 2 s 2 v2    Equation R – Mg = Ma R = Mg + Ma = Mg + M         s h 1 Mg s gh Hence B R Mg a 15. A rocket, set for vertical launching, has a mass of 50 kg and contains 450 kg of fuel. It can have a maximum exhaust speed of 2 km s–1. If g = 10 ms–2, what should be the minimum rate of fuel consumption to just lift it off the launching pad? (A) 2.5 kg s–1 (B) 5 kg s–1 (C) 7.5 kg s–1 (D) 10 kg s–1 Sol: (A) When the rocket is at the launching pad, M = M0 = 50 + 450 = 500 kg. Also u = 2kms–1 = 2000 ms–1. M F u t d = d
• 13. But F = Mg. Therefore M u Mg t d = d or 2 1 1 M 500 kg 10 ms 2.5 kg s t 2000 ms - - - d ´ = = d 16. In above Q. what should be the minimum rate of fuel consumption to give an initial acceleration of 20 ms–2 to the rocket ? (A) 2.5 kg s–1 (B) 5 kg s–1 (C) 7.5 kg s–1 (D) 10 kg s–1 Sol: (C) M F u Mg t d = - d But F = Ma, where a is the initial acceleration. Thus M Ma u Mg t d = - d or 2 1 M M(a g) 500 kg (20 10) ms t u 2000 ms - - d + ´ + = = d = 7.5 kg s–1 17. A block can slide on a smooth inclined plane of inclination  kept on the floor of a lift. When the lift is descending with a retardation a. the acceleration of the block relative to the incline is: (A) (g + a) sin  (B) (g – a) (C) g sin  (D) (g – a) sin  Solution: N mg ma  m(g a)cos   m(g a)sin   net accel. = m(g a)sin (g a)sin m      Ans. (A) 18. A force of 5 N acts on a body of weight of 10 N. What is the acceleration in m/s2 (A) 50 (B) 5 (C) 0.5 (D) 2 Solution: 5N mass = 10 1 10  A = 2 5 5m/ s (10/10)  Ans. (B)
• 14. 19. A man of mass m stands on a crate of mass M. He pulls on a light rope passing over a smooth light pulley. The other end of the rope is attached to the crate. For the system to be is equilibrium, the force exerted by the man on the rope will be: (A) mg (B) Mg (C) 1 (M m)g 2  (D) (m + M)g Solution: T m M T 2T = (M + m)g T = (M m)g 2  Ans. (C) 20. A uniform rope of mass M and length L is pulled along a frictionless inclined plane of angle  by applying a force F (> Mg) parallel to incline. Tension in the chain at a distance x from the end at which force is applied is (A) x x F Mgsin L L   (B) x F 1- L       (C) x x F 1- Mgsin L L         (D) None of these Solution: Mgsin Mgcos  N F F – Mg sin = MA (1) F – M M x gsin T x A L L                F– m M F Mgsin xgsin T x L L M                  T = F x 1 N L        Ans. (D) 21. In the system shown, the acceleration of the wedge of mass 5M is (there is no friction anywhere) (A) zero (B) g/2 (C) g/3 (D) g/4 Solution: 2 1 2 1 m g (m m )g A m m m       M m 5M M 2M 600
• 15. (6 0.4(m 4))g A 0 m 6 4       60 m 4 0.4    44 m 11kg 4   Ans. (C) 22. What is the maximum value of the force F such that the block shown in the arrangement, does not move? (A) 20 N (B) 10 N (C) 12 N (D) 15 N Solution: 60 F 1 r F mg Fsin60 Fcos60 N N = F sin60 + Mg (1) F cos60 = 1 r F = N = (F sin60 + Mg) F = Mg cos60 sin60    = 1 3 1 5 4 2 3 20N 1 1 1 3 2 2 2 3        = 5 4 20N 1   Ans. (A) 23. A force vector F = 6 i  – 8 j  + 10 k  newton applied to a body accelerates it by 1 ms–2. What is the mass of the body? (A) 10 2 kg (B) 2 10 kg (C) 10 kg (D) 20 kg Solution: F = m a |F| = 10 2 10 2 = m × A = m × 1  m = 10 2 kg. Ans. (A) 24. 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 end point. (A) W cosec 2  (B) W sec 2  (C) W cos (D) W sin 3  Solution: 600 F m 3kg  1 2 3   A W B  
• 16. T F   T 2T sin = w  T = w w cosec 2sin 2    Ans. (A) 25. Two blocks A and B of masses m1 and m2 are placed in contact with each other on a horizontal platform. The coefficient of friction between the platform and the block A is 1 and that between block B is 2. The platform moves with a constant acceleration a as shown in fig. The force of interaction between the blocks exists if (A) 1 = 2 (B) 1  2 (C) 1 < 2 (D) none of these Solution: Form of indication will exhibit By 1 2 r r 1 2 F mg F mg m m    1 2 r r 1 2 F F m m   1 1 1 2 1 2 m g m g m m     1  2 Ans. (D) 26. For the system shown in the figure, the pulleys are light and frictionless. The tension in the string will be (A) 2 mgsin 3  (B) 3 mgsin 2  (C) 1 mgsin 2  (D) 2mgsin Solution: mg sin – T = mA (1) T = mA (2) (1) + (2)  mg sin = 2mA  A = gsin 2  T = mgsin 2  Ans. (C) 27. Two masses M and m are connected by the arrangement shown in figure. What is the downward acceleration of mass M. a m1 m2 2 r F 1 r F a 1 m 2 m m  m m M
• 17. (A) 2M m g 3M m         (B) 2M m 2g 4M m         (C) M m g M m         (D) M g 2M m        Solution: Mg – T = MA (1) 2T – mg = mA 2 (2) 2T m T A M 2 × (1) + (2) (2M – m) g = A(2M + m 2 ) A = 2(2M m)g 4M m   Ans. (B) 28. In a rocket, the mass of the fuel is 90% of the total mass. The rocket is blasted from the launching pad. If the exhaust gases are ejected at a speed of 2 km s–1, what is the maximum speed attained by the rocket ? (Neglect the effects of gravity and air resistance). (A) 2 ln (10) kms–1 (B) 2 ln (30) kms–1 (C) 2 ln (60) kms–1 (D) 2 ln (90) kms–1 Sol: (A) The mass of the container = 10% of the total mass of the rocket, 90% being the mass of the fuel. Hence M0/Mc = 10. The rocket will attain the maximum speed vb when all its fuel is burnt. This speed is given by 1 1 0 b c M v u ln 2 kms ln(10) 2ln(10)kms M - - æ ö = = = ç ÷ è ø . 29. Blocks A and C start from rest and move to the right with acceleration aA = 12t m/s2 and aC = 3 m/s2. Here t is in seconds. The time when block B again comes to rest is (A) 2 s (B) 1 s (C) 3/2 s (D) ½ s Solution: B A C
• 18. 12t 3 A 12t 3 A = 12t 3 2  And A = 0, t = 3 12 = 1 4 Ans. (E) 30. At some instant, the 10 kg mass has acceleration of 12 m/s2, what is the acceleration of 20 kg mass (A) 2.5 m/s2 (B) 4.0 m/s2 (C) 3.6 m/s2 (D) 12 m/s2 Solution: 200N (1) (2) 10 20 1 r max F = 0.1 × 10 ×10 = 10 N 2 r max F = 20 × 0.1 × 10 = 20 N For 10 kg block Kx – 10 = 10A = 10 ×12 Kx = 120 + 10 = 130 N. For 20 kg block = 200 – 130 = –20 = 20A  A = 50 2.5m/ s 20  Ans. (A) 31. A block slides down an inclined plane of slope  with constant velocity. It is then projected up the plane with an initial speed v0. How far up the incline will it move before coming to rest (A) 2 0 v 4gsin (B) 2 0 v gsin (C) 2 0 v 2gsin (D) 2 0 v 2g Solution: mgsin  r F For constant velocity the net force acting on the block will be O. 10 kg 20 kg F=200 N 0.1 
• 19.  Fr = mg sin o V r F mgsin   mA = mg sin + mg sin = 2mg sin  A = 2g sin O = V0 2 – 2As s = 1 2 0 o V V 2A 2gsin   32. A force of 100N is applied on a block of mass 3kg as shown in the figure. The coefficient of friction between the wall and block is 0.6. The magnitude of the force exerted by the wall on the block is (A) 15N downwards (B) 25N upwards (C) 20N downwards (D) 30N upwards Solution: F cos30 = N.  N = 6 3 10.39  F sin 30 = 6N < sin So. Fr will act upward And Fr max = N = 0.6 × 10.39= 6.23 N. So force exerted by wall = 2 2 (6.23) (10.4)  = 15N down ward Ans. (A) 33. A force acts for 8 seconds on a body of mass 10 kg initially at rest. The force then stops acting and the body moves 80 m in next 5s. Calculate the force. (A) 10 N (B) 20 N (C) 30 N (D) 40 N Solution : After the force stops acting, the body moves with a uniform velocity given by v = taken time travelled distance = 5 80 = 16 m/s Initial velocity of the body, u = 0 From v = u + at 16 = 0 + at or a = 2 m/s2 As F = ma  F = 10  2 = 20N 12N 100N 30
• 20. Ans. (B) 34. A block is gently placed on a conveyor belt moving horizontally with constant speed. After t = 4s, the velocity of the block becomes equal to the velocity of the belt. If the coefficient of friction between the block and the belt is  = 0.2, then the velocity of the conveyor belt is (A) 2ms–1 (B) 4ms–1 (C) 64ms–1 (D) 8ms–1 Solution: 0 V Fr = mg = 0.2 mg. A = 0.2g = 2m/s Let. velocity of belt = V0 So, vel. of block with respect belt will be V0 U = V0, A = 2, t = 4, v = 0 0 = V0 – 2 × 4 V0 = 8 m/s Ans. (D) 35. A block of mass M is connected to a massless pulley and massless spring of stiffness k. The pulley is frictionless. The spring connecting the block and spring is massless. Initially the spring is untstretched when the block is released. When the spring is maximum stretched, then tension in the rope is (A) zero (B) Mg (C) 2Mg (D) Mg/2 Solution: m k by consternation of energy mgx = 1 kx 2 x = 2mg k T = kx = 2mg Ans. (c) 36. A balloon has a mass of 10 gram in air. The air escapes from the balloon at a uniform rate with a velocity of 5 cm/s and the balloon shrinks completely in 2.5s. Calculate the average force acting on the balloon. (A) 10 Dyne (B) 20 Dyne (C) 30 Dyne (D) 40 Dyne M K
• 21. Solution : Here, m = 10 gm, v = 5 cm/s 5 . 2 10 dt dm  = 4 gm/s.F = ? F =   mv dt d dt dp  = v dt dm       F = 4  5 = 20 dyne. Ans. (B) 37. The weights W1 and W2 are suspended from the ends of a light string passing over a smooth fixed pulley. If pulley is pulled up at an acceleration g, the tension in the string will be (A) 1 2 1 2 4W W W W  (B) 1 2 1 2 2W W W W  (C) 1 2 1 2 W W W W  (D)   1 2 1 2 W W 2 W W  Solution: 1 w 2 w g consider w2 and w1 with respect to bully 2w2 – T = 2 w A g       (1) T – 2w1 = 1 w A g (2) (1) × w1 – (2) × w2  T(– w1 – w2) = – 2gw1w2 T = 1 2 1 2 4w w w w  Ans. (A) 38. A block of mass m is placed on a smooth wedge of inclination . The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude (A) mgtan (B) mgsin (C) mgcos (D) mgsec Solution:  m A consider F. B. D of m with respect to m
• 22. mAcos N mA sin mgsin mgcos mg mA mg sin = mA cos  A = g tan and N = mgcos + mAsin = m(gcos + 2 gsin cos   ) = mg sin Ans. (D) 39. A cube is resting on an inclined plane. What must be the value of coefficient of friction between cube and plane so that cube topples before sliding (A) 1 2  (B)  < 1 (C)  > 1 (D) 1 2  Solution: The given condition on only take place when line of force of mg passes troughs  = 45° …(i) and mg sin  fr max =  mg cos    tan    1 (C) 40. A block is placed on a rough horizontal plane. A time dependent horizontal force F = K t acts on the block. Here K is a positive constant. Acceleration-time graph of the block is (A) a t (B) a t 10 kg 20 kg F=200 N 0.1 
• 23. (C) a t (D) a t Solution: a t F = kt Fnet = F – fr = 0 up to time Kt = mg = 1 r max F And for t > mg kt  Fnet = kt – mg Ans. (C) For Q. no. 41 to 44: A 20 kg box is dragged across a rough level floor (k = 0.3) by means of a rope which is pulled upward at an angle of 30° to the horizontal. The pulling force has a magnitude 80 N. 41. What is the normal force exerted by the floor on the block? (A) 160 N (B) 140 N (C) 180 N (D) 40 N 42. What is the frictional exerted by the floor on the block force? (A) 48 N (B) 50 N (C) 52 N (D) 60 N 43. What is the acceleration of the box? (A) 1.06 m/s2 (B) 2.06 m/s2 (C) 2.08 m/s2 (D) 3.09 m/s2 44. If the force is reduced until the acceleration becomes zero, what is the tension in the rope? (A) 54.42 N (B) 55.42 N (C) 58.42 N (D) 60.42 N Solution for Q. 41 to 44: Let the normal contact force be N (41)  F sin  + N = mg  N = mg – F sin  = 20  10 – 80 sin 300 = 160 N. Ans (A) (42) fk = k N = (0.3 ) (160) = 48 N. Ans (A) F  mg N m fk (43) a = (Fcos   fk)/m = ) 20 ( 48 3 40  = 1.06 m/s2. Ans (A)
• 24. (44) F' cos  = fk  F' = fk sec  = 48  (2 / 3) = (96/3) = 55.42 N. Ans (B) 45. An elevator accelerates upward at a constat rate. A uniform spring of length L and mass m supports a small block of mass M that hangs from the ceiling of the elevator. The tension at distance 1 from the ceiling is T. The acceleration of the elevator is (A) T g M m ml/L    (B) T g 2M m ml/L    (C) T g M ml   (D) T g 2M m ml/L    Solution: m system mass of system m (L ) M g L         m m T (L ) M g (L ) M A L L                   A = T g m (L ) M L    46. A particle of mass 70 g, moving at 50 cm/s, is acted upon by a variable force opposite to its direction of motion. The force F is shown as a function of time t. (A) Its speed will be –50 cm/s after the force stops acting. (B) Its direction of motion will reverse. (C) Its average acceleration will be 1 m/s2 during the interval in which the force acts. (D) Its average acceleration will be 10 m/s2 during the interval in which the force acts. Solution: Total area under curve = (10 – 2 + 4  10–3) × 10 2 = (10–2 + 0.4 × 10–2) × 5 – 2 1.4 10 5 10           = 0.07(V – 0.5) V = –0.5 m/s = – 50 cm/s Ans. (A) 10 F(N) t(s) O 4 x 10-3 10-2 8 x 10-3
• 25. 47. Two masses M and m(M>m) are joined by a light string passing over a smooth light pulley then what is the tension on block (A) The acceleration of each block is M m g. M m         (B) The tension in the string is 2Mmg . M m  (C) The centre of mass of the ‘M plus m’ system moves down with an acceleration of 2 M m g . M m         (D) The tension in the string by which the pulley is attached to the roof is (M + m)g. Solution: (M m) A g M m    m M And for m: Mg – T = MA T = Mg – MA = Mg – Mg (M m) M m   = Mg(2m) 2mMg M m (M m)    Ans. (A, B) 48. In the figure, the blocks A, B and C of mass m each have acceleration a1, a2 and a3 respectively. F1 and F2 are external forces of magnitudes 2mg and mg respectively. (A) a1 = a2 = a3 (B) a1 > a3 > a2 (C) a1 = a2, a2 > a3 (D) a1 > a2, a2 = a3 Solution: 2mg m T mg 2mg 2mg 2mg T m T m T mg mg I II III 2mg 2mg – T = 2mA1  2mg –2mA1 = T m M m m 2m m m A B F1 = 2mg C F2 = mg
• 26. Case II and Case III are identical to each other. So a2 = a3 < a1 Ans. (D) 49. Block A is placed on block B, whose mass is greater than that of A. There is friction between the blocks, while the ground is smooth. A horizontal force P, increasing linearly with time, begins to act on A. The accelerations a1 and a2 of A and B respectively are plotted against time (t). Choose the correct graph. (A) t a2 a1 (B) t a2 a1 a1 a2 (C) t a2 a1 a1 a2 (D) t a1 a1 a2 a2 Solution: Ans. will be B because upto some moment both will move with each other and after that B will move with constant accel. due to kinetic friction and aceel. Of A will increase 50. In the figure, the ball A is released from rest when the spring is at its natural (unstretched) length. For the block B of mass M to leave contact with the ground at some stage, the minimum mass of A must be (A) 2M (B) M (C) M 2 (D) a function of M and the force constant of the spring Solution: M x m o mg x = 2 1 kx 2  x = 2mg k (1) and kx = Mg x = Mg k Mg 2mg k k   m = (M/2) Ans. (c) B A P B M A
• 27. 51. Neglecting friction and mass of pulley, what is the acceleration of mass B (A) g/3 (B) 5g/2 (C) 2g/3 (D) 2g/5 Solution: mg – T = mA (1) 2T – mg = mA 2 (2) A 2 A B 2T T A 2 A (1) × 2 + (2) mg = 5mA 2 A = 2g 5 Ans. (D) 52. In the arrangement shown in figure, pulley is smooth and massless and all the strings are light. Let F1 be the force exerted on the pulley in case (i) and F2 the force in case (ii). Then (A) F1>F2 (B) F1<F2 (C) F1 = F2 (D) F1 = 2F2 Solution: (c) F1 = F2 because cases are identical. 53. A block of mass 2kg is pushed against a rough vertical wall with a force of 40N, the coefficient of static friction being 0.5. Another horizontal force of 15N, is applied on the block in a direction parallel to the wall. Will the block move? If yes, in which direction and with what minimum acceleration ? If no, find the frictional force exerted by the wall on the block. ? (A) 2 m/s2 (B) 2.5 m/s2 (C) 3 m/s2 (D) 4 m/s2 4m 2m (i) 4m m (ii) m A B m m
• 28. Solution : The force which may cause the tendency of motion or motion in the body is its own weight and the applied horizontal force of 15N. The resultant of the forces F = N 25 15 20 2 2   In a direction tan-1       20 15 = 370 with the vertical. The friction will, by its very virtue of opposing the tendency will act in a direction opposite to the resultant. 40N 15N mg The acceleration is minimum when the resultant force is minimum = F –N (as, N is the maximum frictional force) = 25 – 0.5 x 40 = 5 N  Minimum acceleration is 2 s / m 5 . 2 2 5  . Ans (B) 54. An inclined plane makes an angle 30 with the horizontal. A groove OA = 5m cut in the plane makes an angle 30 with OX. A short smooth cylinder is free to slide down the influence of gravity. The time taken by the cylinder to reach from A to O is (g = 10 m/s2) (A) 4 s (B) 2 s (C) 2 2s (D) 1 s Solution: Accel. of 2 2 mgsin 30 gsin 30 m  s = 5 = 2 1 4 t 2    10 4 t 2sec 10    55. Two blocks of masses m1 = 4kg and m2 = 6 kg are connected by a string of negligible mass passing over a frictionless pulley as shown in Fig. The coefficient of friction between block m1 and the horizontal surface is 0.4. When the system is released, the masses m1 and m2 start accelerating. What additional mass m should be placed over mass m1 so that the masses (m1 + m) slide with a uniform speed? (A) 9 kg (B) 10 kg (C) 11 kg (D) 12 kg Solution: For m2 T = m2g. For (m1 + m) T = (m1 + m)g = 0.4(m1 + m) g = m2g 300 3 0 0 O X A cylinder m m2 m1
• 29.  2 1 m m m 0.4    6 10 m 4 11kg 0.4     Ans. (C) 56. A block of mass 4 kg is suspended through two light spring balances A and B as shown in the figure. Then balance A and B will respectively read (A) 4 kg and zero kg (B) zero kg and 4 kg (C) 4 kg and 4 kg (D) 2 kg and 2 kg Solution: (c) If we consider 4 kg block as our system then it will be the reading of B and if we consider (spring B 4kg block) as our system then spring A will give 4kg reading. 57. A block of mass m is placed on another block of mass M lying on smooth horizontal surface, as shown in the figure. The co-efficient of friction between the blocks is . What is the maximum horizontal force F that can be applied to the block M so that the blocks move together ? (A) (M + m) g (B) (M – m) g (C) (M – mg)  (D) None of these F m M Solution: If there is no relative motion between the blocks then acceleration of the blocks is a = m M F  F.B.D. of m relative to M For vertical equilibrium N = mg (1) For horizontal equilibrium fl  ma  mg  m   m M F   F  (M + m)g  Fmax = (M + m) g. Ans (A) ma mg f N 58. When a force F acts on a body of mass m, the acceleration produced in the body is a. If three equal forces F1 = F2 = F3 = F act on the same body as shown in Figure the acceleration produced is (A)   2 1 a  (B)   2 1 a  (C) 2 a (D) None of these Solution: Fnet = 2 2 2 2 2 F F (1 sin45) F cos 45    = F 1 1 1 2sin45    = F 3 2  A B 4 kg F2 F1 F3 m 1350 900
• 30. a = F 3 2 m  Ans. (D) 59. A body is moving down a long inclined plane of angle of inclination . The coefficient of friction between the body and the plane varies as  = 0.5 x, where x is the distance moved down the plane. The body will have the maximum velocity when it has travelled a distance x given by (A) x = 2 tan  (B) 2 x= tanθ (C) x= 2cot (D) 2 x= cot  Solution: mgcos mgsin N r F  N = mg cos Velocity will be maximum when Fr max  N  mg sin 0.5x – mg cos = mg sin x = 2tan Ans. (A) 60. A man of mass 60 kg is standing on a horizontal conveyer belt. When the belt is given an acceleration of 1 ms–2, the man remains stationary with respect of the moving belt. If g = 10 ms–2, the net force acting on the man is (A) 0.6 N (B) 6 N (C) 60 N (D) 600 N Solution: 1 r F 60A 60 1    Fnet = 1 r F = 60N Ans. (C) a = 1 ms-2