Title 05. NEWTONS LAWS OF MOTION med Ans.pdf ✅

1. (d): Inertia means resistance to change. It is the property of the body by virtue of which it cannot change by itself its state of rest or of uniform motion 2. (c) : In uniform translatory motion, all parts of the ball have the same velocity in magnitude and direction and this velocity is constant. 3. (d) : As the rain drop is falling down with a constant speed, its acceleration, a=0 Hence, net force on the drop = ma=0. As the cork is floating on the surface of water, its weight is balanced by the upthrust. Hence net force on the cork is zero. The force exerted by the engine is balanced by the friction due to rough road. As the car is moving with constant velocity, it’s acceleration a=0 Hence. Net force on the car, F=ma=0 Whenever a body is thrown vertically upwards gravitational pull of earth gives it a uniform acceleration a=g in downward direction. Hence, net force on the pebble =mg =0.0510= 0.5N vertically downwards 4. (d) : The second law of motion is a local law which means that force F at a point in space (location of the particle) at a certain instant of time is related to a at that point at that instant. Acceleration here and now is determined by the force here and now, not by any history or the motion of the particle. 5. (c) : Here 90 , 0 1 = = − u ms v m g kg 0.04kg 1000 40 =40 = = s=60cm = 0.6m Using (0) (90) 2 0.6 2 2 2 2 2  − =  − = a v u as 2 2 6750 2 0.6 (90) − =−  a=− ms -ve sing shows the retardation  The average resistive force exerted by block on the bullet is F m a (0.04kg)(6750ms ) 270N 2 =  = = − 6. (d) : The situation is as shown in the figure. pinitial =mu,pfinal =−mu Impulse imparted to the ball = Change in momentum = pfinal − pinital =− mu − mu = −2mu 7. (a) : Given 2 6 8 , 5 −   F = i − j N a= ms F (6) ( 8) 10N 2 2  = + − = Mass of body is kg ms N a F m 2 5 10 2 = = = − 8. (b) : If a large force F acts for a short time dt the impulse imparted I is dt dt dp I = Fdt = I =dp= Change in momentum 9. (a) : Here 2 2 1 y =ut + gt  velocity u gt dt dy v= = + Acceleration, g dt dv a = = The force acting on the particle is F =ma =mg 10.(a) : Tension, thrust, weight are all common forces in mechanics whereas impulse is not a force Impulse = Force × Time duration 11.(a) : Here, F = -50 N (-ve sing for retardation m 10kg,u 10ms ,v 0 1 = = = − As F = ma+ +++++ 2 5ms 10kg 50 m F a − = − −  = = Using 2s 5ms 0 10ms a v u t v u at 2 1 = − − = −  = = + − − 12.(c) : Here m 5kg,u 5ms ,v 10ms ,t 10s 1 1 = = = = − − Using v=u + at 2 1 0.5 10 (10 5) − − = − = − = ms s ms t v u a As F =ma F (5kg)(0.5ms ) 2.5N 2  = = − 13.(d) : The relation F =ma can only be deduced from Newton’s second law, if mass remains constant with time. If mass depends on time then this relation cannot be deduced. 14.(c) : [Re ] [ ] [ ] −2 action = Force = MLT 15.(d): When the cork is floating, its weight is balanced by the upthrust. Therefore, net force on the cork is zero.
16.(b) : Here, mass of the stone , m = 1kg As the stone is lying on the floor of the train, its acceleration is same as that of train  Force acting on the stone, F ma (1kg)(1ms ) 1N 2 = = = − 17.(c) : It is easier to pull lawn mower than to push it. All other statements are true. 18.(d) : The greater the change in the momentum in a given time, the greater is the force that needs to be applied. 19.(d): For equilibrium under the effect of the three concurrent force all the properties mentioned are required 20.(d) : For uniform velocity, acceleration is zero. Hence resultant force will be zero 0 3 2 F1 + F + F = ( 5 9 ) (2 5 ) (3 4 ) 0 3 3       = − + − + − + = or F i j i j i i F 21.(c) : If 1 N and 2 N act in the same direction and 3N acts in opposite direction, then equilibrium is possible. 22.(a): The coefficient of friction between a given pair of substance is independent of the area of contact between tem.  1 for copper on cast iron. Rolling friction is much smaller than sliding friction; Irregularity of the surface in contact is not a main contributor of friction. 23.(b): Friction forces are always parallel to the surface in contact, which in this case are the wall and the cover of the book, this tells us that the friction forces ions either upwards or downwards. Because the tendency of the book is to fall due to gravity, the friction force must be in the upwards direction. 24.(d): A car accelerates on a horizontal road due to the force exerted by the road on the car 25.(b) : Limiting friction f =mg =0.619.8 =5.88N Applied force, F =ma=15 = 5N As F  f , so force of friction =5N 26.(a): Static friction is a self adjusting force 27.(d) In the machines, the flow of compressed and purified air lowers the friction, hence it acts as a lubricant. 28.(a): The various forces acting on the block are as shown in the figure. From figure mg sin= f ...... (i) mg cos= N ..... (ii) Divide (i) by (ii) we get tan tan ( ) 1 =   = = − or N N N f 29.(d): Both static and kinetic friction are independent on the area of contact. Coefficient of static friction depends on the surfaces in constant k  s 30.(a) The net force on the particle directed towards the centre is T 31.(b) : Turing means motion on a curved path, which required centripetal force. Bending of cyclist with respect to vertical direction provides the necessary centripetal force. 32.(a) The object will slip if centripetal force  force of friction mr mg 2 r g 2 mr constant 2   Or 2 1 21 2 1           =        r r r cm r cm 1 4 2 2 2 2   =        = 33.(c): Here 1 2 10 , 5 , 2 − − r = m v= ms a= ms 2 2 2.5 10 5 5 − =  = = ms r v ar The net acceleration is 2 2 2 2 a= ar + at = (2.5) +2 1 10.25 3.2 − = = ms 34.(d) 35.(c): If a is downward acceleration of 4 kg block, the upward acceleration of 1 kg block must be 2a Let T be tension in each part of string. The equation of motion on 4 kg block is
4g − 2T =4a ..... (i) The equation of motion or 1 kg block is T −1g =12a ...(ii) Or 2T − 2g =4a ....(iii) Adding (i) and (iii) we get 2g =8a or 4 g a =  Acceleration of 1 kg block 2 2 g = a= upwards. 36.(c) In uniform translator motion, all parts of the ball have the same velocity in magnitude and direction, and this velocity is constant. 37.(b) When a metre scale Is moving with uniform velocity. The force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero 38.(d): Conservation of momentum in a collision between particles can be understood from both Newton’s second and third law 39.(c) in figure OA= p1 = initial momentum of player northward OB = p2 = final momentum of player westward According to triangle law of vectors, OA+ AB =OB AB =OB−OA= p2 − p1 = change in momentum The change in momentum of player is along south west As motion is due to frictional force of reaction of the ground therefore force that acts on the player is frictional force along south west 40.(a) : Since there are no nearby stars to exert gravitational force on him and the small spaceship exerts negligible gravitational attraction on him, the net force acting on the astronaut, once he is out of the spaceship, is zero. In accordance with the first law of motion the acceleration of the astronaut is zero. 41.(c): Change in momentum =0.1512− (−0.1512) =3.6Ns Impulse = 3.6Ns in the direction from the batsman to the bowler 42.(a) : figure shows that slope of x-t graph changes from positive to negative at t = 2 s and it changes from negative to positive at t = 4s and so on. Thus direction of velocity is reversed after every two seconds. Hence, the body must be receiving consecutive impulses after every two seconds. 43.(c) : Here , m=5kg u (30i 40 j)ms and F ( 6i 5 j)N 1   −   = + = − − uy 40ms and Fy 5N 1  = =− − 2 1 5 5 − =− − = = ms m F a y y vy = 0 As t v u a t y y y  = −  = + 0 40 1 Or t =40s 44.(a): Here Mass of the particle m=0.4kg F = -8N (minus sign for direction of force  Acceleration , 2 20 0.4 8 − =− − = = ms kg N m F a The position of the body at any time t is given by 2 0 2 1 x= x + ut+ at The position of the body at t=0 is 0, therefore 0 x0 = 2 2 1 x=ut + at Position of the body at t =25s Here 10 , 20 , 25 1 2 = = − = − − u ms a ms t 2 ( 20)(25) 2 1 x=10 25+ − =250−6250=−6000m 45.(d) : At t=4s the body has constant velocity 1 4 3 − u = ms After t =4s the body is at rest i.e. v= 0 Impulse =m(v −u)
1 1 4 3 ) 4 3 2 (0 − − = kg − ms =− kgms 46.(a) : Since the graph between x and t is a straight line and passing through the origin x = t Since the graph between y and t is a parabola 2 y =t  = =1 = = 0 dt dv and a dt dx v x x x And 2 2 2 − = = t and a = ms dt dy vy y The force acting on the particle is F may (0.5kg)(2ms ) 1N 2 = = = − along y-axis 47.(a) : For t  0 and t  4s the position of the particle is not chaining i.e. the particle is at rest so no force is acting on the particle at these intervals. For 0  t  4s the position of the particle is continuously chaining. As the position – time graph is s straight line, the motion of the particle is uniform, so acceleration, a=0 Hence no force acts on the particle during this interval also. 48.(b) : As the weight of man increased by 5 times , so acceleration of the rocket, 2 5 5 10 50 − a= g =  = ms Force applied by rocket engine is N F ma 5 4 5 10 0.1 10 50 =  = =   49.(c) : When the speeding bus stops suddenly, the lower part of the passengers’s body in contact with the seat remains at rest whereas the upper part of the body of the passengers continues to be in state of motion due to inertia. Hence m the passengers are thrown forward. 50.(a) : The sixth coin is under the weight of four coins above it. Hence Reaction of the th 6 coin on the th 7 coin = Force on the th 6 coin due to th 7 coin =4mg 51.(d): Acceleration of the rocket at any instant t is dt dm M t dt dm v a r − = Here, 1 6000 , 16 − = = kgs dt dm M kg 2 2 1 1 35 6000 60 16 11000 16 1min 60 11 11000 − − − −        −    = = = = = a ms ms t s v kms ms r 52.(b): Here 1 15 − v= ms Area of cross section 2 2 A 10 m − = Density of water, 3 3 10 − = kgm Mass of water hitting the wall per second 3 3 2 2 1 1 10 10 15 150 − − − − =   = =  kgm m ms kgs A v Force exerted on the wall = Momentum loss of water per second kgs ms N N 1 1 3 =150 15 =2250 =2.2510 − − 53.(c) : Force dt d = (momentum)         =      = = dt dm dt dm mv v dt d ( ) 210 300 Rate of combustion, , 1 0.7 300 210 − = = kgs dt dm 54. Initial momentum of ball) 1 1 (0.05 ) (5 ) 0.25 − − A= kg ms = ms As the speed is reversed on collision Final momentum of the ball 1 1 (0.05 )( 5 ) 0.25 − − A= kg − ms =− kgms Impulse imparted to the ball A = Change in momentum of ball A = 1 1 1 0.25 0.25 0.5 − − − − kgms − kgms =− kgms Similarly, Initial momentum of ball B = 1 1 (0.05 )( 5 ) 0.25 − − kg − ms =− kgms Final momentum of ball B = 1 1 (0.05 ) (5 ) 0.25 − − kg ms = ms Impulse imparted to the ball B = 1 1 1 (0.25 ) ( 0.25 ) 0.5 − − − kgms − − kgms = kgms Impulse imparted to each ball is 1 0.5 − kgms in magnitude. The two impulses are opposite in direction 55.(b) : Here 4 m=210 kg Initial acceleration, 2 5 − a= ms Initial thrust = upthrust required to impart acceleration a + upthrust required to overcome gravitational pull