NCERT Solutions for Class 11

Physics

Chapter 5 Law of Motion

Q.1 A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for ω≤ g R . What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω= 2g R Neglect friction. 

Answer 

Let the radius vector joining the bead with the centre of wire makes an angle θ, with the vertical downward direction as shown in the figure. In the given figure, OP = R = Radius of the circle N = Normal reaction, l = PQ As clear from the figure, the respective vertical and horizontal equations of forces can be written as: mg=Ncosθ … (i) ml ω 2 =Nsinθ … (ii) From ΔOPQ, we have: sinθ= l R l=Rsinθ … (iii) Putting equation iii in equation ii , we obtain: m Rsinθ ω 2 =Nsinθ mR ω 2 =N … (iv) Putting equation iv in equation i , we obtain: mg=mR ω 2 cosθ= g R ω 2 … v As cosθ≤1, the bead will remain at its lowermost point when   g R ω 2 ≤1, i.e., for ω≤ g R For ω= 2g R or  ω 2 = 2g R …(vi) Equating equations v and vi , we obtain:  2g R = g Rcosθ cosθ= 1 2 ∴θ= cos −1 0.5 =60°

Q.2 A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev min-1. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed

Answer 

Here, mass of the man, m=70 kg Radius of the drum, r=3 m Frequency of rotation of the drum, ν=200 revmin -1 = 200 60   = 10 3 rev sec -1 Coefficient of friction, μ=0.15 The wall provides the normal force ( F N ),which provides the necessary centripetal force required for the rotation of the man. As the floor revolves, the man remains stuck to the wall of drum. ∴Weight of the man ( mg ) acting downwards is balanced by the frictional force ( f=μFN ) acting in the upwards direction. ∴Man will remains stuck until: mg<f  i.e., mg<μFN=μmr ω 2 or g<μr ω 2 ∴Minimum angular speed of rotation of the cylinder is given by,  ω min = g μr = 10 0.15×3 =4.71  radsec -1

Q.3 You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m

Answer

In a death – well, a motorcyclist does not fall at the uppermost point, as both the force of normal reaction and the weight  of the motorcyclist acting downwards are balanced by the centrifugal force. This situation is shown in the given figure. The total force acting on the motorcyclist is the sum of the normal force ( F N ) and the force due to gravity ( F g =mg ).The equation of motion for the centripetal acceleration a c  is given as: F net = ma c F net = ma c F N + F g = ma c F N +mg= m v 2 r Normal reaction force is provided by the speed of motorcyclist. At the minimum speed ( v min ), F N =0 mg= m v 2 min r ∴ v min = rg = 25×10 =15.8  ms -1

Q.4 A disc revolves with a speed of 33 1 3 rev/min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record  

Answer 

Frequency of revolution of disc, ν=33 1 3   revmin -1 Coefficient of friction, μ=0.15 Force of friction is given as: f=μmg Required centripetal force, F cent = m v 2 r =mr ω 2 To prevent slipping f≥ F cent For coin placed at 4 cm: Radius of revolution, r’=4 cm =0.04 m Angular frequency, ω=2πυ =2π× 100 180 =3.49  s -1 Frictional force, f=0.15×m×10 =1.5m N Centripetal force on the coin, F cent =mr’ ω 2 =m×0.04× ( 3.49 ) 2 =0.49 mN As,  f > F cent the coin will revolve with the record. For coin placed at 14 cm: Radius, r’’=14 cm =0.14 m Angular frequency, ω=3. 49 s –1 Frictional force, f=1.5m N Centripetal force is given by, F cent =m×0.14× ( 3.49 ) 2 =1.7m N As f < F cent , the coin will slip from the record.

Q.5 The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and  accelerates with 2 ms –2 .At what distance from the starting point does the box fall off the truck (Ignore the size of the box)

Answer 

Given, mass of the box, m=40 kg Coefficient of friction, μ=0.15 Initial velocity of the truck, u=0 Acceleration of the truck, a= 2 ms -2 Distance of the box from the open end of the truck,s’ = 5 m According to Newton’s second law of motion, the force on the box due to the accelerated motion of the truck is given as: F=ma=40×2=80 N This force is in forward direction. According to Newton’s third law of motion, Reaction force on the box, F’ = 80 N, backwards This force is resisted by the force of limiting friction f, present between the box and the floor of the truck. It is given as: f=μmg =0.15×40×10 =60 N in the forward direction ∴Total force acting on the box is given as: F total =80–60 =20 N, in the backward direction The backward acceleration produced in the box is given as: a’= P m = 20 40 =0.5  ms -2 According to the second equation of motion, we have: s=ut+ 1 2 a’ t 2 5=0×t+ 1 2 ×0.5 t 2 ∴t= 5×2 0.5 =4.47 s. Therefore, the box will fall from the truck after 4.47 s from start. The distance (x) covered by the truck during this time is given by equation of motion, s=ut+ 1 2 a t 2 ∴x=0×4.47+ 1 2 ×2 ( 4.47 ) 2 =19.98 m

Q.6 A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0 .5m s –2  for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by ( a ) a stationary observer on the ground, ( b ) an observer moving with the trolley.  

Answer 

( a ) Here, mass of the block, m = 15 kg Coefficient of static friction, μ = 0.18 Acceleration of the trolley, a = 0. 5 ms -2 According to Newton’s second law of motion, the force ( F )on the block due to the motion of the trolley is given as: F = ma = 15 × 0.5 = 7.5 N This force acts towards the direction of motion of the trolley. Force of static friction between the block and the trolley is given as: f = μmg = 0.18 × 15 × 10 = 27 N Relative motion between the block and the trolley is not possible as the frictional force here is less than the maximum frcitional force possible. When the trolley will move with uniform velocity after 20 s, there will be no frictional force on the block.

(b) For the observer moving with the trolley the box will be at rest relative to him. When the trolley will move with uniform velocity after 20 s, there will be no frictional force on the block. Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall Fig. 5.21 . The coefficient of friction between the bodies and the table is 0.15.

Q.7 A force of 200 N is applied horizontally to A. What are  a the reaction of the partition b the action-reaction forces between A and B What happens when the wall is removed Does the answer to b change, when the bodies are in motion Ignore the difference between μ s and μ k . 

Answer 

Given,mass of body A, m A =5 kg Mass of body B, m B =10 kg Applied horizontal force, F = 200 N Coefficient of friction between bodies and table, μ s = 0.15 The force of limiting friction is given as: f s =μ m A + m B g =0.15 5+10 ×10=1.5×15=22.5 N towards left Total force exerted on the partition = 200 – 22.5 = 177.5 N towards right According to Newton’s third law of motion, the force of reaction of the partition will be opposite to dirction of the total applied force. ∴Reaction of the partition  =  177.5 N, towards the left b Force of limiting friction on mass A is given as: f A =μ m A g =0.15×5×10 =7.5 N , to the left Total force applied by mass A on mass B F”=200–7.5 = 192.5 N , towards the right According to Newton’s third law of motion, an equal force of reaction will be applied by mass B on mass A, i.e., 192.5 N acting towards left. When the partition is removed, the two bodies will move in the direction of the force exerted. Total force on the moving system of bodies=177.5 N The equation of motion for the system of acceleration a is given as: Total force = m A + m B a ∴a= Net force m A + m B = 177.5 5+10 = 177.5 15 =11.83  ms -2 Total force producing motion in body A, F A = m A a ∴ F A =5×11.83 =59.1 N ∴Total force applied by mass A on mass B, when partition is removed =F”− F A =192.5−59.15=133.35 N It will act towards the direction of motion. According to Newton’s third law of motion, Equal reaction force of mass B on mass A = 133.3 N , acting opposite to the direction of motion.

Q.8 A monkey of mass 40 kg climbs on a rope ( Fig. 5.20 ) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey 

( a ) climbs up with an acceleration of 6 m s –2 

( b ) climbs down with an acceleration of 4 m s –2 

( c ) climbs up with a uniform speed of 5 m s –1 

( d ) falls down the rope nearly freely under gravity ( Ignore the mass of the rope ). 

Answer

(a) Here, mass of the monkey, m=40 kg Acceleration due to gravity, g=10 ms -1 Maximum tension that the rope can stand,T max = 600 N Acceleration of the monkey, a = 6 ms -2 upwards According to Newton’s second law of motion, we have: T–mg=ma ∴T=m( g+a ) =40( 10+6 ) =640 N In this case, T > T max , therefore, the rope will break.

( b ) In this case, Acceleration of the monkey, a = 4 ms -2 downwards According to Newton’s second law of motion, we have: mg – T = ma ∴T=m( g–a ) =40( 10–4 ) =240 N In this case, T < T max , therefore, the rope will not break.

( c )In this case,  the monkey is climbing with a uniform speed of 5 ms -1 . ∴Acceleration of monkey, a = 0. According to Newton’s second law of motion, we have: T–mg=ma      T–mg=0 ( ∵a=0 ) ∴T = mg = 40 × 10 = 400 N In this case, T < T max , therefore, the rope will not break.

( d )When the monkey falls down nearly freely under the effect of gravity, its acceleration= acceleration due to gravity, i.e., a = g According to Newton’s second law of motion, we have: mg – T = mg ∴T=m( g – g )=0 In this case, T< T max , therefore, the rope will not break.

Q.9 A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding

Answer 

Here, mass of the , m=25 kg Mass of the man, M=50 kg Acceleration due to gravity, g=10 ms -2 Force exerted on the block, F=mg =25×10 =250 N Weight of man, W=Mg =50×10 =500 N ( a ) When the man lifts the block directly: In this case, upward force is applied by the man.            This increases the apparent weight of the man. ∴Action on the floor  =  W + F  = 500+250 N =750 N ( b ) When the man lifts the block by using a pulley: In this case, downward force is applied by the man. This decreases apparent weight of the man.∴Action force on the floor = W – F = 500 – 250 = 250 N As the floor yields to a normal force of 700 N, the man should adopt the second method to easily raise the block by applying comparatively lesser force.

∴Action force on the floor = W – F = 500 – 250 = 250 N As the floor yields to a normal force of 700 N, the man should adopt the second method to easily raise the block by applying comparatively lesser force.

Q.10 A train runs along an unbanked circular track of radius 30 m at a speed of 54 kmh-1.The mass of the train is 106 kg. What provides the centripetal force required for this purpose – The engine or the rails What is the angle of banking required to prevent wearing out of the rail

Answer 

Here, Radius of the circular bend, r=30 m Mass of the train, m=106kg Speed of the train, v= 54 kmh -1 = 15 ms -1 Acceleration due to gravity, g=10  ms -2 The lateral thrust of the rail on the wheel provides the necessary centripetal force. According to Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail causing      its wear and tear.  Using the relation, tanθ= v 2 rg tanθ= ( 15 ) 2 30×10 = 225 300 ∴θ= tan −1 ( 0.75 ) =36. 87 0 ∴Angle of banking= 36. 87 ∘

Q.11 An aircraft executes a horizontal loop speed of 720 kmh-1 with its wings banked at 15°. What is the radius of the loop

Answer 

Given, speed of the aircraft, v=720 kmh -1 =720× 5 18  =200  ms -1 Angle of banking of the wings, θ= 15 0 Acceleration due to gravity, g=10 ms -2 Let radius of the loop = r From the relation, tanθ = v 2 rg r= v 2 g tanθ = 200×200 10×tan 15 0 ∴r=14925.37 m =14.92 km

Q.12  Ten one – rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of  

( a ) the force on the 7 th coin (counted from the bottom) due to all the coins on its top, 

( b ) the force on the 7 th coin by the eighth coin, 

( c ) the reaction of the 6 th coin on the 7 th coin.  

Answer 

( a ) Force on the seventh coin is because of weight of the three coins on the top of it. Given, weight of one coin = mg ∴Weight of three coins = 3 × mg = 3mg Here, g = Acceleration due to gravity ∴ Force on the 7 th coin because of three coins on its top = Weight of three coins = 3mg; vertically downwards.

( b ) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins ( ninth and tenth ) on the top of it. Here, weight of eighth coin = mg Weight of ninth coin = mg Weight of tenth coin = mg ∴Net weight of these three coins = 3mg ∴ Force on the 7 th coin by the eighth coin = Net weight of these three coins = 3mg; vertically downwards.

( c ) The 6th coin experiences a downward force due to the weight of the four coins ( 7 th , 8 th , 9 th , and 10 th ) on the top of it. ∴ Total downward force on 6 th coin = 4mg According to Newton’s third law of motion,the 6th coin will exert an equal and opposite reaction force on the 7th coin. ∴Reaction force exerted by 6th coin on the 7th coin = 4mg; It is directed upwards.

Q.13 A stream of water flowing horizontally with a speed of 15  ms –1 pushes out of a tube of cross -sectional area 10 –2   m 2 , and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound

Answer 

Given,  speed of the water stream, v = 15 ms -1 Area of  cross-section of the tube, A = 10 –2 m 2 Volume of water pushing out from the pipe per second:    V = A × v = 10 –2 × 15 m 3 s -1 As, density of water, ρ= 10 3   kgm -3 ∴Mass of water striking the wall per second,m=ρ×V = 150 kgs -1 Since the water hits the wall and does not rebound, ∴According to Newton’s second law of motion: Force exerted by water on the wall, F = Rate of change of momentum = Change in linear momentum Time = mv t = 150 × 15 1 = 2250 N

Q.14 A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s –2 . The crew and the passengers weigh 300 kg. Give the magnitude and direction of the ( a ) force on the floor by the crew and passengers, ( b ) action of the rotor of the helicopter on the surrounding air, ( c ) force on the helicopter due to the surrounding air.  

Answer 

( a ) Here, mass of the helicopter, m H =1000 kg Mass of the crew and passengers, m P =300 kg ∴Total mass of the system, M=1300 kg Upward acceleration of the helicopter, a= 15 ms -2 Acceleration due to gravity, g=10  ms -2 According to Newton’s second law of motion, the reaction force R, on the system by the floor can be obtained as: R – m P g = ma ∴R= m P ( g+a ) =300( 10+15 ) =300×25  N =7500 N As the helicopter is accelerating in the vertically upward direction, the reaction force will also act upwards. ∴According Newton’s third law of motion, force on the floor by the crew and passengers = 7500 N,directed downwards

( b ) As per Newton’s second law of motion, the reaction force , experienced by the helicopter is given as: R’ = M ( g + a ) =1300( 10+15 ) =1300×25 N =32500 N, directed upwards According to Newton’s third law of motion, Action of the rotor on the surrounding =  32500 N, directed downwards

( c ) Force on the helicopter because of the surrounding air is 32500 N, directed upwards.

Q.15 A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: Choose the correct alternative Lowest Point Highest Point a mg– T 1 mg+ T 2 b mg+ T 1 mg– T 2 c mg+ T 1 – m v 1 2 /R mg– T 2 + m v 1 2 /R d mg– T 1 – m v 1 2 /R mg+ T 2 + m v 1 2 /R T 1 and v 1 denote the tension and speed at the lowest point. T 2 and v 2 denote corresponding values at the highest point.  

Answer

The free body diagrams of stone at the lowest and highest point are shown in figure (c) and (d) respectively. As per Newton’s second law of motion, the net force acting on the stone at the lowest point is given as: F net = T – mg = m v 1 2 R →(i) Here,  v 1 = velocity of the stone at the lowest point According to Newton’s second law of motion, we have:n T+mg= m v 2 2 R →(ii) Here,  v 2 = velocity of the stone at the highest point From equations (i) and (ii) it can be observed that the net force acting at the lowest and highest points are  T – mg  and  T + mg  respectively. Thus, (a) is the correct alternative. Figure 5.18 shows a man standing stationary withrespect to a horizontal conveyor belt that is accelerating  with 1 m s -2 .

Q.16 What is the net force on the man If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt (Mass of the man = 65 kg.) 

Answer 

Given, Mass of the man, m = 65 kg Acceleration of the belt, a = 1 ms -2 According to Newton’s second law of motion, we obtain: Net force acting on the man, F net = 65 × 1 = 65 N As the man is standing stationary w.r.t. the belt, Acceleration of the man = Acceleration of the belt =1  ms -2 As, Coefficient of static friction, μ = 0.2 ∴Force of limiting friction, F = μmg The man will remain stationary with respect to the conveyor belt until the net  force on the man is less than or equal to the frictional force f s , exerted by the belt, i.e., F’ net = f s ma’=μmg ∴a’=0.2×10 = 2 ms -2 ∴The maximum acceleration of the belt up to which the man can remain stationary  = 2 ms -2

Q.17 Figure 5.17 shows the position-time graph of a particle of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the particle What is the magnitude of each impulse

Answer

As physical context to this motion, the graph can be related to a ball between two parallel walls undergoing perfectly elastic collision between t = 0 to 16 s

Given, mass of the particle, m = 0.04 kg The given graph shows that the particle moves from x=0 to x=2 in 2 sec and the body changes its direction of motion after every 2 sec. As the slope of the given graph reverses after every 2 sec, ∴The ball undergoes collision with a wall after every 2 s. ∴The ball experiences an impulse after every 2 s. The slope of the given x – t graph gives the velocity of theball, From the given graph, we obtain: Velocity of the ball, u= ( 2−0 )× 10 -2 ( 2−0 ) = 10 −2   ms -1 Now, Velocity of the ball before collision, u = 10 -2 ms -1 ∴Velocity of the ball after collision, v = – 10 -2 ms -1 The negative sign shows the reversal of direction of motion of the ball. ∴Magnitude of impulse = Total change in linear momentum =| mv−mu |=| m( v−u ) | =| 0.04( − 10 −2 − 10 −2 ) | = 0 .08 × 10 -2   kgms -1

Q.18 Explain why

(a) a horse cannot pull a cart and run in empty space,

(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,

(c) it is easier to pull a lawn mower than to push it,

(d) a cricketer moves his hands backwards while holding a catch.

Answer 

(a) While trying to pull a cart, a horse pushes the ground backward with some force at an angle. As a result, theground exerts an equal and opposite reaction force upon the feet of the horse. The forward component of this reaction force is responsible for the forward motion of the horse and the cart. However, in an empty space, there is no such reaction force. Hence, the horse cannot pull the cart and run in an empty space.

(b) When a speeding bus stops suddenly, the lower part of the body of the passengers in contact with the seat,  suddenly comes to rest. But, the upper part has the tendency to remain in uniform motion. Therefore, the  passenger’s upper part of body is thrown forward.

(c) In order to pull a lawn mower, a force at an angle θ is exerted on it, as shown in the given figure

(A). The vertical component of this applied force is upwards and reduces the effective weight of the mower. On the contrary, in order to push a lawn mower, a force at an angle θ is exerted on it ,as shown in figure

(B). In this case, the vertical component of the applied force acts downwards and increases the effective weight of the mower.  As the effective weight is lesser in the first case, therefore, pulling the lawn mower is easier than pushing it. (d) According to Newton’s second law of motion, we have F=ma=m Δv Δt →(i) From the above equation, we observe that the force of impact is inversely proportional to the impact time. i.e., F∝ 1 Δt →(ii) Equation (ii) indicates that the force experienced by the cricketer decreases if the time of impact increases. By moving his hands backwards, the cricketer increases the time of impact. As time of impact increases, the  force of impact decreases and as a result, his hands are not hurt severely.

Q.19 If in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:

(a) the stone moves radially outwards,

(b) the stone flies off tangentially from the instant the string breaks,

(c) The stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle

Answer

Correct option is (b). As the string breaks, the stone moves in the direction of velocity at that instant of time. Applying Newton’s first law of motion where the direction of velocity is tangential to the path of the stone at that instant; when the string breaks, the stone flies off tangentially.

A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev. min-1 in a horizontal plane. What is the tension in the string

Q.20 What is the maximum speed with which the stone can be whirled around if the string can with stand a maximum tension of 200 N

Answer 

Given, mass of stone, m = 0.25 kg Radius of circle, r = 1.5 m Number of revolutions per second, n = 40 60 = 2 3  rps Angular velocity,ω = v r = 2πn Tension in the string provides the necessary centripetal force i.e., T = F Centripetal  = m v 2 r =mr ω 2 =mr ( 2πn ) 2 =0.25×1.5× ( 2×3.14× 2 3 ) 2 =6.57 N Here, maximum tension in the string,  T max = 200 N But,  T max = m v 2 max r ∴ v max = T max ×r m = 200×1.5 0.25 = 1200 =34.64  ms -1 ∴Maximum speed of the stone = 34.64  ms -1

Q.21 A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 kmh-1. What is the impulse imparted to the ball (Mass of the ball is 0.15 kg.)

Answer

The situation of the problem is shown in the given figure. Here, AO = Incident path followed by the ball OB = Path followed by the ball after reflection ∠AOB = Angle between incident and deflected  paths = 45 o ∠AOP = ∠BOP = 22 .5 o = θ Given, magnitude of initial and final velocity, u = 54  kmh -1 Horizontal component of initial velocity = u cosθ along OS Vertical component of initial velocity = u sinθ along PO Horizontal component of final velocity = u cosθ along OR Vertical component of final velocity = u sinθ along PO There is no change in the horizontal components of velocity, however the vertical components of velocity are in opposite directions. ∴Impulse imparted to the ball = Change in the linear momentum of the ball = mu cosθ -( – mu cosθ ) = 2mu cosθ Given, mass of the ball, m = 0.15 kg Velocity of the ball,  v = 54 kmh -1 = 54×1000 60×60 ms -1 =15  ms -1 ∴Impulse imparted=2×0.15×15cos 22.5 o =4.16  kgms -1

Q.22 A shell of mass 0.22 kg is fired by a gun of mass 100 kg. if the muzzle speed of the shell is 80 ms-1, What is the recoil speed of the gun

Answer 

Here, mass of shell, m=0.020 kg Mass of gun, M = 100 kg Muzzle speed of the shell, v=80  ms -1 Let recoil speed of gun = V As, both the shell and the gun were at rest initially ∴Initial momentum of the system of gun and shell = 0 Final momentum of the system of gun and shell = mv – MV Here, the negative sign indicates that the directions of the shell and the gun are opposite to each other. As per the law of conservation of momentum: Final momentum of the system = Initial momentum of the system mv – MV = 0 ∴V= mv M = 0.020×80 100×1000 =0.016  ms -1

Q.23 Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms-1 collide and rebound with the same speed. What is the impulse imparted to each ball due to the other

Answer 

Given, mass of each ball=0.05 kg Initial velocity of each ball=6  ms -1 ∴Magnitude of initial momentum of each ball,  p i =0.3  kgms -1 As the speed of each ball is reversed on collision, ∴Final momentum of each ball,  p f =-0.3  kgms -1 Impulse of each ball=Change in momentum of the  system=p f – p i =-0.6  kgms -1 The negative sign shows that the impulses acting on the balls are in opposite directions.

Q.24 A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Answer 

As, the nucleus is at rest before disintegration, ∴ Initial momentum of the system = 0 Suppose  m 1 ,  m 2  be the masses of products and  v 1 ,v 2  be their respective velocities. ∴Total linear momentum after disintegration= m 1 v 1 + m 2 v 2 According to the law of conservation of energy,  Total initial momentum =Total final momentum ∴0=  m 1 v 1 + m 2 v 2 ∴ v 1 =− m 2 v 2 m 1 The negative sign shows that the products move in opposite directions.

Q.25 Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Answer

The given system of two masses and a pulley is represented by the given figure. Here, smaller mass, m 1 =8 kg Larger mass, m 2 =12 kg Due to its larger weight, mass  m 2  moves downward and mass  m 1  moves upward. According to Newton’s second law for the system of each mass: For mass  m 1 , the equation of motion is given as: T− m 1 g= m 1 a→(i) For mass  m 2 , the equation of motion is given as: m 2 g−T= m 2 a→(ii) Adding equations (i) and (ii), we obtain: ( m 2 − m 1 )g=( m 1 + m 2 )a ∴a=( m 2 − m 1 m 1 + m 2 )g→(iii) =( 12−8 12+8 )×10 = 4 20 ×10 =2  ms −2 Hence, the acceleration of each mass is 2  ms −2 . Putting the value of a from equation (iii) in equation (i), we obtain: m 2 g−T= m 2 ( m 2 − m 1 m 1 + m 2 )g ∴T=( m 2 − m 2 2 − m 1 m 2 m 1 + m 2 )g =( 2 m 1 m 2 m 1 + m 2 )g =( 2×12×8 12+8 )×10 = 2×12×8 20 ×10 =96 N ∴Tension in the string = 96 N

Q.26 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F=600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case

Answer

Given, horizontal force applied, F=600 N Mass of body A,  m 1 =10 kg Mass of body B,  m 2 =20 kg ∴Total mass of the system, m= m 1 + m 2 =10+20 kg =30 kg According to Newton’s second law of motion, we obtain:    F = ma ∴a= F m = 600 30   ms −2 =20  ms −2

(a) When force F is applied on body A: The equation of motion becomes F−T= m 1 a ∴T=F− m 1 a =600−10×20 N =400 N

(b) When force F is applied on body B: The equation of motion becomes F−T= m 2 a ∴T=F− m 2 a =600−20×20 N =200 N

Q.27 Figure 5.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t<0, t>4, 0

Answer

(a) For t < 0, from the given graph, we observe that the position of the particle coincides with the time axis. It shows that the displacement of the particle is zero in this case. Therefore, in this case, the force acting on the particle is zero.

For 0 < t < 4 s, the position time graph has a constant slope. Therefore, the acceleration produced in the particle is zero in this interval. Hence, force acting on the particle is zero.

For t > 4 s, it can be observed that the position time graph is parallel to time axis. Therefore, the particle remains at the state of rest at a distance of 3 m from the origin. Therefore, force acting on the particle is zero.

(b) At t = 0 Here, mass of the particle, m=4 kg Initial velocity of the particle, u=0 Final velocity of the particle, v= 3 4   ms −1 As, impulse = Change in momentum = mv – mu ∴Impulse = m( v – u ) =4( 3 4 −0 ) =3  kgms −1 At t=4 s Initial velocity, u= 3 4   ms −1 Final velocity, v=0 ∴Impulse=m( v−u ) =4( 0− 3 4 ) =−3  kgms −1

Q.28 A man of mass 70 kg stands on a weighing scale in a lift which is moving

(a) Upward with a uniform speed of 10 ms-1

(b) Downwards with a uniform acceleration of 5 ms-2

(c) Downwards with a uniform acceleration of 5 ms-2

What would be the reading on the scale in each case

(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity

Answer 

Given, mass of the man, m=70 kg Acceleration, a=0 In each case, the weighing machine measures the apparent weight (R).

(a) When the lift is moving upwards with uniform speed, acceleration a=0 According to Newton’s second law of motion, we obtain:    R=mg =70×10 =700 N ∴Reading on the weighing scale= 700 g = 700 10 =70 kg

(b) When the lift is moving downwards with acceleration     a=5  ms −2 According to Newton’s second law of motion, we obtain: R−mg=ma ∴R=m( g−a ) =70( 10−5 ) =350 N ∴Reading on the weighing scale= 350 g = 350 10  kg =35 kg

(c) When the lift is moving upwards with acceleration  a=5m s −2 According to Newton’s second law of motion, we obtain: R – mg = ma R = m( g + a ) = 70( 10 + 5 ) = 1050 N ∴Reading on the weighing scale= 1050 g = 1050 10  kg =105 kg

(d) When the lift moves downward freely under the effect of gravity alone, downward acceleration, a=g

According to Newton’s second law of motion, we obtain: R=m( g−a ) =m( g−g ) =0 ∴Reading on the weighing scale= 0 g =0 kg In this case, the man is in the state of weightlessness .

Q.29 A bob of mass 0.1 kg hung from the ceiling of a room by a starting 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms-1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.

Answer

(a) The velocity of the bob becomes zero at the extreme position. If we cut the string at the extreme position, the bob will fall vertically on the ground.

(b) At the mean position, the velocity of the bob is 1 ms-1 along the tangent to the arc formed by the oscillating bob. If we cut the string at the mean position, then the bob will trace a projectile path having the horizontal component of velocity only. Therefore, the bob will follow a parabolic path.

Q.30 A truck starts from rest and accelerates uniformly with 2 ms-2. At t = 10s, a stone is dropped by a person standing on the top of the truck (6 m high from ground). What are the (a) velocity and (b) acceleration of the stone at t = 11 s Neglect air resistance.

Answer

Given, initial velocity of the truck, u = 0 Acceleration, a=2  ms −2 Time, t=10 s According to the first equation of motion, the final velocity of the truck is given by, v=u+at ∴v=0+2×10 =20  ms −1 The final velocity of the truck and therefore, that of the stone is equal to 20  ms -1 . At t=11 s As the air resistance is neglected, the horizontal component of velocity, v x ( =v=20  ms −1 )remains constant. In the vertical direction,  initial velocity of the stone, u = 0 Acceleration, a = g = 10  ms -2 Time, t’=11−10 s =1 s According to the first equation of motion, the vertical component ( v y ) of velocity of stone is given as: v y =u+ a y t’ =0+10×1  ms −1 =10  ms −1 Resultant velocity of the stone,v is given as: v= v x 2 + v y 2 = ( 20 ) 2 + ( 10 ) 2 =22.4  ms −1 Let θ be the angle made by the resultant velocity with the horizontal component of  velocity,v x ∴tanθ= v y v x = 10 20 =0.5 ∴θ= 26.6 o (b) When the stone is dropped from the car, horizontal force on the stone=0. The stone continues to move  vertically downwards under the effect of gravity. But,the path followed by the stone is parabolic in nature.

Q.31 A body of mass 0.40 kg moving initially with a constant speed of 10 ms-1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, and the position of the particle at that time to be x = 0, predict its position at t = -5 s, 52 s, 100 s

Answer 

Given, mass of the body, m = 0.40 kg Initial speed of the body, u = 10  ms -1  due N Force exerted on the body, F = -8 N According to Newton’s second law of motion, we have: F=ma ∴Acceleration, a= F m = −8 0.40 =−20 m s −2 (i) At t=−5 s Acceleration, a’ = 0; u = 10  ms -1 From the equation of motion: x=ut+ 1 2 a’ t 2 =10×( −5 )  m =−50 m

(ii) At t = 25 s Acceleration,  a”=-20ms -2 ; u=10  ms -1 From the equation of motion: x’=ut’+ 1 2 a” t 2 =10×( 25 )+ 1 2 ( −20 )× ( 25 ) 2 =250−6250 m =−6000 m (iii) At t = 100 E The problem is divided into two parts For 0≤t≤30 s a=−20  ms −2 u=10  ms −1 ∴ x 1 =ut+ 1 2 a” t 2 =10×30+ 1 2 ( −20 )× ( 30 ) 2 =−8700 m

For 30 s≤t≤100 s According to the first equation of motion, final velocity of the body is given as: v=u+at =10+( −20 )×30 =−590  ms −1 For motion between 30 s to 100 s, t = 70 s ∴ x 2 =vt+ 1 2 a” t 2 =−590×70 =−41300 m ∴ Total distance, x= x 1 + x 2 =−8700−41300 m ∴x=−50000 m

Q.32 A rocket with a lift off mass 20000 kg is blasted upwards with a net initial acceleration of 5 ms-2. Calculate the initial thrust (force) of the blast.

Answer

Here, mass of the rocket, m=20000 kg

Initial acceleration of the rocket, a=5  ms −2 Acceleration due to gravity, g=10  ms −2

The thrust must overcome the force of gravity so as to give it an upward acceleration of 5  ms -2 ∴Thrust must produce a net acceleration: a’= 9.8+5 =14.8 m s −2 According to Newton’s second law of motion, the thrust (force) acting on the rocket is given as: F=ma’ ∴F=20000×14.8 kgm s −2 =2.96× 10 5  N

Q.33 The driver of a three wheeler moving with a speed of 36 kmh-1 sees a child standing in the middle of the road and brings his vehicle to rest in 4 s just in time to save the child. What is the average retarding force on the vehicle The mass of the three wheeler is 400 kg and the mass of the driver is 65 kg.

Answer

Here, initial speed of three wheeler, u=36  kmh -1 Final speed of three wheeler, v = 0  ms -1 Time taken, t = 4 s Mass of the three wheeler, m = 400 kg Mass of the driver,  m 1 = 65 kg ∴Total mass of the system,  M = m + m 1 = 400 + 65 = 465 kg According to the first law of motion, we have: v=u+at ∴a= v−u t = 0−10 4 =−2.5 m s −2 The negative sign of acceleration shows the retarding motion of the three wheeler. According to the second law of motion, we have: F=Ma =465×( −2.5 ) =−1162.5 N The negative sign shows that the force is acting opposite to the direction of motion of the three wheeler.

Q.34 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body.

Answer

Here, mass of the body, m = 5 kg The given situation can be represented by the given figure Let,  F → 1 = OS → =8 N  and  F → 2 = OT → =6 N The resultant of two perpendicular forces is given as: R= F 1 2 + F 2 2 ∴Resultant force, R= ( 8 ) 2 + ( 6 ) 2 = 64+36 =10 N Let θ be angle made by R with  F 1 ∴tanθ= SC OS = OT OS = 6 8 ∴θ= tan −1 ( 6 8 ) = 36.87 o This gives the direction of the resultant force and therefore the direction of the acceleration of the body. According to Newton’s second law of motion, we have: F=ma ∴Acceleration, a= F m = 10 5 =2 m s −2

Q.35 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 ms-1 in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force

Answer

Here, mass of body, m = 3.0 kg initial speed of body, u = 2.0  ms -1 Final speed of body, v = 0 According to the first equation of motion: v=u+at ∴a= v−u t = 3.5−2 25 = 1.5 25 =0.06 m s −2 According to Newton’s second law of motion: F = ma =3×0.06 =0.18 N The force is along the direction of motion of the body.

Q.36 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms-1. How long does the body take to stop

Answer 

Here, retarding force applied to the body, F = -50 N Mass of the body, m = 20 kg Initial speed of the body, u = 15  ms -1 Final speed of the body, v = 0 According to Newton’s second law of motion, F = ma −50=20×a ∴a= −50 20 =−2.5 m s −2 According to the first equation of motion, v=u+at ∴t= −u a = −15 −2.5 =6 s One end of a string of length r is connected to a particle of mass m and the other end to a small peg on a smooth horizontal surface. If the particle moves in a circle with speed v, the net force on the particle (directed towards centre) is  (i) T (ii) T – mv 2 r  (ii) T + mv 2 r  (iv) 0 Answer When a particle connected to a string revolves in a circular path ,tension in the string provides the necessary centripetal force.In this case, the net force on the particle directed towards the centre is the force of tension,T. F = T Here, F is the total force acting on the particle.  ∴(i) T is the correct option.

Q.37 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,

(a) Just after it is dropped from the window of a stationary train,

(b) Just after it is dropped from the window of a train running at a constant velocity of 36 kmh-1,

(c) Just after it is dropped from the window of a train accelerating with 1 ms-2,

(d) Lying on the floor of a train which is acceleration with 1, the stone being at rest relative to the train. Neglect the resistance of air throughout.

Answer 

(a) Here, mass of the stone, m=0.1 kg Acceleration of the stone, a=g =10 m s −2 According to Newton’s second law of motion, Net force acting on the stone, F = ma =mg =0.1×10 =1 N ∴F =1 N, vertically downwards

(b) As the train is moving with constant velocity,  ∴Acceleration of the train = 0 ∴Force on the stone, F = weight of stone = mg = 0.1×10 = 1 N; vertically downwards

(c) As the train is accelerating with the rate of 1  ms -2 , an additional force  F’=ma =0.1×10 =0.1 N  acts on the stone in the horizontal direction. But, when the stone is dropped from the train, F’ =0 and the net force acting on the stone is due to its weight only. It is given as: F = mg = 0.1×10 = 1 N; vertically downwards

(d) As the stone is lying on the train, its weight is balanced by the normal reaction of the train.   ∴Acceleration of the stone = Acceleration of the train = 1  ms -2 ∴Force acting on the stone, F = ma = 0.1×1 = 0.1 N This force acts in the direction of motion of the train.

Q.38 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

(a) during its upward motion,

(b) during its downward motion,

(c) at the highest point where it is momentarily at rest.

Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction Ignore air resistance.

Answer

Acceleration due to gravity always acts in the downward direction, whatever may be the direction of motion of the object.

In all the three cases, the only force acting on the pebble is the gravitational force in vertically downward direction.

According to Newton’s second law of motion, its magnitude is given by F=ma Here, F=Net force m = Mass of the pebble =0.05 kg a=g =10  ms −2 ∴F=0.05×10 =0.5 N In all three cases, the net force acting on the pebble is 0.5 in vertically downward direction.

The answer remains unchanged even if pebble was thrown at an angle of 45˚with the horizontal direction as the direction of force will not change.

Q.39 Give the magnitude and direction of the net force acting on

(a) a drop of rain falling down with a constant speed,

(b) a cork of mass 10 g floating on water,

(c) a kite skillfully held stationary in the sky,

(d) a car moving with a constant velocity of 30 km/h on a rough road,

(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Answer

(a) As the rain drop is falling with a constant speed, its acceleration is zero. Thus, according to Newton’s second law of motion, the total force acting on the rain drop is zero.

(b) As the cork is floating on water, its weight is balanced by the buoyant force exerted by the water in the upward direction. Therefore, net force acting on the cork is zero.

(c) As the kite is stationary in the sky, therefore according to Newton’s first law of motion, no net force is acting on the kite.

(d) As the car is moving on a rough road with a constant velocity, its acceleration is zero. Therefore, according to Newton’s second law of motion, no net force is acting on the car.

(e) As the high speed electron is free from the effect of all fields, therefore, net force acting on the electron is zero.