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# Chapter 9 Force and Laws of Motion

We have all heard about the great scientist Newton, he’s proposed different laws to explain motion. In this chapter we will discuss momentum and its effect with example. Conservation of momentum that means momentum can neither be created nor be destroyed.

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### Exercise 1

•  Q1 Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupees coin and a one-rupee coin? Ans: Inertia is the tendency of a body to oppose the changes in their state of rest or motion unless acted upon by an external force. Inertia depends on the mass of the body, higher the mass higher will be the inertia. a. Mass of a stone is greater than the mass of a rubber ball of the same size. Hence, the inertia of the stone is greater than that of a rubber ball. b. Mass of a train is greater than the mass of a bicycle. Hence, the inertia of the train is greater than that of the bicycle. c. Mass of a five rupee coin is greater than that of a one-rupee coin. Hence, inertia of the five rupee coin is greater than that of the one-rupee coin. Q2 In the following example, try to identify the number of times the velocity of the ball changes: “A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”. Also identify the agent supplying the force in each case. Ans: The velocity of the football changes four times . (i) When the first player kicks the football, its speed changes from zero to a certain value.  (ii) When another player kicks the ball towards the goal post , the direction of the ball gets changed. Therefore, its velocity also changes. (iii) When the goalkeeper collects the ball , the speed of the ball reduces to zero from a certain value . Hence , the velocity of the ball has changed. (iv) The goalkeeper kicks the ball towards his team players. Hence, the speed of the ball increases from zero to a certain value. Hence, its velocity changes once again.  Agent supplying the force In first case - first player In second case - second player In third case - goalkeeper In fourth case - goalkeeper Q3 Explain why some of the leaves may get detached from a tree if we vigorously shake its branch. Ans: Some leaves of the tree may get detached from a tree, if we vigorously shake its branch due to inertia of rest. Leaves are in the state of rest with respect to the tree so they possess inertia of rest. Due to this reason, the leaves fall down from the tree when shaken vigorously. Q4 Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest? Ans: When a moving bus stops suddenly, the passengers are jerked forward because of inertia the passengers tend to remain in their state of motion even though the bus has come to rest and we fall backwards when bus starts suddenly from rest because of inertia, passengers tend to remain in state of rest though bus starts moving. Hence, the passenger tends to fall backwards when the bus accelerates forward.

### Exercise 2

•  Q1 If action is always equal to the reaction, explain how a horse can pull a cart. Ans: To pull the cart horse pushes the ground with its foot in the backward direction by pressing the ground. As a reaction of this force, the ground pushes the horse in forward direction. Due to which the horse pulls the cart. Q2 Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity. Ans: When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then equal and opposite reaction forces act on the fireman according to Newton’s third law. Due to this reaction force ,stability of the fireman decreases and fireman faces difficulty to hold the hose. Q3 From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s-1. Calculate the initial recoil velocity of the rifle. Ans: Mass of the rifle, m1 = 4 kg Mass of the bullet, m2 = 50g = 0.05 kg Recoil velocity of the rifle = v1 Bullet is fired with an initial velocity, v2 = 35 m/s Total mass = m1 + m2 = 4 + 0.05 = 4.05 kg Since the system was at rest , velocity (v) = 0 Momentum ( before firing ) = mass x velocity = ( m1 + m2 )v = ( 4.05 ) 0 = 0 kg m/s Total momentum = momentum of bullet + momentum of rifle = m1 v1 + m2 v2  = 0.05 x 35 + 4 x v2 = 7/4 + 4 v2 Since the total momentum remain conserved  Total momentum before firing = Total momentum after firing 0 = 7/4 + 4 v2 4 v2 = - 7/4 v2 = - 7/16 v2 = -04375 m/s Thus , the recoil velocity of the pistol is -0.4375 m/s . The negative sign of velocity means that the rifle will recoil in the opposite direction of the bullet. Q4 Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m s-1 and 1 m s-1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m s-1. Determine the velocity of the second object. Ans: Mass of one of the objects, m1 = 100 g = 100/1000 = 0.1 kg  Mass of the other object, m2 = 200 g = 200/1000 = 0.2 kg Velocity of m1 before collision, v1 = 2 m/s Velocity of m2 before collision, v2 = 1 m/s Velocity of m1 after collision, v3 = 1.67 m/s Velocity of m2 after collision = v4 = ? By law of conservation of momentum : Total momentum before collision = Total momentum after collision m1v1 + m2v2 = m1v3 + m2v4 (0.1)2 + (0.2)1 = (0.1)1.67 + (0.2)v4 0.4 = 0.167 + 0.2v4 v4 = 1.165 m/s The velocity of the second object after collision 1.165 m/s.

### Exercise 3

•  Q1 An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the magnitude and direction of the velocity. If no, provide a reason. Ans: Yes. It is possible. (i) The object should already be moving at uniform speed in a straight line path. (ii) There is no change in speed. (iii) No direction change. (iv) Friction force must be zero. (v) Air resistance must be zero. Q2 When a carpet is beaten with a stick, dust comes out of it. Explain. Ans: When a carpet is beaten with a stick, the dust comes out of it because of the law of inertia. Initially the dust particles are at rest along with the carpet but, the dust particles try to resist their state of rest. According to Newton’s first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet. Q3 Why is it advised to tie any luggage kept on the roof of a bus with a rope? Ans: When a moving bus suddenly stops, the luggage on the roof tends to continue its state of motion and may fall. Also, when the bus suddenly starts from rest, luggage maintains its rest position and may fall backward. So, it is advised to tie any luggage kept on the roof of a bus with a rope. Q4 A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because (a) the batsman did not hit the ball hard enough. (b) velocity is proportional to the force exerted on the ball. (c) there is a force on the ball opposing the motion. (d) there is no unbalanced force on the ball, so the ball would want to come to rest. Ans: (c) there is a force on the ball opposing the motion. Q5 A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.) Ans: Initial velocity, u = 0 Distance travelled, s = 400 m Time taken, t = 20 s According to equation of motion:  a = 2 m/s2 1 metric tonne = 1000 kg (Given) ∴ 7 metric tonnes = 7000 kg Mass of truck, m = 7000 kg Force, F = Mass × Acceleration F = ma = 7000 × 2 = 14000 N The acceleration of the truck is 2 m/s2 and the force acting on the truck is 14000 N. Q6 A stone of 1 kg is thrown with a velocity of 20 m s-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice? Ans: Mass of the stone = 1 kg Initial velocity of the stone, u = 20 m/s Final velocity of the stone, v = 0 Distance travelled by the stone, s = 50 m Using the third equation of motion: v2 = u2 + 2as Where, Acceleration, a (0)2 = (20)2 + 2 × a × 50 a = −4 m/s2 Force, F = Mass × Acceleration F = ma F = 1 × (− 4) = −4 N The negative sign shows that force of friction is in the opposite direction of motion , the force of friction between the stone and the ice is −4 N. Q7 A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force; (b) the acceleration of the train; and (c) the force of wagon 1 on wagon 2. Ans: Mass of engine = 8000 kg Mass of 1 wagon = 2000 kg Mass of 5 wagons = 2000 x 5 = 10000 kg Force exerted by the engine, F = 40000 N Frictional force offered by the track, Ff = 5000 N (a) Net force, Fa = Force of engine - Force of friction = F − Ff       = 40000 − 5000 = 35000 N Hence, the net accelerating force is 35000 N. (b) Total mass of the train, M = mass of 5 wagons + mass of engine       = 10000 + 8000 = 18000 kg From Newton’s second law of motion: Force = Mass x acceleration 35000 = 18000 x a a = 35000 / 18000 = 1.94 m/s2 Hence, the acceleration of the train is 1.94 m/s2. Q8 An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s-2? Ans: Mass of the automobile vehicle, m = 1500 kg Acceleration of the automobile, a = −1.7 ms−2 We know that Force = Mass × Acceleration = 1500 × (−1.7) = −2550 N The negative sign shows that force is in the opposite direction of motion, the force between the automobile and the road is −2550 N. Q9 What is the momentum of an object of mass m, moving with a velocity v? (a) (mv)2      (b) mv2      (c) 1⁄2 mv2      (d) mv Ans: (d) mv Q10 Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet? Ans: Since, a horizontal force of 200 N is used to move a wooden cabinet. Thus, from Newton’s third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the frictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet. Q11 Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m s-1 before the collision during which they stick together. What will be the velocity of the combined object after collision? Ans: Mass of first objects, m1 = 1.5 kg Mass of second object, m2 = 1.5 kg Velocity of m1 before collision, v1 = 2.5 m/s Velocity of m2, moving in opposite direction before collision, v2 = −2.5 m/s (Negative sign arises because mass m2 is moving in an opposite direction) We know that Total momentum before collision = Total momentum after collision m1v1 + m2 v2 = (m1 + m2) v 1.5(2.5) + 1.5 (−2.5) = (1.5 + 1.5) v 3.75 − 3.75 = 3 v v = 0 / 3 = 0 Hence, the velocity of the combined object after collision is 0 m/s. Q12 According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move. Ans: Because of the huge mass of the truck, the force of static friction is very high. To move the car / truck, one has to apply a force more than the static friction. Therefore, when someone pushes the truck and the truck does not move, then it can be said that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction. Hence , the rationale given by the students is correct. Q13 A hockey ball of mass 200 g travelling at 10 m s-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 m s-1. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick. Ans: Mass of the hockey ball, m = 200 g = 200 / 1000 = 0.2 kg Initial velocity, v1 = 10 m/s Initial momentum = mv1 Final velocity, v2 = −5 m/s Final momentum = mv2 Change in momentum = mv1 − mv2 = m(v1 - v2) = 0.2 [10 − (−5)] = 0.2 (15) = 3.0 kg m/s Hence, the change in momentum of the hockey ball is 3 kg m/s. Q14 A bullet of mass 10 g travelling horizontally with a velocity of 150 m s-1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet. Ans: Mass of bullet, m = 10 g = 10 / 1000 = 0.01 kg Initial velocity, u = 150 m/s Final velocity, v = 0 Time taken, t = 0.03 s Using the first equation of motion  v = u + at 0 = 150 + (a ×0.03 s) (Negative sign indicates that the velocity of the bullet is decreasing.) Now, using the third equation of motion: v2 = u2 + 2as 0 = (150)2 + 2 (−5000) s Hence, the distance of penetration of the bullet is 2.25 m. We know that Force, F = Mass × Acceleration F = ma = 0.01 × 5000 = 50 N Hence, the magnitude of force is 50 N. Q15 An object of mass 1 kg travelling in a straight line with a velocity of 10 m s-1 collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object. Ans: Mass of the object, m1 = 1 kg Mass of the wooden block, m2 = 5 kg Velocity of the object before collision, v1 = 10 m/s Velocity of the wooden block before collision, v2 = 0 m/s Total mass of the combined system = m1 + m2 = 1 + 5 = 6 kg ∴ Total momentum before collision = m1 v1 + m2 v2 = 1 (10) + 5 (0) = 10 kg m s−1 Velocity of the combined object = v Momentum = total mass x velocity of combined object = (m1 + m2) v = 6v According to the law of conservation of momentum: Total momentum before collision = Total momentum after collision m1 v1 + m2 v2 = (m1 + m2) v 10 = 6 v The total momentum after collision is also 10 kg m/s. Hence, velocity of the combined object after collision = Q16 An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s-1 to 8 m s-1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object. Ans: Mass of the object, m = 100 kg Initial velocity of the object, u = 5 m/s Final velocity of the object, v = 8 m/s Time take by the object to accelerate, t = 6 s Initial momentum = mu = 100 × 5 = 500 kg m/s Final momentum = mv = 100 × 8 = 800 kg m/s By using Newton’s second law of motion, F = change in momentum / time taken =  Force exerted on the object is 50 N. Q17 Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions Ans: According to the law of conservation of momentum, the momentum of a system before collision is equal to the momentum of the system after collision. Hence, the change in momentum of the car and insect system is zero. Thus, Kiran is wrong with the logic as momentum is always conserved, i.e change in momentum of insect must be equal to that of a motorcar. Akhtar’s logic is incorrect because the mass of the car is very large as compared to the mass of the insect. Rahul gave the correct logic as insects and motorcar experience the same force and change in momentum. Q18 How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s-2. Ans: Mass of the dumbbell, m = 10 kg Distance, s = 80 cm = 80 / 100 = 0.8 m Acceleration, a = 10 m/s2 Initial velocity of the dumbbell, u = 0 Final velocity of the dumbbell = v Using the third equation of motion: v2 = u2 + 2as v2 = 0 + 2 (10) 0.8 v = 4 m/s Hence, the momentum with which the dumbbell hits the floor is = mv = 10 × 4 = 40 kg m/s

### Exercise 4

•  Q1 The following is the distance-time table of an object in motion:    Time in seconds       Distance in metres              0                                    0              1                                    1              2                                    8              3                                   27              4                                   64              5                                 125              6                                 216              7                                 343 (a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero? (b) What do you infer about the forces acting on the object? Ans: (a) From the above table, an unequal change of distance in an equal interval of time. Thus, the acceleration is non − uniform. Since the velocity of the object increases with time, the acceleration is increasing non-uniformly. (b) According to Newton’s second law of motion,  Force = mass x acceleration The force acting on an object is directly proportional to the acceleration produced in the object. Q2 Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.) Ans: Mass of the motor car = 1200 kg Only two persons are able to move the car with constant velocity. So, the acceleration acquired by the car is given by the third person. Thus the third person is responsible for the acceleration generated. Acceleration, a = 0.2 m/s2 From Newton’s second law of motion: Force = Mass × Acceleration F = 1200 × 0.2 = 240 N Hence, each person applies a force of 240 N to push the motor car. Q3 A hammer of mass 500 g, moving at 50 m s-1, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer? Ans: Mass of the hammer, m = 500 g = 500 / 1000 = 0.5 kg Initial velocity of the hammer, u = 50 m/s Time taken, t = 0.01 Since the hammer comes to rest, Velocity of the hammer, v = 0 Using Newton’s second law of motion: The hammer strikes the nail with a force of −2500 N. By Newton’s third law of motion, nail exerts an equal force in the opposite direction on the hammer, thus  the force of the nail on the hammer is equal and opposite, i.e., +2500 N. Q4 A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required. Ans: Mass of the motor car, m = 1200 kg Initial velocity of the motor car, u = 90 km/h = 90 x 5/18 = 25 m/s Final velocity of the motor car, v = 18 km/h = 18 x 5/18 = 5 m/s Time taken, t = 4 s Using the first equation of motion: v = u + at 5 = 25 + a (4) a = -5 m/s2 Change in momentum = mv − mu = m (v−u) = 1200 (5 − 25) = -24000 kg m s−1 Force = Mass × Acceleration = 1200 × (-5) = -6000 N Acceleration of the motor car = -5 m/s2 Change in momentum of the motor car = -24000 kg m s−1 Hence, the force required to decrease the velocity is -6000 N. (Negative sign indicates the retardation, decrease in momentum and retarding force respectively) Q5 A large truck and a car, both moving with a velocity of magnitude v, have a head-on collision and both of them come to a halt after that. If the collision lasts for 1 s: (a) Which vehicle experiences the greater force of impact? (b) Which vehicle experiences the greater change in momentum? (c) Which vehicle experiences the greater acceleration? (d) Why is the car likely to suffer more damage than the truck? Ans: (a)  Both vehicles experience equal forces of action and reaction because both of the objects are moving. (b)  Since both truck and car are moving with the same magnitude of velocity v, momentum change for both vehicles is the same. (c)  Mass and acceleration are inversely proportional to each other. Therefore, car experience greater acceleration due to their smaller mass. (d)  Due to its smaller mass or opposition to the force exerted on it, the car is likely to suffer more damage than the truck.

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