Force and laws of motion activities

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Published Date:28-07-2017
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C hapter 9 F F FORCE ORCE ORCE ORCE ORCE AND AND AND AND AND L L LAWS AWS AWS AWS AWS OF OF OF OF OF M M MOTION OTION OTION OTION OTION F F L L M M In the previous chapter, we described the In our everyday life we observe that some motion of an object along a straight line in effort is required to put a stationary object terms of its position, velocity and acceleration. into motion or to stop a moving object. We We saw that such a motion can be uniform ordinarily experience this as a muscular effort and say that we must push or hit or pull on or non-uniform. We have not yet discovered what causes the motion. Why does the speed an object to change its state of motion. The of an object change with time? Do all motions concept of force is based on this push, hit or require a cause? If so, what is the nature of pull. Let us now ponder about a ‘force’. What this cause? In this chapter we shall make an is it? In fact, no one has seen, tasted or felt a force. However, we always see or feel the effect attempt to quench all such curiosities. For many centuries, the problem of of a force. It can only be explained by motion and its causes had puzzled scientists describing what happens when a force is and philosophers. A ball on the ground, when applied to an object. Pushing, hitting and given a small hit, does not move forever. Such pulling of objects are all ways of bringing objects in motion (Fig. 9.1). They move observations suggest that rest is the “natural state” of an object. This remained the belief because we make a force act on them. until Galileo Galilei and Isaac Newton From your studies in earlier classes, you developed an entirely different approach to are also familiar with the fact that a force can understand motion. be used to change the magnitude of velocity of an object (that is, to make the object move faster or slower) or to change its direction of motion. We also know that a force can change the shape and size of objects (Fig. 9.2). (a) The trolley moves along the (b) The drawer is pulled. direction we push it. (a) (b) (c) The hockey stick hits the ball forward Fig. 9.2: (a) A spring expands on application of force; (b) A spherical rubber ball becomes oblong Fig. 9.1: Pushing, pulling, or hitting objects change their state of motion. as we apply force on with a small force, the box does not move 9.1 Balanced and Unbalanced because of friction acting in a direction Forces opposite to the push Fig. 9.4(a). This friction force arises between two surfaces in contact; Fig. 9.3 shows a wooden block on a horizontal in this case, between the bottom of the box table. Two strings X and Y are tied to the two and floor’s rough surface. It balances the opposite faces of the block as shown. If we pushing force and therefore the box does not apply a force by pulling the string X, the block move. In Fig. 9.4(b), the children push the begins to move to the right. Similarly, if we box harder but the box still does not move. pull the string Y, the block moves to the left. This is because the friction force still balances But, if the block is pulled from both the sides the pushing force. If the children push the with equal forces, the block will not move. box harder still, the pushing force becomes Such forces are called balanced forces and bigger than the friction force Fig. 9.4(c). do not change the state of rest or of motion of There is an unbalanced force. So the box an object. Now, let us consider a situation in starts moving. which two opposite forces of different What happens when we ride a bicycle? magnitudes pull the block. In this case, the When we stop pedalling, the bicycle begins block would begin to move in the direction of to slow down. This is again because of the the greater force. Thus, the two forces are friction forces acting opposite to the direction not balanced and the unbalanced force acts of motion. In order to keep the bicycle moving, in the direction the block moves. This we have to start pedalling again. It thus suggests that an unbalanced force acting on appears that an object maintains its motion an object brings it in motion. under the continuous application of an unbalanced force. However, it is quite incorrect. An object moves with a uniform velocity when the forces (pushing force and frictional force) acting on the object are balanced and there is no net external force on it. If an unbalanced force is applied on the object, there will be a change either in its speed or in the direction of its motion. Thus, to accelerate the motion of an object, an unbalanced force is required. And the change Fig. 9.3: Two forces acting on a wooden block in its speed (or in the direction of motion) would continue as long as this unbalanced What happens when some children try to force is applied. However, if this force is push a box on a rough floor? If they push the (a)(b)(c) Fig. 9.4 FORCE AND LAWS OF MOTION 115removed completely, the object would continue to move with the velocity it has acquired till then. 9.2 First Law of Motion By observing the motion of objects on an inclined plane Galileo deduced that objects move with a constant speed when no force acts on them. He observed that when a marble rolls down an inclined plane, its velocity increases Fig. 9.5(a). In the next chapter, you will learn that the marble falls under the unbalanced force of gravity as it rolls down and attains a definite velocity by the time it reaches the bottom. Its velocity decreases when it climbs up as shown in Fig. 9.5(b). Fig. 9.5(c) shows a marble resting on an ideal frictionless plane inclined on both sides. Fig. 9.5: (a) the downward motion; (b) the upward motion of a marble on an inclined plane; Galileo argued that when the marble is and (c) on a double inclined plane. released from left, it would roll down the slope and go up on the opposite side to the same Newton further studied Galileo’s ideas on height from which it was released. If the force and motion and presented three inclinations of the planes on both sides are fundamental laws that govern the motion of equal then the marble will climb the same objects. These three laws are known as distance that it covered while rolling down. If Newton’s laws of motion. The first law of the angle of inclination of the right-side plane motion is stated as: were gradually decreased, then the marble An object remains in a state of rest or of would travel further distances till it reaches uniform motion in a straight line unless the original height. If the right-side plane were compelled to change that state by an applied ultimately made horizontal (that is, the slope force. is reduced to zero), the marble would continue In other words, all objects resist a change to travel forever trying to reach the same in their state of motion. In a qualitative way, height that it was released from. The the tendency of undisturbed objects to stay unbalanced forces on the marble in this case at rest or to keep moving with the same are zero. It thus suggests that an unbalanced velocity is called inertia. This is why, the first (external) force is required to change the law of motion is also known as the law of motion of the marble but no net force is inertia. needed to sustain the uniform motion of the Certain experiences that we come across marble. In practical situations it is difficult while travelling in a motorcar can be to achieve a zero unbalanced force. This is explained on the basis of the law of inertia. because of the presence of the frictional force We tend to remain at rest with respect to the acting opposite to the direction of motion. seat until the driver applies a braking force Thus, in practice the marble stops after to stop the motorcar. With the application of travelling some distance. The effect of the brakes, the car slows down but our body frictional force may be minimised by using a tends to continue in the same state of motion smooth marble and a smooth plane and because of its inertia. A sudden application providing a lubricant on top of the planes. of brakes may thus cause injury to us by 116 SCIENCEimpact or collision with the panels in front. Galileo Galilei was born Safety belts are worn to prevent such on 15 February 1564 in accidents. Safety belts exert a force on our Pisa, Italy. Galileo, right body to make the forward motion slower. An from his childhood, had opposite experience is encountered when we interest in mathematics are standing in a bus and the bus begins to and natural philosophy. move suddenly. Now we tend to fall But his father backwards. This is because the sudden start Vincenzo Galilei wanted him to become a medical of the bus brings motion to the bus as well doctor. Accordingly, as to our feet in contact with the floor of the Galileo Galilei Galileo enrolled himself bus. But the rest of our body opposes this (1564 – 1642) for a medical degree at the motion because of its inertia. University of Pisa in 1581 which he never When a motorcar makes a sharp turn at completed because of his real interest in a high speed, we tend to get thrown to one mathematics. In 1586, he wrote his first side. This can again be explained on the basis scientific book ‘The Little Balance La of the law of inertia. We tend to continue in Balancitta’, in which he described our straight-line motion. When an Archimedes’ method of finding the relative unbalanced force is applied by the engine to densities (or specific gravities) of substances change the direction of motion of the using a balance. In 1589, in his series of motorcar, we slip to one side of the seat due essays – De Motu, he presented his theories to the inertia of our body. about falling objects using an inclined plane to slow down the rate of descent. The fact that a body will remain at rest In 1592, he was appointed professor of unless acted upon by an unbalanced force mathematics at the University of Padua in can be illustrated through the following the Republic of Venice. Here he continued his activities: observations on the theory of motion and through his study of inclined planes and the Activity ______________9.1 pendulum, formulated the correct law for • Make a pile of similar carom coins on uniformly accelerated objects that the a table, as shown in Fig. 9.6. distance the object moves is proportional to • Attempt a sharp horizontal hit at the the square of the time taken. bottom of the pile using another carom Galileo was also a remarkable craftsman. coin or the striker. If the hit is strong He developed a series of telescopes whose enough, the bottom coin moves out optical performance was much better than quickly. Once the lowest coin is that of other telescopes available during those removed, the inertia of the other coins days. Around 1640, he designed the first makes them ‘fall’ vertically on the table. pendulum clock. In his book ‘Starry Messenger’ on his astronomical discoveries, Galileo claimed to have seen mountains on the moon, the milky way made up of tiny stars, and four small bodies orbiting Jupiter. In his books ‘Discourse on Floating Bodies’ and ‘Letters on the Sunspots’, he disclosed his observations of sunspots. Using his own telescopes and through his observations on Saturn and Venus, Galileo argued that all the planets must orbit the Sun Fig. 9.6: Only the carom coin at the bottom of a and not the earth, contrary to what was pile is removed when a fast moving carom believed at that time. coin (or striker) hits it. FORCE AND LAWS OF MOTION 117five-rupees coin if we use a one-rupee coin, we Activity ______________9.2 find that a lesser force is required to perform • Set a five-rupee coin on a stiff card the activity. A force that is just enough to covering an empty glass tumbler cause a small cart to pick up a large velocity standing on a table as shown in will produce a negligible change in the motion Fig. 9.7. of a train. This is because, in comparison to • Give the card a sharp horizontal flick the cart the train has a much lesser tendency with a finger. If we do it fast then the to change its state of motion. Accordingly, we card shoots away, allowing the coin to say that the train has more inertia than the fall vertically into the glass tumbler due cart. Clearly, heavier or more massive objects to its inertia. offer larger inertia. Quantitatively, the inertia • The inertia of the coin tries to maintain its state of rest even when the card of an object is measured by its mass. We may flows off. thus relate inertia and mass as follows: Inertia is the natural tendency of an object to resist a change in its state of motion or of rest. The mass of an object is a measure of its inertia. uestions Fig. 9.7: When the card is flicked with the 1. Which of the following has more finger the coin placed over it falls in the inertia: (a) a rubber ball and a tumbler. stone of the same size? (b) a bicycle and a train? (c) a five- Activity ______________9.3 rupees coin and a one-rupee coin? Q 2. In the following example, try to • Place a water-filled tumbler on a tray. identify the number of times the • Hold the tray and turn around as fast as you can. velocity of the ball changes: • We observe that the water spills. Why? “A football player kicks a football to another player of his team who Observe that a groove is provided in a kicks the football towards the saucer for placing the tea cup. It prevents goal. The goalkeeper of the the cup from toppling over in case of sudden opposite team collects the football jerks. and kicks it towards a player of his own team”. Also identify the agent supplying 9.3 Inertia and Mass the force in each case. All the examples and activities given so far 3. Explain why some of the leaves illustrate that there is a resistance offered by may get detached from a tree if an object to change its state of motion. If it is we vigorously shake its branch. at rest it tends to remain at rest; if it is moving 4. Why do you fall in the forward it tends to keep moving. This property of an direction when a moving bus object is called its inertia. Do all bodies have brakes to a stop and fall the same inertia? We know that it is easier to backwards when it accelerates push an empty box than a box full of books. from rest? Similarly, if we kick a football it flies away. But if we kick a stone of the same size with 9.4 Second Law of Motion equal force, it hardly moves. We may, in fact, get an injury in our foot while doing so The first law of motion indicates that when Similarly, in activity 9.2, instead of a an unbalanced external force acts on an 118 SCIENCEobject, its velocity changes, that is, the object change the momentum of an object depends gets an acceleration. We would now like to on the time rate at which the momentum is study how the acceleration of an object changed. depends on the force applied to it and how The second law of motion states that the we measure a force. Let us recount some rate of change of momentum of an object is observations from our everyday life. During proportional to the applied unbalanced force the game of table tennis if the ball hits a player in the direction of force. it does not hurt him. On the other hand, when 9.4.1 MATHEMATICAL FORMULATION OF a fast moving cricket ball hits a spectator, it may hurt him. A truck at rest does not require SECOND LAW OF MOTION any attention when parked along a roadside. Suppose an object of mass, m is moving along But a moving truck, even at speeds as low as –1 a straight line with an initial velocity, u. It is 5 m s , may kill a person standing in its path. uniformly accelerated to velocity, v in time, t A small mass, such as a bullet may kill a by the application of a constant force, F person when fired from a gun. These throughout the time, t. The initial and final observations suggest that the impact momentum of the object will be, p = mu and produced by the objects depends on their 1 p = mv respectively. mass and velocity. Similarly, if an object is to 2 be accelerated, we know that a greater force The change in momentum ∝ p – p 2 1 is required to give a greater velocity. In other ∝ mv – mu words, there appears to exist some quantity ∝ m × (v – u). of importance that combines the object’s mv ×−()u mass and its velocity. One such property The rate of change of momentum ∝ called momentum was introduced by Newton. t The momentum, p of an object is defined as Or, the applied force, the product of its mass, m and velocity, v. mv ×−()u That is, F ∝ t p = mv (9.1) km×−() v u Momentum has both direction and F = (9.2) t magnitude. Its direction is the same as that of velocity, v. The SI unit of momentum is = k m a -1 kilogram-metre per second (kg m s ). Since (9.3) the application of an unbalanced force brings Here a = (v – u)/t is the acceleration, a change in the velocity of the object, it is which is the rate of change of velocity. The therefore clear that a force also produces a quantity, k is a constant of proportionality. The change of momentum. SI units of mass and acceleration are kg and Let us consider a situation in which a car -2 m s respectively. The unit of force is so chosen with a dead battery is to be pushed along a that the value of the constant, k becomes one. -1 straight road to give it a speed of 1 m s , For this, one unit of force is defined as the which is sufficient to start its engine. If one amount that produces an acceleration of 1 m or two persons give a sudden push -2 s in an object of 1 kg mass. That is, (unbalanced force) to it, it hardly starts. But -2 a continuous push over some time results in 1 unit of force = k × (1 kg) × (1 m s ). a gradual acceleration of the car to this speed. Thus, the value of k becomes 1. From Eq. (9.3) It means that the change of momentum of the car is not only determined by the F = ma (9.4) magnitude of the force but also by the time -2 The unit of force is kg m s or newton, during which the force is exerted. It may then which has the symbol N. The second law of also be concluded that the force necessary to FORCE AND LAWS OF MOTION 119motion gives us a method to measure the force The first law of motion can be acting on an object as a product of its mass mathematically stated from the mathematical and acceleration. expression for the second law of motion. Eq. The second law of motion is often seen in (9.4) is action in our everyday life. Have you noticed F = ma that while catching a fast moving cricket ball, a fielder in the ground gradually pulls his mv() −u = or F hands backwards with the moving ball? In t doing so, the fielder increases the time during (9.5) which the high velocity of the moving ball or Ft = mv – mu decreases to zero. Thus, the acceleration of the ball is decreased and therefore the impact That is, when F = 0, v = u for whatever time, t of catching the fast moving ball (Fig. 9.8) is is taken. This means that the object will also reduced. If the ball is stopped suddenly continue moving with uniform velocity, u then its high velocity decreases to zero in a throughout the time, t. If u is zero then v will very short interval of time. Thus, the rate of also be zero. That is, the object will remain change of momentum of the ball will be large. at rest. Therefore, a large force would have to be applied for holding the catch that may hurt Example 9.1 A constant force acts on an the palm of the fielder. In a high jump athletic event, the athletes are made to fall either on object of mass 5 kg for a duration of a cushioned bed or on a sand bed. This is to 2 s. It increases the object’s velocity –1 -1 increase the time of the athlete’s fall to stop from 3 m s to 7 m s . Find the after making the jump. This decreases the magnitude of the applied force. Now, if rate of change of momentum and hence the the force was applied for a duration of force. Try to ponder how a karate player 5 s, what would be the final velocity of breaks a slab of ice with a single blow. the object? Solution: –1 We have been given that u = 3 m s -1 and v = 7 m s , t = 2 s and m = 5 kg. From Eq. (9.5) we have, mv() −u = F t Substitution of values in this relation gives -1 -1 F = 5 kg (7 m s – 3 m s )/2 s = 10 N. Now, if this force is applied for a duration of 5 s (t = 5 s), then the final velocity can be calculated by rewriting Eq. (9.5) as Ft vu=+ m On substituting the values of u, F, m and Fig. 9.8: A fielder pulls his hands gradually with the t, we get the final velocity, moving ball while holding a catch. -1 v = 13 m s . 120 SCIENCESolution: Example 9.2 Which would require a greater force –– accelerating a 2 kg mass at 5 m From Eq. (9.4) we have m = F/a ; and 1 1 –2 -2 -2 s or a 4 kg mass at 2 m s ? m = F/a . Here, a = 10 m s ; 2 2 1 -2 a = 20 m s and F = 5 N. 2 Solution: -2 Thus, m = 5 N/10 m s = 0.50 kg; and 1 -2 From Eq. (9.4), we have F = ma. m = 5 N/20 m s = 0.25 kg. 2 -2 Here we have m = 2 kg; a = 5 m s If the two masses were tied together, 1 1 -2 and m = 4 kg; a = 2 m s . the total mass, m would be 2 2 -2 m = 0.50 kg + 0.25 kg = 0.75 kg. Thus, F = m a = 2 kg × 5 m s = 10 N; 1 1 1 -2 and F = m a = 4 kg × 2 m s = 8 N. The acceleration, a produced in the 2 2 2 combined mass by the 5 N force would ⇒ F F . 1 2 -2 be, a = F/m = 5 N/0.75 kg = 6.67 m s . Thus, accelerating a 2 kg mass at -2 5 m s would require a greater force. Example 9.5 The velocity-time graph of a ball of mass 20 g moving along a Example 9.3 A motorcar is moving with a straight line on a long table is given in velocity of 108 km/h and it takes 4 s to Fig. 9.9. stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg. Solution: The initial velocity of the motorcar u = 108 km/h = 108 × 1000 m/(60 × 60 s) -1 = 30 m s and the final velocity of the motorcar -1 v = 0 m s . The total mass of the motorcar along Fig. 9.9 with its passengers = 1000 kg and the How much force does the table exert on time taken to stop the motorcar, t = 4 s. the ball to bring it to rest? From Eq. (9.5) we have the magnitude of the force (F) applied by the brakes as Solution: m(v – u)/t. On substituting the values, we get -1 The initial velocity of the ball is 20 cm s . -1 F = 1000 kg × (0 – 30) m s /4 s Due to the friction force exerted by the -2 = – 7500 kg m s or – 7500 N. table, the velocity of the ball decreases –1 The negative sign tells us that the force down to zero in 10 s. Thus, u = 20 cm s ; -1 exerted by the brakes is opposite to the v = 0 cm s and t = 10 s. Since the direction of motion of the motorcar. velocity-time graph is a straight line, it is clear that the ball moves with a constant acceleration. The acceleration a is Example 9.4 A force of 5 N gives a mass vu − –2 m , an acceleration of 10 m s and a a = 1 -2 t mass m ,an acceleration of 20 m s . 2 -1 -1 = (0 cm s – 20 cm s )/10 s What acceleration would it give if both -2 -2 = –2 cm s = –0.02 m s . the masses were tied together? FORCE AND LAWS OF MOTION 121The force exerted on the ball F is, -2 F = ma = (20/1000) kg × (– 0.02 m s ) = – 0.0004 N. The negative sign implies that the frictional force exerted by the table is opposite to the direction of motion of Fig. 9.10: Action and reaction forces are equal and the ball. opposite. Suppose you are standing at rest and intend to start walking on a road. You must 9.5 Third Law of Motion accelerate, and this requires a force in The first two laws of motion tell us how an accordance with the second law of motion. applied force changes the motion and provide Which is this force? Is it the muscular effort us with a method of determining the force. you exert on the road? Is it in the direction The third law of motion states that when one we intend to move? No, you push the road object exerts a force on another object, the below backwards. The road exerts an equal second object instantaneously exerts a force and opposite reaction force on your feet to back on the first. These two forces are always make you move forward. equal in magnitude but opposite in direction. It is important to note that even though These forces act on different objects and never the action and reaction forces are always on the same object. In the game of football equal in magnitude, these forces may not sometimes we, while looking at the football produce accelerations of equal magnitudes. and trying to kick it with a greater force, This is because each force acts on a different collide with a player of the opposite team. object that may have a different mass. Both feel hurt because each applies a force When a gun is fired, it exerts a forward to the other. In other words, there is a pair of force on the bullet. The bullet exerts an equal forces and not just one force. The two and opposite reaction force on the gun. This opposing forces are also known as action and results in the recoil of the gun (Fig. 9.11). reaction forces. Since the gun has a much greater mass than Let us consider two spring balances the bullet, the acceleration of the gun is much connected together as shown in Fig. 9.10. The less than the acceleration of the bullet. The fixed end of balance B is attached with a rigid third law of motion can also be illustrated support, like a wall. When a force is applied when a sailor jumps out of a rowing boat. As through the free end of spring balance A, it is the sailor jumps forward, the force on the boat observed that both the spring balances show moves it backwards (Fig. 9.12). the same readings on their scales. It means that the force exerted by spring balance A on balance B is equal but opposite in direction to the force exerted by the balance B on balance A. The force which balance A exerts on balance B is called the action and the force of balance B on balance A is called the reaction. This gives us an alternative statement of the third law of motion i.e., to every action there is an equal and opposite reaction. However, it must be remembered Fig. 9.11: A forward force on the bullet and recoil of that the action and reaction always act on two the gun. different objects. 122 SCIENCEThe cart shown in this activity can be constructed by using a 12 mm or 18 mm thick plywood board of about 50 cm × 100 cm with two pairs of hard ball-bearing wheels (skate wheels are good to use). Skateboards are not as effective because it is difficult to maintain straight-line motion. 9.6 Conservation of Momentum Suppose two objects (two balls A and B, say) of masses m and m are travelling in the same A B Fig. 9.12: As the sailor jumps in forward direction, direction along a straight line at different the boat moves backwards. velocities u and u , respectively Fig. 9.14(a). A B And there are no other external unbalanced forces acting on them. Let u u and the Activity ______________9.4 A B two balls collide with each other as shown in • Request two children to stand on two Fig. 9.14(b). During collision which lasts for separate carts as shown in Fig. 9.13. a time t, the ball A exerts a force F on ball B AB • Give them a bag full of sand or some and the ball B exerts a force F on ball A. BA other heavy object. Ask them to play a Suppose v and v are the velocities of the two A B game of catch with the bag. balls A and B after the collision, respectively • Does each of them receive an Fig. 9.14(c). instantaneous reaction as a result of throwing the sand bag (action)? • You can paint a white line on cartwheels to observe the motion of the two carts when the children throw the bag towards each other. Fig. 9.14: Conservation of momentum in collision of two balls. From Eq. (9.1), the momenta (plural of momentum) of ball A before and after the collision are m u and m v , respectively. The A A A A rate of change of its momentum (or F , action) AB () vu − AA m during the collision will be . A t Similarly, the rate of change of momentum of Fig. 9.13 ball B (= F or reaction) during the collision BA () vu − BB Now, place two children on one cart and m will be . B t one on another cart. The second law of motion According to the third law of motion, the can be seen, as this arrangement would show force F exerted by ball A on ball B (action) different accelerations for the same force. AB FORCE AND LAWS OF MOTION 123and the force F exerted by the ball B on ball BA Activity ______________9.6 A (reaction) must be equal and opposite to • Take a test tube of good quality glass each other. Therefore, material and put a small amount of F = – F (9.6) AB BA water in it. Place a stop cork at the () vu − () vu − AA BB mouth of it. m m or = – . A B t t • Now suspend the test tube horizontally This gives, by two strings or wires as shown in Fig. 9.16. m u + m u = m v + m v (9.7) A A B B A A B B • Heat the test tube with a burner until water vaporises and the cork blows Since (m u + m u ) is the total momentum A A B B out. of the two balls A and B before the collision • Observe that the test tube recoils in and (m v + m v ) is their total momentum A A B B the direction opposite to the direction after the collision, from Eq. (9.7) we observe that the total momentum of the two balls remains unchanged or conserved provided no other external force acts. As a result of this ideal collision experiment, we say that the sum of momenta of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum. This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision. Activity ______________9.5 • Take a big rubber balloon and inflate of the cork. it fully. Tie its neck using a thread. Fig. 9.16 Also using adhesive tape, fix a straw on the surface of this balloon. • Also, observe the difference in the • Pass a thread through the straw and velocity the cork appears to have and hold one end of the thread in your that of the recoiling test tube. hand or fix it on the wall. • Ask your friend to hold the other end of the thread or fix it on a wall at some Example 9.6 A bullet of mass 20 g is distance. This arrangement is shown horizontally fired with a velocity in Fig. 9.15. -1 150 m s from a pistol of mass 2 kg. • Now remove the thread tied on the What is the recoil velocity of the pistol? neck of balloon. Let the air escape from the mouth of the balloon. Solution: • Observe the direction in which the straw moves. We have the mass of bullet, m = 20 g (= 0.02 kg) and the mass of 1 the pistol, m = 2 kg; initial velocities of 2 the bullet (u ) and pistol (u ) = 0, 1 2 respectively. The final velocity of the -1 bullet, v = + 150 m s . The direction of 1 bullet is taken from left to right (positive, by convention, Fig. 9.17). Let v be the Fig. 9.15 124 SCIENCErecoil velocity of the pistol. Example 9.7 A girl of mass 40 kg jumps Total momenta of the pistol and bullet -1 with a horizontal velocity of 5 m s onto before the fire, when the gun is at rest a stationary cart with frictionless –1 = (2 + 0.02) kg × 0 m s wheels. The mass of the cart is 3 kg. –1 = 0 kg m s What is her velocity as the cart starts Total momenta of the pistol and bullet moving? Assume that there is no after it is fired external unbalanced force working in –1 = 0.02 kg × (+ 150 m s ) the horizontal direction. –1 + 2 kg × v m s –1 = (3 + 2v) kg m s Solution: According to the law of conservation of Let v be the velocity of the girl on the momentum cart as the cart starts moving. Total momenta after the fire = Total The total momenta of the girl and cart momenta before the fire before the interaction 3 + 2v = 0 –1 –1 –1 ⇒ v = − 1.5 m s . = 40 kg × 5 m s + 3 kg × 0 m s –1 Negative sign indicates that the direction = 200 kg m s . in which the pistol would recoil is Total momenta after the interaction opposite to that of bullet, that is, right –1 to left. = (40 + 3) kg × v m s –1 = 43 v kg m s . According to the law of conservation of momentum, the total momentum is conserved during the interaction. That is, 43 v = 200 –1 ⇒ v = 200/43 = + 4.65 m s . The girl on cart would move with a –1 velocity of 4.65 m s in the direction in Fig. 9.17: Recoil of a pistol which the girl jumped (Fig. 9.18). (a) (b) Fig. 9.18: The girl jumps onto the cart. FORCE AND LAWS OF MOTION 125If v is the velocity of the two entangled Example 9.8 Two hockey players of players after the collision, the total opposite teams, while trying to hit a momentum then hockey ball on the ground collide and =(m + m ) × v 1 2 immediately become entangled. One –1 = (60 + 55) kg × v m s has a mass of 60 kg and was moving –1 = 115 × v kg m s . –1 with a velocity 5.0 m s while the other Equating the momenta of the system has a mass of 55 kg and was moving before and after collision, in accordance –1 faster with a velocity 6.0 m s towards with the law of conservation of the first player. In which direction and momentum, we get with what velocity will they move after v= – 30/115 –1 they become entangled? Assume that = – 0.26 m s . the frictional force acting between the feet Thus, the two entangled players would –1 of the two players and ground is move with velocity 0.26 m s from right negligible. to left, that is, in the direction the second player was moving before Solution: the collision. Fig. 9.19: A collision of two hockey players: (a) before collision and (b) after collision. Let the first player be moving from left uestions to right. By convention left to right is taken as the positive direction and thus 1. If action is always equal to the right to left is the negative direction (Fig. reaction, explain how a horse 9.19). If symbols m and u represent the can pull a cart. mass and initial velocity of the two 2. Explain, why is it difficult for a players, respectively. Subscripts 1 and Q fireman to hold a hose, which 2 in these physical quantities refer to the ejects large amounts of water at two hockey players. Thus, -1 m = 60 kg; u = + 5 m s ; and a high velocity. 1 1 -1 m = 55 kg; u = – 6 m s . 3. From a rifle of mass 4 kg, a bullet 2 2 The total momentum of the two players of mass 50 g is fired with an before the collision –1 initial velocity of 35 m s . -1 = 60 kg × (+ 5 m s ) + Calculate the initial recoil -1 55 kg × (– 6 m s ) velocity of the rifle. -1 = – 30 kg m s 126 SCIENCE4. Two objects of masses 100 g and respectively. They collide and 200 g are moving along the same after the collision, the first object –1 line and direction with velocities moves at a velocity of 1.67 m s . –1 –1 of 2 m s and 1 m s , Determine the velocity of the second object. CONSERVATION LAWS All conservation laws such as conservation of momentum, energy, angular momentum, charge etc. are considered to be fundamental laws in physics. These are based on observations and experiments. It is important to remember that a conservation law cannot be proved. It can be verified, or disproved, by experiments. An experiment whose result is in conformity with the law verifies or substantiates the law; it does not prove the law. On the other hand, a single experiment whose result goes against the law is enough to disprove it. The law of conservation of momentum has been deduced from large number of observations and experiments. This law was formulated nearly three centuries ago. It is interesting to note that not a single situation has been realised so far, which contradicts this law. Several experiences of every-day life can be explained on the basis of the law of conservation of momentum. What you have learnt • First law of motion: An object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force. • The natural tendency of objects to resist a change in their state of rest or of uniform motion is called inertia. • The mass of an object is a measure of its inertia. Its SI unit is kilogram (kg). • Force of friction always opposes motion of objects. • Second law of motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force. –2 • The SI unit of force is kg m s . This is also known as newton and represented by the symbol N. A force of one newton –2 produces an acceleration of 1 m s on an object of mass 1 kg. • The momentum of an object is the product of its mass and velocity and has the same direction as that of the velocity. –1 Its SI unit is kg m s . • Third law of motion: To every action, there is an equal and opposite reaction and they act on two different bodies. • In an isolated system (where there is no external force), the total momentum remains conserved. FORCE AND LAWS OF MOTION 127Exercises 1. 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. 2. When a carpet is beaten with a stick, dust comes out of it. Explain. 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope? 4. 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. 5. 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 tonnes (Hint: 1 tonne = 1000 kg.) –1 6. A stone of 1 kg is thrown with a velocity of 20 m s 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? 7. 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. 8. 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 –2 stopped with a negative acceleration of 1.7 m s ? 9. What is the momentum of an object of mass m, moving with a velocity v? 2 2 2 (a) (mv) (b) mv (c) ½ mv (d) mv 10. 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? 11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each 128 SCIENCE-1 object is 2.5 m s before the collision during which they stick together. What will be the velocity of the combined object after collision? 12. 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. –1 13. A hockey ball of mass 200 g travelling at 10 m s is struck by a hockey stick so as to return it along its original path with a –1 velocity at 5 m s . Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick. 14. A bullet of mass 10 g travelling horizontally with a velocity of –1 150 m s 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. 15. An object of mass 1 kg travelling in a straight line with a velocity –1 of 10 m s 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. 16. An object of mass 100 kg is accelerated uniformly from a velocity –1 –1 of 5 m s to 8 m s in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object. 17. 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. 18. 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 –2 acceleration to be 10 m s . FORCE AND LAWS OF MOTION 129Additional Exercises A1. The following is the distance-time table of an object in motion: Time in seconds Distance in metres 00 11 28 327 464 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? A2. 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 -2 0.2 m s . With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.) -1 A3. A hammer of mass 500 g, moving at 50 m s , 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? A4. 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. 130 SCIENCE

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