Force and Laws of Motion

Welcome dear students! Today we are going to learn about Force_and_Laws_of_Motion from Class 9 Science. In the previous chapter, we described the motion of an object along a straight line in terms of its position, velocity and acceleration. We saw that such a motion can be uniform or non-uniform. We have not yet discovered what causes the motion. Why does the speed of an object change with time? Do all motions require a cause? If so, what is the nature of this cause? In this chapter we shall make an attempt to quench all such curiosities. For many centuries, the problem of motion and its causes had puzzled scientists and philosophers. A ball on the ground, when given a small hit, does not move forever. Such observations suggest that rest is the natural state of an object. This remained the belief until Galileo Galilei and Isaac Newton developed an entirely different approach to understand motion. In our everyday life we observe that some effort is required to put a stationary object into motion or to stop a moving object. We ordinarily experience this as a muscular effort and say that we must push or hit or pull on an object to change its state of motion. The concept of force is based on this push, hit or pull. Let us now ponder about a force. What is it? In fact, no one has seen, tasted or felt a force. However, we always see or feel the effect of a force. It can only be explained by describing what happens when a force is applied to an object. Pushing, hitting and pulling of objects are all ways of bringing objects in motion. They move because we make a force act on them. From your studies in earlier classes, you are also familiar with the fact that a force can 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. In one diagram, a trolley moves along the direction we push it. In another, a drawer is pulled. In a third, a hockey stick hits the ball forward. These show how pushing, pulling, or hitting objects change their state of motion. Another diagram shows a spring expanding on application of force, and a spherical rubber ball becoming oblong as we apply force on it. [CHECKPOINT] Now let us move on to balanced and unbalanced forces. Imagine a wooden block on a horizontal table. Two strings, X and Y, are tied to the two opposite faces of the block. If we apply a force by pulling the string X, the block begins to move to the right. Similarly, if we pull the string Y, the block moves to the left. But, if the block is pulled from both the sides with equal forces, the block will not move. Such forces are called balanced forces and do not change the state of rest or of motion of an object. Now, let us consider a situation in which two opposite forces of different magnitudes pull the block. In this case, the block would begin to move in the direction of the greater force. Thus, the two forces are not balanced and the unbalanced force acts in the direction the block moves. This suggests that an unbalanced force acting on an object brings it in motion. What happens when some children try to push a box on a rough floor? If they push the box with a small force, the box does not move because of friction acting in a direction opposite to the push. This friction force arises between two surfaces in contact, in this case, between the bottom of the box and floor's rough surface. It balances the pushing force and therefore the box does not move. If the children push the box harder but the box still does not move, it is because the friction force still balances the pushing force. If the children push the box harder still, the pushing force becomes bigger than the friction force. There is an unbalanced force. So the box starts moving. What happens when we ride a bicycle? When we stop pedalling, the bicycle begins to slow down. This is again because of the friction forces acting opposite to the direction of motion. In order to keep the bicycle moving, we have to start pedalling again. It thus appears that an object maintains its 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 in its speed or in the direction of motion would continue as long as this unbalanced force is applied. However, if this force is removed completely, the object would continue to move with the velocity it has acquired till then. [CHECKPOINT] Let us now learn about the 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. 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. Another diagram shows a marble resting on an ideal frictionless plane inclined on both sides. Galileo argued that when the marble is released from left, it would roll down the slope and go up on the opposite side to the same height from which it was released. If the inclinations of the planes on both sides are equal then the marble will climb the same distance that it covered while rolling down. If the angle of inclination of the right side plane were gradually decreased, then the marble would travel further distances till it reaches the original height. If the right side plane were ultimately made horizontal, that is, the slope is reduced to zero, the marble would continue to travel forever trying to reach the same height that it was released from. The unbalanced forces on the marble in this case are zero. It thus suggests that an unbalanced external force is required to change the motion of the marble but no net force is needed to sustain the uniform motion of the marble. In practical situations it is difficult to achieve a zero unbalanced force. This is because of the presence of the frictional force acting opposite to the direction of motion. Thus, in practice the marble stops after travelling some distance. The effect of the frictional force may be minimised by using a smooth marble and a smooth plane and providing a lubricant on top of the planes. Newton further studied Galileo's ideas on force and motion and presented three fundamental laws that govern the motion of objects. These three laws are known as Newton's laws of motion. The first law of motion is stated as: An object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force. In other words, all objects resist a change in their state of motion. In a qualitative way, the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why, the first law of motion is also known as the law of inertia. Certain experiences that we come across while travelling in a motorcar can be explained on the basis of the law of inertia. We tend to remain at rest with respect to the seat until the driver applies a braking force to stop the motorcar. With the application of brakes, the car slows down but our body tends to continue in the same state of motion because of its inertia. A sudden application of brakes may thus cause injury to us by impact or collision with the panels in front. Safety belts are worn to prevent such accidents. Safety belts exert a force on our body to make the forward motion slower. An opposite experience is encountered when we are standing in a bus and the bus begins to move suddenly. Now we tend to fall backwards. This is because the sudden start of the bus brings motion to the bus as well as to our feet in contact with the floor of the bus. But the rest of our body opposes this motion because of its inertia. When a motorcar makes a sharp turn at a high speed, we tend to get thrown to one side. This can again be explained on the basis of the law of inertia. We tend to continue in our straight line motion. When an unbalanced force is applied by the engine to change the direction of motion of the motorcar, we slip to one side of the seat due to the inertia of our body. [CHECKPOINT] The fact that a body will remain at rest unless acted upon by an unbalanced force can be illustrated through the following activities. Activity 8.1: Make a pile of similar carom coins on a table. Attempt a sharp horizontal hit at the bottom of the pile using another carom coin or the striker. If the hit is strong enough, the bottom coin moves out quickly. Once the lowest coin is removed, the inertia of the other coins makes them fall vertically on the table. Activity 8.2: Set a five rupee coin on a stiff card covering an empty glass tumbler standing on a table. Give the card a sharp horizontal flick with a finger. If we do it fast then the card shoots away, allowing the coin to fall vertically into the glass tumbler due to its inertia. The inertia of the coin tries to maintain its state of rest even when the card flows off. Activity 8.3: Place a water filled tumbler on a tray. Hold the tray and turn around as fast as you can. We observe that the water spills. Why? Observe that a groove is provided in a saucer for placing the tea cup. It prevents the cup from toppling over in case of sudden jerks. Now let us discuss inertia and mass. All the examples and activities given so far illustrate that there is a resistance offered by an object to change its state of motion. If it is at rest it tends to remain at rest, if it is moving it tends to keep moving. This property of an object is called its inertia. Do all bodies have the same inertia? We know that it is easier to push an empty box than a box full of books. Similarly, if we kick a football it flies away. But if we kick a stone of the same size with equal force, it hardly moves. We may, in fact, get an injury in our foot while doing so! Similarly, in activity 8.2, instead of a five rupees coin if we use a one rupee coin, we find that a lesser force is required to perform the activity. A force that is just enough to cause a small cart to pick up a large velocity will produce a negligible change in the motion of a train. This is because, in comparison to the cart the train has a much lesser tendency to change its state of motion. Accordingly, we say that the train has more inertia than the cart. Clearly, heavier or more massive objects offer larger inertia. Quantitatively, the inertia of an object is measured by its mass. We may 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. [CHECKPOINT] Let us answer the questions from this section. Question 1 asks which has more inertia: a rubber ball and a stone of the same size, a bicycle and a train, a five rupees coin and a one rupee coin. The answers are: the stone has more inertia because it has more mass. The train has more inertia because it has more mass. The five rupees coin has more inertia because it has more mass. Question 2 asks to identify the number of times the velocity of the ball changes in the football scenario and the agent supplying the force. The velocity changes four times. First kick by the first player, second kick by the teammate, third stop by the goalkeeper, fourth kick by the goalkeeper. The agents are the respective players and the goalkeeper. Question 3 asks why some leaves may get detached from a tree if we vigorously shake its branch. The leaves tend to remain at rest due to inertia while the branch moves. This relative motion causes the leaves to detach. Question 4 asks why you fall forward when a moving bus brakes and fall backwards when it accelerates. When braking, your body continues forward due to inertia. When accelerating, your feet move forward with the bus but your upper body tends to stay at rest due to inertia, making you fall backward. [CHECKPOINT] Now we will study the second law of motion. The first law of motion indicates that when an unbalanced external force acts on an object, its velocity changes, that is, the object gets an acceleration. We would now like to study how the acceleration of an object depends on the force applied to it and how we measure a force. Let us recount some observations from our everyday life. During the game of table tennis if the ball hits a player it does not hurt him. On the other hand, when a fast moving cricket ball hits a spectator, it may hurt him. A truck at rest does not require any attention when parked along a roadside. But a moving truck, even at speeds as low as 5 m s^-1, may kill a person standing in its path. A small mass, such as a bullet may kill a person when fired from a gun. These observations suggest that the impact produced by the objects depends on their mass and velocity. Similarly, if an object is to be accelerated, we know that a greater force is required to give a greater velocity. In other words, there appears to exist some quantity of importance that combines the object's mass and its velocity. One such property called momentum was introduced by Newton. The momentum, p of an object is defined as the product of its mass, m and velocity, v. That is, p = mv. Momentum has both direction and magnitude. Its direction is the same as that of velocity, v. The SI unit of momentum is kilogram metre per second, written as kg m s^-1. Since the application of an unbalanced force brings a change in the velocity of the object, it is therefore clear that a force also produces a change of momentum. Let us consider a situation in which a car with a dead battery is to be pushed along a straight road to give it a speed of 1 m s^-1, which is sufficient to start its engine. If one or two persons give a sudden push to it, it hardly starts. But a continuous push over some time results in a gradual acceleration of the car to this speed. It means that the change of momentum of the car is not only determined by the magnitude of the force but also by the time during which the force is exerted. It may then also be concluded that the force necessary to change the momentum of an object depends on the time rate at which the momentum is changed. The second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force. [CHECKPOINT] Let us look at the mathematical formulation of the second law of motion. Suppose an object of mass, m is moving along a straight line with an initial velocity, u. It is uniformly accelerated to velocity, v in time, t by the application of a constant force, F throughout the time, t. The initial and final momentum of the object will be, p_1 = mu and p_2 = mv respectively. The change in momentum is proportional to p_2 minus p_1, which is mv minus mu, or m times (v minus u). The rate of change of momentum is proportional to m times (v minus u) divided by t. Or, the applied force, F is proportional to m times (v minus u) divided by t. We write this as F equals k times m times (v minus u) divided by t, which equals k times m times a. Here a, which equals (v minus u) divided by t, is the acceleration, which is the rate of change of velocity. The quantity, k is a constant of proportionality. The SI units of mass and acceleration are kg and m s^-2 respectively. The unit of force is so chosen that the value of the constant, k becomes one. For this, one unit of force is defined as the amount that produces an acceleration of 1 m s^-2 in an object of 1 kg mass. That is, 1 unit of force equals k times 1 kg times 1 m s^-2. Thus, the value of k becomes 1. From this equation, we get F equals m times a. The unit of force is kg m s^-2 or newton, which has the symbol N. The second law of motion gives us a method to measure the force acting on an object as a product of its mass and acceleration. The second law of motion is often seen in action in our everyday life. Have you noticed that while catching a fast moving cricket ball, a fielder in the ground gradually pulls his hands backwards with the moving ball? In doing so, the fielder increases the time during which the high velocity of the moving ball decreases to zero. Thus, the acceleration of the ball is decreased and therefore the impact of catching the fast moving ball is also reduced. If the ball is stopped suddenly then its high velocity decreases to zero in a very short interval of time. Thus, the rate of change of momentum of the ball will be large. Therefore, a large force would have to be applied for holding the catch that may hurt the palm of the fielder. In a high jump athletic event, the athletes are made to fall either on a cushioned bed or on a sand bed. This is to increase the time of the athlete's fall to stop after making the jump. This decreases the rate of change of momentum and hence the force. Try to ponder how a karate player breaks a slab of ice with a single blow. [CHECKPOINT] The first law of motion can be mathematically stated from the mathematical expression for the second law of motion. The equation is F equals m a, or F equals m times (v minus u) divided by t, or F times t equals mv minus mu. That is, when F equals 0, v equals u for whatever time, t is taken. This means that the object will continue moving with uniform velocity, u throughout the time, t. If u is zero then v will also be zero. That is, the object will remain at rest. Now let us solve the worked examples step by step. Example 8.1: A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object's velocity from 3 m s^-1 to 7 m s^-1. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object? Solution: We have u equals 3 m s^-1, v equals 7 m s^-1, t equals 2 s, and m equals 5 kg. From the equation F equals m times (v minus u) divided by t, substitution gives F equals 5 kg times (7 m s^-1 minus 3 m s^-1) divided by 2 s, which equals 10 N. Now, if this force is applied for a duration of 5 s, we rewrite the equation as v equals u plus F times t divided by m. Substituting the values, we get v equals 3 plus 10 times 5 divided by 5, which equals 13 m s^-1. [CHECKPOINT] Example 8.2: Which would require a greater force, accelerating a 2 kg mass at 5 m s^-2 or a 4 kg mass at 2 m s^-2? Solution: From F equals m a, we have m_1 equals 2 kg, a_1 equals 5 m s^-2, and m_2 equals 4 kg, a_2 equals 2 m s^-2. Thus, F_1 equals m_1 times a_1 equals 2 kg times 5 m s^-2 equals 10 N. And F_2 equals m_2 times a_2 equals 4 kg times 2 m s^-2 equals 8 N. Since F_1 is greater than F_2, accelerating a 2 kg mass at 5 m s^-2 requires a greater force. Example 8.3: A motorcar is moving with a velocity of 108 km/h and it takes 4 s to 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 u equals 108 km/h, which converts to 108 times 1000 divided by 3600, giving 30 m s^-1. The final velocity v equals 0 m s^-1. Mass equals 1000 kg, time equals 4 s. Using F equals m times (v minus u) divided by t, we get F equals 1000 kg times (0 minus 30) m s^-1 divided by 4 s, which equals minus 7500 kg m s^-2 or minus 7500 N. The negative sign indicates the braking force is opposite to the direction of motion. Example 8.4: A force of 5 N gives a mass m_1 an acceleration of 10 m s^-2 and a mass m_2 an acceleration of 20 m s^-2. What acceleration would it give if both masses were tied together? Solution: Using m equals F divided by a, we get m_1 equals 5 N divided by 10 m s^-2 equals 0.50 kg. And m_2 equals 5 N divided by 20 m s^-2 equals 0.25 kg. Combined mass m equals 0.50 kg plus 0.25 kg equals 0.75 kg. The acceleration a equals F divided by m equals 5 N divided by 0.75 kg, which equals 6.67 m s^-2. [CHECKPOINT] Example 8.5: The velocity time graph of a ball of mass 20 g moving along a straight line on a long table is given. The graph shows velocity on the y axis from 0 to 25 cm/s and time on the x axis from 0 to 11 s. A straight line starts at 0 seconds and 20 cm/s, and ends at 10 seconds and 0 cm/s. How much force does the table exert on the ball to bring it to rest? Solution: The initial velocity u equals 20 cm s^-1. The final velocity v equals 0 cm s^-1. Time t equals 10 s. Since the graph is a straight line, acceleration is constant. Acceleration a equals (v minus u) divided by t, which equals (0 minus 20) divided by 10, giving minus 2 cm s^-2, or minus 0.02 m s^-2. The force F equals m times a. Mass is 20 g, which is 20 divided by 1000 kg. So F equals (20/1000) kg times minus 0.02 m s^-2, which equals minus 0.0004 N. The negative sign implies the frictional force exerted by the table is opposite to the direction of motion. Now let us move to the third law of motion. The first two laws tell us how an applied force changes motion and how to determine force. The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object. In football, when players collide, both feel hurt because each applies a force to the other. There is a pair of forces, known as action and reaction forces. Consider two spring balances connected together. The fixed end of balance B is attached to a rigid support. When a force is applied through the free end of spring balance A, both show the same readings. The force by A on B is equal and opposite to the force by B on A. Any can be action, the other reaction. This gives the alternative statement: to every action there is an equal and opposite reaction. Remember, action and reaction act on two different objects simultaneously. [CHECKPOINT] Suppose you are standing at rest and intend to start walking. You must accelerate, requiring a force. You push the road backwards. The road exerts an equal and opposite force on your feet to make you move forward. Even though action and reaction are equal, they may not produce equal accelerations because masses differ. When a gun is fired, it exerts a forward force on the bullet. The bullet exerts an equal and opposite force on the gun, causing recoil. Since the gun's mass is greater, its acceleration is much less. The third law is also seen when a sailor jumps out of a rowing boat. As the sailor jumps forward, the boat moves backwards. Activity 8.4: Request two children to stand on two separate carts. Give them a heavy bag. Ask them to play catch. Each experiences an instantaneous force when throwing. Painting a white line on cart wheels shows their motion. Placing two children on one cart and one on another demonstrates the second law, showing different accelerations for the same force. The cart can be made from a 12 mm or 18 mm thick plywood board of about 50 cm by 100 cm with two pairs of hard ball bearing wheels. Now, let us review what you have learnt. First law: 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 to resist change is inertia. Mass measures inertia. SI unit is kilogram. Friction opposes motion. Second law: Rate of change of momentum is proportional to applied unbalanced force. SI unit of force is kg m s^-2, called newton (N). One newton gives 1 kg an acceleration of 1 m s^-2. Momentum is mass times velocity, unit kg m s^-1. Third law: To every action, there is an equal and opposite reaction acting on two different bodies. [CHECKPOINT] Now we will solve all the exercises completely. Exercise 1: An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? Yes. The condition is that the object must already be moving with a constant velocity in a straight line, and the net zero force must continue to act on it. Exercise 2: When a carpet is beaten with a stick, dust comes out. Explain. The stick applies a sudden force to the carpet, putting it in motion. The dust particles tend to remain at rest due to inertia, so they separate from the carpet and fall out. Exercise 3: Why tie luggage on a bus roof with a rope? When the bus accelerates, brakes, or turns, the luggage tends to maintain its state of motion or rest due to inertia. Tying it prevents it from falling off. Exercise 4: A cricket ball rolls and stops because (c) there is a force on the ball opposing the motion, which is friction. Exercise 5: A truck starts from rest, rolls down a hill with constant acceleration, travels 400 m in 20 s. Find acceleration and force if mass is 7 tonnes. Using s equals u t plus half a t squared, with u equals 0, s equals 400, t equals 20. 400 equals 0 plus 0.5 times a times 400. So a equals 2 m s^-2. Mass is 7000 kg. Force F equals m a equals 7000 times 2 equals 14000 N. Exercise 6: A 1 kg stone thrown at 20 m s^-1 on ice stops after 50 m. Find friction force. Using v squared equals u squared plus 2 a s. 0 equals 20 squared plus 2 times a times 50. 0 equals 400 plus 100 a. So a equals minus 4 m s^-2. Force F equals m a equals 1 times minus 4 equals minus 4 N. The magnitude is 4 N. [CHECKPOINT] Exercise 7: An 8000 kg engine pulls 5 wagons of 2000 kg each. Engine force 40000 N, friction 5000 N. Find net accelerating force and acceleration. Total mass equals 8000 plus 5 times 2000 equals 18000 kg. Net force equals 40000 minus 5000 equals 35000 N. Acceleration a equals F divided by m equals 35000 divided by 18000, which equals 1.94 m s^-2. Exercise 8: A 1500 kg vehicle stops with negative acceleration 1.7 m s^-2. Force F equals m a equals 1500 times minus 1.7 equals minus 2550 N. The force between vehicle and road is 2550 N opposite to motion. Exercise 9: Momentum of mass m, velocity v is (d) mv. Exercise 10: Moving a cabinet at constant velocity with 200 N horizontal force. Since velocity is constant, net force is zero. Friction force equals applied force, so friction is 200 N. Exercise 11: Pushing a massive truck. The student says forces cancel. This logic is incorrect because action and reaction act on different objects. The truck does not move because the applied force is less than the maximum static friction between the truck and the road. Exercise 12: Hockey ball mass 200 g, initial velocity 10 m s^-1, returns at 5 m s^-1 opposite direction. Change in momentum equals m times (v minus u). u equals 10, v equals minus 5. Change equals 0.2 times (minus 5 minus 10) equals 0.2 times minus 15 equals minus 3 kg m s^-1. Magnitude is 3 kg m s^-1. [CHECKPOINT] Exercise 13: Bullet mass 10 g, velocity 150 m s^-1, stops in 0.03 s. Find penetration distance and force. Acceleration a equals (v minus u) divided by t equals (0 minus 150) divided by 0.03 equals minus 5000 m s^-2. Distance s equals u t plus half a t squared equals 150 times 0.03 plus 0.5 times minus 5000 times 0.03 squared equals 4.5 minus 2.25 equals 2.25 m. Force F equals m a equals 0.01 times minus 5000 equals minus 50 N. Magnitude is 50 N. Exercise 14: 1 kg object at 10 m s^-1 collides with 5 kg stationary block, they stick. Momentum before equals 1 times 10 plus 5 times 0 equals 10 kg m s^-1. Momentum after equals total mass times v equals 6 times v. By conservation, 10 equals 6v, so v equals 1.67 m s^-1. Exercise 15: 100 kg object accelerated from 5 to 8 m s^-1 in 6 s. Initial momentum equals 100 times 5 equals 500 kg m s^-1. Final momentum equals 100 times 8 equals 800 kg m s^-1. Force equals change in momentum divided by time equals (800 minus 500) divided by 6 equals 300 divided by 6 equals 50 N. Exercise 16: Insect hits car. Kiran says insect has greater change in momentum. Akhtar says car exerts larger force. Rahul says both experience same force and change in momentum. Rahul is correct. By Newton's third law, forces are equal and opposite. By conservation of momentum, the change in momentum is equal in magnitude for both. The insect dies due to its small mass causing huge acceleration. Exercise 17: 10 kg dumbbell falls from 80 cm. Downward acceleration 10 m s^-2. Find momentum transferred. Using v squared equals u squared plus 2 a s. v squared equals 0 plus 2 times 10 times 0.8 equals 16. So v equals 4 m s^-1. Momentum equals m v equals 10 times 4 equals 40 kg m s^-1. [CHECKPOINT] Now for the additional exercises. A1: Distance time table shows distance equals t cubed. Acceleration is increasing because distance is proportional to t cubed, meaning velocity is proportional to t squared, and acceleration is proportional to t. Since acceleration is increasing, the net force acting on the object is also increasing. A2: Two persons push 1200 kg car at uniform velocity, so net force is zero. Friction equals force of two persons. Three persons push to get acceleration 0.2 m s^-2. Net force equals m a equals 1200 times 0.2 equals 240 N. This net force is provided by the third person. So each person pushes with 240 N. A3: Hammer mass 500 g, velocity 50 m s^-1, stops in 0.01 s. Force equals m times (v minus u) divided by t equals 0.5 times (0 minus 50) divided by 0.01 equals minus 2500 N. Magnitude is 2500 N. A4: Car mass 1200 kg, velocity 90 km/h slows to 18 km/h in 4 s. Convert velocities: 90 km/h equals 25 m s^-1. 18 km/h equals 5 m s^-1. Acceleration a equals (v minus u) divided by t equals (5 minus 25) divided by 4 equals minus 5 m s^-2. Change in momentum equals m times (v minus u) equals 1200 times (5 minus 25) equals minus 24000 kg m s^-1. Force magnitude equals m times a equals 1200 times 5 equals 6000 N. [CHECKPOINT] We have now covered every concept, activity, example, and exercise from this chapter. Remember to practice the numerical problems regularly and understand the real life applications of Newton's laws. This knowledge is fundamental for your future studies in physics. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]