Force and Laws of Motion

"We can control the magnitude of the applied force and also its direction, so force is a vector quantity, just like velocity and acceleration."

A cricketer strikes the ball with his bat, applying a force and directing the ball into a particular part of the ground. He can hit the ball at different speeds and direct it into different parts of the ground.

1.0Introduction

In common usage, a force is a push or a pull on some object. If you pull hard enough on a trolley [see figure (a)], the trolley moves. If a player kicks the football [see figure (b)], the football moves.

Examples of some contact forces

• There are two types of forces : contact forces and non-contact forces. The types of contact forces are muscular force, tension, friction etc. The types of non-contact force are electrostatic force, magnetic force, gravitational force.
• Frictional force not only opposes motion but also helps a body to move. It is friction that helps us walk and run.
• The earth is acted upon by the gravitational pull of the stars, planets, sun and moon. However, the gravitational pull of the sun is the most dominant.

2.0Effects of force

(1) A force can distort an object i.e. it can change the shape and size of an object. (2) A force can start an object at rest i.e. it can move a stationary object. (3) A force can stop a moving object i.e. it can cause a moving object to come to rest. (4) A force can change the speed or the magnitude of velocity of an object i.e. it can increase or decrease the speed of an object. (5) A force can change the direction of a moving

3.0Balanced and unbalanced forces

Balanced forces

• If the net force exerted on an object is zero, then the forces acting on it are said to be balanced. In such a case, the acceleration of the object is zero and its velocity remains constant. That is, if the net force acting on the object is zero, the object either remains at rest or continues to move with constant velocity.

Unbalanced forces

• If the net force exerted on an object is not zero, then the forces acting on it are said to be unbalanced. In such a case, the acceleration of the object is not zero and its velocity changes. That is, if the net force acting on the object is not zero, then such a force changes state of rest or the state of uniform motion of the object.

If there is an unbalanced force acting on an object, the change in its velocity would continue as long as this unbalanced force persists. If this force is removed completely, the object would continue to move with the velocity it has acquired till then.

An object is in equilibrium when it has zero acceleration. This means, it is the state of an object or system of objects for which there are no changes in motion. It includes the state of rest as well as the state of uniform motion.

• Q. What happens when some children try to push a box on a rough floor? Explanation: 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 [see figure (a)]. This frictional force arises between two surfaces in contact. Here, the friction is 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 [see figure (b)], the box still does not move. This is because the frictional force still balances the pushing force. If the children continue to increase the push force on the box, at some point the pushing force becomes bigger than the frictional force [see figure (c)]. That is, there is an unbalanced force due to which the box starts moving (accelerating).
  

4.0Galileo's inclined planes

Galileo studied motion of objects on an inclined plane. He noted that balls rolling down [see figure (a)] the inclined planes picked up speed (i.e. acceleration), while balls rolling on [see figure (b)] up the inclined planes lost speed (i.e. retardation). From this he reasoned that balls rolling on [see figure (c)] a horizontal plane would neither speed up nor slow down. The ball would finally come to rest not because of its 'nature' but because of friction. This idea was supported by Galileo's observation of motion along smoother surfaces. When there was less friction, the motion of objects persisted for a longer time. The smaller the friction, the more the motion approached constant velocity. Galileo concluded that an object moving on a frictionless horizontal plane must neither have acceleration nor retardation, i.e. it should move with constant velocity.

Galileo concluded that 'a ball rolling down the first inclined plane on the left tends to roll up to its initial height on the second plane on the right, thus, the ball must roll a greater distance as the angle of the second inclined plane on the right is reduced' [(see figure (b)]. He argued that when the slope of the second plane is made zero i.e. it becomes a horizontal plane, the ball must travel an infinite distance since it can never reach its initial height on first plane [(see figure (c)]. In other words, its motion never ceases. This is, of course, an idealised situation.

• Galileo arrived at a new insight that 'the state of rest and the state of uniform motion (motion with constant velocity) are equivalent'. In both cases, there is no net force acting on the body. It is incorrect to assume that a net force is needed to keep a body in uniform motion. To maintain a body in uniform motion, we need to apply an external force to counter the frictional force, so that the two forces sum up to zero net external force. A body does not change its state of rest or uniform motion, unless an external force compels it to change that state. The tendency of things to resist changes in motion was what Galileo called inertia.
  
  (a)
  
  (b)
  
  (c) Galileo's observations on motion of a ball on a double inclined plane

5.0Newton's first law of motion

Newton built on Galileo's ideas and laid the foundation of mechanics in terms of three laws of motion that go by his name. Galileo's law of inertia was his starting point which he formulated as the first law of motion : 'Every object continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it'.

Concept of Inertia

• It should be noted that Newton's first law of motion is also called Galileo's law of inertia.

• Q. Do all bodies have same inertia? Is there any relation between inertia and mass of body? Explanation: 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 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.

• The mass of an object is a quantitative measure of inertia. More the mass, more will be the inertia of an object and vice-versa. Inertia of an object can be of three types : (1) Inertia of rest (2) Inertia of motion (3) Inertia of direction

Inertia of rest

It is the tendency of an object to remain at rest. This means an object at rest remains at rest until a sufficiently large external force is applied on it.

Examples of inertia of rest

Inertia of motion

It is the tendency of an object to remain in the state of uniform motion. This means an object in uniform motion continues to move uniformly until an external force is applied on it.

Examples of inertia of motion

Inertia of direction

It is the tendency of an object to maintain its direction. This means an object moving in a particular direction continues to move in that direction until an external force is applied to change it.

Examples of inertia of direction

Air bags in car

• a safety device: If a head-on collision of a car is violent enough, sodium azide undergoes a rapid chemical reaction to produce non-toxic nitrogen gas, which inflates an airbag. The inflated airbag provides a protective cushion to slow down the head and body of a motorist.
  
  Airbag systems in cars are designed to safeguard the motorist during vehicle collisions

6.0Linear momentum (or momentum)

The linear momentum of a particle or an object that has a mass ' $m$ ' moving with a velocity ' $v$ ' is defined to be the product of the mass and velocity.

• Linear momentum is a vector quantity, because it equals the product of a scalar quantity ' $m$ ' and a vector quantity ' $v$ '. The direction of linear momentum is 'the direction along the velocity'. $\mathbf{p}=\mathbf{m}\times \mathbf{v}$
• If there is an unbalanced force acting on an object, its velocity changes, hence, its momentum also changes. If the forces acting on an object are balanced, its velocity is constant, hence, its momentum is also constant.
• The linear momentum of a particle is (i) directly proportional to its mass (ii) directly proportional to its velocity

Units of linear momentum

• Q. (i) A car and a truck both are moving with same velocities. Which one has more momentum? (ii) A car and a truck have same momentum. Which one has more velocity? Explanation: (i) For a given velocity, the momentum is directly proportional to the mass of the object. This means more the mass, more will be the momentum and vice-versa. Here, velocity of both truck and car is same. Since, the mass of a truck is greater than the mass of a car, therefore, the momentum of truck is more than the momentum of car. (ii) For a given momentum, the velocity is inversely proportional to the mass of the object. This means smaller the mass, more will be the velocity of an object and viceversa. Here, momentum of both truck and car is same. Since, the mass of a car is smaller than the mass of a truck, therefore, the velocity of car is more than the velocity of truck.

Numerical Ability

Decode the problem To find the momentum, we should know the mass and velocity of the object. Apply the formula $\mathrm{p}=\mathrm{mv}$

• Momentum of an object is $20{\mathbf{kg}\mathbf{m}\mathbf{s}}^{-1}$. What will be its momentum if (a) its mass is doubled but the velocity remains the same? (b) the velocity is reduced to $(1/3)$ of its original magnitude but mass remains the same? Solution: Let the mass of the object be $m$ and its velocity be $v$. Initial value of momentum, $p=mv=20\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$ (a) New mass, ${\mathrm{m}}^{\prime }=2\mathrm{m}$, velocity remains the same i.e. ${\mathrm{v}}^{\prime }=\mathrm{v}$ New value of momentum, ${\mathrm{p}}^{\prime }={\mathrm{m}}^{\prime }{\mathrm{v}}^{\prime }=(2\mathrm{m})(\mathrm{v})$ $=2\mathrm{mv}=2\times 20=40\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$ (b) New velocity, ${\mathrm{v}}^{\prime }=\mathrm{v}/3$, mass remains the same i.e., ${\mathrm{m}}^{\prime }=\mathrm{m}$ New value of momentum, ${\mathrm{p}}^{\prime }={\mathrm{m}}^{\prime }{\mathrm{v}}^{\prime }=\mathrm{m}(\mathrm{v}/3)=(\mathrm{mv})/3$ $=20/3=6.67{\mathbf{kg}\mathbf{m}\mathbf{s}}^{-1}$
• Calculate the momentum of a motor car of mass 800 kg moving with a velocity of $10\mathrm{m}/\mathrm{s}$. Solution: Given, mass $=800\mathrm{kg}$; velocity, $\mathrm{v}=10\mathrm{m}/\mathrm{s}$ $\mathrm{p}=\mathrm{mv}=800\times 10$ $=8000\mathbf{kg}\mathbf{m}/\mathrm{s}$
• A $65-\mathbf{kg}$ girl is driving a $535-\mathrm{kg}$ car at a constant velocity of $11.5\mathbf{m}/\mathrm{s}$. Calculate the momentum of the girl-car system. Solution: Since, we have to find the momentum of the girl-car system, the total mass of the system, $\mathrm{m}=$ mass of girl + mass of car $=65+535=600\mathrm{kg}$ Velocity of car, $\mathrm{v}=11.5\mathrm{m}/\mathrm{s}$ Now, momentum, $\mathrm{p}=\mathrm{mv}=600\times 11.5$ $=6900{\mathbf{kg}\mathbf{m}\mathbf{s}}^{-1}$
• The momentum of a $75-\mathrm{g}$ bullet is $9\mathrm{kg}\mathrm{m}/\mathrm{s}$. What is the velocity of the bullet? Solution: Given, mass of bullet, $\mathrm{m}=75\mathrm{g}=(75/1000)\mathrm{kg}=0.075\mathrm{kg}$ momentum, $\mathrm{p}=9\mathrm{kg}\mathrm{m}/\mathrm{s}$; velocity, $\mathrm{v}=$ ? Now, momentum, $\mathrm{p}=\mathrm{mv}$ $\text{ or }\begin{array}{rl}\mathrm{v}& =\frac{\mathrm{p}}{\mathrm{m}}=\frac{9}{0.075}=\frac{9\times 1000}{75}\\ & =120\mathrm{m}/\mathrm{s}\end{array}$

7.0Newton's second law of motion

According to the second law of motion, 'the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts'.

Mathematically, the force, $\mathrm{F}\propto \frac{\Delta \mathrm{p}}{\mathrm{t}}$ where, $\Delta \mathrm{p}=$ change in momentum, $\mathrm{t}=$ time interval.

Mathematical formulation of second law of motion

Let 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, ${\mathrm{p}}_1=\mathrm{mu}$ and $p_2=mv$, respectively.

The change in momentum, $\Delta \mathrm{p}={\mathrm{p}}_2-{\mathrm{p}}_1=\mathrm{mv}-\mathrm{mu}=\mathrm{m}(\mathrm{v}-\mathrm{u})$ The force F is proportional to the rate of change of momentum, that is,

where, k is a constant of proportionality. Now, acceleration, $\mathrm{a}=\frac{(\mathrm{v}-\mathrm{u})}{\mathrm{t}}$ From eq. (1) & eq. (2), we get, $\mathrm{F}=\mathrm{km}$ a The SI units of mass and acceleration are kg and $\mathrm{m}{\mathrm{s}}^{-2}$ respectively. The unit of force is so chosen that the value of the constant ' $k$ ' becomes one. That is, 1 unit of force $=\mathrm{k}\times (1\mathrm{kg})\times \left(1\mathrm{m}{\mathrm{s}}^{-2}\right)$ or $\mathrm{k}=1$ Thus, the value of k becomes 1 . Therefore, the eq.(3) reduces to, $\mathrm{F}=\mathbf{ma}$

Unit of force

• Q. Equal forces are applied to a car and a truck which produce acceleration in both of them. Which one has smaller acceleration? Explanation: For a given force, the acceleration produced in the object is inversely proportional to the mass of the object $(a\propto 1/m)$. This means more the mass of an object, less will be the acceleration produced in it. Here, force applied on the truck as well as the car is same. Since, the mass of a truck is greater than the mass of a car, therefore, the acceleration of the truck is smaller than the acceleration of the car.

Applications/results of Newton's second law

(i) Suppose a light-weight vehicle (say a small car) and a heavy-weight vehicle (say a loaded truck) are parked on a horizontal road. A much greater force is needed to push the truck than the car to bring them to the same speed in same time. This is because, for a given time interval, the force is directly proportional to the change in momentum. Here, the change in momentum of truck is quite large than that of the car, therefore, force required for truck is quite large as compared to that required for car. (ii) An experienced cricketer catches a cricket ball coming in with great speed far more easily than a beginner, who can hurt his hands in the act. One reason is that the cricketer allows a longer time for his hands to stop the ball. He draws in the hands backward in the act of catching the ball (see figure). As the time for catching increases, the force with which the ball hurts his hand decreases. As a result, his hands are not injured. A beginner, on the other hand, keeps his hands fixed and tries to catch the ball almost instantly (i.e. in a very small time interval). Thus, a much greater force is exerted by the ball on his hands and this hurts a lot. (iii) If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. This is because the force is directly proportional to the mass of an object. (iv) Speed/velocity is another important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. The same bullet fired with moderate speed will not cause much damage. Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time. (v) When an athlete goes for a high jump, he is made to fall on a cushioned bed. This increases the time of falling of the athlete, thereby reducing the force exerted on him, causing no injury.

Numerical Ability

• How much force should be applied to a body of mass 250 g to produce an acceleration of $12\mathrm{m}/{\mathrm{s}}^2$ in it? Solution: Apply the formula $\mathrm{F}=\mathrm{ma}$ Given mass, $\mathrm{m}=250\mathrm{g}=\frac{250}{1000}\mathrm{kg}=0.25\mathrm{kg}$ Acceleration, $\mathrm{a}=12\mathrm{m}/{\mathrm{s}}^2$ force, $\mathrm{F}=\mathrm{ma}$ $=0.25\times 12=3\mathbf{N}$
• A cricket ball of 0.2 kg moving with a speed of $40\mathrm{m}/\mathrm{s}$ is brought to rest by player in 0.05 s . Find the average force applied by the player. Solution: Apply the formula $\mathrm{F}=\frac{{\mathrm{p}}_2-{\mathrm{p}}_1}{\mathrm{t}}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}$ Given, mass of ball, $\mathrm{m}=0.2\mathrm{kg}$; initial velocity, $\mathrm{u}=40\mathrm{m}/\mathrm{s}$; final velocity, $\mathrm{v}=0$; time, $\mathrm{t}=0.05\mathrm{s}$ Now, initial momentum, ${\mathrm{p}}_1=\mathrm{mu}=0.2\times 40=8\mathrm{kg}\mathrm{m}/\mathrm{s}$ Final momentum, ${\mathrm{p}}_2=\mathrm{mv}=0.2\times 0=0\mathrm{kg}\mathrm{m}/\mathrm{s}$ Force, $\mathrm{F}=\frac{{\mathrm{p}}_2-{\mathrm{p}}_1}{\mathrm{t}}=\frac{0-8}{0.05}$ $=-\frac{800}{5}=-160\mathbf{N}$
• Estimate the net force needed to accelerate a 1000 kg car at $5\mathrm{m}/{\mathrm{s}}^2$. If same force is applied to a 200 g ball, what will be its acceleration? Solution: Apply the formula $\mathrm{F}=\mathrm{ma}$ Given, mass of car, $\mathrm{m}=1000\mathrm{kg}$; acceleration of car, $\mathrm{a}=5$ $\mathrm{m}/{\mathrm{s}}^2$; force needed, $\mathrm{F}=$ ? Force, $\mathrm{F}=\mathrm{ma}=1000\times 5=5000\mathrm{N}$ Now, same force is applied to a 200 g ball i.e., $\mathrm{F}=5000\mathrm{N}$ mass of ball, ${\mathrm{m}}^{\prime }=200\mathrm{g}=200/1000=0.2\mathrm{kg}$ Thus, acceleration of ball, ${\mathrm{a}}^{\prime }=\frac{\mathrm{F}}{{\mathrm{m}}^{\prime }}=\frac{5000}{0.2}=2.5\times {10}^4\mathbf{m}/{\mathbf{s}}^2$
• A car of mass 800 kg , moving with velocity $10\mathbf{m}/\mathrm{s}$ is brought to rest in $20$ seconds by applying brakes. Calculate the force acting on the wheels. Solution: Apply the formula $\mathrm{F}=\frac{{\mathrm{p}}_2-{\mathrm{p}}_1}{\mathrm{t}}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}$ Given, mass, $\mathrm{m}=800\mathrm{kg}$; initial velocity, $\mathrm{u}=10\mathrm{m}/\mathrm{s}$; final velocity, $v=0$; time, $t=20\mathrm{s}$ $\mathrm{f}=\mathrm{ma}=\frac{\mathrm{m}(\mathrm{v}-\mathrm{u})}{\mathrm{t}}=\frac{800\times (0-10)}{20}\left(\because \mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}\right)$ $=\frac{800\times (-10)}{20}$ $=-400\mathrm{N}$
• A constant force acts on an object of mass 5 kg for a duration of $2s$. It increases the object's velocity from $3\mathrm{m}{\mathrm{s}}^{-1}$ to $7\mathrm{m}{\mathrm{s}}^{-1}$. Find the magnitude of the applied force. If the force was applied for a duration of 5 s , what would be the final velocity of the object? Solution: Apply the formula $\mathrm{F}=\mathrm{ma}$ Identify the formula for final velocity $\mathrm{v}=\mathrm{u}+\mathrm{at}$ Given, initial velocity, $u=3\mathrm{m}{\mathrm{s}}^{-1}$; final velocity, $\mathrm{v}=7{\mathrm{ms}}^{-1}$ time, $\mathrm{t}=2\mathrm{s}$; mass of the object, $\mathrm{m}=5\mathrm{kg}$; force, $\mathrm{F}=$ ? Acceleration, $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{7-3}{2}=\frac{4}{2}=2\mathrm{m}{\mathrm{s}}^{-2}$ Force, $\mathrm{F}=\mathrm{ma}=5\times 2=10\mathrm{N}$ If same force was applied for a duration of 5 s , the acceleration will remain the same i.e., $a=2\mathrm{m}{\mathrm{s}}^{-2}$ Now, from first equation of motion, $\mathrm{v}=\mathrm{u}+\mathrm{at}=3+(2\times 5)=3+10=13{\mathbf{m}\mathbf{s}}^{-1}$
• A force of 5 N gives a mass ${\mathrm{m}}_1$ an acceleration of $10\mathrm{m}{\mathrm{s}}^{-2}$ and a mass $m_2$, an acceleration of $20\mathrm{m}{\mathrm{s}}^{-2}$. What acceleration would it give if both the masses were tied together? Solution: Apply the formula $\mathrm{F}=\mathrm{ma}$ Given, ${\mathrm{m}}_1=$ ? ; force on ${\mathrm{m}}_1,\mathrm{F}=5\mathrm{N}$ acceleration on ${\mathrm{m}}_1,{\mathrm{a}}_1=10\mathrm{m}{\mathrm{s}}^{-2};{\mathrm{m}}_2=$ ? force on ${\mathrm{m}}_2,\mathrm{F}=5\mathrm{N}$; acceleration on ${\mathrm{m}}_2,{\mathrm{a}}_2=20\mathrm{m}{\mathrm{s}}^{-2}$ We know that, $\mathrm{F}=\mathrm{ma}$ or $\mathrm{m}=\frac{\mathrm{F}}{\mathrm{a}}$ Thus, ${\mathrm{m}}_1=\frac{\mathrm{F}}{{\mathrm{a}}_1}=\frac{5}{10}=0.5\mathrm{kg}$. Similarly, ${\mathrm{m}}_2=\frac{\mathrm{F}}{{\mathrm{a}}_2}=\frac{5}{20}=0.25\mathrm{kg}$ Now, if both the masses are tied together and same force $F$ is applied on them, then, $F=\left(m_1+m_2\right)a^{\prime }$, or $a^{\prime }=\frac{F}{m_1+m_2}=\frac{5}{0.5+0.25}=\frac{5}{0.75}=6.67{\mathbf{m}\mathbf{s}}^{-2}$
• The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in figure. How much force does the table exert on the ball to bring it to rest?
  
  Solution: Apply the formula $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}$ $\mathrm{F}=\mathrm{ma}$ The initial velocity of the ball is $20\mathrm{cm}{\mathrm{s}}^{-1}$. Due to the friction force exerted by the table, the velocity of the ball decreases down to zero in 10 s (as per graph). Initial velocity, $u=20\mathrm{cm}{\mathrm{s}}^{-1}=(20/100)\mathrm{m}{\mathrm{s}}^{-1}=0.2\mathrm{m}{\mathrm{s}}^{-1}$; final velocity, $\mathrm{v}=0$; time, $\mathrm{t}=10\mathrm{s}$; mass of the ball, $\mathrm{m}=20\mathrm{g}=(20/1000)\mathrm{kg}=0.02\mathrm{kg}$ Since the velocity-time graph is a straight line, it is clear that the ball comes to rest with a constant acceleration. Acceleration, $\mathrm{a}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{t}}=\frac{0-0.2}{10}=-0.02\mathrm{m}{\mathrm{s}}^{-2}$ Force exerted on the ball, $\mathrm{F}=\mathrm{ma}=0.02\times (-0.02)=-4\times 10-4\mathbf{N}$ The negative sign implies that the frictional force exerted by the table is opposite to the direction of motion of the ball.

8.0Newton's third law of motion

The second law relates the external force on a body to its acceleration. What is the origin of the external force on the body? What agency provides the external force? The simple answer is that the external force on a body always arises due to some other body.

Forces always exist in pairs

When two objects interact, two forces will always be involved. One force is the action force and the other is the reaction force.

• According to Newton's third law, 'whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body'. In other words, 'to every action, there is always an equal and opposite reaction'. Consider a pair of bodies A and B. According to Newton's third law, ${\mathrm{F}}_{\mathrm{AB}}=-{\mathrm{F}}_{\mathrm{BA}}$, where, ${\mathrm{F}}_{\mathrm{AB}}$ $=$ force on $A$ due to $B$, and $F_{BA}=$ force on $B$ due to $A$.
• An important aspect of Newton's third law is that the action and reaction forces act on different bodies. This means that action-reaction pairs will never be added together i.e., they can not cancel out each other.

• Q. If the action force is equal in magnitude to the reaction force, how can there ever be an acceleration? Explain using an example. Explanation: Though the action-reaction pair are equal in magnitude and opposite in direction but the reaction force always acts on a different object than the action force. Thus, these forces do not cancel out each other. Hence, there can be an acceleration in an object. Example : Let us consider a volleyball player bumping the ball (see figure). At the instant when both the ball and the player's arms are in contact, the action force is the upward force that the player exerts on the ball. The reaction force is the downward force that the ball exerts on the player's arms. During the collision, the ball accelerates upward and the player's arms accelerate downward. We hardly notice the acceleration of player's arms since his mass is quite large as compared to the ball, and the effect of the force on his motion is negligible.
  

Examples/applications of Newton's third law

• Newton's third law is also applicable to non-contact forces. For example, the Earth pulls an object downwards due to gravity (see figure). The object's force also exerts the same force on the Earth but in upward direction. But, we hardly see the effect of the object on the Earth because the Earth is very massive and the effect of a small force on its state of rest or motion is negligible. That is, the acceleration of Earth is negligible due to its huge mass.
  
• If boy exerts a small force on the wall, the wall will exert a small force on boy [see figure (a)]. When girl pushes hard against the wall, it pushes back just as hard [see figure (b)].
• When a sailor jumps out of a rowing boat. As the sailor jumps forward, the force on the boat moves it backwards [see figure (c)].
• When a gun is fired, it exerts a forward force on the bullet. The bullet exerts an equal and opposite reaction force on the gun. This results in the recoil of the gun [see figure (d)]. Since the gun has a much greater mass than the bullet, the acceleration of the gun is much less than the acceleration of the bullet.
• Suppose you are standing at rest and intend to start walking (or running) on a road [see figure (e)]. While walking, you push the road backwards. Thus, according to Newton's third law, the road exerts an equal and opposite reaction force on your feet to make you move forward.
  
• Rocket Propulsion : In a rocket engine, the highly combustible fuel burns at a tremendous rate [see figure (f) ]. The rocket exerts a downward (or backward) force on the exhaust gas and thus, according to Newton's third law, the exhaust gas exerts an upward (or forward) force on the rocket, these forces are equal in magnitude. It is the reaction force of the exhaust gas that accelerates the rocket forward. That is why a rocket can accelerate even in outer space.

9.0Conservation of momentum

The second and the third laws of motion lead to an important result : the law of conservation of momentum. Let us understand it by taking an example : When a bullet is fired from a gun, according to Newton's third law, if the force on the bullet by the gun is F, then the force on the gun by the bullet is -F . The two forces act for a common interval of time t .

According to Newton's second law, force F can be written as, $\mathrm{F}=\frac{\Delta \mathrm{p}}{\mathrm{t}}$ where, $\Delta \mathrm{p}$ is the change in momentum of the object. or $\Delta \mathrm{p}=\mathrm{F}\times \mathrm{t}$ This means, $\mathrm{F}\times \mathrm{t}$ is the change in momentum of the bullet and $-\mathrm{F}\times \mathrm{t}$ is the change in momentum of the gun. Since initially, both are at rest, the change in momentum equals the final momentum for each. Let ${\mathrm{p}}_{\mathrm{b}}$ be the momentum of the bullet after firing and ${\mathrm{p}}_{\mathrm{g}}$ is the recoil momentum of the gun, then,

Adding eq. (1) and eq. (2), we get, ${\mathrm{p}}_{\mathrm{b}}+{\mathrm{p}}_{\mathrm{g}}=0$. That is, the final momentum of the system (bullet plus gun) is zero. But, initial momentum of the system is also zero. This means initial momentum is equal to the final momentum i.e., total momentum is conserved. Thus, in an isolated system (a system with no external force), mutual forces between pairs of particles in the system can cause momentum change in individual particles, but since the mutual forces for each pair are equal and opposite, the momentum changes cancel in pairs and the total momentum remains unchanged. This fact is known as the law of conservation of momentum.

• According to the law of conservation of momentum, 'when the net external force on a system of objects is zero, the total momentum of the system remains constant'.

The term 'collision' is used to represent the event of two particles coming together for a short time and thereby producing 'impulsive forces' on each other. These forces are assumed to be much greater than any external forces present because they act for a very short time interval.

Law of conservation of momentum : proof

Let us consider two balls $A$ and $B$ having masses $m_A$ and $m_B$, travelling in the same direction along a straight line with different velocities $u_A$ and $u_B$, respectively [see figure (a)]. No other external unbalanced forces are acting on them. Let $u_A>u_B$ and the two balls collide with each other as shown in figure (b). During collision which lasts for a very short time $t$, the ball A exerts a force $F_{BA}$ on ball B, and the ball B exerts a force $F_{AB}$ on ball A. Suppose va and vв are the velocities of the two balls A and B after the collision, respectively [see figure(c)].

Initial momentum of ball $A=m_A{u}_A$, and final momentum of ball $A=m_{AVA}$ Force on $A$ due to $B,F_{AB}=$ rate of change of momentum of ball $A$ or $F_{AB}=\frac{m_A{v}_A-m_A{u}_A}{t}=\frac{m_A\left(v_A-u_A\right)}{t}$ Initial momentum of ball $B=m_B{U}_B$, Final momentum of ball $B=m_{BV}$ Force on B due to $\mathrm{A},{\mathrm{F}}_{\mathrm{BA}}=$ rate of change of momentum of ball B or $F_{BA}=\frac{m_B{v}_B-m_B{u}_B}{t}=\frac{m_B\left(v_B-u_B\right)}{t}$ According to Newton's third law of motion, the force ${\mathrm{F}}_{\mathrm{BA}}$ exerted by ball A on ball B (action) and the force $F_{AB}$ exerted by the ball $B$ on ball $A$ (reaction) must be equal and opposite to each other. That is, ${\mathrm{F}}_{\mathrm{BA}}=-{\mathrm{F}}_{\mathrm{AB}}$

or $m_B\left(v_B-u_B\right)=-m_A\left(v_A-u_A\right)$ or $m_B{V}_B-m_B{u}_B=-m_A{V}_A+m_A{u}_A$ or ${\mathbf{m}}_{\mathrm{B}}{\mathbf{v}}_{\mathrm{B}}+{\mathbf{m}}_{\mathrm{A}}{\mathbf{v}}_{\mathrm{A}}={\mathbf{m}}_{\mathrm{B}}{\mathbf{u}}_{\mathrm{B}}+{\mathbf{m}}_{\mathrm{A}}{\mathbf{u}}_{\mathrm{A}}$ Since, $\left(m_{B{V}_B}+m_{AVA}\right)$ represent the total final momentum and (mBuB $+m_A{u}_A)$ represent the total initial momentum, from eq. (3), we can conclude that Total final momentum = Total initial momentum This is law of conservation of momentum.

Numerical Ability

• A bullet of mass 20 g is horizontally fired with a velocity $150\mathrm{m}{\mathrm{s}}^{-1}$ from a pistol of mass 2 kg . What is the recoil velocity of the pistol? Solution: Apply the formula $m_B{v}_B+m_A{v}_A=m_B{u}_B+m_A{u}_A$ Given, mass of bullet, ${\mathrm{m}}_1=20\mathrm{g}=0.02\mathrm{kg}$; mass of pistol, ${\mathrm{m}}_2=2\mathrm{kg}$; initial velocity of bullet, ${\mathrm{u}}_1=0$; initial velocity of pistol, ${\mathrm{u}}_2=0$. Let the direction of bullet is taken left to right (see figure). Now, the final velocity of the bullet, ${\mathrm{v}}_1=+150\mathrm{m}{\mathrm{s}}^{-1}$ (by sign convention, left to right is taken positive).
  
  Let ' $v$ ' be the recoil velocity of the pistol. Initial momentum of the pistol-bullet system, ${\mathrm{p}}_{\mathrm{i}}=$ initial momentum of the pistol + initial momentum of the bullet $=2\times 0+0.02\times 0=0\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$ Final momentum of the pistol-bullet system, ${\mathrm{p}}_{\mathrm{f}}=$ final momentum of the pistol + final momentum of the bullet $=2\times v+0.02\times 150=(2v+3)\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-1}$ According to the law of conservation of momentum, final momentum of the system = initial momentum of the system $\therefore 2\mathrm{v}+3=0$ or $2\mathrm{v}=-3$ or $\mathrm{v}=-3/2=-1.5\mathbf{m}/\mathbf{s}$ (Negative sign indicates that the direction in which the pistol would recoil is opposite to that of bullet, that is, right to left).
• A boy of mass 40 kg jumps with a horizontal velocity of $5{\mathbf{m}\mathbf{s}}^{-1}$ onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg . What is his velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction. Solution: Apply the formula $m_B{v}_B+m_A{v}_A=m_B{u}_B+m_A{u}_A$ Given, mass of boy, ${\mathrm{m}}_1=40\mathrm{kg}$; initial velocity of boy, ${\mathrm{u}}_1=+5\mathrm{m}/\mathrm{s}$; mass of cart, ${\mathrm{m}}_2=3\mathrm{kg}$; initial velocity of cart, ${\mathrm{u}}_2=0$ [see figure (a)]. Initial momentum of the boy-cart system, ${\mathrm{p}}_{\mathrm{i}}=$ Initial momentum of the boy + Initial momentum of the cart

Let ' $v$ ' be the common velocity of the boy-cart system when the cart starts moving along with the boy [see figure (b)].

• Consider a large fish that swims towards and swallows a small fish at rest (see figure). If the large fish has a mass of 5 kg and swims at $1\mathrm{m}/\mathrm{s}$ toward the small fish having mass of 1 kg , what is the velocity of the large fish immediately after the lunch? Neglect the effects of water resistance.
  
  Solution: Apply the formula $m_B{v}_B+m_A{v}_A=m_B{u}_B+m_A{u}_A$ Given, mass of large fish, ${\mathrm{m}}_1=5\mathrm{kg}$; initial velocity of large fish, ${\mathrm{u}}_1=1\mathrm{m}/\mathrm{s}$; mass of small fish, ${\mathrm{m}}_2=1\mathrm{kg}$; initial velocity of small fish, ${\mathrm{u}}_2=0$. Let ' $v$ ' be the velocity of the large fish-small fish system. Applying law of conservation of momentum, we get, total momentum after lunch = total momentum before lunch or $m_{1}v+m_{2}v=m_1{u}_1+m_2{u}_2$ or $\left(m_1+m_2\right)v=m_1{u}_1+m_2{u}_2$ or $(5+1)v=(5)\times (+1)+(1)\times (0)$ or $6\mathrm{v}=5$, or $v=(5/6)\mathrm{m}/\mathrm{s}$ or $v=+0.833\mathbf{m}/\mathrm{s}$ The large fish will move with a velocity of $0.833\mathbf{m}/\mathrm{s}$ in the direction in which it was moving initially.
• Two ice hockey players of opposite teams, while trying to hit a hockey ball on the ice collide and immediately become entangled. One has a mass of 60 kg and was moving with a velocity $5.0\mathrm{m}/\mathrm{s}$ while the other has a mass of 55 kg and was moving faster with a velocity $6.0\mathrm{m}/\mathrm{s}$ towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and ice is negligible. Solution: Apply the formula ${\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}+{\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}={\mathrm{m}}_{\mathrm{B}}{\mathrm{u}}_{\mathrm{B}}+{\mathrm{m}}_{\mathrm{A}}$ Let the first player be moving towards right and the second player moving towards left [see figure (a)]. mass of $1^{\text{st }}$ player, ${\mathrm{m}}_1=60\mathrm{kg}$; initial velocity of $1^{\text{st }}$ player, ${\mathrm{u}}_1=+5\mathrm{m}/\mathrm{s}$; mass of $2^{\text{nd }}$ player, ${\mathrm{m}}_2=55\mathrm{kg}$; initial velocity of $2^{\text{nd }}$ player, ${\mathrm{u}}_2=-6\mathrm{m}/\mathrm{s}$. Let the common velocity of both the players after collision be 'v' [see figure(b)]. According to the law of conservation of momentum, final momentum of the system = initial momentum of the system or $m_{1}v+m_{2}v=m_1{u}_1+m_2{u}_2$ or $\left(m_1+m_2\right)v=m_1{u}_1+m_2{u}_2$ or $(60+55)v=60\times (+5)+(55)\times (-6)$ or $115\mathrm{v}=300-330$, or $115\mathrm{v}=-30$ or $v=-30/115=-0.26\mathbf{m}/\mathbf{s}$ Negative sign shows that the two entangled players will move towards left, i.e., in the direction the second player was moving before the collision.
  

10.0Some Basic Terms

• Muscular force - The contact force applied to an object with the help of muscles is known as muscular force.
• Tension - It is define as the force transmitted through a rope, string or wire when pulled by forces acting from opposite sides.
• Friction - It is the force generated by two surfaces that contact slide against each other.
• Electrostatic force - The force exerted by a charged body on another charged or uncharged body by virtue of its charged is known as electrostatic force.
• Magnetic force - Magnetic force can be defined as the attractive or repulsive force that is exerted between the poles of a magnet and electrically charged moving particles.
• Gravitational force - The force of attraction that arises between two bodies having definite masses.
• Distort- Distort is to change the physical shape of something.
• Stationary - Stationary means not moving, not appearing to move, stable of fixed.
• Frictionless - In physics, frictionless means that there is no resistance between a surface or substance and something that is moving along or through it. In other words, without any friction.
• Mechanics - It is a branch of science which deals with the motion of an object and the forces acting on it.
• Jerk - A sudden, quick, sharp movement.
• Tumbler - A tall glass for drinking out of with straight side and no handle.
• Collision - A brief event in which two or more bodies come together. For example : Two billiards ball, a golf club and a ball, a hammer and a nail head, or a falling object and a floor
• Head on collision - A crash of two objects that are moving directly towards each other.
• Inflates - It means to add air or gas to something such as a tire or a balloon and make it larger.
• External force - It is a force that act on a structure from the outside. For example : wind pushing on a tree is an external force.
• Internal force - Any force that acts on a structure from with in known as a internal force. Internal forces not change the mechanical energy of an object.
• Resultant force - The resultant force is described as the total amount of force acting on the object or body along with a direction of the body.
• Recoil - Recoil is a movement backwards, usually from some force or impact. For example : The recoil of a gun is backward movement caused by momentum.