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Chapter 8 Force And Laws Of Motion
In the previous chapter, we studied the description of motion in terms of position, velocity, and acceleration. However, we did not explore the fundamental question of what causes motion or changes in motion. Why does the speed of an object increase or decrease? Does every change in motion require a cause, and what is the nature of this cause?
Historically, the prevailing belief for centuries was influenced by Aristotelian physics, suggesting that the 'natural state' of an object was rest, and a continuous force was required to keep something moving. This view was challenged and ultimately overturned by the work of Galileo Galilei and Isaac Newton, who developed a completely new understanding of motion and its causes.
In our everyday experience, we know that some effort is needed to make a stationary object move, or to stop a moving one. This effort is commonly described as a **push**, a **pull**, or a **hit**. The concept of **force** arises from these actions. While we cannot see, taste, or feel force itself, we can observe and experience its effects when it is applied to an object. Forces can initiate motion in a stationary object, change the speed or direction of a moving object (causing it to accelerate or decelerate), or even change the shape and size of an object (like compressing a spring or deforming a rubber ball).
Balanced And Unbalanced Forces
Forces acting on an object can be classified as balanced or unbalanced, depending on their net effect.
- **Balanced Forces:** When two or more forces acting on an object are equal in magnitude and opposite in direction, they cancel each other out. The net force acting on the object is zero. Balanced forces **do not cause any change in the state of motion** of an object. If the object is at rest, it remains at rest. If it is moving with uniform velocity, it continues to move with the same uniform velocity. For example, if you push a box with a certain force, and friction opposes your push with an equal force, the box will not move.
- **Unbalanced Forces:** When the forces acting on an object are not equal in magnitude or are not directly opposite, they do not cancel out. There is a non-zero **net force** acting on the object. An unbalanced force is necessary to **change the state of motion** of an object. This means an unbalanced force can start a stationary object moving, stop a moving object, change its speed (accelerate or decelerate), or change its direction of motion. In the box example, if you push harder than the friction force, there is a net unbalanced force, and the box will start moving in the direction of your push.
Consider children pushing a box on a rough floor. Initially, a small push might not move the box because the static friction force between the box and the floor is equal and opposite to the push (balanced forces). As they push harder, the static friction increases to oppose it. Only when the pushing force exceeds the maximum static friction does an unbalanced force act on the box, causing it to accelerate and move.
The observation that a rolling ball or a bicyclist stops suggests that some force (like friction or air resistance) is acting to oppose the motion. To maintain motion at a constant velocity, the applied force (like pedalling) must be equal to the opposing force (friction and air resistance), resulting in a net force of zero. An unbalanced force is required to cause a change in velocity (acceleration or deceleration), not to sustain motion at a constant velocity.
First Law Of Motion
Based on experiments and observations like the behaviour of objects on inclined planes, Galileo Galilei concluded that an object moving on a frictionless horizontal surface would continue to move indefinitely at a constant speed. He reasoned that motion changes only when an unbalanced force is present, but no net force is needed to maintain uniform motion.
Isaac Newton built upon Galileo's insights and formulated three fundamental laws of motion. The **First Law of Motion** states:
An object will remain at rest or in a state of uniform motion along a straight line unless acted upon by an external unbalanced force.
This law essentially says that objects resist changes to their state of motion. A stationary object tends to stay stationary, and a moving object tends to keep moving with the same speed and in the same direction. This inherent property of objects to resist changes in their state of rest or uniform motion is called **inertia**.
For this reason, the First Law of Motion is also known as the **Law of Inertia**.
Everyday examples of inertia:
- When you are standing in a bus at rest and it suddenly starts moving, you tend to fall backwards. Your feet, in contact with the bus floor, start moving, but the rest of your body tends to remain at rest due to inertia.
- When you are travelling in a moving car and the driver suddenly applies the brakes, you tend to move forward. The car and your feet (touching the floor) decelerate, but your upper body tends to continue moving at the original velocity due to inertia. Safety belts work by exerting a force to counteract this forward motion due to inertia, preventing injury.
- When a car takes a sharp turn at high speed, you tend to be thrown outwards or sideways. Your body tends to continue moving in a straight line (inertia of direction), while the car changes direction.
These effects can be demonstrated by simple activities:
- Stack carom coins and strike the bottom coin quickly with a striker (Activity 8.1). The bottom coin is removed, but the inertia of rest of the coins above causes them to fall vertically.
- Place a coin on a card over a glass tumbler and flick the card away (Activity 8.2). The card moves, but the coin's inertia of rest keeps it in place momentarily, and it falls into the tumbler.
- Carry a tray with a water-filled tumbler and turn quickly (Activity 8.3). The water spills because its inertia causes it to resist the change in direction and continue moving in its original direction.
Inertia And Mass
While all objects possess inertia, they do not all have the same amount of inertia. The resistance to changing motion is greater for some objects than for others. For instance, it's much easier to push an empty box than a heavy box filled with books. Kicking a small stone of the same size as a football with the same force will result in very different outcomes; the stone will barely move, and you might get hurt, while the football will fly away. Similarly, it takes less effort to move a lighter coin than a heavier one.
These observations show that heavier objects (those with more mass) resist changes to their state of motion more strongly than lighter objects. This resistance is a measure of inertia.
Quantitatively, the **mass** of an object is a measure of its inertia. An object with a larger mass has greater inertia, meaning it is harder to start it moving if it's at rest, and harder to stop it or change its direction if it's already moving.
The SI unit of mass is the **kilogram (kg)**.
Question 1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupees coin and a one-rupee coin?
Answer:
Inertia is measured by mass. The object with greater mass has more inertia.
(a) A **stone** of the same size as a rubber ball typically has more mass than the rubber ball, so it has more inertia.
(b) A **train** has significantly more mass than a bicycle, so it has more inertia.
(c) A **five-rupees coin** generally has more mass than a one-rupee coin, so it has more inertia.
Question 2. In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.
Also identify the agent supplying the force in each case.
Answer:
Let's trace the changes in the ball's velocity and the agent applying the force:
- The ball is initially likely at rest or moving slowly. When the first football player kicks it, its velocity changes from rest (or slow motion) to a high velocity in a certain direction.
Agent of force: The **first football player's foot**. - The ball is moving with some velocity. The second player kicks it towards the goal. The ball's velocity changes (likely magnitude and/or direction).
Agent of force: The **second football player's foot**. - The ball is moving towards the goal. The goalkeeper collects it, bringing its velocity to zero.
Agent of force: The **goalkeeper's hands/body**. - The ball is at rest (or momentarily held by the keeper). The goalkeeper kicks it towards his teammate. The ball's velocity changes from zero to a high velocity in a new direction.
Agent of force: The **goalkeeper's foot**.
The velocity of the ball changes a total of **four** times in this sequence.
Question 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
This is due to the principle of inertia (Newton's First Law). When you vigorously shake a tree branch, the branch is set into motion. The leaves, due to their inertia of rest, tend to remain in their stationary state. This sudden, rapid movement of the branch while the leaves resist motion causes the stem holding the leaf to the branch to break, and the leaves get detached and fall off.
Question 4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer:
This is explained by Newton's First Law (Inertia):
- When a moving bus brakes suddenly, the bus and your feet (which are in contact with the bus floor) quickly slow down. However, your upper body, due to its inertia of motion, tends to continue moving forward at the bus's original velocity. This causes you to fall forward relative to the bus.
- When a stationary bus accelerates from rest, the bus and your feet suddenly start moving forward. However, your upper body, due to its inertia of rest, tends to remain in its original stationary position. This causes you to fall backward relative to the bus.
Second Law Of Motion
The first law of motion tells us that an unbalanced force is required to change an object's state of motion (cause acceleration). The second law of motion quantifies this relationship and provides a way to measure force.
Everyday observations suggest that the 'impact' of a moving object depends on both its mass and its velocity. For example, a slow-moving table tennis ball has little impact, but a fast-moving cricket ball can cause injury. A stationary truck is harmless, but a moving truck, even at low speed, can be dangerous. This implies there is a quantity that combines mass and velocity that is important in describing motion and the effect of forces.
This quantity is called **momentum**. The **momentum ($\mathbf{p}$) of an object** is defined as the product of its mass ($m$) and its velocity ($v$).
$$ \mathbf{p} = m\mathbf{v} $$
Momentum is a **vector quantity**; its direction is the same as the direction of velocity. The SI unit of momentum is **kilogram-metre per second (kg m/s or kg m s⁻¹)**.
Since an unbalanced force changes an object's velocity, it also changes its momentum. The effectiveness of a force in changing momentum depends not only on the force's magnitude but also on the time for which it acts. A small force applied over a long time can produce a larger change in momentum than a large force applied for a very short time (related to impulse, $Ft = \Delta p$).
Newton's **Second Law of Motion** formally states the relationship between force and the change in momentum:
The rate of change of momentum of an object is proportional to the applied unbalanced force and takes place in the direction of the force.
Mathematical Formulation Of Second Law Of Motion
Let's consider an object of mass $m$ moving in a straight line with initial velocity $u$. A constant unbalanced force $F$ acts on it for time $t$, causing its velocity to change uniformly to $v$.
- Initial momentum, $p_1 = mu$.
- Final momentum, $p_2 = mv$.
The change in momentum is $p_2 - p_1 = mv - mu = m(v-u)$.
The rate of change of momentum is the change in momentum divided by the time taken:
$$ \text{Rate of change of momentum} = \frac{\text{Change in momentum}}{\text{Time taken}} = \frac{m(v-u)}{t} $$
According to the Second Law, the applied force $F$ is proportional to the rate of change of momentum:
$$ F \propto \frac{m(v-u)}{t} $$
We can introduce a constant of proportionality, $k$:
$$ F = k \frac{m(v-u)}{t} $$
Since acceleration $a = \frac{v-u}{t}$, we can write:
$$ F = kma $$
The unit of force, called the **Newton (N)**, is defined such that the constant $k$ equals 1. One Newton is the force required to produce an acceleration of 1 m/s² in an object of mass 1 kg.
$$ 1 \text{ N} = 1 \times (1 \text{ kg}) \times (1 \text{ m/s}^2) $$
Thus, $k=1$, and the mathematical form of the Second Law of Motion is:
$$ \mathbf{F} = m\mathbf{a} $$
Where $\mathbf{F}$ is the net external unbalanced force, $m$ is the mass, and $\mathbf{a}$ is the acceleration produced. Force is a vector quantity, and its direction is the same as the direction of the acceleration.
The SI unit of force is $\text{kg} \cdot \text{m}/\text{s}^2$ or Newton (N).
The Second Law explains many everyday phenomena:
- Catching a cricket ball: A fielder pulls their hands back while catching a fast ball. This increases the time ($t$) over which the momentum changes from its initial high value to zero. Since $F \propto \frac{\Delta p}{t}$ (or $F \times t = \Delta p$), increasing $t$ for a given change in momentum ($\Delta p$) reduces the average force ($F$) exerted on the fielder's hands, making the catch less painful.
- High jump: Athletes land on a cushioned bed or sand pit. This increases the time taken for their body's momentum to become zero upon landing. By increasing the time of impact, the force experienced by the athlete is reduced.
- Karate chop: A karate expert applies force on a stack of tiles or ice slabs for a very short duration. By reducing the time of impact, a large force is produced, sufficient to break the slabs ($\text{Force} \propto 1/\text{time}$ for a given momentum change).
The First Law of motion can be derived from the Second Law:
From $F = ma$, we have $F = m \frac{v-u}{t}$. Rearranging, $Ft = m(v-u) = mv - mu$.
If the net external force $F=0$, then $0 \times t = mv - mu$. This means $0 = mv - mu$, or $mv = mu$. Since the mass $m$ is non-zero, we must have $v = u$. This shows that if no unbalanced force acts on an object, its velocity remains constant ($v=u$). If the initial velocity $u$ was zero (object at rest), the final velocity $v$ will also be zero, meaning it remains at rest. If $u$ was non-zero, $v$ remains equal to $u$, meaning it continues moving with uniform velocity.
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?
Answer:
Given: mass $m = 5$ kg, initial velocity $u = 3$ m/s, final velocity $v = 7$ m/s, time $t = 2$ s.
First, calculate the acceleration: $a = \frac{v - u}{t} = \frac{7 \text{ m/s} - 3 \text{ m/s}}{2 \text{ s}} = \frac{4 \text{ m/s}}{2 \text{ s}} = 2 \text{ m/s}^2$.
Now, find the magnitude of the force using $F = ma$:
$F = (5 \text{ kg}) \times (2 \text{ m/s}^2) = 10 \text{ N}$.
The magnitude of the applied force is 10 N.
Next, if this force (10 N) is applied for a duration of $t = 5$ s, we need to find the new final velocity ($v'$). The initial velocity remains $u = 3$ m/s.
Using the first equation of motion, $v' = u + at'$. We first need the acceleration produced by the 10 N force on the 5 kg mass, which we found to be $a = 2$ m/s².
So, $v' = 3 \text{ m/s} + (2 \text{ m/s}^2) \times (5 \text{ s})$
$v' = 3 \text{ m/s} + 10 \text{ m/s} = 13 \text{ m/s}$.
If the force was applied for a duration of 5 s, the final velocity of the object would be 13 m/s.
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 ?
Answer:
Using Newton's Second Law, $F = ma$, we calculate the force required in each case:
Case 1: Mass $m_1 = 2$ kg, acceleration $a_1 = 5$ m/s².
Force required, $F_1 = m_1 a_1 = (2 \text{ kg}) \times (5 \text{ m/s}^2) = 10 \text{ N}$.
Case 2: Mass $m_2 = 4$ kg, acceleration $a_2 = 2$ m/s².
Force required, $F_2 = m_2 a_2 = (4 \text{ kg}) \times (2 \text{ m/s}^2) = 8 \text{ N}$.
Comparing the forces, $F_1 = 10$ N and $F_2 = 8$ N. Since $F_1 > F_2$, accelerating a 2 kg mass at 5 m s⁻² would require 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.
Answer:
Given:
Initial velocity, $u = 108$ km/h. Convert to m/s: $u = 108 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 108 \times \frac{5}{18} \text{ m/s} = 6 \times 5 \text{ m/s} = 30 \text{ m/s}$.
Final velocity, $v = 0$ m/s (car stops).
Time taken, $t = 4$ s.
Mass of the motorcar with passengers, $m = 1000$ kg.
First, calculate the acceleration produced by the brakes: $a = \frac{v - u}{t} = \frac{0 \text{ m/s} - 30 \text{ m/s}}{4 \text{ s}} = \frac{-30 \text{ m/s}}{4 \text{ s}} = -7.5 \text{ m/s}^2$.
The negative sign indicates that the acceleration (and thus the force) is in the direction opposite to the motion.
Now, calculate the force exerted by the brakes using $F = ma$:
$F = (1000 \text{ kg}) \times (-7.5 \text{ m/s}^2) = -7500 \text{ N}$.
The magnitude of the force exerted by the brakes is 7500 N. The negative sign confirms that this force is acting in the direction opposite to the car's motion.
Example 8.4. A force of 5 N gives a mass m₁, an acceleration of 10 m s⁻² and a mass m₂, an acceleration of 20 m s⁻². What acceleration would it give if both the masses were tied together?
Answer:
Given force, $F = 5$ N.
For mass $m_1$, acceleration $a_1 = 10$ m/s².
Using $F = m_1 a_1$, we find $m_1 = \frac{F}{a_1} = \frac{5 \text{ N}}{10 \text{ m/s}^2} = 0.5 \text{ kg}$.
For mass $m_2$, acceleration $a_2 = 20$ m/s².
Using $F = m_2 a_2$, we find $m_2 = \frac{F}{a_2} = \frac{5 \text{ N}}{20 \text{ m/s}^2} = 0.25 \text{ kg}$.
If the two masses are tied together, the total mass $m_{total} = m_1 + m_2 = 0.5 \text{ kg} + 0.25 \text{ kg} = 0.75 \text{ kg}$.
Now, the same force $F = 5$ N is applied to the combined mass $m_{total}$. We need to find the acceleration $a$ produced.
Using $F = m_{total} a$:
$5 \text{ N} = (0.75 \text{ kg}) \times a$
$a = \frac{5 \text{ N}}{0.75 \text{ kg}} = \frac{5}{3/4} \text{ m/s}^2 = 5 \times \frac{4}{3} \text{ m/s}^2 = \frac{20}{3} \text{ m/s}^2$.
$a \approx 6.67 \text{ m/s}^2$.
If both masses were tied together, a force of 5 N would give them an acceleration of $\frac{20}{3}$ m/s² (approximately 6.67 m/s²).
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 in Fig. 8.9.
How much force does the table exert on the ball to bring it to rest?
Answer:
From the velocity-time graph (Fig 8.9):
Initial velocity, $u = 20$ cm/s = $20 \times 10^{-2}$ m/s = 0.2 m/s (at t=0).
Final velocity, $v = 0$ cm/s = 0 m/s (at t=10 s).
Time taken, $t = 10$ s (time when velocity becomes zero).
Mass of the ball, $m = 20$ g = $20/1000$ kg = 0.02 kg.
Since the velocity-time graph is a straight line sloping downwards, the ball is moving with uniform deceleration (negative acceleration).
Calculate the acceleration: $a = \frac{v - u}{t} = \frac{0 \text{ m/s} - 0.2 \text{ m/s}}{10 \text{ s}} = \frac{-0.2 \text{ m/s}}{10 \text{ s}} = -0.02 \text{ m/s}^2$.
The negative sign indicates the acceleration is opposite to the direction of initial velocity.
Now, calculate the force exerted by the table (frictional force) using $F = ma$:
$F = (0.02 \text{ kg}) \times (-0.02 \text{ m/s}^2) = -0.0004 \text{ N}$.
The magnitude of the force exerted by the table (friction) is 0.0004 N. The negative sign indicates that this force acts in the direction opposite to the ball's motion, bringing it to rest.
Question 1. A bus starting from rest moves with a uniform acceleration of 0.1 m s⁻² for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Answer:
Given:
Initial velocity, $u = 0$ m/s (starting from rest)
Uniform acceleration, $a = 0.1$ m/s²
Time, $t = 2$ minutes. Convert to seconds: $t = 2 \times 60 = 120$ s.
(a) To find the speed acquired (final velocity, $v$), use the first equation of motion, $v = u + at$:
$v = 0 \text{ m/s} + (0.1 \text{ m/s}^2) \times (120 \text{ s})$
$v = 12 \text{ m/s}$.
The speed acquired by the bus is 12 m/s.
(b) To find the distance travelled ($s$), use the second equation of motion, $s = ut + \frac{1}{2}at^2$:
$s = (0 \text{ m/s}) \times (120 \text{ s}) + \frac{1}{2} \times (0.1 \text{ m/s}^2) \times (120 \text{ s})^2$
$s = 0 + \frac{1}{2} \times 0.1 \times (14400) \text{ m}$
$s = 0.05 \times 14400 \text{ m} = 720 \text{ m}$.
The distance travelled by the bus is 720 m.
Question 2. A train is travelling at a speed of 90 km h⁻¹. Brakes are applied so as to produce a uniform acceleration of – 0.5 m s⁻². Find how far the train will go before it is brought to rest.
Answer:
Given:
Initial velocity, $u = 90$ km/h. Convert to m/s: $u = 90 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 90 \times \frac{5}{18} \text{ m/s} = 5 \times 5 \text{ m/s} = 25 \text{ m/s}$.
Final velocity, $v = 0$ m/s (train comes to rest).
Uniform acceleration (deceleration), $a = -0.5$ m/s².
To find the distance travelled ($s$), use the third equation of motion, $v^2 - u^2 = 2as$:
$(0 \text{ m/s})^2 - (25 \text{ m/s})^2 = 2 \times (-0.5 \text{ m/s}^2) \times s$
$0 - 625 \text{ m}^2/\text{s}^2 = -1 \text{ m/s}^2 \times s$
$-625 \text{ m}^2/\text{s}^2 = -s \text{ m/s}^2$
$s = 625 \text{ m}$.
The train will go 625 metres before it is brought to rest.
Question 3. A trolley, while going down an inclined plane, has an acceleration of 2 cm s⁻². What will be its velocity 3 s after the start?
Answer:
Given:
Initial velocity, $u = 0$ cm/s (assuming it starts from rest on the inclined plane).
Acceleration, $a = 2$ cm/s².
Time, $t = 3$ s.
To find the velocity ($v$) after 3 s, use the first equation of motion, $v = u + at$:
$v = 0 \text{ cm/s} + (2 \text{ cm/s}^2) \times (3 \text{ s})$
$v = 6 \text{ cm/s}$.
The velocity of the trolley 3 s after the start will be 6 cm/s.
Question 4. A racing car has a uniform acceleration of 4 m s⁻². What distance will it cover in 10 s after start?
Answer:
Given:
Initial velocity, $u = 0$ m/s (assuming it starts from rest).
Uniform acceleration, $a = 4$ m/s².
Time, $t = 10$ s.
To find the distance covered ($s$), use the second equation of motion, $s = ut + \frac{1}{2}at^2$:
$s = (0 \text{ m/s}) \times (10 \text{ s}) + \frac{1}{2} \times (4 \text{ m/s}^2) \times (10 \text{ s})^2$
$s = 0 + \frac{1}{2} \times 4 \times 100 \text{ m}$
$s = 2 \times 100 \text{ m} = 200 \text{ m}$.
The racing car will cover a distance of 200 m in 10 s after starting.
Question 5. A stone is thrown in a vertically upward direction with a velocity of 5 m s⁻¹. If the acceleration of the stone during its motion is 10 m s⁻² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer:
Given:
Initial velocity (upward), $u = 5$ m/s.
Acceleration (downward), $a = -10$ m/s² (negative because it opposes the upward motion).
At the highest point, the final velocity $v = 0$ m/s.
To find the height attained ($s$), use the third equation of motion, $v^2 - u^2 = 2as$:
$(0 \text{ m/s})^2 - (5 \text{ m/s})^2 = 2 \times (-10 \text{ m/s}^2) \times s$
$0 - 25 \text{ m}^2/\text{s}^2 = -20 \text{ m/s}^2 \times s$
$-25 \text{ m}^2/\text{s}^2 = -20 \text{ m/s}^2 \times s$
$s = \frac{-25 \text{ m}^2/\text{s}^2}{-20 \text{ m/s}^2} = \frac{25}{20} \text{ m} = 1.25 \text{ m}$.
The maximum height attained by the stone is 1.25 m.
To find the time taken ($t$) to reach the highest point, use the first equation of motion, $v = u + at$:
$0 \text{ m/s} = 5 \text{ m/s} + (-10 \text{ m/s}^2) \times t$
$-5 \text{ m/s} = -10 \text{ m/s}^2 \times t$
$t = \frac{-5 \text{ m/s}}{-10 \text{ m/s}^2} = 0.5 \text{ s}$.
The stone will take 0.5 seconds to reach the highest point.
Third Law Of Motion
The first two laws describe how forces affect the motion of a single object. The Third Law of Motion addresses the fundamental nature of force itself: forces always occur in pairs as a result of interaction between two objects.
Newton's **Third Law of Motion** states:
To every action, there is always an equal and opposite reaction. These forces act on two different objects.
This means that whenever one object exerts a force on a second object (the **action force**), the second object simultaneously exerts a force back on the first object (the **reaction force**). The action and reaction forces have **equal magnitude** and are **opposite in direction**. Critically, they **always act on different objects**, never on the same object.
Illustrations of the Third Law:
- When you push against a wall, the wall pushes back on you with an equal and opposite force.
- When you walk, you push the ground backward with your feet (action). The ground simultaneously pushes forward on your feet with an equal and opposite force (reaction), propelling you forward.
- Consider two spring balances connected and pulled. Balance A pulls B (action), and balance B pulls A back with an equal force (reaction), as shown by equal readings.
- When a gun is fired, the expanding gases push the bullet forward (action). The bullet simultaneously pushes backward on the gun with an equal and opposite force (reaction), causing the gun to recoil. Even though the forces are equal, the gun has much greater mass than the bullet, so its backward acceleration (recoil) is much less than the bullet's forward acceleration ($a = F/m$).
- When a sailor jumps forward from a boat, the sailor exerts a force on the boat pushing it backward (action). The boat exerts an equal and opposite force forward on the sailor, enabling the jump (reaction). Because the boat's mass is much less than the sailor's mass (in this relative motion context), the boat experiences a noticeable backward acceleration (recoil).
Activities like children throwing a sandbag between carts (Activity 8.4) demonstrate that when one child throws the bag (exerting force on the bag), they experience a backward reaction force on their cart. The child catching the bag experiences a forward force on their cart as they bring the bag to rest or throw it back.
It is crucial to remember that because action and reaction forces act on different objects, they **do not cancel each other out**. They are responsible for the changes in momentum and state of motion of the *respective* objects they act upon.
Intext Questions
Page No. 91
Question 1. Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupees coin and a one-rupee coin?
Answer:
Question 2. In the following example, try to identify the number of times the velocity of the ball changes:
“A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team”.
Also identify the agent supplying the force in each case.
Answer:
Question 3. Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Answer:
Question 4. Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Answer:
Exercises
Question 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.
Answer:
Question 2. When a carpet is beaten with a stick, dust comes out of it, Explain.
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Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
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Question 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.
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Question 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 = $1000\ kg$.)
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Question 6. A stone of $1\ kg$ is thrown with a velocity of $20\ m\ s^{-1}$ across the frozen surface of a lake and comes to rest after travelling a distance of $50\ m$. What is the force of friction between the stone and the ice?
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Question 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 and
(b) the acceleration of the train.
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Question 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 stopped with a negative acceleration of $1.7\ m\ s^{-2}$?
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Question 9. What is the momentum of an object of mass $m$, moving with a velocity $v$?
(a) $(mv)^2$
(b) $mv^2$
(c) $\frac{1}{2} mv^2$
(d) $mv$
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Question 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?
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Question 11. 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.
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Question 12. A hockey ball of mass $200\ g$ travelling at $10\ m\ s^{-1}$ is struck by a hockey stick so as to return it along its original path with a velocity at $5\ m\ s^{-1}$. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
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Question 13. A bullet of mass $10\ g$ travelling horizontally with a velocity of $150\ m\ s^{-1}$ strikes a stationary wooden block and comes to rest in $0.03\ s$. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
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Question 14. An object of mass $1\ kg$ travelling in a straight line with a velocity of $10\ m\ s^{-1}$ collides with, and sticks to, a stationary wooden block of mass $5\ kg$. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
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Question 15. An object of mass $100\ kg$ is accelerated uniformly from a velocity of $5\ m\ s^{-1}$ to $8\ m\ s^{-1}$ in $6\ s$. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
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Question 16. 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.
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Question 17. How much momentum will a dumb-bell of mass $10\ kg$ transfer to the floor if it falls from a height of $80\ cm$? Take its downward acceleration to be $10\ m\ s^{-2}$.
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Additional Exercises
Question A1. The following is the distance-time table of an object in motion:
| Time in seconds | Distance in metres |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 8 |
| 3 | 27 |
| 4 | 64 |
| 5 | 125 |
| 6 | 216 |
| 7 | 343 |
(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?
(b) What do you infer about the forces acting on the object?
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Question 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 $0.2\ m\ s^{-2}$. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)
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Question A3. A hammer of mass $500\ g$, moving at $50\ m\ s^{-1}$, strikes a nail. The nail stops the hammer in a very short time of $0.01\ s$. What is the force of the nail on the hammer?
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Question 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.
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