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Chapter 9: Force And Laws Of Motion
In the previous chapter, we studied how to describe motion in terms of position, velocity, and acceleration. However, we did not explore what causes motion or changes in motion. This chapter delves into the concept of force and the fundamental laws that govern motion.
Historically, it was believed that an object required a continuous force to remain in motion, and that 'rest' was the natural state of objects. This idea was challenged by the work of scientists like Galileo Galilei and Isaac Newton.
We commonly observe that some effort (a push, pull, or hit) is needed to start a stationary object moving or to stop a moving object. This effort is associated with the concept of force. Although we cannot directly see or feel a force itself, we experience and observe its effects.
Effects of Force: Applying a force can:
- Put a stationary object into motion.
- Stop a moving object.
- Change the speed of a moving object (make it move faster or slower).
- Change the direction of motion of an object.
- Change the shape and size of an object (e.g., stretching a spring, deforming a rubber ball).
Balanced And Unbalanced Forces
Consider a situation where multiple forces act on an object.
Balanced Forces: If two or more forces acting on an object are equal in magnitude and opposite in direction, they are called balanced forces. The net force (vector sum of all forces) is zero. Balanced forces do not change the state of rest or uniform 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 that uniform velocity.
Unbalanced Forces: If the forces acting on an object are not balanced (i.e., their vector sum is not zero), there is a net unbalanced force acting on the object. An unbalanced force changes the state of rest or uniform motion of an object. It can start a stationary object moving, stop a moving object, or change its speed or direction of motion, causing acceleration.
Example: Pushing a box on a floor. Friction between the box and the floor opposes the push. If the pushing force is balanced by friction, the box remains at rest or moves with uniform velocity. If the pushing force is greater than friction, there is a net unbalanced force, and the box accelerates.
To accelerate an object, an unbalanced force is required. The object will continue to accelerate as long as the unbalanced force is applied. If the unbalanced force is removed, the object will continue to move with the constant velocity it had at that instant (if there are no other unbalanced forces like friction). This contradicts the earlier belief that continuous force is needed for motion.
First Law Of Motion
Building on Galileo's experiments with objects on inclined planes, which suggested that objects would continue moving at a constant speed indefinitely in the absence of friction, Isaac Newton formulated his three laws of motion. The First Law of Motion describes the behaviour of objects when no unbalanced force acts on them.
Statement of Newton's First Law of Motion: 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.
This law essentially states that objects resist changes in their state of motion. This inherent property of objects to resist changes in their state of rest or motion is called inertia. Because it describes inertia, the First Law of Motion is also known as the Law of Inertia.
Examples illustrating the Law of Inertia:
- In a moving bus, applying brakes: When a bus moving at a speed suddenly stops, our body tends to continue moving forward due to inertia, even though the bus and our feet (in contact with the floor) have stopped. Safety belts help counter this by exerting a force to slow us down.
- In a stationary bus, starting suddenly: When a bus starts moving from rest, our body tends to remain at rest due to inertia, while the bus and our feet move forward. This makes us feel as if we are pushed backwards.
- Taking a sharp turn in a car: When a car turns sharply, our body tends to continue moving in a straight line due to inertia. The force from the car's engine and the road changes the car's direction, but our body resists this change, making us lean or get thrown to the side.
- Coin and Card Experiment: Placing a coin on a card over a glass and flicking the card quickly horizontally makes the card shoot away. The coin, due to its inertia of rest, resists the horizontal motion and falls vertically into the glass.
- Water in a Tumbler on a Tray: Rapidly rotating a tray with a water-filled tumbler causes the water to spill outwards. The water, due to its inertia, tends to continue moving in a straight line tangential to the circular path of the tray.
Inertia And Mass
Inertia is the resistance of an object to a change in its state of motion (or rest). Do all objects have the same amount of inertia? Experience tells us that it is harder to change the state of motion of heavier objects.
For example, pushing an empty box is easier than pushing a full box. Kicking a football makes it fly, but kicking a stone of the same size can hurt your foot and the stone moves very little. Similarly, in the coin-card experiment, using a heavier coin makes it harder to move with the card.
This suggests that heavier objects offer greater resistance to changes in motion, meaning they have greater inertia.
Quantitatively, the inertia of an object is measured by its mass. A more massive object has more inertia.
Inertia is the natural tendency of an object to resist a change in its state of motion or rest. The mass of an object is a measure of its inertia.
The SI unit of mass is the kilogram (kg).
Second Law Of Motion
The First Law explains what happens in the absence of an unbalanced force. The Second Law of Motion quantifies the relationship between the unbalanced force applied to an object and the resulting change in its motion (acceleration).
Observations in everyday life show that the impact produced by a moving object depends on both its mass and its velocity. For example, a table tennis ball is harmless, but a fast-moving cricket ball can be painful. A stationary truck is not a threat, but a slow-moving truck is dangerous. This suggests that a quantity combining mass and velocity is important in describing the effect of motion.
This quantity is called momentum.
Momentum (p) of an object is defined as the product of its mass (m) and velocity (v).
$$ p = mv $$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⁻¹).
An unbalanced force changes the velocity of an object, and therefore it also changes its momentum.
The rate at which momentum changes depends not only on the magnitude of the force but also on the duration over which it acts. Applying a force for a longer time results in a greater change in momentum.
Statement of Newton's Second Law of Motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Mathematical Formulation Of Second Law Of Motion
Consider an object of mass $m$ moving with initial velocity $u$. A constant force $F$ is applied for time $t$, causing its velocity to uniformly change to $v$.
- Initial momentum, $p_1 = mu$.
- Final momentum, $p_2 = mv$.
Change in momentum $= p_2 - p_1 = mv - mu = m(v - u)$.
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:
$F \propto \frac{m(v - u)}{t}$
Introducing a constant of proportionality $k$:
$F = k \frac{m(v - u)}{t}$
Since $\frac{v - u}{t}$ is the acceleration $a$, we have:
$F = kma$
The unit of force (Newton, N) is defined such that a force of 1 N produces an acceleration of 1 m s⁻² in an object of 1 kg mass. Setting $F=1$, $m=1$, and $a=1$, we find $1 = k \times 1 \times 1$, so $k=1$.
Thus, the magnitude of force is given by:
$$ F = ma $$The SI unit of force is kg m s⁻² or Newton (N). 1 N = 1 kg m s⁻².
The Second Law provides a way to measure force: Force is the product of mass and acceleration.
Examples illustrating the Second Law:
- Catching a cricket ball: A fielder pulls their hands back while catching a fast ball. This increases the time taken to reduce the ball's momentum to zero ($t$ increases). Since Force $\propto \frac{\text{Change in momentum}}{\text{Time}}$, increasing the time interval reduces the force exerted on the fielder's hands, preventing injury. Stopping the ball abruptly ($t$ is very small) would require a much larger force.
- High jumpers falling on cushioned/sand bed: The bed increases the time over which the athlete's momentum changes to zero upon landing, reducing the impact force.
- Karate player breaking ice slab: A quick, forceful blow applies a large force over a very short time interval, causing a rapid change in momentum and sufficient force to break the slab.
The First Law of Motion can be derived from the Second Law. From $F = m \frac{(v - u)}{t}$, we have $Ft = mv - mu$. If the net external force $F = 0$, then $0 \times t = mv - mu$, which means $mv - mu = 0$, or $mv = mu$. If $m \neq 0$, then $v = u$. This shows that if no unbalanced force acts ($F=0$), the velocity remains constant ($v=u$), which is the statement of the First Law.
Example 9.1. 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: $m = 5$ kg, $u = 3$ m s⁻¹, $v = 7$ m s⁻¹, $t_1 = 2$ s.
Using $F = m \frac{(v - u)}{t_1}$ to find the force:
$F = 5 \text{ kg} \times \frac{(7 \text{ m s}^{-1} - 3 \text{ m s}^{-1})}{2 \text{ s}} = 5 \text{ kg} \times \frac{4 \text{ m s}^{-1}}{2 \text{ s}} = 5 \text{ kg} \times 2 \text{ m s}^{-2} = 10 \text{ N}$.
The magnitude of the applied force is 10 N.
Now, if this force ($F = 10$ N) is applied for $t_2 = 5$ s, starting from the initial velocity $u = 3$ m s⁻¹, we need to find the new final velocity $v'$.
Using $v' = u + at_2$. We need to find the acceleration $a$ first, using $F = ma$.
$10 \text{ N} = 5 \text{ kg} \times a$
$a = \frac{10 \text{ N}}{5 \text{ kg}} = 2 \text{ m s}^{-2}$.
Now calculate $v'$:
$v' = 3 \text{ m s}^{-1} + (2 \text{ m s}^{-2}) \times (5 \text{ s})$
$v' = 3 \text{ m s}^{-1} + 10 \text{ m s}^{-1} = 13 \text{ m s}^{-1}$.
If the force was applied for 5 s, the final velocity would be 13 m s⁻¹.
Example 9.2. 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:
Case 1: $m_1 = 2$ kg, $a_1 = 5$ m s⁻².
Force required, $F_1 = m_1 a_1 = 2 \text{ kg} \times 5 \text{ m s}^{-2} = 10 \text{ N}$.
Case 2: $m_2 = 4$ kg, $a_2 = 2$ m s⁻².
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 the 2 kg mass at 5 m s⁻² requires a greater force.
Example 9.3. 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: Mass of motorcar + passengers, $m = 1000$ kg.
Initial velocity, $u = 108 \text{ km/h}$. Convert to m s⁻¹: $u = 108 \times \frac{5}{18} \text{ m s}^{-1} = 6 \times 5 \text{ m s}^{-1} = 30 \text{ m s}^{-1}$.
Final velocity, $v = 0 \text{ m s}^{-1}$ (car stops).
Time taken, $t = 4 \text{ s}$.
Using $F = m \frac{(v - u)}{t}$:
$F = 1000 \text{ kg} \times \frac{(0 \text{ m s}^{-1} - 30 \text{ m s}^{-1})}{4 \text{ s}} = 1000 \text{ kg} \times \frac{-30 \text{ m s}^{-1}}{4 \text{ s}}$
$F = 1000 \times (-7.5) \text{ kg m s}^{-2} = -7500 \text{ N}$.
The magnitude of the force exerted by the brakes is 7500 N. The negative sign indicates that the force is acting in the direction opposite to the motion of the car (retarding force).
Example 9.4. 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 s⁻². Using $F = m_1 a_1$, $5 \text{ N} = m_1 \times 10 \text{ m s}^{-2}$. So, $m_1 = \frac{5 \text{ N}}{10 \text{ m s}^{-2}} = 0.5 \text{ kg}$.
For mass $m_2$: acceleration $a_2 = 20$ m s⁻². Using $F = m_2 a_2$, $5 \text{ N} = m_2 \times 20 \text{ m s}^{-2}$. So, $m_2 = \frac{5 \text{ N}}{20 \text{ m s}^{-2}} = 0.25 \text{ kg}$.
When masses are tied together, the total mass $m = m_1 + m_2 = 0.5 \text{ kg} + 0.25 \text{ kg} = 0.75 \text{ kg}$.
If the same force $F = 5$ N acts on the combined mass $m = 0.75$ kg, the acceleration $a$ is given by $F = ma$.
$5 \text{ N} = 0.75 \text{ kg} \times a$
$a = \frac{5 \text{ N}}{0.75 \text{ kg}} = \frac{5}{3/4} \text{ m s}^{-2} = 5 \times \frac{4}{3} \text{ m s}^{-2} = \frac{20}{3} \text{ m s}^{-2} \approx 6.67 \text{ m s}^{-2}$.
The acceleration of the combined mass would be approximately 6.67 m s⁻².
Example 9.5. The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in Fig. 9.9.
How much force does the table exert on the ball to bring it to rest?
Answer:
From the velocity-time graph:
- Initial velocity at $t=0$, $u = 20$ cm s⁻¹.
- Final velocity at $t=10$ s, $v = 0$ cm s⁻¹ (comes to rest).
- Time interval, $t = 10$ s.
Convert mass to kg and velocities to m s⁻¹:
- Mass, $m = 20 \text{ g} = 20/1000 \text{ kg} = 0.02 \text{ kg}$.
- Initial velocity, $u = 20 \text{ cm s}^{-1} = 20/100 \text{ m s}^{-1} = 0.2 \text{ m s}^{-1}$.
- Final velocity, $v = 0 \text{ m s}^{-1}$.
Calculate the acceleration ($a$). Since the graph is a straight line, acceleration is uniform.
$a = \frac{v - u}{t} = \frac{0 \text{ m s}^{-1} - 0.2 \text{ m s}^{-1}}{10 \text{ s}} = \frac{-0.2}{10} \text{ m s}^{-2} = -0.02 \text{ m s}^{-2}$.
The force exerted by the table on the ball ($F$) is given by $F = ma$.
$F = 0.02 \text{ kg} \times (-0.02 \text{ m s}^{-2})$
$F = -0.0004 \text{ kg m s}^{-2} = -0.0004 \text{ N}$.
The magnitude of the force is 0.0004 N. The negative sign indicates that the force (friction) acts in the direction opposite to the ball's motion, causing it to slow down and stop.
Third Law Of Motion
Newton's First and Second Laws describe how a force affects the motion of a single object. The Third Law of Motion describes how forces arise from the interaction between two objects.
Statement of Newton's Third Law of Motion: When one object exerts a force on a second object, the second object simultaneously exerts a force of equal magnitude and opposite direction on the first object.
In simpler terms, forces always occur in pairs. These pairs of forces are often referred to as action and reaction forces.
Important points about Action and Reaction Forces:
- They are always equal in magnitude.
- They are always opposite in direction.
- They always act on two different objects. They never act on the same object, so they cannot cancel each other out (unless you consider the system as a whole, in which case the net internal force is zero).
- They occur simultaneously.
Examples illustrating the Third Law:
- Walking: When you walk, you push the ground backwards with your feet (action). The ground simultaneously pushes you forwards with an equal and opposite force (reaction), propelling you forward.
- Firing a Gun: When a gun fires a bullet, the gun exerts a forward force on the bullet (action). The bullet exerts an equal and opposite force backwards on the gun (reaction). This causes the gun to recoil. Since the gun's mass is much larger than the bullet's mass ($a=F/m$), the gun's backward acceleration (recoil) is much less than the bullet's forward acceleration.
- Jumping from a Boat: When a sailor jumps forward out of a rowing boat, the sailor exerts a force backwards on the boat (action). The boat exerts an equal and opposite force forwards on the sailor (reaction). This force moves the sailor forward, but also pushes the boat backward.
- Spring Balances: Connecting two spring balances and pulling one demonstrates that both show the same reading, indicating equal and opposite forces between them.
Conservation Of Momentum
Newton's Third Law leads directly to a very important principle in physics: the Law of Conservation of Momentum.
Consider a system of two objects, say two balls A and B, of masses $m_A$ and $m_B$, moving with initial velocities $u_A$ and $u_B$ respectively. Assume no external unbalanced forces are acting on this system.
Suppose the balls collide. During the collision, which lasts for a short time $t$, ball A exerts a force $F_{AB}$ on ball B, and ball B exerts a force $F_{BA}$ on ball A.
According to Newton's Third Law, $F_{AB} = -F_{BA}$.
Let $v_A$ and $v_B$ be the velocities of the balls after the collision.
Initial momentum of ball A = $m_A u_A$. Final momentum of ball A = $m_A v_A$.
Rate of change of momentum of ball A = $\frac{m_A v_A - m_A u_A}{t} = \frac{m_A (v_A - u_A)}{t}$. According to the Second Law, this is equal to the force $F_{BA}$ exerted on A by B.
$F_{BA} = \frac{m_A (v_A - u_A)}{t}$.
Initial momentum of ball B = $m_B u_B$. Final momentum of ball B = $m_B v_B$.
Rate of change of momentum of ball B = $\frac{m_B v_B - m_B u_B}{t} = \frac{m_B (v_B - u_B)}{t}$. According to the Second Law, this is equal to the force $F_{AB}$ exerted on B by A.
$F_{AB} = \frac{m_B (v_B - u_B)}{t}$.
Using Newton's Third Law, $F_{AB} = -F_{BA}$:
$\frac{m_B (v_B - u_B)}{t} = - \frac{m_A (v_A - u_A)}{t}$
$m_B (v_B - u_B) = - m_A (v_A - u_A)$
$m_B v_B - m_B u_B = - m_A v_A + m_A u_A$
Rearranging the terms (bringing initial momentum terms to one side and final momentum terms to the other):
$$ m_A u_A + m_B u_B = m_A v_A + m_B v_B $$The term $(m_A u_A + m_B u_B)$ is the total momentum of the system (ball A + ball B) before the collision.
The term $(m_A v_A + m_B v_B)$ is the total momentum of the system after the collision.
This equation shows that the total momentum of the system before the collision is equal to the total momentum after the collision, provided there are no external unbalanced forces acting on the system.
Statement of the Law of Conservation of Momentum: The total momentum of an isolated system of interacting objects remains constant (conserved) if no external unbalanced force acts on the system.
This law is a fundamental principle in physics, valid for all types of interactions (collisions, explosions, etc.). It is based on extensive experimental observations and is considered a conservation law, meaning the total quantity is conserved within the system under specific conditions.
Examples illustrating Conservation of Momentum:
- Recoil of a Gun: Before firing, the total momentum of the gun-bullet system is zero (both are at rest). After firing, the bullet moves forward (positive momentum), and the gun recoils backward (negative momentum). According to conservation of momentum, the total momentum after firing must also be zero. The magnitude of the bullet's forward momentum equals the magnitude of the gun's backward momentum.
- Balloon Rocket: Air escaping backward from an inflated balloon creates momentum in the backward direction. By conservation of momentum, the balloon gains an equal amount of momentum in the forward direction, causing it to move forward.
- Jumping onto a Cart: When a person jumps onto a stationary cart, their initial momentum before the jump is transferred to the combined system of the person and the cart. The total momentum of the person+cart system before the jump (person's momentum + cart's momentum, assuming the cart is initially at rest) is equal to the total momentum of the combined system (person and cart moving together) after the jump.
Example 9.6. A bullet of mass 20 g is horizontally fired with a velocity 150 m s⁻¹ from a pistol of mass 2 kg. What is the recoil velocity of the pistol?
Answer:
Let $m_1$ be the mass of the bullet and $m_2$ be the mass of the pistol.
$m_1 = 20 \text{ g} = 0.02 \text{ kg}$
$m_2 = 2 \text{ kg}$
Before firing, both the bullet and the pistol are at rest. So, their initial velocities are $u_1 = 0 \text{ m s}^{-1}$ and $u_2 = 0 \text{ m s}^{-1}$.
Total momentum before firing = $m_1 u_1 + m_2 u_2 = (0.02 \text{ kg} \times 0 \text{ m s}^{-1}) + (2 \text{ kg} \times 0 \text{ m s}^{-1}) = 0$.
After firing, the bullet's velocity is $v_1 = +150 \text{ m s}^{-1}$ (taking the forward direction as positive). Let the recoil velocity of the pistol be $v_2$.
Total momentum after firing = $m_1 v_1 + m_2 v_2 = (0.02 \text{ kg} \times 150 \text{ m s}^{-1}) + (2 \text{ kg} \times v_2)$
Total momentum after firing = $(0.02 \times 150) \text{ kg m s}^{-1} + 2v_2 \text{ kg m s}^{-1} = 3 \text{ kg m s}^{-1} + 2v_2 \text{ kg m s}^{-1}$.
According to the law of conservation of momentum, total momentum before firing = total momentum after firing.
$0 = 3 + 2v_2$
$2v_2 = -3$
$v_2 = \frac{-3}{2} \text{ m s}^{-1} = -1.5 \text{ m s}^{-1}$.
The recoil velocity of the pistol is 1.5 m s⁻¹. The negative sign indicates that the pistol moves in the opposite direction to the bullet (backward).
Example 9.7. A girl of mass 40 kg jumps with a horizontal velocity of 5 m s⁻¹ onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction.
Answer:
Let $m_1$ be the mass of the girl and $m_2$ be the mass of the cart.
$m_1 = 40 \text{ kg}$
$m_2 = 3 \text{ kg}$
Before the girl jumps onto the cart:
- Girl's velocity, $u_1 = 5 \text{ m s}^{-1}$ (horizontal velocity)
- Cart's velocity, $u_2 = 0 \text{ m s}^{-1}$ (stationary)
Total momentum before jump = $m_1 u_1 + m_2 u_2 = (40 \text{ kg} \times 5 \text{ m s}^{-1}) + (3 \text{ kg} \times 0 \text{ m s}^{-1}) = 200 \text{ kg m s}^{-1}$.
After the girl jumps onto the cart, they move together as a combined object with mass $(m_1 + m_2)$ and a common velocity, let's call it $v$.
Total momentum after jump = $(m_1 + m_2) \times v = (40 \text{ kg} + 3 \text{ kg}) \times v = 43 \text{ kg} \times v$.
According to the law of conservation of momentum, total momentum before = total momentum after.
$200 \text{ kg m s}^{-1} = 43v \text{ kg}$
$v = \frac{200}{43} \text{ m s}^{-1} \approx 4.65 \text{ m s}^{-1}$.
The velocity of the girl as the cart starts moving (i.e., the velocity of the combined system) is approximately 4.65 m s⁻¹ in the horizontal direction the girl was jumping.
Example 9.8. Two hockey players of opposite teams, while trying to hit a hockey ball on the ground collide and immediately become entangled. One has a mass of 60 kg and was moving with a velocity 5.0 m s⁻¹ while the other has a mass of 55 kg and was moving faster with a velocity 6.0 m s⁻¹ 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 ground is negligible.
Answer:
Let $m_1$ and $m_2$ be the masses of the two hockey players, and $u_1$ and $u_2$ be their initial velocities.
$m_1 = 60 \text{ kg}$
$m_2 = 55 \text{ kg}$
Let's choose the direction of the first player's motion as positive. So, $u_1 = +5.0 \text{ m s}^{-1}$.
The second player is moving towards the first, so their direction is opposite. $u_2 = -6.0 \text{ m s}^{-1}$.
Total momentum before collision = $m_1 u_1 + m_2 u_2 = (60 \text{ kg} \times 5.0 \text{ m s}^{-1}) + (55 \text{ kg} \times -6.0 \text{ m s}^{-1})$
Total momentum before collision = $300 \text{ kg m s}^{-1} - 330 \text{ kg m s}^{-1} = -30 \text{ kg m s}^{-1}$.
After the collision, the players become entangled and move together as a single combined mass $(m_1 + m_2)$ with a common velocity, let's call it $v$.
Total momentum after collision = $(m_1 + m_2) \times v = (60 \text{ kg} + 55 \text{ kg}) \times v = 115 \text{ kg} \times v$.
According to the law of conservation of momentum, total momentum before = total momentum after.
$-30 \text{ kg m s}^{-1} = 115v \text{ kg}$
$v = \frac{-30}{115} \text{ m s}^{-1} \approx -0.261 \text{ m s}^{-1}$.
The velocity of the entangled players is approximately -0.261 m s⁻¹. The negative sign indicates that they will move in the negative direction, which is the direction the second player was initially moving (towards the first player, from right to left). So, they move with a velocity of about 0.261 m s⁻¹ from right to left.
Intext Questions
Page No. 118
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:
Page No. 126 - 127
Question 1. If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:
Question 2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer:
Question 3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 $m s^{-1}$. Calculate the initial recoil velocity of the rifle.
Answer:
Question 4. Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 $m s^{-1}$ and 1 $m s^{-1}$, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 $m s^{-1}$. Determine the velocity of the second object.
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 the magnitude and direction of the velocity. If no, provide a reason.
Answer:
Question 2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
Question 3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
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.
Answer:
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.)
Answer:
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. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m $s^{-1}$ before the collision during which they stick together. What will be the velocity of the combined object after collision?
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Question 12. According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
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Question 13. 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 14. 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 15. 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 16. 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 17. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
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Question 18. How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward 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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