CBSE Class 9 Science Force and laws of motion Notes

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FORCE AND LAWS OF MOTION

FORCE AND LAWS OF MOTION

FORCE

A force is anything that can cause a change to objects. Forces can change the shape of an object, move or stop an object, and change the direction of a moving object. A force can be classified as either a contact force or a non-contact force.

A contact force must touch, or be in contact with, an object to cause a change. Examples: the force used to push or pull things, like on a door to open or close it; the force a sculptor uses to turn clay into a pot; the force of the wind that turns a windmill.

A non-contact force does not have to touch an object to cause a change. Examples: the force due to gravity, like the Earth pulling the Moon towards itself; the force due to electricity, like a proton and an electron attracting each other; the force due to magnetism, like a magnet pulling a paper clip towards itself.

The unit of force is the newton (symbol N), named after Sir Isaac Newton, who first defined force. Force is a vector quantity: it has a magnitude and a direction.

EFFECT OF FORCE

1. Force can make a stationary body move. For example, a football is set in motion by kicking it, i.e. by applying a force.
2. Force can stop a moving body. For example, by applying brakes a running cycle or vehicle can be stopped.
3. Force can change the direction of a moving object. For example, by turning the handle the direction of a running bicycle is changed; similarly, by turning the steering wheel the direction of a running vehicle is changed.
4. Force can change the speed of a moving body. By accelerating, the speed of a running vehicle can be increased, and by applying brakes it can be decreased.
5. Force can change the shape and size of an object. For example, by hammering, a block of metal can be turned into a thin sheet, and a stone can be broken into pieces.

Forces can also be divided into two types: balanced forces and unbalanced forces.

Balanced forces. If the resultant of the applied forces is equal to zero, they are called balanced forces. Example: in a tug of war, if both teams apply forces of the same magnitude in opposite directions, the rope does not move to either side, because the resultant of the applied forces is zero. Balanced forces do not cause any change in the state of motion of an object; they are equal in magnitude and opposite in direction. Balanced forces can, however, change the shape and size of an object: for example, when forces are applied from both sides on a balloon, its size and shape change.

Unbalanced forces. If the resultant of the applied forces is greater than zero, the forces are called unbalanced forces. An object at rest can be moved by applying unbalanced forces. Unbalanced forces can: move a stationary object; increase the speed of a moving object; decrease the speed of a moving object; stop a moving object; change the shape and size of an object.

NEWTON’S LAWS OF MOTION

• First law: any object remains in the state of rest or in uniform motion along a straight line until it is compelled to change that state by an applied external force.
• Second law: the rate of change of momentum is directly proportional to the force applied, and takes place in the direction of the force.
• Third law: there is an equal and opposite reaction for every action.

NEWTON’S FIRST LAW OF MOTION

Explanation. If an object is at rest, it remains at rest until an external force is applied to change its state. Similarly, an object remains in motion until an external force is applied to change its state. This means all objects resist a change in their state. The state of any object can be changed only by applying external forces.

Newton’s first law in everyday life

1. A person standing in a bus falls backward when the bus starts moving suddenly. The person and the bus are both at rest, but as the bus starts moving, the legs of the person move along with the bus while the rest of his body tends to remain at rest. Because of this the person falls backward if he is not alert.
2. A person standing in a moving bus falls forward if the driver applies the brakes suddenly. While the bus is moving, the person is also in motion along with the bus. When the driver applies the brakes, the speed of the bus decreases suddenly; the legs of the person in contact with the bus come to rest, while the rest of his body tends to remain in motion. Because of this the person falls forward if he is not alert.

MASS AND INERTIA

The property of an object because of which it resists any change in its state is called inertia. In other words, the natural tendency of an object to resist a change in its state of rest or of motion is called inertia. The inertia of an object is measured by its mass: inertia is directly proportional to mass. A heavy object has more inertia than a lighter one. Since a heavy object has more inertia, it is more difficult to push or pull a heavy box over the ground than a lighter one.

INTEXT QUESTIONS (PAGE 118)

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-rupee coin and a one-rupee coin?

Answer: Inertia is the measure of the mass of a body. The greater the mass of the body, the greater its inertia, and vice versa.

(a) The mass of a stone is more than the mass of a rubber ball of the same size, so the stone has greater inertia.

(b) The mass of a train is more than the mass of a bicycle, so the train has greater inertia.

(c) The mass of a five-rupee coin is more than that of a one-rupee coin, so the five-rupee coin has greater inertia.

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: The velocity of the ball changes four times.

First, the football player kicks the football: its speed changes from zero to a certain value, so its velocity changes. The agent supplying the force is the first player.

Second, another player kicks the ball towards the goal post: the direction of the ball changes, so its velocity changes. The agent is the second player.

Third, the goalkeeper collects the ball: the ball comes to rest, so its speed reduces from a certain value to zero and its velocity changes. The agent is the goalkeeper, who applies an opposing force to stop the ball.

Fourth, the goalkeeper kicks the ball towards his team players: the speed increases from zero to a certain value, so the velocity changes once again. The agent is the goalkeeper.

Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer: When the branches of a tree are shaken, they move to and fro, but the leaves tend to remain at rest because of their inertia, which resists the to-and-fro motion. Due to this, the leaves get detached and fall down.

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: It is due to the inertia of the passenger. Every body tries to maintain its state of motion or of rest. In a moving bus, a passenger moves with the bus. As the driver applies the brakes, the bus comes to rest, but the passenger tries to maintain his state of motion and so falls forward. Similarly, when the bus accelerates from rest, the passenger’s inertia opposes the forward motion of the bus (his body tends to stay at rest), so he falls backwards.

MOMENTUM

Momentum is the quantity of motion of an object. The product of mass and velocity is called momentum. It is denoted by p:

\[p=m\times v\]

where p = momentum, m = mass of the object, v = velocity of the object.

NEWTON’S SECOND LAW OF MOTION

Newton’s second law states that the rate of change of momentum is directly proportional to the force applied, in the direction of the force. For example, when a vehicle accelerates, its momentum increases, and the increase is in the direction of motion because the force is applied in that direction. When the brake is applied on a moving vehicle, its momentum decreases, and the decrease is in the direction opposite to the motion because the force is applied in the opposite direction.

Mathematical formulation. Let the mass of a moving object be m, and let its velocity change from u to v in time t. Initial momentum \(=mu\); final momentum \(=mv\). Change in momentum \(=mv-mu\), so

\[\text{Rate of change of momentum}=\dfrac{mv-mu}{t}=\dfrac{m(v-u)}{t}\qquad\ldots(i)\]

According to the second law, force is directly proportional to the rate of change of momentum:

\[F\propto\dfrac{m(v-u)}t\;\Rightarrow\;F\propto ma\quad\left(a=\dfrac{v-u}t\right)\;\Rightarrow\;F=k\,ma\qquad\ldots(ii)\]

where k is a constant of proportionality. One unit of force is defined as the force which produces an acceleration of 1 m/s² in an object of mass 1 kg. So 1 unit of force \(=k\times1\text{ kg}\times1\text{ m/s}^2\), thus \(k=1\). Substituting k = 1 in (ii):

\[F=ma\qquad\ldots(iii)\]

Force = mass × acceleration. Thus Newton’s second law gives the relation between force, mass and acceleration of an object. The product of mass and acceleration is the force acting on the object.

SI unit of force. Since force = mass × acceleration, with mass in kg and acceleration in m/s², 1 newton = 1 kg × 1 m/s², i.e. N = kg m/s².

Equation (iii) can also be written as

\[a=\dfrac Fm\qquad\ldots(iv)\]

So Newton’s second law can also be stated as: the acceleration produced in an object is directly proportional to the force applied on it and inversely proportional to its mass. Acceleration increases with increase in force and vice versa. Acceleration decreases with increase in mass and vice versa. That is why a small vehicle requires less force to attain a given acceleration, while a heavy vehicle requires more force to get the same acceleration.

Newton’s second law in everyday life

1. A fielder pulls his hand backward while catching a cricket ball coming with great speed, so that the momentum of the ball is reduced to zero over a longer time.
2. While catching a cricket ball, the momentum of the ball is reduced to zero when it is stopped by the fielder’s hands. If the ball is stopped suddenly, its momentum becomes zero instantly; the rate of change of momentum is very large, and the player’s hand may get injured. By pulling the hand backward, the fielder gives more time for the momentum to become zero, so the rate of change of momentum (the force) is smaller. This prevents the hands from getting hurt.

NUMERICALS (SET 1)

What is the acceleration produced by a force of 12 N exerted on an object of mass 3 kg?

Answer: 4 m/s².

\(a=\dfrac Fm=\dfrac{12}3=4\) m/s².

What force would be needed to produce an acceleration of 4 m/s² on a ball of mass 6 kg?

Answer: 24 N.

\(F=ma=6\times4=24\) N.

A force of 5 N gives a mass \(m_1\) an acceleration of 8 m/s², and a mass \(m_2\) an acceleration of 24 m/s². What acceleration would it give if both the masses are tied together?

Answer: 6 m/s².

\(m_1=\dfrac58=0.625\) kg and \(m_2=\dfrac5{24}\) kg. Together \(m_1+m_2=\dfrac58+\dfrac5{24}=\dfrac56\) kg. \(a=\dfrac F{m_1+m_2}=\dfrac5{5/6}=6\) m/s².

Calculate the force required to impart to a car a velocity of 30 m/s in 10 seconds. The mass of the car is 1500 kg.

Answer: 4500 N.

Starting from rest, \(a=\dfrac{30-0}{10}=3\) m/s², so \(F=ma=1500\times3=4500\) N.

A motorcycle is moving with a velocity of 90 km/h and it takes 5 seconds to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcycle if its mass along with the rider is 200 kg.

Answer: 1000 N (opposite to the motion).

\(u=90\) km/h \(=25\) m/s, \(v=0\), \(t=5\) s. \(a=\dfrac{0-25}5=-5\) m/s². \(F=ma=200\times(-5)=-1000\) N. The magnitude of the braking force is 1000 N.

What is the momentum of a man of mass 75 kg when he walks with a velocity of 2 m/s?

Answer: 150 kg m/s.

\(p=mv=75\times2=150\) kg m/s.

What would be the force required to produce an acceleration of 2 m/s² in a body of mass 12 kg? What would be the acceleration if the force were doubled?

Answer: 24 N; 4 m/s².

\(F=ma=12\times2=24\) N. If the force is doubled to 48 N, \(a=\dfrac{48}{12}=4\) m/s².

A man pushes a box of mass 50 kg with a force of 80 N. What will be the acceleration of the box? What would be the acceleration if the mass were doubled?

Answer: 1.6 m/s²; 0.8 m/s².

\(a=\dfrac Fm=\dfrac{80}{50}=1.6\) m/s². If the mass is doubled to 100 kg, \(a=\dfrac{80}{100}=0.8\) m/s².

A certain force exerted for 1.2 s raises the speed of an object from 1.8 m/s to 4.2 m/s. Later, the same force is applied for 2 s. How much does the speed change in 2 s?

Answer: The speed changes by 4 m/s.

\(a=\dfrac{4.2-1.8}{1.2}=2\) m/s². The same force on the same object gives the same acceleration, so in 2 s the change in speed is \(a\times t=2\times2=4\) m/s.

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 to 7 m/s. Find the magnitude of the applied force. Now if the force were applied for a duration of 5 s, what would be the final velocity of the object?

Answer: 10 N; 13 m/s.

\(a=\dfrac{7-3}2=2\) m/s², so \(F=ma=5\times2=10\) N. If the force acts for 5 s, \(v=u+at=3+2\times5=13\) m/s.

(The source gives the initial velocity as “3 cm/s”, which looks like a misprint for 3 m/s; the answer is for 3 m/s.)

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: 7500 N (opposite to the motion).

\(u=108\) km/h \(=30\) m/s, \(v=0\), \(t=4\) s. \(a=\dfrac{0-30}4=-7.5\) m/s². \(F=ma=1000\times(-7.5)=-7500\) N; the magnitude is 7500 N.

A force of 5 N gives a mass \(m_1\) an acceleration of 10 m/s², and a mass \(m_2\) an acceleration of 20 m/s². What acceleration would it give if both the masses were tied together?

Answer: 6.67 m/s² (\(\dfrac{20}3\) m/s²).

\(m_1=\dfrac5{10}=0.5\) kg, \(m_2=\dfrac5{20}=0.25\) kg. Together 0.75 kg, so \(a=\dfrac5{0.75}=6.67\) m/s².

For how long should a force of 100 N act on a body of mass 20 kg so that it acquires a velocity of 100 m/s?

Answer: 20 s.

\(a=\dfrac Fm=\dfrac{100}{20}=5\) m/s². Starting from rest, \(t=\dfrac{v}a=\dfrac{100}5=20\) s.

A 150 g ball travelling at 30 m/s strikes the palm of a player’s hand and is stopped in 0.06 s. Calculate the force exerted by the ball on the hand.

Answer: 75 N.

\(m=0.15\) kg. Change in momentum \(=0.15\times30=4.5\) kg m/s. \(F=\dfrac{4.5}{0.06}=75\) N.

A body of mass 1 kg is kept at rest. A constant force of 6.0 N starts acting on it. Find the time taken by the body to move through a distance of 12 m.

Answer: 2 s.

\(a=\dfrac61=6\) m/s². \(s=\tfrac12at^2\Rightarrow12=\tfrac12\times6\times t^2\Rightarrow t^2=4\Rightarrow t=2\) s.

A force of 4 N acts on a body of mass 2 kg for 4 s. Assuming that the body is initially at rest, find (i) its velocity when the force stops acting, (ii) the distance covered in 10 s after the force starts acting.

Answer: (i) 8 m/s; (ii) 64 m.

\(a=\dfrac42=2\) m/s². (i) \(v=at=2\times4=8\) m/s.

(ii) In the first 4 s: \(s_1=\tfrac12\times2\times4^2=16\) m. After that the force stops and the body moves with constant velocity 8 m/s for the remaining 6 s: \(s_2=8\times6=48\) m. Total distance \(=16+48=64\) m.

A feather of mass 10 g is dropped from a height. It is found to fall down with a constant velocity. What is the net force acting on it?

Answer: Zero.

Constant velocity means zero acceleration, so by \(F=ma\) the net force is zero: the weight of the feather (about 0.098 N, downward) is balanced by the air resistance (upward).

A hockey ball of mass 200 g travelling from west to east at 10 m/s is struck by a hockey stick. As a result, the ball gets turned back and now has a speed of 5 m/s. If the ball and the hockey stick were in contact for 0.2 s, calculate (i) the initial and final momentum of the ball, (ii) the rate of change of momentum of the ball, (iii) the force exerted by the hockey stick on the ball.

Answer: (i) 2 kg m/s and –1 kg m/s; (ii) –15 kg m/s²; (iii) 15 N towards the west.

Taking west to east as positive: \(m=0.2\) kg. (i) Initial momentum \(=0.2\times10=2\) kg m/s; final momentum \(=0.2\times(-5)=-1\) kg m/s.

(ii) Rate of change of momentum \(=\dfrac{-1-2}{0.2}=-15\) kg m/s².

(iii) Force \(=\) rate of change of momentum \(=-15\) N, i.e. 15 N directed from east to west.

A stone of mass 500 g is thrown with a velocity of 20 m/s across the frozen surface of a lake. It comes to rest after travelling a distance of 0.1 km. Calculate the force of friction between the stone and the frozen surface of the lake.

Answer: 1 N.

\(v^2=u^2+2as\Rightarrow0=20^2+2a(100)\Rightarrow a=-2\) m/s². \(F=ma=0.5\times(-2)=-1\) N, so the force of friction has magnitude 1 N, opposite to the motion.

A body starts from rest and rolls down a hill with a constant acceleration. If it travels 400 m in 20 s, calculate the force acting on the body if its mass is 10 kg.

Answer: 20 N.

\(s=\tfrac12at^2\Rightarrow400=\tfrac12a(20)^2\Rightarrow a=2\) m/s². \(F=ma=10\times2=20\) N.

The velocity–time graph of a ball of mass 20 g moving along a straight line on a long table is given in the figure below. How much force does the table exert on the ball to bring it to rest?

Answer: 0.04 N.

From the graph the speed falls from 20 m/s to 0 in 10 s, so \(a=\dfrac{0-20}{10}=-2\) m/s². \(m=20\) g \(=0.02\) kg. \(F=ma=0.02\times(-2)=-0.04\) N; the table exerts a force of 0.04 N opposing the motion.

The speed–time graph of a ball of mass 30 g moving along a straight line is shown in the figure below. Calculate the opposing force that brings the ball to rest.

Answer: 0.3 N.

From the graph the speed falls from 50 m/s to 0 in 5 s, so \(a=\dfrac{0-50}5=-10\) m/s². \(m=30\) g \(=0.03\) kg. \(F=ma=0.03\times(-10)=-0.3\) N; the opposing force is 0.3 N.

What will be the percentage change in momentum of a body when both its mass and velocity are doubled?

Answer: 300% increase.

Initial momentum \(p=mv\). New momentum \(=(2m)(2v)=4mv=4p\). Change \(=4p-p=3p\), i.e. \(\dfrac{3p}{p}\times100=300\%\) increase.

A force of 2 N gives a mass \(m_1\) an acceleration of 5 m/s², and a mass \(m_2\) an acceleration of 7 m/s². What acceleration would be produced if both the masses are tied together?

Answer: 2.92 m/s² (\(\dfrac{35}{12}\) m/s²).

\(m_1=\dfrac25=0.4\) kg, \(m_2=\dfrac27\) kg. Together \(\dfrac25+\dfrac27=\dfrac{24}{35}\) kg, so \(a=\dfrac{2}{24/35}=\dfrac{35}{12}=2.92\) m/s².

A body of mass 2 kg moving with a velocity of 10 m/s is brought to rest in 5 s. Calculate the stopping force applied.

Answer: 4 N.

\(a=\dfrac{0-10}5=-2\) m/s², \(F=ma=2\times(-2)=-4\) N; the stopping force has magnitude 4 N.

NEWTON’S THIRD LAW OF MOTION

Newton’s third law states that there is always a reaction for every action, in the opposite direction and of equal magnitude. Explanation: whenever a force is applied on a body, that body also applies a force of equal magnitude on the first, in the opposite direction.

Examples

1. Walking of a person: a person is able to walk because of Newton’s third law. While walking, a person pushes the ground backward, and as a reaction the ground pushes the person with an equal force in the opposite direction. This enables him to move forward.
2. Recoil of a gun: when a bullet is fired from a gun, the bullet also pushes the gun backward with an equal force. The gunman feels a backward push from the butt of the gun.
3. Propulsion of a boat: the sailor pushes the water backward with an oar, and the water pushes the oar forward. Consequently the boat is pushed forward. The forces applied by the oar and the water are of equal magnitude but opposite in direction.

CONSERVATION OF MOMENTUM

Law of conservation of momentum. The sum of the momenta of two objects remains the same even after a collision. In other words, the sum of the momenta of two objects before the collision is equal to the sum of their momenta after the collision, provided no external unbalanced force acts on them.

Mathematical formulation. Suppose two objects A and B are moving along a straight line in the same direction, and the velocity of A is greater than that of B. Let the masses be \(m_A\) and \(m_B\), the initial velocities \(u_A\) and \(u_B\), and the velocities after the collision \(v_A\) and \(v_B\). The collision lasts for t seconds.

Momentum of A before collision \(=m_Au_A\); momentum of B before collision \(=m_Bu_B\); momentum of A after collision \(=m_Av_A\); momentum of B after collision \(=m_Bv_B\). The rate of change of momentum of A during the collision is \(F_{AB}=m_A\dfrac{v_A-u_A}t\), and that of B is \(F_{BA}=m_B\dfrac{v_B-u_B}t\).

By Newton’s third law the force exerted by A on B equals the force exerted by B on A, but in the opposite direction: \(F_{AB}=-F_{BA}\). Therefore

\[m_A\dfrac{v_A-u_A}t=-m_B\dfrac{v_B-u_B}t\;\Rightarrow\;m_A(v_A-u_A)=-m_B(v_B-u_B)\]

\[m_Au_A+m_Bu_B=m_Av_A+m_Bv_B\qquad\ldots(i)\]

This says that the total momentum of A and B before the collision is equal to the total momentum after the collision. The total momentum of the two objects remains unchanged, or conserved, provided no external force acts. This is the law of conservation of momentum.

Practical applications

• Bullet and gun: when a bullet is fired from a gun, the gun recoils in the direction opposite to the bullet. The momentum of the bullet is equal in magnitude to the momentum of the gun. Since the bullet has a very small mass compared to the gun, the velocity of the bullet is very high compared to the recoil velocity of the gun.
• When a cricket ball is hit by a bat, the law of conservation of momentum applies.

NUMERICALS (SET 2)

The velocity of a body of mass 10 kg increases from 4 m/s to 8 m/s when a force acts on it for 2 s. (a) What is the momentum before and after the force acts? (b) What is the gain in momentum per second? (c) What is the value of the force?

Answer: (a) 40 kg m/s and 80 kg m/s; (b) 20 kg m/s²; (c) 20 N.

(a) Momentum before \(=10\times4=40\) kg m/s; after \(=10\times8=80\) kg m/s.

(b) Gain in momentum \(=80-40=40\) kg m/s in 2 s, i.e. \(\dfrac{40}2=20\) kg m/s per second.

(c) Force = rate of change of momentum = 20 N.

A boy pushes a wall with a force of 20 N. What is the magnitude and direction of the force experienced by the boy?

Answer: 20 N, in the direction opposite to his push (from the wall towards the boy).

By Newton’s third law, the wall exerts an equal and opposite reaction on the boy.

A 20 g bullet is shot from a 5 kg gun with a velocity of 500 m/s. What is the speed of the recoil of the gun?

Answer: 2 m/s.

\(0=0.02\times500+5v\Rightarrow v=-\dfrac{10}5=-2\) m/s. The gun recoils at 2 m/s, opposite to the bullet.

A 10 g bullet is shot from a 5 kg gun with a velocity of 400 m/s. What is the speed of the recoil of the gun?

Answer: 0.8 m/s.

\(0=0.01\times400+5v\Rightarrow v=-\dfrac45=-0.8\) m/s.

When two bodies A and B interact with each other, A exerts a force of 10 N on B towards the east. What is the force exerted by B on A?

Answer: 10 N towards the west.

By Newton’s third law the reaction is equal in magnitude and opposite in direction.

A man weighing 60 kg runs along the rails with a velocity of 18 km/h and jumps into a car of mass 1 quintal standing on the rails. Calculate the velocity with which the car will start travelling along the rails.

Answer: 1.875 m/s.

18 km/h \(=5\) m/s; 1 quintal \(=100\) kg. \(60\times5+100\times0=(60+100)v\Rightarrow v=\dfrac{300}{160}=1.875\) m/s.

Car A of mass 1500 kg, travelling at 25 m/s, collides with another car B of mass 1000 kg travelling at 15 m/s in the same direction. After the collision, the velocity of car A becomes 20 m/s. Calculate the velocity of car B after the collision.

Answer: 22.5 m/s.

\(1500\times25+1000\times15=1500\times20+1000v_B\Rightarrow52500=30000+1000v_B\Rightarrow v_B=22.5\) m/s.

A bullet of mass 10 g is fired from a gun of mass 6 kg with a velocity of 300 m/s. Calculate the recoil velocity of the gun.

Answer: 0.5 m/s.

\(0=0.01\times300+6v\Rightarrow v=-0.5\) m/s.

A bullet of mass 50 g is fired from a gun of mass 6 kg with a velocity of 400 m/s. Calculate the recoil velocity of the gun.

Answer: 3.33 m/s.

\(0=0.05\times400+6v\Rightarrow v=-\dfrac{20}6=-3.33\) m/s.

A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900 g. What is the velocity acquired by the block?

Answer: About 4.4 m/s.

Momentum before \(=0.01\times400=4\) kg m/s. After the bullet is embedded, the total mass is \(0.9+0.01=0.91\) kg. \(v=\dfrac4{0.91}=4.4\) m/s.

A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 s to move through the barrel of the gun and acquires a velocity of 100 m/s. Calculate (i) the velocity with which the gun recoils, (ii) the force exerted on the gunman due to the recoil of the gun.

Answer: (i) 1 m/s; (ii) 1000 N.

(i) \(0=0.03\times100+3v\Rightarrow v=-1\) m/s.

(ii) Change in momentum of the gun \(=3\times1=3\) kg m/s in 0.003 s, so \(F=\dfrac3{0.003}=1000\) N.

A heavy car A of mass 200 kg travelling at 10 m/s has a head-on collision with a sports car B of mass 500 kg. If both cars stop dead on colliding, what was the velocity of car B?

Answer: 4 m/s, in the direction opposite to car A.

Total momentum before = total momentum after = 0. Taking the direction of A as positive: \(200\times10+500\times v_B=0\Rightarrow v_B=-4\) m/s.

A machine gun fires 25 g bullets at the rate of 600 bullets per minute with a speed of 200 m/s. Calculate the force required to keep the gun in position.

Answer: 50 N.

600 bullets per minute \(=10\) bullets per second. Momentum of each bullet \(=0.025\times200=5\) kg m/s. Momentum given to bullets per second \(=10\times5=50\) kg m/s², so the force required is 50 N.

(The source reads “25h bullet”; it is taken as 25 g.)

A bullet of mass 20 g moving with a velocity of 300 m/s gets embedded in a freely suspended wooden block of mass 880 g. What is the velocity acquired by the block?

Answer: 6.67 m/s.

Momentum before \(=0.02\times300=6\) kg m/s. Total mass after \(=0.88+0.02=0.9\) kg. \(v=\dfrac6{0.9}=6.67\) m/s.

A girl of mass 50 kg jumps out of a rowing boat of mass 300 kg onto the bank with a horizontal velocity of 3 m/s. With what velocity does the boat begin to move backwards?

Answer: 0.5 m/s.

\(0=50\times3+300v\Rightarrow v=-0.5\) m/s.

A truck of mass 2500 kg moving at 15 m/s collides with a car of mass 1000 kg moving at 5 m/s in the opposite direction. What is their common velocity?

Answer: 9.29 m/s in the direction of the truck’s motion.

Taking the truck’s direction as positive: \(2500\times15+1000\times(-5)=(2500+1000)v\Rightarrow32500=3500v\Rightarrow v=9.29\) m/s.

A bullet of mass 20 g is fired horizontally with a velocity of 150 m/s from a pistol of mass 2 kg. What is the recoil velocity of the pistol?

Answer: 1.5 m/s.

\(0=0.02\times150+2v\Rightarrow v=-1.5\) m/s.

A body of mass 60 kg running at 3 m/s jumps on to a trolley of mass 140 kg moving with a velocity of 1.5 m/s in the same direction. What is their common velocity?

Answer: 1.95 m/s.

\(60\times3+140\times1.5=(60+140)v\Rightarrow390=200v\Rightarrow v=1.95\) m/s.

A girl of mass 40 kg jumps with a horizontal velocity of 5 m/s 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: 4.65 m/s.

\(40\times5+3\times0=(40+3)v\Rightarrow v=\dfrac{200}{43}=4.65\) m/s.

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 of 5 m/s, while the other has a mass of 55 kg and was moving faster with a velocity of 6 m/s towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force between the feet of the players and the ground is negligible.

Answer: 0.26 m/s in the direction in which the 55 kg player was moving.

Taking the first player’s direction as positive: \(60\times5+55\times(-6)=(60+55)v\Rightarrow300-330=115v\Rightarrow v=-\dfrac{30}{115}=-0.26\) m/s. The negative sign means they move in the direction of the second (55 kg) player.

INTEXT QUESTIONS (PAGE 126)

If action is always equal to the reaction, explain how a horse can pull a cart.

Answer: The horse pushes the ground in the backward direction. According to Newton’s third law, a reaction force is exerted by the Earth on the horse in the forward direction. As a result the cart moves forward.

Explain why it is difficult for a fireman to hold a hose which ejects large amounts of water at a high velocity.

Answer: Because of the backward reaction of the water being ejected. When water is ejected at high velocity, by Newton’s third law a reaction force acts on the hose, and so on the fireman, in the backward direction. This decreases his stability, so it is difficult for him to remain stable while holding the hose.

From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m/s. Calculate the initial recoil velocity of the rifle.

Answer: 0.4375 m/s, backwards.

Mass of rifle \(m_1=4\) kg; mass of bullet \(m_2=50\) g \(=0.05\) kg; velocity of bullet \(v_2=35\) m/s; initially both are at rest, so the total initial momentum is 0. After firing the total momentum is \(4v_1+0.05\times35=4v_1+1.75\). By conservation of momentum, \(4v_1+1.75=0\Rightarrow v_1=-\dfrac{1.75}4=-0.4375\) m/s. The negative sign indicates that the rifle recoils backwards.

Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m/s and 1 m/s, respectively. They collide and after the collision the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.

Answer: 1.165 m/s.

\(m_1=0.1\) kg, \(m_2=0.2\) kg. By conservation of momentum: \(0.1\times2+0.2\times1=0.1\times1.67+0.2v\Rightarrow0.4=0.167+0.2v\Rightarrow v=1.165\) m/s.

EXERCISE QUESTIONS (PAGE 128)

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: Yes. This is possible if the object has been moving with a constant velocity (constant magnitude and a constant direction). Then there is no net unbalanced force on it and it keeps moving with that non-zero velocity. To change the state of motion, a net non-zero external unbalanced force must be applied.

When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer: Inertia of an object resists any change in its state of rest or of motion. When the carpet is beaten with a stick, the carpet is set in motion, but the dust particles tend to remain at rest. According to Newton’s first law the dust particles stay at rest while the carpet moves, so the dust comes out of the carpet.

Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer: When the bus accelerates, it acquires a state of motion, but the luggage on the roof, owing to its inertia, tends to remain at rest. So the luggage tends to stay at its original position and may fall off the back of the roof. (Similarly, when the bus turns or stops suddenly, the luggage tends to continue in its earlier motion.) To avoid this, luggage on the roof is tied with a rope.

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: (c) There is a force on the ball opposing the motion.

The ball comes to rest because of the frictional force on it, which always acts in the direction opposite to the motion.

A truck starts from rest and rolls down a hill with 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 metric tonnes. (Hint: 1 metric tonne = 1000 kg.)

Answer: Acceleration 2 m/s²; force 14000 N.

\(u=0\), \(s=400\) m, \(t=20\) s. \(s=ut+\tfrac12at^2\Rightarrow400=\tfrac12a(20)^2\Rightarrow a=2\) m/s². Mass \(=7\times1000=7000\) kg. \(F=ma=7000\times2=14000\) N.

A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Answer: 4 N (opposite to the motion).

\(u=20\) m/s, \(v=0\), \(s=50\) m. \(v^2=u^2+2as\Rightarrow0=400+100a\Rightarrow a=-4\) m/s². \(F=ma=1\times(-4)=-4\) N; the negative sign shows that the force acts against the motion.

An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate (a) the net accelerating force, (b) the acceleration of the train, and (c) the force of wagon 1 on wagon 2.

Answer: (a) 35000 N; (b) 1.944 m/s²; (c) about 15556 N.

(a) \(F_a=F-F_f=40000-5000=35000\) N.

(b) Mass of the 5 wagons \(=5\times2000=10000\) kg; with the engine, the total mass \(M=10000+8000=18000\) kg. \(a=\dfrac{F_a}M=\dfrac{35000}{18000}=1.944\) m/s².

(c) The force of wagon 1 on wagon 2 is the force needed to accelerate the other four wagons (wagons 2 to 5): mass \(=4\times2000=8000\) kg, acceleration \(=1.944\) m/s². Force \(=8000\times1.944\approx15556\) N.

(The source used an acceleration of 3.5 m/s² here, which is not the acceleration of the train, and got 28000 N; it was corrected.)

An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and the road if the vehicle is to be stopped with a negative acceleration of 1.7 m/s²?

Answer: 2550 N, opposite to the direction of motion.

\(F=ma=1500\times(-1.7)=-2550\) N. The force between the vehicle and the road is 2550 N, directed opposite to the motion.

What is the momentum of an object of mass m moving with a velocity v? (a) \((mv)^2\), (b) \(mv^2\), (c) \(\tfrac12mv^2\), (d) \(mv\)

Answer: (d) \(mv\).

Momentum = mass × velocity = \(mv\).

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?

Answer: 200 N, opposite to the direction of the applied force.

The cabinet moves with constant velocity, so its acceleration is zero and the net force on it is zero. The applied force of 200 N is therefore balanced by an equal and opposite force of friction, 200 N.

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 before the collision, during which they stick together. What will be the velocity of the combined object after the collision?

Answer: Zero.

Take the direction of the first object as positive: \(v_1=2.5\) m/s, \(v_2=-2.5\) m/s. By conservation of momentum: \(1.5(2.5)+1.5(-2.5)=(1.5+1.5)v\Rightarrow3.75-3.75=3v\Rightarrow v=0\).

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.

Answer: The student’s logic is not correct. The action and reaction forces act on different bodies (one on the truck, the other on us), so they can never cancel each other.

The truck does not move because the force we apply is smaller than the maximum force of static friction between the truck and the road. The truck’s large mass makes this friction large. The push on the truck is balanced by the force of friction acting on the truck in the opposite direction, so the net force on the truck is zero. To move it, one must apply a force greater than the static friction.

(The source said that the student was right; this was corrected.)

A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity of 5 m/s. Calculate the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Answer: 3 kg m/s in magnitude, directed opposite to the original motion (–3 kg m/s).

\(m=0.2\) kg. Initial momentum \(=0.2\times10=2\) kg m/s. Final momentum \(=0.2\times(-5)=-1\) kg m/s. Change in momentum \(=-1-2=-3\) kg m/s, i.e. 3 kg m/s in the direction opposite to the original motion.

A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer: Distance 2.25 m; force 50 N.

\(u=150\) m/s, \(v=0\), \(t=0.03\) s. From \(v=u+at\): \(0=150+a(0.03)\Rightarrow a=-5000\) m/s². From \(v^2=u^2+2as\): \(0=150^2-2(5000)s\Rightarrow s=\dfrac{22500}{10000}=2.25\) m. Force \(=ma=0.01\times5000=50\) N (magnitude).

An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also calculate the velocity of the combined object.

Answer: 10 kg m/s before and after; velocity \(\dfrac53\) m/s \(\approx1.67\) m/s.

Momentum before \(=1\times10+5\times0=10\) kg m/s. By conservation of momentum: \(1(10)+5(0)=(1+5)v\Rightarrow v=\dfrac{10}6=\dfrac53\) m/s. Momentum just after \(=(1+5)\times\dfrac53=10\) kg m/s.

An object of mass 100 kg is accelerated uniformly from a velocity of 5 m/s to 8 m/s in 6 s. Calculate the initial and final momentum of the object. Also find the magnitude of the force exerted on the object.

Answer: 500 kg m/s and 800 kg m/s; force 50 N.

Initial momentum \(=100\times5=500\) kg m/s; final momentum \(=100\times8=800\) kg m/s. \(F=\dfrac{800-500}6=\dfrac{300}6=50\) N.

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.

Answer: Rahul’s explanation is the correct one.

By Newton’s third law the car and the insect exert equal and opposite forces on each other for the same time, so each experiences the same magnitude of force. The total momentum of the car–insect system is conserved, so the momentum gained by the insect is equal in magnitude to the momentum lost by the car; the changes in momentum are equal and opposite.

Kiran is not right: the insect’s change in velocity is much larger, but its mass is very small, so its change in momentum is only the same as the car’s. Akhtar is not right either: the car does not exert a larger force on the insect. The insect dies because the same force acts on its tiny mass and gives it a huge acceleration, whereas the same force on the very heavy car produces an almost negligible change in its velocity.

(The source judged Kiran and Akhtar to be correct; this was corrected.)

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/s².

Answer: 40 kg m/s.

\(m=10\) kg, \(s=0.8\) m, \(a=10\) m/s², \(u=0\). \(v^2=u^2+2as=2\times10\times0.8=16\Rightarrow v=4\) m/s. Momentum transferred to the floor \(=mv=10\times4=40\) kg m/s.

ADDITIONAL EXERCISE QUESTIONS (PAGE 128)

The following is the distance–time table of an object in motion. (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?

Answer: (a) The acceleration is increasing. (b) The force acting on the object is also increasing.

(a) The distance covered in equal intervals of time is not equal, so the motion is non-uniform. The distance follows \(s=t^3\), so the velocity (\(\propto t^2\)) keeps increasing at an increasing rate, i.e. the acceleration (\(\propto t\)) increases with time. (b) According to Newton’s second law, the force acting on an object is directly proportional to the acceleration produced. Since the acceleration is increasing, the force acting on the object is also increasing.

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/s². With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)

Answer: 240 N.

Two persons push the car at uniform velocity, so their push just balances the friction. The acceleration of 0.2 m/s² is produced by the push of the third person alone: \(F=ma=1200\times0.2=240\) N. Hence each person pushes the car with a force of 240 N.

A hammer of mass 500 g, moving at 50 m/s, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Answer: 2500 N, opposite to the motion of the hammer.

\(m=0.5\) kg, \(u=50\) m/s, \(v=0\), \(t=0.01\) s. \(F=\dfrac{m(v-u)}t=\dfrac{0.5(0-50)}{0.01}=-2500\) N. The negative sign shows that the force of the nail on the hammer acts opposite to the hammer’s motion; by Newton’s third law the hammer exerts an equal force, 2500 N, on the nail in the direction of its motion.

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 the change in momentum. Also calculate the magnitude of the force required.

Answer: Acceleration –5 m/s²; change in momentum –24000 kg m/s; force 6000 N.

\(u=90\) km/h \(=25\) m/s, \(v=18\) km/h \(=5\) m/s, \(t=4\) s. From \(v=u+at\): \(5=25+4a\Rightarrow a=-5\) m/s² (retardation). Change in momentum \(=m(v-u)=1200(5-25)=-24000\) kg m/s. Force \(=ma=1200\times(-5)=-6000\) N; the magnitude is 6000 N (the negative signs show retardation, decrease in momentum and a retarding force).

A large truck and a car, both moving with a velocity of magnitude v, have a head-on collision and both of them come to a halt after that. If the collision lasts for 1 s: (a) Which vehicle experiences the greater force of impact? (b) Which vehicle experiences the greater change in momentum? (c) Which vehicle experiences the greater acceleration? (d) Why is the car likely to suffer more damage than the truck?

Answer: (a) The truck; (b) the truck; (c) both the same; (d) the car, being lighter and structurally weaker, suffers more damage.

Let the mass of the truck be M and that of the car be m, with \(M>m\). Initial velocity of both \(=v\); final velocity \(=0\); time of impact \(t=1\) s.

(a) \(F_{car}=\dfrac{m(0-v)}1=-mv\) and \(F_{truck}=\dfrac{M(0-v)}1=-Mv\). Since \(M>m\), the truck experiences the greater force of impact (by this calculation).

(b) Change in momentum of the car \(=0-mv=-mv\); of the truck \(=-Mv\). Since \(M>m\), the truck experiences the greater change in momentum.

(c) The acceleration \(=\dfrac{0-v}{1}=-v\) is the same for both, since their initial velocity, final velocity and time of impact are the same.

(d) The same acceleration acts on the smaller, lighter car, whose body is also weaker, and it is thrown back more strongly by the interaction (by Newton’s third law the truck and car exert equal and opposite forces on each other); hence the car suffers more damage than the truck.

1. Balanced Forces

2. Unbalanced Forces