Laws of Motion

The diagram below shows the forces acting on the ball and the string.

The diagram below shows the forces acting on the spring when it is compressed against a rigid wall.

The diagram below shows the point of application of force and the direction of these forces when a wooden block is placed on a table top.

The forces acting are —

(i) The block exerts a force ( = weight ) on the table top, downwards.

(ii) The table top exerts an equal reaction force upwards normal to the top of the table.

All of the above

Reason — A force when applied on a body can produce the following two main effects :

1. It can change the state of rest or of motion of the body.
2. It can change the size or shape of the body i.e., it can change the dimensions of the body.

force due to gravity

Reason — The forces experienced by bodies even without being physically in touch, are called non-contact forces. Hence, out of the given options, force due to gravity is a non-contact force.

non-contact force

Reason — The force on a body due to earth's attraction is called the force of gravity or the weight of the body and it is a non-contact force.

electrostatic force

Reason — When a comb is rubbed on dry hair, it gets charged. If this comb is brought near the small bits of paper, opposite charges are induced on the bits of paper and they begin to move towards the comb. The motion of paper bits is due to the electrostatic force of attraction exerted between the unlike charges on the comb and the paper bits.

attractive

Reason — In the universe each particle attracts the other particle due to its mass. This force of attraction between them is called gravitational force.

decreases, increases

Reason — The magnitude of non-contact force between two bodies depends on the distance of separation between them. It decreases with the increase in separation and increases as the separation decreases.

one-fourth

Reason — Magnitude of non-contact force varies inversely as the square of distance of separation i.e., on doubling the separation, the force becomes one-fourth.

restoring elastic force

Reason — In a spring balance, a spring is fixed at one end with a hook attached to an object at the other end. When the spring is stretched, the spring exerts a force F opposite to the direction of displacement of its free end, the magnitude of this force is directly proportional to the magnitude of displacement. This force is restoring force.

pushes, pulls

Reason — As the spring has the tendency to regain its original form or shape due to restoring force, therefore, if the spring is compressed, it pushes away each object and if it is stretched, it pulls in each object.

(a) Contact Forces — The forces which are applied on bodies by making a physical contact with them, are called contact forces.

Examples of contact force:

1. Frictional force — When a body slides (or rolls) over a rough surface, a force starts acting on the body in a direction opposite to the motion of the body, along the surface in contact. This is called frictional force or force of friction.
2. Normal reaction force — When a body is placed on a surface, the body exerts a force downwards, equal to it's weight, on the surface, but the body does not move (or fall) because the surface exerts an equal and opposite force on the body normal to the surface which is called normal reaction force.

(b) Non-Contact forces — The forces experienced by bodies even without being physically in touch, are called non-contact forces or forces at a distance.

Examples of non-contact force:

1. Gravitational force — In the universe, each particle attracts the other particle due to it's mass. This force of attraction between them is called gravitational force.
2. Electrostatic Force — Two like charges repel, while two unlike charges attract each other. The force between the charges is called electrostatic force.

(i) Non-rigid force — A force when applied on a non-rigid object, changes the inter - spacing between it's constituent particles and therefore causes a change in it's dimensions and can also produce motion in it.

(ii) Rigid - force — A force when applied on a rigid object does not change the inter-spacing between it's constituent particles and therefore it does not change the dimensions of the object, but causes only motion in it.

When a body slides (or rolls) over a rough surface, a force starts acting on the body in direction opposite to the motion of the body, along the surface in contact. This is called frictional force or the force of friction.

When a body is placed on a surface, the body exerts a force downwards, equal to its weight, on the surface, but the body does not move (or fall) because the surface exerts an equal and opposite force on the body normal to the surface which is called normal reaction force.

The force which are applied on bodies by making a physical contact with them, are called contact forces.

The force experienced by bodies even without being physically in touch, are called non-contact force or forces at a distance.

(a) frictional force — Contact force

(b) normal reaction force — Contact force

(c) force of tension in a string — Contact force

(d) gravitational force — Non-contact force

(e) electrostatic force — Non-contact force

(f) magnetic force — Non-contact force

(a) The force of contact — When a body slides over a rough surface, a force starts acting on the body in a direction opposite to the motion of the body, along the surface in contact. This is frictional force and is an example of contact force.

(b) The force at a distance or non contact force — In the universe, each particle attracts the other particle due to it's mass. This force of attraction between them is called gravitational force and it is an example of a non contact force.

The magnitude of non-contact forces between two bodies depends on the distance of separation between them.

It decreases with the increase in separation and increases as the separation decreases. It's magnitude varies inversely as the square of distance of separation i.e., on doubling the separation, the force becomes one fourth.

As we know that, force of attraction acting between two bodies is inversely proportional to the square of the distance between them.

Hence, magnitude of gravitational force will become four times.

The examples are as follows —

(a) Stops a moving body — A fielder on the ground stops a moving ball by applying force with his hands.

(b) Moves a stationary body — A ball lying on the ground moves when it is kicked.

(c) Changes the size of a body — By loading a spring hanging from a rigid support, the length of the spring increases.

(d) Changes the shape of a body — On pressing a piece of rubber, it's shape changes.

The reason is that the lower part of the passenger's body is in close contact with the train. As the train starts moving, his lower part shares the motion at once, but the upper part due to inertia of rest cannot share the motion simultaneously and so it tends to remain at the same place.
Consequently, the lower part of the body moves ahead and the upper part is left behind, so the passenger tends to fall backwards.

The reason is that the frame of sliding door being in contact with the floor of the train also comes in motion on start of train, but the sliding door remains in it's position due to inertia.
Thus, the frame moves ahead with the train while door slides opposite to the direction of motion of the train. Thus, the door may open.

One needs to run for some distance in the direction of the bus after alighting from it because inside the bus, his whole body was in a state of motion with the bus. On jumping out of the moving bus, as soon as his feet touch the ground, the lower part still remains in motion due to inertia of motion. As a result he falls in the direction of motion of the bus.
Hence in order to avoid falling, as soon as the passenger's feet touch the ground he should start running on the ground in the direction of motion of the bus for some distance.

The reason is that the part of the carpet where the stick strikes, comes in motion at once, while the dust particles settled on it's fur, remain in position due to inertia of rest. Thus, the part of the carpet moves ahead with the stick, leaving behind the dust particles which fall down due to the earth's pull.

It is advantageous to run before taking a long jump because by running a person brings his entire body in the state of motion. When the body is in motion, it becomes easier to take a long jump.

both (a) and (b)

Reason — Newton's first law can be understood in two parts:

1. Definition of inertia — In the first part, Newton's first law gives the definition of inertia, according to which an object cannot change its state by itself. If the object is in state of rest, it will remain in the state of rest and if it is moving it will continue to move with same speed and in same direction, unless an external force is applied.

2. Definition of force — The second part of Newton's first law defines force, according to which force is that external cause which can move a stationary object or which can change the state of motion of a moving object.

vector

Reason — Force is a vector quantity as it has magnitude as well as direction.

inertia

Reason — If the object is in the state of rest, it will remain in the state of rest and if it is moving in some direction, it will continue to move with the same speed in the same direction unless an external force is applied on it. This property is called inertia.

table tennis ball, tennis ball, football, cricket ball

Reason — As the property of inertia is because of the mass of the body. Greater the mass, greater the inertia. Ascending order of mass is table tennis ball < tennis ball < football < cricket ball, hence ascending order of inertia is table tennis ball < tennis ball < football < cricket ball.

When a passenger jumps out of a moving train, he falls down.

Reason — When a passenger jumps out of a moving train, he falls down because inside the train, his whole body was in a state of motion. On jumping out of the moving train, as soon as his feet touches the ground, the lower part of the body comes to rest, while the upper part still remains in motion due to inertia of motion and not inertia of rest. Hence, the person falls down.

both (a) and (c)

Reason — An athlete often runs before taking a long jump. The reason is that by running he brings his entire body in the state of motion. When the body is in motion, it becomes easier to take a long jump. Hence, it is an example of Newton's first law and inertia of motion.

same, same

Reason — According to inertia of motion, a body in a state of motion, continues to be in the state of motion with the same speed in the same direction in a straight line unless an external force is applied on it to change its state.

As the ball moves on the table top, force of friction comes into play and it opposes the motion of the ball. Hence, the ball stops.

A ball moving on a table top stops, as the force of friction between the moving ball and the table top opposes the motion.

The property of an object by virtue of which it tends to retain its state of rest or of motion is called inertia.

If the object is in the state of rest, it will remain in the state of rest and if it is moving in some direction, it will continue to move with the same speed in the same direction unless an external force is applied on it.

Example — A book lying on a table top will remain placed at it's place unless it is displaced. Similarly, a ball rolling on a horizontal plane keeps on rolling unless the force of friction between the ball and the plane stops it.

Below examples illustrate that mass is a measure of inertia i.e., greater the mass, greater is the inertia of the body :

1. A cricket ball is more massive then a tennis ball. The cricket ball acquires much smaller velocity than a tennis ball when the two balls are pushed with equal force for the same duration.
   In case when they are moving with the same velocity, it is more difficult to stop the cricket ball (which has more mass) in comparison to the tennis ball (which has less mass).

2. It is difficult (i.e., larger force is required) to set a loaded trolley (which has more mass) in motion than an unloaded trolley (which has less mass). Similarly, it is difficult to stop a loaded trolley than an unloaded one, if both are moving initially with the same speed.

As we know, more the mass, more is the inertia of the body. So mass is a measure of inertia.

If we take the example of a loaded trolley, then we observe that, it is difficult (i.e., larger force is required) to set a loaded trolley (which has more mass) in motion than an unloaded trolley (which has less mass).

No, the body will not move when two equal and opposite forces act on a stationary body. As the net force on the body is zero, so the body will remain stationary due to inertia of rest.

When two equal and opposite forces are acting on a moving object, the motion remains unaffected because the net force on the object is zero.

A person falls when he jumps out from a moving train, because inside the train, his whole body was in a state of motion with the train. On jumping out of the moving train, as soon as his feet touch the ground, the lower part of the body comes to rest, while the upper part still remains in motion.

As a result, he falls in the direction of motion of the train and gets hurt.

To avoid falling, as soon as the passenger's feet touch the ground he should start running on the ground in the direction of motion of the train for some distance.

A coin placed on a card, drop into the tumbler when the card is rapidly flicked with the finger, because when the card is flicked the momentary forces acts on the card, so it moves away. But the coin kept on it does not share the motion at once and it remain at it's place due to inertia of rest. The coin then falls down into the tumbler due to the pull of gravity.

A ball thrown vertically upwards in a moving train, comes back to the thrower’s hand because when ball was thrown, it was in motion along with the person and the train. It remains in the same state of forward motion even during the time the ball remains in air.

The person, the inside air and the ball all move ahead by the same distance due to inertia and so the ball falls back into his palm on it's return.

The reason is that when the stem (or branches) of the tree are shaken, they come in motion, while the fruits due to inertia, remain in the state of rest. Thus, the massive and weakly attached fruits get detached from the branches and the fall down due to the pull of gravity.

Force is the physical quantity which causes motion in a body.

No, force is not needed to keep a moving body in motion.
Reason — If a body is set in motion, it will remain in motion even when the force applied to set the body in motion is withdrawn, provided that there is no other force such as friction etc., to oppose the motion.

It's speed will remain unchanged.

Reason — According to Newton's first law of motion, if a body is in a state of rest, it will remain in the state of rest and if it is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.

According to Galileo’s law of inertia — "An object, if once set in motion, moves with uniform velocity if no force acts on it."

Thus, a body continues to be in a state of rest or in a state of uniform motion unless an external force is applied on it.

According to Newton's first law of motion, if a body is in a state of rest, it will remain in the state of rest and if it is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.

The qualitative definition of force on the basis of Newton's first law of motion is —

Force is that external cause which tends to change the state of rest or the state of motion of an object.

Example — A book lying on a table gets displaced from it's place when it is pushed.

The factor on which inertia of a body depends is mass.

More the mass, more is the inertia of the body. Thus, a lighter body has less inertia than a heavier body. In other words, more the mass of a body, more difficult it is to move the body from rest (or to stop the body if it initially in motion).

Hence, mass is a measure of inertia.

The two kinds of inertia are —

1. Inertia of rest
2. Inertia of motion

(a) Example of inertia of rest — When a train suddenly starts moving forward, the passenger standing in the compartment tends to fall backwards. The reason is that the lower part of the passenger's body is in close contact with the train. As the train starts moving, the person's lower part shares the motion at once. However, the upper part, due to the inertia of rest cannot share the motion simultaneously and so it tends to remain at the same place.
Hence, the lower part of the body moves ahead and the upper part is left behind, so the passenger tends to fall backwards.

(b) Example of inertia of motion — A cyclist riding along a level road does not come to rest immediately after he stops pedalling. The reason is that the bicycle continues to move due to inertia of motion even after the cyclist stops applying the force on the pedal.

When the aeroplane is acted upon by two opposing forces then the net force acting on the aeroplane is zero as the two forces are in opposite direction and they cancel each other.

(i) The acceleration produced in a body of given mass is directly proportional to the force applied on it. i.e.,

a ∝ F (if mass remains constant)

The graph plotted for acceleration on force for a constant mass is as shown below —

(ii) The force needed to produce a given acceleration in a body is proportional to the mass of the body. i.e.,

F ∝ m (if acceleration remains the same)

The graph plotted for force on mass for a constant acceleration is shown below —

When a force is applied on bodies of different masses, the acceleration produced in them is inversely proportional to their masses

i.e.,

a ∝ $\dfrac{1}{m}$ (for a given force F).

The graph plotted for acceleration against mass m is a hyperbola and is shown below,

Acceleration produced in a body of given mass is directly proportional to the force applied on it. i.e., a ∝ F (if mass remains constant)

The graph plotted for acceleration a against force F is a straight line as shown below:

both (a) and (b)

Reason — The force needed to stop a moving body in a definite time depends on the product of mass and velocity, which is called linear momentum of the moving body. Thus, p = mv.

vector quantity

Reason — Linear momentum is a vector quantity in the direction of the motion of the body (i.e., along the velocity of the body).

qualitatively

Reason — Newton's first law of motion defines force only qualitatively. A force is that physical cause which changes the state of motion of a body when it is applied on it. It means that force produces acceleration in the body.

directly proportional to the force applied on it

Reason — Experimentally, Newton found that the acceleration produced in a body is directly proportional to the force applied on it.

F = $\dfrac{Δ\text p}{Δ\text t}$

Reason — As we know,

Force = Rate of change of momentum

F = $\dfrac{Δ\text p}{Δ\text t}$

= $\dfrac{Δ\text {(mv)}}{Δ\text t}$

= $\dfrac{\text m(Δ\text v)}{Δ\text t}$

= ma

Thus, Newton’s second law will be F = $\dfrac{Δ\text p}{Δ\text t}$

Reason — If a given force is applied on bodies of different masses, the acceleration produced in them is inversely proportional to their masses i.e., a ∝ $\dfrac{1}{\text{m}}$ (for a given force F). A graph plotted for acceleration a against mass m is a curve (hyperbola).

mass of the body

Reason — As we know,

Force (f) = mass (m) x acceleration (a)

Hence, acceleration produced in a body by a force of given magnitude depends on mass of the body.

10⁵ dyne

Reason — 1 newton = 1 kg x 1 ms^-2

= 1000 g x 100 cms^-2

= 10⁵ g x cms^-2

= 10⁵ dyne.

Thus, 1 N = 10⁵ dyne.

10 N

Reason — Given, a = 5 ms^-2 and m = 2 kg

Using, F = ma

we get, F = 5 x 2 = 10 N

Hence, force required = 10 N

similar to the direction of the force applied

Reason — According to Newton's second law of motion, the rate of change of momentum of a body is directly proportional to the force applied on it and the change in momentum takes place in the direction in which the force is applied.

both (b) and (c)

Reason — It is observed that the mass of a particle increases with increase in velocity but it becomes perceptible only when the velocity v of the particle is comparable with the speed of light c. At velocities v << c, the change in mass is not perceptible. At such velocities (v << c ), mass m can be considered to be constant. Then Newton's second law takes the form F = $\dfrac{\text m(Δ\text v)}{Δ\text t}$ = ma.

Thus, for F = $\dfrac{\text m(Δ\text v)}{Δ\text t}$ = ma to hold true, two conditions are required :

1. velocities are much smaller than the velocity of light
2. mass remains constant.

1 : 1

Reason — Given,

m₁ = 4m m₂ = 8m
v₁ = 2v
v₂ = v

As, p = mv

Hence, $\dfrac{\text{p}_1}{\text{p}_2}$ = $\dfrac{\text{(4\text{m})(2\text{v})}}{\text{(8\text{m)}\text{(v)}}}$ = $\dfrac{\text{8\text{mv}}}{\text{8\text{m}\text{v}}}$ = 1 : 1

Hence, ratio of their momentum = 1 : 1

As we know,

linear momentum (p) = mass (m) x velocity (v)

Given,

v = 2 m s^-1

m = 5 kg

Substituting the values in the formula above we get,

$\text p = 5 \times 2 \\[0.5em] \text p = 10 \\[0.5em]$

Hence, linear momentum = 10 kg m s^-1

As we know,

linear momentum (p) = mass (m) x velocity (v)

Given,

p = 0.5 kg m s^-1

m = 50 g

Converting g to kg,

$1000\ \text g = 1\ \text {kg} \\[0.5em] 50\ \text g = (\dfrac{1}{1000}) \times 50 \\[0.5em] \Rightarrow 50\ \text g = 0.05\ \text {kg} \\[0.5em]$

Hence, m = 0.05 kg

Substituting the values in the formula above we get,

$0.5 = 0.05 \times \text v \\[0.5em] \Rightarrow \text v = \dfrac{0.5}{0.05} \\[0.5em] \Rightarrow \text v = \dfrac{0.1}{0.01} \\[0.5em] \Rightarrow \text v = 10\ \text {m s}^{-1}\\[0.5em]$

Hence, velocity = 10 m s^-1

As we know,

Force (f) = mass (m) x acceleration (a)

Given,

F = 15 N

m = 2 kg

Substituting the values in the formula above we get,

$15 = 2 \times \text a \\[0.5em] \Rightarrow \text a = \dfrac{15}{2} \\[0.5em] \Rightarrow \text a = 7.5\ \text {m s}^{-2} \\[0.5em]$

Hence,

Acceleration produced by the body = 7.5 m s^-2

As we know,

Force (f) = mass (m) x acceleration (a)

Given,

F = 10 N

m = 5 kg

Substituting the values in the formula above we get,

$10 = 5 \times \text a \\[0.5em] \Rightarrow \text a = \dfrac{10}{5} \\[0.5em] \Rightarrow \text a = 2\ \text {m s} ^{-2} \\[0.5em]$

Hence,

Acceleration produced by the body = 2 m s^-2

As we know,

Force (f) = mass (m) x acceleration (a)

Given,

a = 5 m s ^-2

m = 0.5 kg

Substituting the values in the formula above we get,

$\text F = 0.5 \times 5 \\[0.5em] \Rightarrow \text F = 2.5 \text N \\[0.5em]$

Hence,

Magnitude of force = 2.5 N

As we know,

Force (f) = mass (m) x acceleration (a)

Given,

f = 10 N

m = 2 kg

t = 3 s

u = 0

Substituting the values in the formula above we get,

$10 = 2 \times \text a \\[0.5em] \Rightarrow \text a = \dfrac{10}{2} \\[0.5em] \Rightarrow \text a = 5\ \text {m s}^{-2} \\[0.5em]$

Hence,

Acceleration produced by the body = 5 m s^-2

The 1^st equation of motion states that;

v = u + at

Substituting the values in the formula above we get,

$\text v = 0 + (5 \times 3) \\[0.5em] \Rightarrow v = 15\ \text {m s}^{-1}$

Hence, velocity of the body is 15 m s^-1

(ii) As we know,

Change in momentum = Final momentum - Initial momentum

Initial momentum = mu = 2 x 0 = 0

Final momentum = mv = 2 x 15 = 30 kg m s^-1

Substituting the values in the formula above we get,

Change in momentum = 30 - 0 = 30 kg m s^-1

(i) Velocity (v) = $\dfrac{\text {distance (S)}}{\text {time (t)}}$

Given,

s = 100 m in 5 s

Substituting the values in the formula above we get,

$\text v = \dfrac{100}{5} \\[0.5em] \text v = 20\ \text {m s}^{-1}\\[0.5em]$

Hence, velocity acquired by the body = 20 m s^-1

(ii) As we know,

v² - u² = 2as

Given,

s = 100 m

u = 0

v = 20 m s^-1

Substituting the values in the formula above we get,

$20^2 - 0^2 = 2 \times \text a \times 100 \\[0.5em] 400 = 200 \times \text a \\[0.5em] \Rightarrow \text a = 2\ \text {m s}^{-2} \\[0.5em]$

Hence, a = 2 m s^-2

(iii) As we know,

Force (f) = mass (m) x acceleration (a)

Given,

m = 100 kg

a = 2 m s ^-2

Substituting the values in the formula above we get,

$\text F = 100 \times 2 \\[0.5em] \Rightarrow \text F = 200\ \text N \\[0.5em]$

Hence,

Magnitude of force = 200 N

As we know,

Force (f) = mass (m) x acceleration (a)

Given,

m = 100 g

Converting g to kg,

1000 g = 1 kg

100 g = ($\dfrac{1}{1000}$) x 100 kg

100 g = 0.1 Kg

Hence, m = 0.1 Kg

As,

acceleration = slope of v-t graph = $\dfrac{\text v}{\text t}$

= $\dfrac{20}{5}=4\ \text {m s}^{-2}$

Hence, a = 4 m s^-2

Substituting the values in the formula above we get,

$\text F = 0.1 \times 4 \\[0.5em] \text F = 0.4\ \text N \\[0.5em]$

Hence, force acting on the particle = 0.4 N

As we know,

Force (f) = mass (m) x acceleration (a)

Given,

m = 500 g

Converting g to kg,

1000 g = 1 kg

500 g = ($\dfrac{1}{1000}$) x 500 kg

500 g = 0.5 Kg

Hence, m = 0.5 Kg

a = 10 m s ^-2

Substituting the values in the formula above we get,

$\text F = 0.5 \times 10 \\[0.5em] \Rightarrow F = 5\ \text {N} \\[0.5em]$

Hence,

Magnitude of force = 5 N

If, m = 5 Kg , f = 5 N

Substituting the values in the formula above we get,

$5 = 5 \times \text a \\[0.5em] \Rightarrow \text a = \dfrac{5}{5} \\[0.5em] \Rightarrow \text a = 1\ \text {m s}^{-2} \\[0.5em]$

Hence, a = 1 m s ^-2

The 1^st equation of motion states that;

v = u + at

Given,

m = 150 g

Converting g to kg,

1000 g = 1 kg

150 g = ($\dfrac{1}{1000}$) x 150

150 g = 0.15 kg

Hence, m = 0.15 kg

u = 25 m s^-1

v = 0

t = 0.03 s

Substituting the values in the formula above we get,

$0 = 25 + (\text a \times 0.03) \\[0.5em] \Rightarrow \text a = -\dfrac{25}{0.03} \\[0.5em] \Rightarrow \text a = - 833.33\ \text {m s}^{-2} \\[0.5em]$

Hence, acceleration of the ball = - 833.33 m s^-2`

Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

$\text F = 0.15 \times (- 833.33) \\[0.5em] \Rightarrow \text F = -125\ \text N \\[0.5em]$

Hence,

Magnitude of force = -125 N

The 1^st equation of motion states that;

v = u + at

Given,

t = 0.1 s

m = 2.0 Kg

u = 0

v = 2 m s^-1

Substituting the values in the formula above we get,

$2 = 0 + (\text a \times 0.1) \\[0.5em] \Rightarrow \text a = \dfrac{2}{0.1} \\[0.5em] \Rightarrow \text a = 20\ \text {m s}^{-2} \\[0.5em]$

Hence, acceleration of the body = 20 m s^-2`

Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

$\text F = 2 \times 20 \\[0.5em] \Rightarrow \text F = 40\ \text N \\[0.5em]$

Hence,

Magnitude of force = 40 N

As we know,

S = ut + ($\dfrac{1}{2}$) at²

Given,

m = 500 g

Converting g to kg,

1000 g = 1 kg

500 g = ($\dfrac{1}{1000}$) x 500 kg

100 g = 0.5 Kg

Hence,

m = 0.5 Kg

u = 0

s = 4 m

t = 2 s

Substituting the values in the formula above we get,

$4 = (0 \times 2) + (\dfrac{1}{2} \times \text a \times 2^2) \\[0.5em] 4 = 0 + (\dfrac{1}{2} \times \text a \times 4) \\[0.5em] 4 = 2\text a \\[0.5em] \Rightarrow \text a = \dfrac{4}{2} \\[0.5em] \Rightarrow \text a = 2\ \text {m s}^{-2} \\[0.5em]$

Hence,

a = 2 m s ^-2

Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

$\text F = 0.5 \times 2 \\[0.5em] \Rightarrow \text F = 1\ \text N \\[0.5em]$

Hence,

Magnitude of force = 1 N

We know that,

acceleration = $\dfrac{\text v - \text u}{\text t}$

Given,

m = 480 kg

u = 54 km h^-1

converting km h^-1 to m s^-1

$54 \text {km h}^{-1} = 54 \times (\dfrac{1000 \text {m}}{60 \times 60 \text {s}}) \\[0.5em] 54 \text {km h}^{-1} = 54 \times (\dfrac{10 \text {m}}{36 \text {s}}) \\[0.5em] 54 \text {km h}^{-1} = \dfrac{540 \text {m}}{36 \text {s}} \\[0.5em] 54 \text {km h}^{-1} = 15 {\text {m s}^{-1}} \\[0.5em]$

Hence, u = 15 m s^-1

v = 0

t = 10 s

Substituting the values in the formula above we get,

$\text a = \dfrac{0 - 15}{10} \\[0.5em] \Rightarrow \text a = -1.5\ \text{m s}^{-2} \\[0.5em]$

Hence, a = - 1.5 m s ^-2

The negative sign shows that it is retardation.

Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

$\text F = 480 \times (1.5) \\[0.5em] \Rightarrow \text F = 720\ \text N \\[0.5em]$

Hence,

Magnitude of force applied by the brakes = 720 N

(i) We know that,

acceleration = $\dfrac{\text v - \text u}{\text t}$

Given,

v = 0

u = 30 m s^-1

t = 2 s

f = 1500 N

Substituting the values in the formula above we get,

$\text a = \dfrac{0 - 30}{2} \\[0.5em] \Rightarrow \text a = - 15\ \text{m s}^{-2} \\[0.5em]$

The negative sign shows that it is retardation. So retardation = 15 m s ^-2

Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

$1500 = \text m \times (15) \\[0.5em] \Rightarrow \text m = 100 \text{ kg} \\[0.5em]$

Hence,

Mass of the car = 100 kg

As we know,

Change in momentum = Final momentum - Initial momentum

Initial momentum = mu = 100 x 30 = 3000 kg m s^-1

Final momentum = mv = 100 x 0 = 0

Substituting the values in the formula above we get,

Change in momentum = 0 - 3000 = - 3000 kg m s^-1

Hence,

(i) change in momentum = 3000 kg m s^-1

(ii) The retardation produced in the car = 15 m s ^-2

(iii) The mass of the car = 100 Kg

Given,

m = 50 g

Converting g to Kg

1000 g = 1 Kg

50 g = ($\dfrac{1}{1000}$) x 50 kg

50 g = 0.05 Kg

Hence, m = 0.05 Kg

u = 100 m s^-1

S = 2 cm

Converting cm to m

100 cm = 1 m

2 cm = ($\dfrac{1}{100}$) x 2 = 0.02 m

(i) Initial momentum of the bullet = mass (m) x initial velocity (u)

Substituting the values in the formula above we get,

Initial momentum = 0.05 x 100 = 5 kg m s^-1

Hence, Initial momentum = 5 kg m s^-1

(ii) Final momentum = mass (m) x final velocity (u)

Substituting the values in the formula above we get,

Final momentum = 0.05 x 0 = 0 kg m s^-1

Hence, Final momentum = Zero

(iii) As we know,

v² - u² = 2as

Substituting the values in the formula above we get,

$0^2 - 100^2 = 2 \times \text a \times 0.02 \\[0.5em] - 10000 = 0.04 \times \text a \\[0.5em] \Rightarrow \text a = - \dfrac {10000}{0.04} \\[0.5em] \Rightarrow \text a = - \dfrac {1000000}{4} \\[0.5em] \Rightarrow \text a = - 250000\ \text{m s}^{-2}\\[0.5em]$

Hence,

Retardation caused by wooden block = 2.5 x 10⁵ m s^-2

(iv) Now,

Force (f) = mass (m) x acceleration (a)

Substituting the values in the formula above we get,

$\text F = 0.05 \times (2.5 \times 10^5) \\[0.5em] \Rightarrow \text F = 0.125 \times 10^5\ \text{N} \\[0.5em]$

Hence,

Resistive force exerted by wooden block = 12500 N

The two factors on which the force needed to stop a moving body in a given time depends are:

1. Mass
2. It's Velocity

When a force is applied on a moving body, it's velocity changes. Due to change in velocity of the motion, it's momentum also changes.

Let a force (F) be applied on a body having mass (m) for time (t) due to which it's velocity changes from u to v.

Then,

Initial momentum of the body = mu
Final momentum of the body = mv

Change in momentum of the body in (t) seconds

= mv - mu = m (v - u)

Then, rate of change of momentum

= $\dfrac{\text {Change in velocity}}{\text {Time}}$

= $\dfrac{\text {m(v - u)}}{\text {t}}$

And we know,

a = $\dfrac{\text {(v - u)}}{\text {t}}$

Therefore,

Rate of change of momentum = ma = mass x acceleration.

This relation holds true when mass of the body remains constant.

Newton’s second law of motion states that —

The rate of change of momentum of a body is directly proportional to the force applied on it and the change in momentum takes place in the direction in which the force is applied.

Newton’s second law of motion provides the quantitative value of force, i.e., it relates force to the measurable quantities like acceleration and mass.

Newton's first law of motion defines force only qualitatively. A force is that physical cause which changes the state of motion of a body when it is applied on it. It means that the force produces acceleration in the body i.e., the force is the cause of acceleration.

Whereas, Newton's second law of motion gives the quantitative value of force, i.e., it relates force to the measurable quantities like acceleration and mass.

The mathematical form of Newton’s second law of motion is —

F = ma

where,

F = force
m = mass and
a = acceleration

The conditions for the relation to hold true are —

(i) velocities must be much smaller than the velocity of light and

(ii) mass of the body should remain constant.

Newton’s second law of motion states that —

The rate of change of momentum of a body is directly proportional to the force applied on it and the change in momentum takes place in the direction in which the force is applied.

The mathematical form of Newton’s second law of motion is —

F = ma

where,

F = force
m = mass and
a = acceleration

The conditions for the relation to hold true are —

1. velocities must be much smaller than the velocity of light.
2. Mass of the body should remain constant.

To obtain Newton’s first law of motion from second law of motion —

From Newton’s second law, F=ma

If F = 0, then a = 0

This means that if no force is applied on the body, it's acceleration will be zero. If the body is at rest, it will remain at rest and if it is moving, it will remain moving in the same direction with the same speed.

Thus, a body not acted upon by any external force, does not change it's state of rest or of motion.

This statement is Newton's first law of motion.

When a glass vessel falls from a height on a hard floor, it comes to rest almost instantaneously (i.e., in a very short time) so the floor exerts a large force on the vessel and it breaks.

But, if the glass vessel falls on a carpet, the time duration in which the vessel comes to rest, increases and so the carpet exerts a less force on the vessel and it does not break.

(a) Let u be the velocity of the ball of mass m, when it reaches the hands of the player catching it.

The initial momentum of the ball = mu

If the cricketer does not pull his hands and stops the ball as soon as it touches his hands, he uses very little time (t₁) to stop the ball.

Then, the force exerted by the ball on the hands of the cricketer is —

F = $\dfrac{\text {Change in momentum}}{\text {Time interval}}$ = $\dfrac{\text {mu}}{\text t_1}$

But if the cricketer pulls back his hands along with the ball, he takes a much longer time t₂ to stop the ball.

The force now exerted by the ball on his hands is F₂ = $\dfrac{\text {mu}}{\text t_2}$

Since, t₂ > t₁, therefore, F₂ < F ₁ or force exerted on the hands of cricketer by the fast moving ball is less when he withdraws his hands. Thus cricketer avoids the chances of injury to his palms by withdrawing his hands along with the ball while catching it.

(b) When an athlete lands from a height on a hard floor, he may hurt his feet because his feet comes to rest almost instantaneously (i.e., in a very short time) so a very large force is exerted by the floor of his feet.

On the other hand, when he lands on sand, his feet push the sand for some distance, therefore the time duration in which his feet comes to rest, increases.

As a result, the force exerted on his feet decreases and he is saved from getting hurt.

Linear momentum of a body can be defined as the product of its mass and velocity.

For a body of mass m, moving with velocity v, linear momentum p is expressed as

p = mv

The S.I. unit of linear momentum is kg ms^-1

(i) When v << c

i.e., if the velocity of the moving particle is much smaller than the velocity of light (c) , then —

Change in momentum (Δp) = m Δv

It happens when the velocity of particle is of the order of 10⁶ m s^-1 or less than this, then the variation in mass with velocity is small enough and mass can be considered to be constant.

(ii) when v → c

Change in momentum (Δp) = Δ(mv)

(iii) when v << c but m does not remain constant.

In case of atomic particles moving with velocity comparable to the velocity of light c, it was observed that mass of the particle does not remain constant, but it increases with increase in velocity, according to the relation

m = $\dfrac{\text m_0}{\sqrt {1-\big(\dfrac{\text v}{\text c}\big)^2}}$

where, m₀ is the mass of the particle when it is at rest (i.e., v = 0)

Then,

Change in momentum (Δp) = Δ(mv)

(i) The factor on which inertia of a body depends is mass.

More the mass, more is the inertia of the body. Thus, if mass of two bodies is same their inertia will also be same.

Hence, inertia of A and B will be same as mass of both A and B are same.

Hence, ratio of their inertia is 1 : 1

(ii) As we know, momentum of a body (p) = mass (m) x velocity (v)

Given, mass of A and B are equal.

For A,

P_A = mv_A

For B,

P_B = mv_B

Ratio between the two is —

$\dfrac{{\text P}<em>\text A}{\text {P}</em>\text B} = \dfrac{\text m{\text v}<em>\text A}{\text m{\text v}</em>\text B} \\[0.5em]$

As, mass of A = mass of B , hence,

$\dfrac{{\text P}<em>\text A}{{\text P}</em>\text B} = \dfrac{{\text v}<em>\text A}{\text {v}</em>\text B} \\[0.5em]$

Substituting the values, we get,

$\dfrac{{\text P}<em>\text A}{{\text P}</em>\text B} = \dfrac{\text v}{2\text v} \\[0.5em] \Rightarrow \dfrac{{\text P}<em>\text A}{\text {P}</em>\text B} = \dfrac{1}{2} \\[0.5em]$

Hence, ratio between the momentum of A and B is 1 : 2

(i) Given,

Mass of A = m

Mass of B = 2m

The factor on which inertia of a body depends is mass.

More the mass, more is the inertia of the body.

Therefore,

$\dfrac{\text {Inertia of A}}{\text {Inertia of B}} = \dfrac{\text {mass of A}}{\text {mass of B}} \\[0.5em]$

Substituting the values, we get,

$\dfrac{\text {Inertia of A}}{\text {Inertia of B}} = \dfrac{\text {m}}{\text {2m}} \\[0.5em] \dfrac{\text {Inertia of A}}{\text {Inertia of B}} = \dfrac{\text {1}}{\text {2}} \\[0.5em]$

Hence, ratio of their inertia = 1 : 2

(ii) As we know, momentum of a body (p) = mass (m) x velocity (v)

Given,

v_A = 2v

v_B = v

Ratio between the two is —

$\dfrac{{\text P}<em>\text A}{\text {P}</em>\text B} = \dfrac{\text {(mv)}<em>\text A}{\text {(mv)}</em>\text B} \\[0.5em]$

Substituting the values, we get,

$\dfrac{\text {P}<em>\text A}{{\text P}</em>\text B} = \dfrac{\text m \times 2\text v}{2\text m \times \text v} \\[0.5em] \dfrac{{\text P}<em>\text A}{\text {P}</em>\text B} = \dfrac{2\text {mv}}{2\text {mv}} \\[0.5em] \Rightarrow \dfrac{{\text P}<em>\text A}{{\text P}</em>\text B} = \dfrac{1}{1} \\[0.5em]$

Hence, ratio between the momentum of A and B is 1 : 1

(iii) According to Newton's second law of motion, the rate of change of momentum of a body is directly proportional to the force applied on it and as the ratio of momentum between A and B is 1 : 1, hence, ratio of force needed to stop A and B is also 1 : 1

The S.I. unit of force is newton.

One newton is the force which when acts on a body of mass 1 kg, produces an acceleration of 1 m s^-2.

Hence, 1 newton = 1 kg x 1 m s^-2

The standard symbol of newton is N.

In C.G.S system, unit of force is dyne.

One dyne is the force which when acts on a body of mass 1 g, produces an acceleration of 1 cm s^-2

i.e.,

Hence, 1 dyne = 1 g x 1 cm s^-2

The S.I. unit of force is newton (N) and the C.G.S. unit of force is dyne.

Relationship between newton and dyne —

1 newton = 1 kg x 1 m s^-2

= 1000 g x 100 cm s^-2

= 10⁵ g cm s^-2

= 10⁵ dyne.

Thus, 1 newton = 10⁵ dyne.

explains the way a force acts on a body

Reason — Newton's third law explains the way a force acts on a body and states that, to every action there is always an equal and opposite reaction.

Newton's third law

Reason — To move a boat ahead in water, the boatman pushes (action) the water backwards with his oar and at the same time, the water exerts an equal and opposite force (reaction) in the forward direction on the boat due to which the boat moves ahead. Hence, it is an example of Newton's third law

$\overrightarrow{\text{F}}<em>{\text{AB}} = -\overrightarrow{\text{F}}</em>{\text{BA}}$

Reason — According to Newton’s third law, to every action there is always an equal and opposite reaction. In magnitude, F_AB = F_BA but they are in opposite directions. Therefore, $\overrightarrow{\text{F}}<em>{\text{AB}} = -\overrightarrow{\text{F}}</em>{\text{BA}}$.

10 N

Reason — According to Newton’s third law to every action there is always an equal and opposite reaction. Hence, the footballer will also experience a reaction force of 10 N.

action and reaction act on different bodies in opposite direction.

Reason — The action and reaction never act on the same body, but they always act simultaneously on two different bodies i.e., the forces on interaction are always present in a pair.

(iii)

Reason — When a man applies a force F (action) backward by his foot on the ground against the force of friction, the ground exerts an equal and opposite force R (reaction) forward on his foot. The horizontal component of the force of reaction enables the man to move forward.

Obviously, it will be difficult to move on a slippery road where friction is less.

To prevent passengers from being thrown forward due to their inertia when the car stops suddenly.

Reason — When a car suddenly stops, the car experiences an external force (brakes), but the passengers tend to keep moving forward due to their inertia. If not restrained by a seat belt, the passenger can be thrown forward, possibly leading to injury so seat belts apply an external force to the body, reducing motion safely and keeping the person in place.

The forward push of the road on the rear wheel.

Reason — When the rear wheel of the bicycle pushes backward on the road (action), the road pushes forward on the rear wheel (reaction) and this forward force from the road is what actually propels the bicycle forward.

The force needed to produce a given acceleration in a body is inversely proportional to the mass of the body.

Reason — From Newton’s second law of motion,

$\text F = \text {ma}$

Here, force is directly proportional to both mass and acceleration so if acceleration is constant then force is directly proportional to mass, not inversely.

As we know, Newton’s third law states that for every action there is always an equal and opposite reaction.

Hence, when the boy pushes the wall with a force of 10 N towards east then the wall will also push the boy with a force of 10 N towards west.

(a) The force exerted by the block on the string is 15 N acting downwards due to the weight of the block.

(b) The force exerted by the string on the block is 15 N acting upwards because of the tension generated.

Newton's first law and second law does not explain how the force acts on the object. This question is answered by Newton's third law, which states —

"To every action there is always an equal and opposite reaction".

Example — While moving on the ground, we exert a force by our feet to push the ground backwards, the ground exerts a force of the same magnitude on our feet forward which makes us to move forward.

Hence, the force exerted by our feet on the ground is the force of action and the force exerted by the ground on our feet is the force of reaction.

The Newton’s third law or the law of action and reaction states —

“To every action there is always an equal and opposite reaction”.

Examples

1. Motion of boat in water — To move a boat ahead in water, the boatman pushes (action) the water backwards with his oar and the water exerts an equal and opposite force (reaction) in the forward direction on the boat due to which the boat moves ahead.
2. Catching a ball — While catching a ball, the ball exerts a force (action) on the palm of cricketer and the cricketer exerts an equal force (reaction) on the ball to stop it.

Newton’s third law states that for every action there is always an equal and opposite reaction.

In a rocket, fuel burning inside the rocket and the burnt gases at high pressure and high temperature are expelled out of the rocket through a nozzle. Thus, rocket exerts a force (action) on gases to expel them through a nozzle backwards.

The outgoing gases exert an equal and opposite force R (reaction) on the rocket due to which it moves in the forward direction.

Hence, Newton's third law is obeyed.

The Newton’s third law states —

“To every action there is always an equal and opposite reaction”.

So, when a bullet is fired from a gun, a force F is exerted on the bullet (action) and the gun experiences an equal recoil R (reaction) as shown.

Hence, Newton's third law is obeyed.

In order to get out of the boat we exert a force (action) on the board of the boat. This creates a force (reaction) which enables us to step out of the boat.

At the same instant, the boat tends to leave the shore due to the force exerted by us (i.e., action).

For the safety of the passengers the boatman ties the boat to the pole on the shore so that it does not move away.

The spring of balance A pulls the spring of balance B due to which we get some reading in balance B. The same reading is seen in balance A because the spring of balance B also pulls the spring of balance A by the same force.

The pull on the spring B by the spring A is the action F_BA and the pull on the spring A by the spring B is the reaction F_AB.

This demonstrates that "to every action, there is an equal and opposite reaction" (i.e., in magnitude F_AB = F_BA but they are in opposite directions. )

To move a boat ahead in water, the boatman pushes (action) the water backwards with his oar and the water exerts an equal and opposite force (reaction) in the forward direction on the boat due to which the boat moves ahead.

When a person exerts a force (action) on a wall by pushing the palm of his hand against it, he will experience a force (reaction) exerted by the wall on his palm and hence, he may fall back.

When a light ball strikes the ground it exerts a force on the ground (action) and the ground in turn exerts an equal amount of force R (reaction) on the ball, due to which the ball rises up.

Newton's third law of motion states that to every action there is always an equal and opposite reaction.

(a) Firing a bullet from a gun — When a man fires a bullet from a gun, a force F is exerted on the bullet (action) and the gun experiences an equal recoil R (reaction).

(b) Hammering a nail — When we hammer a nail, the hammer exerts a force F (action) on the nail and the nail in turn also exerts an equal and opposite force R (reaction) on the hammer.

(c) A book lying on a table — When a book is placed on a table top, the book exerts a force equal to it's weight (action) on the table in downward direction and the table balances it by an equal force called the (reaction) acting upwards on the book.

(d) A moving rocket — The rocket exerts a force F (action) on gases (produced by burning of fuel) to expel them through a nozzle backwards. The outgoing gases exert an equal and opposite force R (reaction) on the rocket due to which it moves in the forward direction.

(e) A person walking on the floor — When a man applies a force F (action) backward by his foot on the ground against the force of friction, the ground exerts an equal and opposite force R (reaction) forward on his foot.

(f) A moving train colliding with a stationary train — The moving train exerts a force F (action) on the stationary train and the stationary train in turn exerts a force R (reaction) on the moving train.

Yes, action and reaction both act simultaneously.

Example — To move a boat ahead in water, the boatman pushes (action) the water backwards with his oar and at the same time, the water exerts an equal and opposite force (reaction) in the forward direction on the boat due to which the boat moves ahead.

Yes, action and reaction are equal in magnitude.

Example — The spring of balance A pulls the spring of balance B due to which we get some reading in balance B. The same reading is seen in balance A because the spring of balance B also pulls the spring of balance A by the same force.

The pull on the spring B by the spring A is the action F_BA and the pull on the spring A by the spring B is the reaction F_AB.

This demonstrates that "to every action, there is an equal and opposite reaction" (i.e., in magnitude F_AB = F_BA but they are in opposite directions. )

The statement 'the sum of action and reaction on a body is zero' is wrong.

As per the Newton’s third law of motion —

In an interaction of two bodies A and B, the magnitude of reaction (i.e., the force F_AB applied by the body A) is equal in magnitude to the action (i.e., the force F_BA applied by the body A on the body B), but they are in directions opposite to each other.

Hence, the two forces can't add up to zero.

both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because non-contact forces like gravitational, electrostatic, and magnetic forces show same trend.

Reason (R) is true because these forces follow the inverse square law :

$\text F ∝ \dfrac {1}{\text r^2}$

Since the reason correctly explains why the force decreases with increasing distance, the assertion and reason are both true, and the reason is the correct explanation.

assertion is false but reason is true

Explanation

Assertion (A) is false because inertia is the tendency of a body to resist any change in its state, whether it is rest or uniform motion in a straight line. So, it does not apply only to rest, but also to motion.

Reason (R) is true because this is the definition of force as per Newton’s first law as a force causes a change in a body’s state of rest or motion.

both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because when equal force is applied for the same duration, the lighter object (tennis ball) gains more velocity than the heavier object (cricket ball). This is because the cricket ball has more mass, so it accelerates less.

Reason (R) is true because inertia is the resistance to change in motion, and mass quantifies this resistance. A body with more mass (like a cricket ball) resists acceleration more than a lighter one (tennis ball). Since the Reason (R) correctly explains Assertion (A) (i.e., the cricket ball's greater mass/inertia is the reason it acquires less velocity), the statement is fully correct.

both A and R are true and R is not the correct explanation of A

Explanation

Assertion (A) is true because the cricketer pulls his hands back to increase the time over which the ball’s momentum is reduced to zero. This reduces the force on his hands as per Newton's second law of motion which is :

$\text F=\dfrac {Δ\text p}{Δ\text t}$

So, on increasing time (Δt) reduces the force (F) experienced.

​Reason (R) is true because this is Newton’s third Law, which is correct in itself, but it is not the reason why the cricketer withdraws his hands as the reason involves Newton’s second law.

assertion is false but reason is true

Explanation

Assertion (A) is false because momentum (p) is defined as :

$\text p=\text {mv}\text {(mass × velocity)}$

​Reason (R) is true because acccording to Newton’s second law :

$\text F=\dfrac {Δ\text p}{Δ\text t}=\text {ma (if mass is constant)}$

assertion is true but reason is false

Explanation

Assertion (A) is true because once pedalling stops, no external force is applied to keep the bicycle moving then frictional forces cause the bicycle to gradually slow down.

​Reason (R) is false because friction always acts opposite to the direction of motion, resisting it.

assertion is false but reason is true

Explanation

Assertion (A) is false as on the Moon, a person actually feels lighter, not heavier because the gravitational acceleration (g) on the Moon is about 1/6th of that on Earth, so weight is also 1/6th.

​Reason (R) is true because the Moon’s gravitational acceleration is much lower than that of Earth.

both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because according to Newton’s law of gravitation :

$\text F=\text G\dfrac{\text M_1\text M_2}{\text r^2}$

If distance r becomes 4 times, then :

$\text F'=\dfrac {\text F}{4^2}=\dfrac {\text F}{16}$

So, the force becomes 1/16th of the original.

​Reason (R) is true because this is exactly what Newton’s law states :

$\text F\propto\dfrac{1}{\text r^2}$

Since the reason correctly explains the assertion, both are true and reason is the correct explanation.

assertion is true but reason is false

Explanation

Assertion (A) is true because weight is defined as the gravitational force exerted by the Earth on a body :

$\text W=\text {m×g}$

where m is the mass and g is the acceleration due to gravity.

​Reason (R) is false because mass is the measure of the quantity of matter and weight is the force due to gravity acting on that mass.

assertion is true but reason is false

Explanation

Assertion (A) is true because when a swimmer wants to move forward in water, he pushes the water backward using his hands and feet which is an action force so according to Newton’s third law of motion : "for every action, there is an equal and opposite reaction", the water pushes back on the swimmer with an equal force in the opposite (forward) direction. This reaction force is what actually propels the swimmer forward.

​Reason (R) is false because water exerts an equal and opposite force (not greater), according to Newton’s third law of motion.

always attractive

Reason — The force of attraction between between two particles because of their masses, is called the gravitational force of attraction. Hence, the gravitational force between two bodies is always attractive.

their masses

Reason — Each mass particle of the universe attracts other mass particles. The force of attraction between two particles because of their masses, is called gravitational force of attraction

Sir Isaac Newton

Reason — For the magnitude of gravitational force of attraction, Sir Isaac Newton gave a law, known as the law of gravitation.

both (a) and (c)

Reason — According to Newton, the force of attraction acting between two bodies is :

1. directly proportional to the product of their masses.
2. inversely proportional to the square of the distance between them.

Hence, we get, F = $\dfrac{\text {Gm}_1\text m_2}{\text r^2}$

none of these

Reason — The value of gravitational constant G remains same at all places, and is independent of the nature of particle, temperature, medium, etc.

it remains unchanged

Reason — The value of gravitational constant G remains same at all places, and is independent of height, place, distance, etc.

all of the above

Reason — The gravitational force between two masses is :

1. is always attractive
2. directly proportional to the product of the masses
3. inversely proportional to the square of the separation between them
4. significant between heavenly bodies, but is insignificant between atomic bodies because of small magnitude of G.

1.60 x 10^-9 N

Reason — As we know,

$\text{F} = \text{G}\dfrac{\text{m}_1\text{m}_2}{\text{R}^2}$

Given,

masses 60 kg and 40 kg, when separation = 10 m,

Hence,

$\text{F} = \dfrac{{6.7 \times 10^{-11} \times 60 \times 40}}{\text{10}^2} \\[0.5em] \text{F} = 1.60 \times 10^{-9}\ \text N$

Hence, F = 1.60 x 10^-9 N

both (a) and (b)

Reason — As, g = $\dfrac{\text {GM}}{\text R^2}$ hence, the value of g on a planet depends on the mass and radius of that planet (or satellite).

2u/g

Reason — At the maximum height, v = 0

maximum height = $\dfrac{\text u^2}{2\text g}$ (from equation v² = u² - 2gh)

Time taken by the body to reach the maximum height (t) = $\dfrac{\text u}{\text g}$ (from equation v = u - gt).

Hence, the same will be the time it takes to get back to the initial point from the highest point. So, total time of journey = 2t = $\dfrac{2\text u}{\text g}$.

∴ The time for which the body remains in air = $\dfrac{2\text u}{\text g}$

19.6 m s^-1

Reason — As we know,

Velocity of the object = g x t

Given,

u = 0,

g = 9.8 m s^-2

t = 2 s

v = u + gt

Substituting the values in the formula above, we get,

Velocity of the object = 0 + 9.8 x 2 = 19.6 m s^-1

40 N

Reason — As we know,

Weight (W) = mass (m) x acceleration due to gravity (g)

Given,

mass = 25 kg

g on moon = 1.6 ms^-2

W_m = ?

Substituting the values, we get,

W_m = 25 x 1.6 = 40 N

Hence, weight of body on moon = 40 N.

W = mg

Reason — The weight of the body is related to the mass as follows :

Weight = mass x acceleration due to gravity or W = mg

Newton

Reason — The weight of a body is the force with which the earth attracts it. In other words, weight of a body is the force of gravity on it.

Weight is a vector quantity. It's direction is downwards towards the centre of the earth.

S.I. unit of weight is Newton (N).

changes when the velocity of the body is close to the velocity of light

Reason — The mass of a body increases with its velocity but this change is perceptible only when the velocity of the body v becomes more than 10⁶ ms^-1 i.e., reaches close to the speed of light c = (3 x 10⁸ ms^-1), so for a body moving with velocity less than 10⁶ ms^-1, its mass is taken to be constant.

Mass remains constant, but weight changes with gravity.

Reason — Mass is the amount of matter in a body so it does not change regardless of location but weight is the force with which a body is attracted towards the Earth (or any other celestial body), and it depends on gravity and is given by :

$\text {Weight (W)} = \text {Mass (m)} \times \text {Acceleration due to gravity (g)}$

So, weight changes when gravity (g) changes (e.g., on the Moon, where gravity is less, weight is less).

As we know,

$\text{F} = \text{G}\dfrac{\text{M}\text{m}}{\text{R}^2}$

Given,

F = 10 N, when separation = R,

Hence,

F₁ = $\dfrac{\text {G M m}}{\text R^2}$ = 10 N     [Equation 1]

If separation = $\dfrac{\text R}{2}$ then,

Substituting the values in the formula above, we get,

$\text F_2 = \dfrac{\text {G M m}}{\big(\dfrac{\text R}{2}\big)^2} \\[0.5em] \text F_2 = \dfrac{4 \text {G M m}}{\text R^2} \\[0.5em]$

$\text F_2 = 4 \times \big(\dfrac{\text {G M m}}{\text R^2}\big)$    [Equation 2]

Substituting the value of equation 1 in equation 2 we get,

F₂ = 4 x 10 N = 40 N

As we know,

Weight = mg

Given,

m = 5 kg,

Assumption : g = 10 m s ^-2

Substituting the values in the formula above, we get,

$\text W = 5 \times 10 \\[0.5em] \text W = 50\ \text N \\[0.5em]$

Hence, weight of the body = 50 N.

(i) As we know,

Weight = mg

Given,

m = 10 kg

g = 9.8 m s^-2

As m s^-2 can be written as N kg^-1,

So, g = 9.8 N kg^-1

Substituting the values in the formula above, we get,

$\text W = 10 \times 9.8 \\[0.5em] \text W = 98\ \text N\\[0.5em]$

As 9.8 N = 1 kgf,

Hence, 98 N = 10 kgf

Therefore, weight in kgf = 10 kgf

(ii) Weight in newton = 9.8 N

As we know,

Force of gravity (F) = mass (m) x acceleration due to gravity (g)

Given,

m = 5 kg

g = 9.8 m s^-2

Substituting the values in the formula above, we get,

$\text F = 5 \times 9.8 \\[0.5em] \Rightarrow \text F = 49\ \text N \\[0.5em]$

Hence, the force of gravity = 49 N, acting vertically downwards.

As we know,

Weight (W) = mass (m) x acceleration due to gravity (g)

Given,

w = 2 N

g = 10 m s^-2

Substituting the values in the formula above, we get,

$2 = \text m \times 10 \\[0.5em] \Rightarrow \text m = \dfrac{2}{10} \\[0.5em] \Rightarrow \text m = 0.2 \text{ kg} \\[0.5em]$

Hence, the mass of the body = 0.2 kg.

(a) As we know,

Weight (W) = mass (m) x acceleration due to gravity (g)

Given,

w = 98 N

g = 9.8 m s^-2

Substituting the values in the formula above, we get,

$98 = \text m \times 9.8 \\[0.5em] \Rightarrow \text m = \dfrac{98}{9.8} \\[0.5em] \Rightarrow \text m = 10 \text { kg} \\[0.5em]$

Hence, the mass of the body = 10 kg.

(b) As mass is constant and it's value remains same on moon (m) as on earth (e), hence,

m_e = m_m = 10 kg

g_moon = 1.6 m s^-2

Substituting the values in the formula above, we get,

$\text W = 10 \times 1.6 \\[0.5em] \Rightarrow \text W = 16\ \text N \\[0.5em]$

Hence, the weight on moon = 16 N.

As we know,

Weight (W) = mass (m) x acceleration due to gravity (g)

Given,

W_e = 600 N

W_m = ?

and

g_m = $\dfrac{1}{6}$g_e

Hence, W_m = $\dfrac{1}{6}$W_e

Substituting the values, we get,

W_m = $\dfrac{1}{6}$ x 600 = 100 N

Hence, weight of the man on the moon = 100 N.

As we know,

Force of gravity (F) = mass (m) x acceleration due to gravity (g)

Given,

m = 10.5 kg

g = 10 m s^-2

Substituting the values in the formula above, we get,

$\text F = 10.5 \times 10 \\[0.5em] \Rightarrow \text F = 105\ \text N \\[0.5em]$

Hence, the force of gravity = 105 N

(b) As we know,

Weight (W) = mass (m) x acceleration due to gravity (g)

Given,

m = 10.5 kg

g = 10 m s^-2

Substituting the values, we get,

$\text W = 10.5 \times 10 \\[0.5em] \Rightarrow \text F = 105\ \text N \\[0.5em]$

Hence, weight = 105 N.

(a) As we know from the equation of motion;

s = ut + $\dfrac{1}{2}$gt²

where, s = height

Given,

t = 3 s

g = 9.8 m s^-2

initial velocity (u) = 0

Substituting the values in the formula above, we get,

$\text S = (0 \times t) + (\dfrac{1}{2} \times 9.8 \times 3^2) \\[0.5em] \text S = 0 + (\dfrac{1}{2} \times 9.8 \times 9) \\[0.5em] \text S = 44.1\ \text m \\[0.5em]$

Hence, the height from which the ball was released = 44.1 m

(b) From the equation of motion,

v² = u² - 2gs

where, v = final velocity

Substituting the values in the formula, we get,

$\text v^2 = 0^2 - (2 \times 9.8 \times 44.1) \\[0.5em] \text v^2 = 2 \times 9.8 \times 44.1 \\[0.5em] \Rightarrow \text v^2 = 864.36 \\[0.5em] \Rightarrow \text v = 29.4 \text{ m s}^{-1} \\[0.5em]$

Hence, velocity with which the ball strikes the ground = 29.4 m s^-1

As we know,

Force (F) = mass (m) x acceleration due to gravity (g)

Given,

m = 5 kg

Assumption : g = 9.8 N kg^-1

Substituting the values in the formula above, we get,

$\text F = 5 \times 9.8 \\[0.5em] \Rightarrow \text F = 49\ \text N \\[0.5em]$

Hence, force = 49 N

(a) As we know, from the equation of motion,

v² = u² - 2gs (a = - g as movement is against gravity )

Given,

s = 20 m

g = 10 m s^-2

v = 0

Substituting the values in the formula, we get,

$0 = \text u^2 - 2 \times 10 \times 20 \\[0.5em] \text u^2 = 400 \\[0.5em] \Rightarrow \text u = 20\ \text{m s}^{-1} \\[0.5em]$

Hence, initial velocity of the ball = 20 m s^-1

(b) As we know, from the equation of motion,

v² = u² + 2gs

When the ball starts falling after reaching the maximum height, it's velocity (u) = 0

s = 20 m

g = 10 m s^-2

Substituting the values in the formula, we get,

$\text v^2 = 0^2 + (2 \times 10 \times 20) \\[0.5em] \text v^2 = 400 \\[0.5em] \Rightarrow \text v = 20\ \text{m s}^{-1} \\[0.5em]$

Hence, final velocity of the ball on reaching the ground = 20 m s^-1

(c) As we know,

total time of of journey of ball (t) = $\dfrac{2\text u}{\text g}$ and

u = 20 m s^-1

g = 10 m s^-2

Substituting the values in the formula, we get,

$\text t = \dfrac{2 \times 20}{10} \\[0.5em] \Rightarrow \text t = 4\ \text s \\[0.5em]$

Hence, total time for which the ball stays in air = 4 s

As we know, from the equation of motion,

v² = u² + 2gs

u = 0

v = 20 m s^-1

g = 10 m s^-2

Substituting the values in the formula, we get,

$20^2 = 0^2 + (2 \times 10 \times \text s) \\[0.5em] 400 = 20 \text s \\[0.5em] \Rightarrow \text s = \dfrac{400}{20} \\[0.5em] \Rightarrow \text s = 20\ \text m \\[0.5em]$

Hence, the height of the tower = 20 m

(i) From the equation of motion,

h = ut + $\dfrac{1}{2}$ gt²

(we consider motion of ball from highest point to the ground)

Since, total time for journey of the ball in air = 6 s.
Therefore, time for reaching maximum height = $\dfrac{6}{2} = 3$ s

g = 10 m s^-2

u = 0

Substituting the values in the formula, we get,

$\text h = (0 \times 3) + (\dfrac{1}{2} \times 10 \times 3^2) \\[0.5em] \text h = 0 + (5 \times 9) \\[0.5em] \text h = 45\ \text m \\[0.5em]$

Hence, greatest height reached by the ball = 45 m

(ii) From the equation of motion,

v² = u² - 2gs

Substituting the values in the formula, we get,

$0^2 = \text u^2 - (2 \times 10 \times 45) \\[0.5em] \text u^2 = (2 \times 10 \times 45) \\[0.5em] \text u^2 = 900 \\[0.5em] \text u = 30 \text { m s}^{-1} \\[0.5em]$

Hence, the initial velocity of the ball = 30 m s^-1

As we know, the equation of motion

h = ut - $\dfrac{1}{2}$ gt² (a = -g, as movement is against gravity)

Given,

t = 2 s

g = 10 m s^-2

u = 20 m s^-2

Substituting the values in the formula, we get,

$\text h = (20 \times 2) - (\dfrac{1}{2} \times 10 \times 2^2) \\[0.5em] \text h = 40 - (5 \times 4) \\[0.5em] \text h = 20\ \text m \\[0.5em]$

Hence, greatest height reached by the ball = 20 m

(i) From the equation of motion,

h = ut + $\dfrac{1}{2}$ gt²

Given,

g = 10 m s^-2

h = 80 m

u = 0

Substituting the values in the formula, we get,

$80 = (0 \times \text t) + (\dfrac{1}{2} \times 10 \times \text t^2) \\[0.5em] 80 = 0 + (5 \times \text t^2) \\[0.5em] 80 = 5 \times \text t^2 \\[0.5em] \text t^2 = \dfrac{80}{5} \\[0.5em] \text t^2 = 16 \\[0.5em] \text t = 4\ \text s \\[0.5em]$

Hence, time taken = 4 s

(b) From the equation of motion,

v² = u² + 2gh

u = 0

g = 10 m s^-2

h = 80 m

Substituting the values in the formula, we get,

$\text v^2 = 0^2 + (2 \times 10 \times 80) \\[0.5em] \text v^2 = 1600 \\[0.5em] \Rightarrow \text v = 40 \text{ m s} ^{-1}\\[0.5em]$

Hence, final velocity of the stone on reaching the ground = 40 m s^-1

From the equation of motion,

h = ut + $\dfrac{1}{2}$ gt²