In this article, we shall study to solve numerical problems to calculate the magnitude of force, momentum, and change in momentum.

Example – 01:

The speed of a tortoise and hare are 1 m/s and 3 m/s respectively. The mass of the hare is 5 kg while that of tortoise is 20 kg. Which of the two has greater momentum?

Solution:

The momentum of hare = Mass of hare x Speed of hare= 5 x 3 = 15 kg m/s

The momentum of tortoise = Mass of tortoise x Speed of tortoise= 20 x 1 = 20 kg m/s

Thus the momentum of tortoise is more than that of the hare.

Example – 02:

What is the magnitude of force exerted by a horse in pulling a cart of mass 600 kg and accelerating at the rate of 1.2 m/s²?

Given: mass of cart = m = 600 kg, acceleration = a = 1.2 m/s².

To Find: F =?

Solution:

By Newton’s second law of motion, magnitude of force

F = ma = 600  x 1.2

∴ F = 720 N

Ans: The required force is 720 N.

Example – 03:

An object of mass 10 kg is moving with the initial velocity of 10 m/s. A constant force acts on it for 4 s giving it a speed of 2 m/s in opposite direction. Find the acceleration and the force.

Given: mass of object = m = 10 kg, initial velocity = u = 10 m/s, Final velocity = v = – 2m/s (negative since it is in opposite direction), time for which force is acting = t = 4 s

To Find: acceleration = a =? Force acting = F =?

Solution:

a = (v – u)/t = (-2 – 10)/4 = -12/4 = -3 m/s²

The negative sign indicated retardation.

By newton’s second law of motion we have

F = ma = 10 x (-3) = -30 N

Negative sign indicated retarding force.

Ans: The acceleration = 3 m/s² and Force = -30 N

Example – 04:

A constant force of 2 N acts on a body for 5 seconds to change its velocity. Calculate the change in the momentum.

Given: Force acting = F = 2 N, time for which the force is acting = t = 5s.

To Find:  Change in momentum =?

Solution:

Change in momentum = F x t = 2  x 5  = 10 kg m/s.

Ans: the change in momentum is 10 kg m/s.

Example – 05:

An empty truck of mass 1000 kg is moving at a speed of 36 km/hr. It is loaded with 500 kg of material on its way and again moves at the same speed. Will the momentum of the truck remain the same after loading? if not, find the momentum of the truck after loading.

Given: Mass of a truck = 1000 kg, Mass of Load = 500 kg, Speed of the vehicle = v = 36 km/hr = 36 x 5/18 = 10 m/s.

To Find: Momentum of truck = p =?

Solution:

The momentum of a body depends on its mass. In this case, the truck is loaded on the way, hence its momentum should change.

Total Mass = m = 1000 + 500 = 1500 kg

New momentum = Total Mass x Velocity = 1500 X 10 = 15000 kg m/s

Ans: The momentum of the truck after loading is 15000 kg m/s.

Example – 06:

A railway wagon of mass 1000 kg is pulled with a force of 10000 N. What is the acceleration?

Given : Force applied = F = 10000 N, mass of wagon = m = 1000 kg.

To Find:  acceleration = a =?,

Solution:

By Newton’s second law of motion, magnitude of force

F = ma

∴ 10000 = 1000 x a

∴ a = 10000/1000 = 10 m/s².

Ans: The acceleration is 10 m/s².

Example – 07:

A car of mass 1000 kg is moving at a certain speed when a constant braking force 1000 N acts on it for 5 s and speed of the car reduced to half the original speed. Find the further time required to stop the car, if the same constant force acts on it.

Given: mass of car = m = 1000 kg, Force acting = F = 1000 N,time taken =  t = 5 s, Final speed (v) = 1/2 Initial speed (u) = u/2

To find: t =? when v = 0

Solution:

By Newton’s second law of motion, magnitude of force

F = ma

∴1000 = 1000 x a

∴ a = 1000/1000 = 1 m/s².

By the first equation of motion

∴ a = (v – u)/t  = (u/2 – u)/5 = (-u/2)/5 = (-u/10)

∴ a = -u/10 = – 1

∴ u = 10 m/s

We have v = u + at

∴ 0 = 10 + (-1)t

∴ -10 = – t

∴ t = 10 s

Ans: The further time required to stop the car is 10 s

Example – 08:

Find the magnitude of the force applied to a block of mass 5 kg at rest, if it moves 36 m, in first 3 seconds. Neglect the force of friction.

Given: mass of block = m = 5 kg, Initial velocity = u = 0, Distance traveled = s = 36 m, time taken = t = 3 s,

To find: F =?

Solution:

s = ut + 1/2 at²

∴   36 = (0)(3) + 1/2 a(3)²

∴   36 = 1/2 a(9)

∴   72 = a (9)

∴ a = 72/9 = 8 m/s².

By Newton’s second law of motion, the magnitude of force

F = ma = 5 x 8 = 40 N

Ans: The force applied = 40 N

Example – 09:

Two spheres of mass 10 g and 100 g each fall on the two pans of a table balance from a height of 40 cm and 10 cm respectively. If both are brought to rest in 0.1 seconds. Determine the force exerted by each sphere on the pans.

Solution:

For the first sphere: 

Given : m₁ = 10 g = 0.01 kg, h = – 40 cm  = – 0.4 m,  t = 0.1 s, g = – 9.8 m/s²

v² = u² + 2gh

∴  v² = (0)² + 2(-9.8)(-0.4)

∴  v² =  7.84

∴  v = 2.8 m/s

F₁ = m₁a

F₁ = m₁ (v – u)/t  = 0.01 x (2.8 – 0)/0.1 = 0.28 N

For the second sphere: 

Given : m₁ = 100 g = 0.1 kg, h = – 10 cm  = – 0.1 m,  t = 0.1 s, g = – 9.8 m/s²

v² = u² + 2gh

∴ v² = (0)² + 2(-9.8)(-0.1)

∴ v² =  1.96

∴ v = 1.4 m/s

F₂ = m₂a

F₂ = m₂ (v – u)/t  = 0.1 x (1.4 – 0)/0.1 = 1.4 N

Ans: The force exerted by the first sphere is 0.28 N and that by second sphere is 1.4 N

Example – 10:

Calculate the density of a cubical ice block of side 50 cm. If a force of 1125 N applied to it produces an acceleration of 10 m/s² in it. Neglect the force of friction. Assume the ice block remains in solid state without melting.

Given: Force applied = F = 1125 N, acceleration = a = 10 ms-1, Side of a block = 50 cm = 0.5 m

To Find:  Density = ρ = ?,

Solution:

By Newton’s second law of motion

F = m.a

∴ 1125 = m x 10

∴ m = 1125/10 = 112.5 kg

Ans: The density of ice block is 900 kg/m³.

Example – 11:

A ball of mass 50 g at rest is hit by a bat and the ball covers a distance of 400 m, in 2 seconds. If the ball was in contact with it for 0.1 s, find the magnitude of the force acting on it. Assuming no other force acts on a ball after it is hit by the bat.

Given: mass of ball = m = 50 g = 0.05 kg,  time of contact = 0.1 s, distance covered  s= 400 m, time taken to cover the distance = t = 2 s.

To find: Force acting =F =?

Solution:

No other force acts on a ball after it is hit by the bat. Thus it is in uniform motion after hit.

v = s/t = 400/2 = 200 m/s

Now Force, F = ma

F = m (v – u)/t  = 0.05 x (200 – 0)/0.1 = 100 N

Ans: The force acting = 100 N

Example – 12:

A force of 500 N acts on a body of mass 1000 kg and the body is brought to rest within a distance of 64 m. Find the initial velocity and time taken by the body to come to rest.

Given: mass of body = m = 1000 kg, Force acting = F = 500 N, Final velocity = v = 0 ms-1, distance traveled = s = 64m.

To find: initial velocity =u =? time taken = t = ?

Solution:

We have     F = ma

∴   500 = 1000 x a

∴ a = 500/1000 = 0.5 m/s²

As the body is brought to rest a = – 0.5 m/s²

v² = u² + 2as

∴ (0)² = u² + 2(-0.5)(64)

∴ (0)² = u² – (64)

∴ u² = 64

∴ u = 8 m/s

By first equation of motion

v = u + at

∴ 0 = 8 + (- 0.5) x t

∴ 8 = – 0.5 x t

∴ t = 8/0.5 = 16 s

Ans: Initial velocity = 8 m/s, time taken to come to rest = 16 s

Example – 13:

A car of mass 1000 kg is moving uniformly with 10 m/s. If the engine of the car develops an extra linear momentum of 1000 kg m/s. Calculate the new velocity with which the car runs.

Solution:

Given : mass of car = m = 100 kg, initial velocity = u = 10 m/s, Extra momentum = 1000 kg m/s,

To find: Final velocity = v = ?

Initial momentum = p₁ = mu = 1000 x 10 = 10000 kg m/s

Final momentum = p₂ = 10000 + 1000 = 11000 kg ms-1.

Now, Final momentum = p₂ = mv

∴ 11000 = 1000 x v

∴ v = 11000/1000 = 11m/s

Ans: New Velocity = 11 m/s

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Newton’s laws of motion have everyday applications. The change in momentum of a body in a small time is a cause of large force. Most of the applications are related to the concept of change in momentum.

Cricket fielder lowers his hand while catching a ball:

If the ball is caught without lowering hands, the duration of the impact is small. As a result, the rate of change of momentum increases producing a large force. Thus the fielder will hurt his hands due to large force. When the ball is caught by moving the hand in the direction of the motion of the ball, the duration of impact increases. As a result, the rate of change of momentum decreases and thus the force exerted by the ball on hands get reduced.

An athlete taking a long jump or a high jump bends his knees before landing:

If the athlete does not bend his knees before landing, the duration of the impact is small. As a result, the rate of change of momentum increases producing a large force. Thus the athlete will hurt his legs due to large force. If the athlete bends his knees before landing, the duration of impact increases. As a result, the rate of change of momentum decreases and thus the force exerted by the ground (Reaction) on the athlete get reduced. Thus the athlete lands safely by bending knees before landing.

A Blacksmith holds the rod on the anvil with his tongs while striking with his hammer:

For reshaping of a rod, the blacksmith has to apply very large force. By holding the rod on the anvil with his tongs while striking with his hammer, the blacksmith reduces the time of impact. As a result, the rate of change of momentum increases drastically producing a very large force.

An athlete runs a certain distance before taking a large jump:

Here Newton’s First law, second law, and Third law is used. As the athlete is running a certain distance he gets additional velocity and hence additional kinetic energy, when he is actually jumping. Due to high velocity at a point where he starts his jump (Point of rising) the time of impact is reduced. As a result the rate of change of momentum increases, producing a very large force on the ground. The ground produces a large reaction and thus a long jump can be taken easily.

A clean hole is made in a glass windowpane when it is struck by a bullet fired from a gun. Whereas the same windowpane will be broken into pieces when struck by a stone of similar size:

When a bullet is fired at a glass window, a clear hole is formed in it. This is because only that part of the glass moves with the bullet, where the bullet hits the glass. The remaining part due to the inertia remains in its position. Thus bullet is able to form a hole in the glass window due to the inertia of window pane. The time for which the force is applied is very small, thus there is very less time for the reaction by the window pane.

When the same window pane is struck with a stone of a similar size, the time for which the force is applied is large compared to that in case of a fired bullet. Thus all the section around the striking point is set into motion. Due to the brittle nature of the glass, this impact causes it to break into pieces.

The engine of a moving car exerts a constant force on the car, but car movers with a constant velocity:

When a car is moving there is friction between different parts of the engine and mechanisms, besides there is friction between the road and the tyres. These frictional forces are opposed to the direction of motion. In the case of a car moving with a constant velocity the forward force exerted by the engine on the car is exactly equal and opposite to the car backward force exerted on the car due to friction and resultant force acting on the car is zero. This is the reason why the case moves with a constant velocity, even though the engine exerts a force on the car.

Apparent Weight of a Person in a Lift:

Let a person of mass ‘m’ standing on a weighing machine placed on the floor of the lift. The actual weight of the person is ‘mg’ which acts vertically down. The reaction offered by the weighing machine is ‘R’ which is also known as the apparent weight of the person.

Case – 1 (When the lift is at rest):

Since the lift is at rest, the net force acting on the person is zero.

R – mg = 0. Thus R = mg.

Thus his apparent weight is equal to his actual weight.

Case – 2 (When the lift moves upward or downward with uniform velocity):

Since lift is moving with uniform velocity, the net acceleration of the lift is zero, thus the net force acting on the person is zero.

R – mg = 0. Thus R = mg.

Thus his apparent weight is equal to his actual weight.

Case – 3 (When the lift moves upward with constant acceleration):

As the lift is moving upward with the acceleration say ‘a’, the net force acting on the person is ‘ma’ acting upward.

R – mg = ma.

Thus R = mg + ma = m(g  + a).

Thus his apparent weight is greater than his actual weight.

Case – 4 (When the lift moves downward with constant acceleration):

As the lift is moving downward with the acceleration, the net force acting on the person is ‘ma’ acting downward.

R – mg = – ma. 

Thus R = mg – ma = m(g  – a).

Thus his apparent weight is less than his actual weight.

Case – 5 (If the lift falls freely):

As the lift falls freely, its acceleration is equal to the acceleration due to gravity. Thus a = g. Substituting this value in equation of case -4 we get

R =  m(g  – g) = m(0) = 0.

Thus his apparent weight is zero and the person feels weightless. In this case, the surface on which the person is standing and the person himself have the same acceleration and the same direction of acceleration. Hence the person feels weightless.

In satellite orbiting around the earth, the astronaut and the surface of the spacecraft have the same acceleration acting towards the centre of the Earth. Hence the astronaut feels weightless.

Weightlessness is feeling and not actually weightless. The feeling of weightlessness can be experienced by a person jumping from a height.

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In this article, we shall study Newton’s third law of motion and the concept of action-reaction pair of forces.

Statement:

To every action, there is always equal, opposite and simultaneous reaction.

Explanation:

Let us consider a block of metal kept on the table. The weight of the block acts vertically downward. Thus the block tries to push the surface of the table downward by a force which is equal to its weight. This is the action. Now the surface of the table pushes the block vertically upward which is the reaction.

Here it should be noted that action and reaction act simultaneously. The action is on the table while the reaction is on the block. Thus action and reaction act on two different bodies. Hence though action and reaction are equal they can not cancel each other’s effect.

Examples to Illustrate Action Reaction Pair:

Example 1: To explain the third law of motion we can take an example of the working of a rocket. When the fuel of the rocket is ignited the combustion of fuels takes place. The hot gases produced in the chemical reaction are pushed out through the small hole at the bottom (tail) of the rocket with very high velocity. Thus the action is downward. Now, these outgoing gases produce an equal and opposite force on the rocket, which constitutes the reaction. Under the effect of upward reaction, the rocket is propelled upward. The jet engine also works on the same principle, the difference is that the rocket has to carry oxygen or an oxidizing agent with it to propel because it has to go out of the atmosphere of the Earth where oxygen is not available. While planes with jet engines fly in Earth,s atmosphere, hence they need not carry oxygen with them.

Example 2: Let us sit on a chair in front of a wall and push the wall with our legs. We will find that the chair is pushed backward. When we are pushing the wall we are applying action force, now the wall will put equal and opposite reaction on us and the chair is pushed backward.

Example 3: When we walk on the ground, as our foot pushes down (action) the ground, the ground pushes up (reaction) which is equal and opposite. This reaction is responsible for our forward movement. If the ground is not able to give a required reaction, then our foot sinks in it as in the case of sand, mud, water etc. If we push the ground harder we can jump into the air.

Example 4: Two-wheeled blocks of the same kind are at rest on the top of a smooth surface of a table. They are tied with a thread with each other. A compressed spring is placed between the two blocks. When the thread is cut the blocks move in the opposite direction.

Example 5: If we attach a spring balance to the hook on the wall and its hook is engaged with a hook of another spring balance and apply a pull on the second spring balance that both of the spring balance show the same reading but in opposite direction.

Example 6: The ball bounces back on hitting the ground. A ball strikes the ground with certain force (Action) and the ground pushes back the ball with equal force(Reaction)

Example 7: A gun recoils when a shot is fired from it. Initially both the bullet and gun are at rest, thus the total momentum of the system is constant. When a shot is fired, the bullet moves forward pushing the gun backward. This phenomenon is called the recoil of the gun. Due to recoiling of the gun, a backward jerk is experienced on the shoulders by the shooter. Hence to avoid injury he has to hold the rifle holding tight on shoulders.

Example 8: When a boy jumps from the boat to the bank. The boat moves in the direction opposite to jump. In order to jump out of the boat, he has to push the ground (surface of the boat) with greater force (action). Now the boat is in water, thus it is representing the non-rigid surface. Thus the boat moves backward under the action of pushing force.

Example 9: While swimming, the swimmer pushes the water backward with his hands. The action of pushing water backward gives rise to a reaction in the opposite direction which results in the swimmer moving forward.

Example 10: While hammering a nail. a force is experienced on the hand holding the hammer. When a nail is hammered a force is applied to it (action) due to which the nail applies equal and opposite reaction on the hammer.

Example 11: When moving train collides with a stationary one, the former experiences a reverse motion. The moving train applies a force on the stationary train and in return, the stationary train puts a reaction which is equal and opposite to the former train.

Example 12: A person falling from a certain height on a hard surface gets hurt more seriously than when he falls on a soft surface. By Newton’s third law of motion “to every action, there is always equal, opposite and simultaneous reaction. Thus when the person falls on the surface its weight (action) acts on the surface in turn the surface puts equal and opposite reaction on the person. Now the hard surface is rigid it transfers complete reaction without absorbing any impact (energy). Thus hard surface puts more reaction on the person. While the soft surface is elastic it transfers part of reaction absorbing some impact (energy). Thus soft surface put less reaction on the person. Thus hard surface is providing greater reaction than the soft surface. Hence, the person falling from a certain height on a hard surface gets hurt more seriously than when he falls on a soft surface.

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