Force And Laws Of Motion

(a) Since, we know Inertia depends on the mass of an object hence, in the case of a rubber ball and a stone of the same size , stone will be having a greater Inertia.

(b) As we know that Inertia is the measure of mass of an object hence, in the case of a bicycle and a train, since, train has a larger mass. Hence, it will have a greater Inertia too.

(c) Since, Inertia is directly proportional to the mass of an object hence, in the case of a five-rupee coin and a one-rupee coin, a five-rupee coin has more inertia (than a one-rupee coin) because it has more mass than a one rupee coin.

The velocity of football changes four times.

First, when a football player kicks to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper stops the football. Fourth when the goalkeeper kicks the football towards a player of his own team.

Agent supplying the force –

First case – First player

Second case – Second player

Third case – Goalkeeper

Fourth case – Goalkeeper

The answer of this cause lies behind the Newton’s First Law of Motion. Initially, leaves and tree both are in rest. But when the tree is shaken vigorously, tree comes in motion while leaves have tendency to be in rest. Thus, because of remaining in the position of rest some of the leaves may get detached from a tree if we vigorously shake its branch.

As we know by newton first law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force

(a) In a moving bus, passengers are in motion along with bus. When brakes are applied to stop a moving bus, bus comes in the position of rest. But because of tendency to be in the motion a person falls in forward direction.

(b) Similarly, when a bus accelerates from rest, we tend to fall backwards because when a bus is accelerated from rest, the tendency to be in rest, a person in the bus falls backwards.

Horse pushes the ground in backward direction. In reaction to this, the horse moves forward and cart moves along with horse in forward direction.

In this case when horse tries to pull the cart in forward direction, cart pulls the horse in backward direction, but to make the cart move, the horse bends forwards and pushes the ground with its feet. When the forward reaction to the backward push of the horse is greater than the opposing frictional forces to the wheels, the cart moves

When large amount of water is ejected from a hose at a high velocity, according to Newton’s Third Law of Motion, water pushes the hose in backward direction with the same force. Therefore, it is difficult for a fireman to hold a hose in which ejects large amount of water at a high velocity.

Firstly, we will calculate the momentum of both rifle and bullet separately.

Momentum of bullet = Mass of bullet × Velocity of bullet

As we know that p= m × v

Putting the values in above equation we get:-

= 1.75 kg m/s (i)

Momentum of rifle = Mass of rifle × Velocity

Putting the value in above equation we get:-

According to the law of conservation of momentum:

1.75 = 4v [From (i) and (ii)]

v =

v = 0.4375 m/s

Therefore, the recoil velocity of the rifle is 0.4375 m/s

For this, firstly we have to calculate the total momentum of both the objects:

Momentum of first object:

= Mass of first object × Velocity of first object

= kg × 2 m/s

= 0.1 kg × 2 m/s

= 0.2 kg m/s

Momentum of second object:

= Mass of second object × velocity of object

= kg × 1 m/s

= 0.2 kg × 1 m/s

= 0.2 kg m/s

Total momentum = Momentum of first object + Momentum of second object

= 0.2 + 0.2

= 0.4 kg m/s

It is given that after the collision, the velocity of first object becomes 1.67 m/s

Momentum after collision (First object):

= kg × 1.67 m/s

= 0.1 kg × 1.67 m/s

= 0.167 kg m/s

Let, the velocity of second object after the collision be v m/s

Momentum after collision (Second object)

= kg × v m/s

= 0.2 kg × v m/s

= 0.2v kg m/s(ii)

Total momentum of the objects after collision = 0.167 + 0.2 v

We know that,

According to the law of conversion of momentum:

Total momentum before collision = Total momentum after collision

0.4 = 0.167 + 0.2 v

0.2 v = 0.4 – 0.167

0.2 v = 0.233

v =

v = 1.165 m/s

Therefore, the velocity of second object will be 1.165 m/s

Yes, when external unbalanced force on an object is zero, the object can be travelling with a non-zero velocity

This can happen under the following conditions:

(a) The object should already be travelling with a uniform speed in a straight line path.

(b) There should be no change in the magnitude of speed.

(c) There should be no change in the direction of motion.

(d) The friction between the object and the ground must be zero.

Beating of a carpet with a stick; makes the carpet come in motion suddenly, while dust particles trapped within the pores of carpet have tendency to remain in rest, and in order to maintain the position of rest they come out of carpet. This happens because of the application of Newton’s First Law of Motion which states that any object remains in its state unless any external force is applied over it.

Luggage kept on the roof of a bus has the tendency to maintain its state of rest when bus is in rest and to maintain the state of motion when bus is in motion according to Newton’s First Law of Motion.

When bus will come in motion from its state of rest, in order to maintain the position of rest, luggage kept over its roof may fall down. Similarly, when a moving bus will come in the state of rest or there is any sudden change in velocity because of applying of brake, luggage may fall down because of its tendency to remain in the state of motion.

This is the cause that it is advised to tie any luggage kept on the roof a bus with a rope so that luggage can be prevented from falling down.

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

Because, when ball moves on the ground, the force of friction opposes its movement and after some time ball comes to the state of rest.

Firstly, we will calculate the acceleration:

Given,

Initial velocity, u = 0

Distance travelled, s = 400 m

Time, t = 20 s

Acceleration, a =?

We know that,

s = ut + at²

400 = 0 × 20 + × a × (20)²

400 = 0 + × a × 400

400 = a × 200

a =

a = 2 m/s²

Therefore, the acceleration of the truck is 2 m/s²

Now, we will calculate the force:

Force, F = m × a

F = 7 × 1000 × 2

= 14000 N

Therefore, the force acting on the truck is of 14000 N

Given,

Initial velocity, u = 20 m/s

Final velocity, v = 0 (As the stone stops)

Acceleration, a =?

Distance travelled, s = 50 m

We know that,

v²= u²+2as

(0)² = (20)² + 2 × a × 50

0 = 400 + 100 a

100 a = - 400

a = -

a = - 4 m/s²

As,

Force, F = m × a

F = 1 × (-4)

F = - 4 N

Therefore, the value of force is – 4 N. Here, the negative sign indicates that the stone opposes the motion of stone.

(a) Given,

Force of engine = 40,000 N

So, let us consider the complete train as single body on which a force of 40000 N is applied by engine of mass 8000 kg on which a opposing force(friction force) is applied in direction opposite to direction of motion of body.

= 40000 – 5000

(b)
given:
Net force exerted by the engine = 35000 N

The mass of 5 wagons of train = 2000 × 5= 10000 kg

The mass of engine = 8000 kg
The mass of train = 10000 +8000 =18000 kg

We know that,

F = m × a

a = m/s² = 1.9444 m/s²

Therefore, the acceleration of the train is 1.9444 m/s²

(c)
There are total 5 wagons behind the engine (which have been marked 1, 2, 3, 4 and 5 in the figure) and there are only 4 wagons behind wagon 1.

Force of wagon 1 = Mass of 4 wagons × Acceleration of train

= 2000 × 4 × 1.9444 (We know that weight of one Wagon is 2000 kg)

= 15555.2 N
Thus, the force of Wagon 1 on Wagon 2 is 15555.2 N

Given,

Mass of vehicle, m = 1500 kg

Acceleration, a = - 1.7 m/s²

We know that,

Force, F = m × a

F = 1500 × (-1.7)

F = - 2550 N

Here, the negative sign shows that the force acts in a direction opposite to the direction of motion.

(d) mv

Given,

Mass = m

Velocity = v

Therefore, Momentum =?

We know that,

Momentum, P = Mass x Velocity

= mv

Therefore, option (d) mv is correct.

Since, a horizontal force of 200N is used to move a wooden cabinet, thus a friction force of 200N will be exerted on the cabinet. Because according to third law of motion, an equal magnitude of force will be applied in the opposite direction.

Given,

Mass of first object, m₁ = 1.5 kg

Velocity of first object, v₁ = 2.5 m/s

Momentum of first object = m₁ × v₁

= 1.5 × 2.5

= 3.75 kg m/s

Mass of second object, m₂ = 1.5 kg

Velocity of first object, v₂ = - 2.5 m/s

So, Momentum of second object, m₂ × v₂

= 1.5 × (-2.5)

= - 3.75 kg m/s

So, Total momentum = 3.75 + (-3.75)

= 0 kg m/s (i)

Combined mass = m₁ + m₂

= 1.5 + 1.5

= 3.0 kg

Let, the velocity of the combined objects after collision will be v m/s.

Total momentum = (m₁ + m₂) × v

= 3.0 × v (ii)

We know that,

According to the principle of conservation of momentum:

Total momentum before collision = Total momentum after collision

0 = 3.0 × v

v =

= 0 m/s

Therefore, the velocity of the combined objects after the collision will be 0 m/s.

Because of the huge mass of the truck, the force of static friction is very high. The force applied by the student is unable to overcome the static friction and hence he is unable to move the truck. In this case, the net unbalanced force in either direction is zero which is the reason of no motion happening here. The force applied by the student and the force because of static friction are cancelling out each other. Hence, the rationale given by the student is correct.

Firstly, we will calculate the initial momentum:

Given,

= kg

Initial velocity, v₁ = 10 m/s (Given)

So,

= 2 kg m/s (i)

Now, we will calculate the final momentum:

Given,

Mass of hockey ball, m₂ = 200 g

(We know that 1kg=1000gram)

Final velocity, v₂= - 5 m/s (As the ball is going in reverse direction)

Final momentum = m₂ × v₂

= - 1 kg m/s (ii)

Now,

= -1 -2

Therefore, the change in momentum of the hockey ball is 3 kg m/s.

Firstly, we have to calculate the acceleration of the bullet:

Initial velocity, u = 150 m/s (Given)

Time, t = 0.03 s(Given)

We know that,

0 = 150 + a × 0.03

a = -

Now, we will calculate the distance of penetration of bullet:

We know that,

v² = u² + 2as

(0)² = (150)² + 2 × (-5000) × s

0 = 22500 – 10000 × s

s =

s = 2.25 m

We know that:-
Force, F = m × a

F = kg × (-5000) m/s (1 gram=1/1000 kg)

= -50 N

Thus the magnitude of retarding force is exerted by the wooden block on the bullet is 50 N

Given,

Mass of object, m₁ = 1 kg

Velocity of object, v₁= 10 m/s

Momentum of object = m₁ × v₁

= 1× 10 kg m/s

= 10 kg m/s (i)

Mass of wooden block, m₂ = 5 kg

Velocity of wooden block, v₂ = 0 (As it is stationary)

Momentum of wooden block = m₂ × v₂

= 5 × 0

= 0 kg m/s (ii)

From (i) and (ii), we get:

Total momentum = 10 + 0

Before impact = 10 kg m/s (iii)

Total momentum (after impact) =10 kg m/s[Law of conservation of momentum]

Total mass of object = 1 kg + 5 kg

= 6 kg

Velocity of object and wooden block = v m/s

10 = 6 × v

v =

v = 1.67 m/s

Therefore, the velocity of the object and wooden block together is 1.67 m/s.

Firstly, we have to calculate the initial momentum:

= Mass × Initial velocity

= 100 × 5

= 500 kg m/s (i)

Now, we will find the final momentum

= Mass × Final velocity

= 100 × 8

= 800 kg m/s (ii)

Now,

Force =

=

=

= 50 N

We know that, as per the Law of Conservation of Momentum; total momentum of a system before collision is equal to the total momentum of the system after collision.

In this case, since the insect experiences a greater change in its velocity so it experiences a greater change in its momentum. From this angle, Kiran’s observation is correct.

Motorcar is moving with a larger velocity and has a bigger mass; as compared to the insect. Moreover, the motorcar continues to move in the same direction even after the collision; which suggests that motorcar experiences minimal change in its momentum, while the insect experiences the maximum change in its momentum. Hence, Akhtar’s observation is also correct.

Rahul’s observation is also correct; because the momentum gained by the insect is equal to the momentum lost by the motorcar. This also happens in accordance to the law of conservation of momentum.

Firstly, we will calculate the final velocity:

Given,

Initial velocity, u = 0 (As it falls from rest)

Final velocity, v =?

Acceleration, a = 10 ms²

And, Distance, s = 80 cm

= m

= 0.8 m

We know that,

v² = u² + 2as

v² = (0)² + 2 × 10 × 0.8

v² = 16

v =

= 4 m/s

Momentum = Mass × Velocity

= 10 ×4

= 40 kg m/s

(a) Given,

Time (t) = 0 s

Distance (s) = 0 m

And at time, (t) = 1 s

Distance (s) = 1 m

When time (t) = 2 s

Distance (s) = 8 m

We know that,

(2)³ = 2×2×2 = 8

When time (t) = 3 s,

Distance (s) = 27 m

(3)³ = 3×3×3 = 27

From above, we can conclude that:

Distance ∝ (Time)³

s ∝ t³

(i) When the distance travelled is proportional to time (s ∝ t), then the object has constant velocity.

Hence, acceleration in that case would be zero.

Since, s ∝ t³ (So the acceleration in this case cannot be zero)

(ii) When the distance travelled is proportional to the square of time (s∝t²), then the object has constant acceleration.

Since, s∝t³ (So the acceleration in this case cannot be constant)

(iii) The data given in this question shows that the distance travelled is proportional to the cube of time (s∝t³), therefore, the conclusion drawn is that the acceleration is increasing uniformly with time.

(b) We know that,

Force = Mass × Acceleration

F= m × a

In other words we can also say that the acceleration of a body is proportional to the force applied to it. Now, since the acceleration of the object in this case is increasing uniformly with respect to time, therefore, the forces acting on the object must also be increasing uniformly with time.

Let, the force applied by each person be F.

So,

Total force applied by two persons = F + F

= 2F (i)

Force of friction, f = 2F (ii)

Now, when three persons apply force to push the motorcar, then:

Total force applied by three persons = F + F + F

= 3F (iii)

Now, the total force applied by three persons is 3F whereas the opposing force of friction is f.

Force that produces acceleration = 3F – f (iv)

As from equation (ii),

f = 2F

Force that produces acceleration = 3F – 2F

= F

Now, Force = Mass × Acceleration

F = m × a

= 1200 × 0.2

= 240 N

Therefore, each person pushes the motorcar with a force of 240 N

Given,

Mass, m = 500 g

= kg

= 0.5 kg

Initial velocity, u = 50 m/s

Final velocity, v = 0 (As the hammer stops)

Acceleration, a =?

Time, t = 0.01 s

We know that,

v = u + at

0 = 50 + a × 0.01

0.01 a = - 50

a = -

= - 5000 m/s²

We know that:

Force, F = m × a

F = 0.5 × (- 5000)

= - 2500 N

Here, the negative sign indicated that this force is opposing motion.

Given,

Initial velocity, u = 90 km/h

=

= 25 m/s

Final velocity, v = 18 km/h

=

= 5 m/s

Acceleration, a =?

Time, t = 4 s

Now, v = u + at

5 = 25 + a × 4

4a = 5 – 25

4a = - 20

a = -

a = - 5 m/s²

Therefore, the acceleration of the motorcar is -5 m/s².

Now we will calculate change in momentum:

Change in momentum = mv – mu

= 1200 × 5 - 1200 × 25

= 6000 – 30000

= - 24000 kg m/s

Therefore, change in momentum is 24000 kg m/s

Now we will calculate the magnitude of force:

Force, F = m × a

= 1200 × (-5)

= - 6000 N

(a) Since, the action force and reaction force are equal and opposite, both the vehicles experience the same impact of force after the collision.

(b) Both the vehicles will experience equal change in momentum because the change in momentum of truck is equal and opposite to the change in momentum of car.

(c) We can see that the force on each vehicle is the same but the mass of car is smaller than that of the truck, therefore, the car experiences the greater acceleration or greater retardation.

(d) Since the mass of the car is less than that of the truck, therefore, the car will be suffering​ greater damage during the collision than the truck (The car has greater retardation than the truck).