10th Science / Chapter - 1 Laws of Motion 2m & word problem Question & Answer Key

Samacheer Kalvi 10th Science Laws of Motion  Text Book Back TWO Mark , FIVE MARK & WORD PROBLEM Questions and Answers :-

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Chapter -1 LAW OF MOTION லில்

1. 2 Mark 

2. 5 Mark 

3. Word Problem  BOOK BACK  QUESTION & ANSWER KEY 

 Answer briefly - 2 Marks

Question 1.

Define inertia. Give its classification.

Answer:

Inertia: The inherent property of a body to resist any change in its state of rest or the state of uniform motion, unless it is influenced upon by an external unbalanced force, is known as ‘inertia’.
Types of Inertia

• Inertia of rest
• Inertia of motion
• Inertia of direction

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Question 2.

Classify the types of force based on their application.

Answer:

1. Like parallel forces
2. Unlike parallel forces

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Question 3.

If a 5 N and a 15 N forces are acting opposite to one another. Find the resultant force and the direction of action of the resultant force.

Solution:

The two forces are unlike parallel forces

Let P = 5N, Q = 15N
Resultant force (R) = P – Q = 5 + (-15) = -10N
R = -10N.
The resultant force acting along the direction of “Q”.

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Question 4.

Differentiate mass and weight.

Answer:

S.NO	Mass	Weight
1	The quantity of matter contained in the body	The gravitation force exerted on it due to the Earth’s gravity alone.
2	Scalar quantity	Vector quantity
3	Unit: Kg	Unit: N
4	Constant at all the places	Variable with respect to gravity.

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Question 5.

Define moment of a couple.

Answer:

Rotating effect of a couple is known as moment of a couple.
Moment of a couple = Force × perpendicular distance between the line of action of forces
M = F × S

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Question 6.

State the principle of moments.

Answer:

When a number of like or unlike parallel forces act on a rigid body and the body is in equilibrium, then the algebraic sum of the moments in the clockwise direction is equal to the algebraic sum of the moments in the anticlockwise direction.

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Question 7.

State Newton’s second law.

Answer:

“The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force”.

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Question 8.

Why a spanner with a long handle is preferred to tighten screws in heavy vehicles?

Answer:

This is because turning effect to tighten the screws depends upon the perpendicular distance of the applied force from the axis of rotation is power arm. Larger the power armless is the force required to turn the screws. So spanner is provided with a long handle.

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Question 9.

While catching a cricket ball the fielder lowers his hands backwards. Why?

Answer:

(i) When the fielder lowers his hands backwards, he increases the value of time of collision and so retardation is decreased.
(ii) Hence retarding force becomes lesser than before and the palm of the fielder is not hurt very much.

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Question 10.

How does an astronaut float in a space shuttle?

Answer:

An astronaut float in a space shuttle because both are in the state of weightlessness. Both are experiencing equal acceleration towards earth as free fall bodies. Astronauts are not floating but falling freely. 

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 Solve the given problems.

Question 1.
Two bodies have a mass ratio of 3 : 4. The force applied to the bigger mass produces an acceleration of 12 ms-2. What could be the acceleration of the other body, if the same force acts on it?

Solution:

Mass ratio of the bodies = 3 : 4 and same force is (m1 : m2) acting on the body and a2 = 12 ms-2
∴ m1a1 = m2a2

m1m2=a2a1⇒34=a2a1

a1=43×12=16ms−2

Question 2.
A ball of mass 1 kg moving with a speed of 10 ms-1 rebounds after a perfectly elastic collision with the floor. Calculate the change in linear momentum of the ball.

Solution:

Mass of a ball = 1 kg
Velocity of the bail before collision,
u = 10 m/s
Velocity of the ball after collision,
v = – u
= -10 m/s
Change in momentum,
P = m(v – u)
P = 1(-10 – 10)
= -20 kg m/s.

Question 3.
A mechanic unscrews a nut by applying a force of 140 N with a spanner of length 40 cm. What should be the length of the spanner if a force of 40 N is applied to unscrew the same nut?

Solution:

Given F1 = 140 N, d1 = 40 cm, F2 = 40 N, d2 = ?
In, both the cases, moment of forces applied are equal
F1d1 = F2d2
d2=(F1F2)d1d2=40×14040=140cm

Question 4.

The ratio of masses of two planets is 2 : 3 and the ratio of their radii are 4 : 7 Find the ratio of their accelerations due to gravity.
Solution:

The ratio of masses of two planets m1 : m2 = 2 : 3
The ratio of radii of two planets R1 : R2 = 4 : 7
Formula:

 Answer in detail.

Question 1.

What are the types of inertia? Give an example for each type.

Answer:
Types of Inertia:
(i) Inertia of rest: The resistance of a body to change its state of rest is called inertia of rest. Eg: When you vigorously shake the branches of a tree, some of the leaves and fruits are detached and they fall down.

(ii) Inertia of motion: The resistance of a body to change its state of motion is called inertia of motion. Eg: An athlete runs some distance before jumping. Because, this will help him jump longer and higher.

(iii) Inertia of direction: The resistance of a body to change its direction of motion is called inertia of direction. Eg: When you make a sharp turn while driving a car, you tend to lean sideways.

Question 2.

State Newton’s laws of motion.
Answer:

Newton’s First Law: Everybody continues to be in its state of rest or the state of uniform motion along a straight line unless it is acted upon by some external force.

Newtons Second law: The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force.

Newtons Third Law: For every action, there is an equal and opposite reaction. They always act on two different bodies.

Question 3.
Deduce the equation of a force using Newton’s second law of motion.

Answer:
“The force acting on a body is directly proportional to the rate of change of linear momentum of the body and the change in momentum takes place in the direction of the force”.
Let, ‘m’ be the mass of a moving body, moving along a straight line with an initial speed ‘u’ After a time interval of ‘t’, the velocity of the body changes to ‘v’ due to the impact of an unbalanced external force F.
Initial momentum of the body (Pi) = mu
Final momentum of the body (Pf) = mv
Change in momentum ∆p = Pf – Pi = mv – mu
By Newton’s second law of motion,
Force, F ∝ rate of change of momentum
F ∝ change in momentum / time
F∝mv−mutF=km(v−u)t
Here, k is the proportionality constant, k = 1 in all systems of units.
Hence, F=m(v−u)t
Since, acceleration = change in velocity / time,
a = (v – u)/t.
Hence, we have F = m × a
Force = mass × acceleration

• No external force is required to maintain the motion of a body moving with uniform velocity.
• When the net force acting on a body is not equal to zero, then definitely the velocity of the body will change.
• Thus, change in momentum takes place in the direction of the force. The change may take place either in magnitude or in direction or in both.

Question 4.
State and prove the law of conservation of linear momentum.

Answer:
(i) There is no change in the linear momentum of a system of bodies as long as no net external force acts on them.
(ii) Let us prove the law of conservation of linear momentum with the following illustration:

(iii) Let two bodies A and B having masses m1 and m2 move with initial velocity u1 and u2 in a straight line.
(iv) Let the velocity of the first body be higher than that of the second body. i.e., u1 > u2.
(v) During an interval of time t second, they tend to have a collision. After the impact, both of them move along the same straight line with a velocity v1 and v2 respectively.
Force on body B due to A,
FB=m2[v2−u2]t
Force on body A due to B,
FA=m1[v1−u1]t
By Newton’s III law of motion, Action force = Reaction force
FA = -FB
=m2[v2−u2]t
The above equation confirms in the absence of an external force, the algebraic sum of the momentum after collision is numerically equal to the algebraic sum of the momentum before collision.
Hence the law of conservation linear momentum is proved.

Question 5.
Describe rocket propulsion.

Answer:
Propulsion of rockets is based on the law of conservation of linear momentum as well as Newton’s III law of motion. Rockets are filled with a fuel (either liquid or solid) in the propellant tank. When the rocket is fired, this fuel is burnt and a hot gas is ejected with a high speed from the nozzle of the rocket, producing a huge momentum. To balance this momentum, an equal and opposite reaction force is produced in the combustion chamber, which makes the rocket project forward.

While in motion, the mass of the rocket gradually decreases, until the fuel is completely burnt out. Since, there is no net external force acting on it, the linear momentum of the system is conserved. The mass of the rocket decreases with altitude, which results in the gradual increase in velocity of the rocket. At one stage, it reaches a velocity, which is sufficient to just escape from the gravitational pull of the Earth. This velocity is called escape velocity.

Question 6.
State the universal law of gravitation and derive its mathematical expression.

Answer:
This law states that every particle of matter in this universe attracts every other particle with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between the centres of these masses. The direction of the force acts along the line joining the masses.

The force between the masses is always attractive and it does not depend on the medium where they are placed.
Let, m1 and m2 be the masses of two bodies A and B placed r metre apart in space.
Force F ∝ m1 × m2
F ∝ 1/r2
On combining the above two expressions
F∝m1×m2r2F=Gm1m2r2
Where G is the universal gravitational constant.
Its value in SI unit is 6.674 × 10-11 Nm2 kg-2.

Question 7.
Give the applications of the universal law of gravitation.

Answer:
Application of Newton’s law of gravitation are:
(i) Dimensions of the heavenly bodies can be measured using the gravitation law. Mass of the Earth, radius of the Earth, acceleration due to gravity, etc., can be calculated with a higher accuracy.
(ii) Helps in discovering new stars and planets.
(Hi) Helps to explain germination of roots is due to the property of geotropism which is the property of a root responding to the gravity.
(iv) One of the irregularities in the motion of stars is called ‘Wobble’ lead to the disturbance in the motion of a planet nearby. In this condition the mass of the star can be calculated using the law of gravitation.
(v) Helps to predict the path of the astronomical bodies.

IX. HOT Questions.

Question 1.

Two blocks of masses 8 kg and 2 kg respectively, lie on a smooth horizontal surface in contact with one other. They are pushed by a horizontally applied force of 15 N. Calculate the force exerted on the 2 kg mass.
Solution:

Given: m1 = 8 kg, m2 = 2 kg, F = 15 N
F = mtotal, F = (m1 + m2) a = (8 + 2) a = 10 a
15 = 10 a
⇒ a = 1510=32 ms-2
Force exerted by mass of 8 kg
F = m1 a = 8×32 = 12 N.

Question 2.
A heavy truck and bike are moving with the same kinetic energy. If the mass of the truck is four times that of the bike, then calculate the ratio of their momenta. (Ratio of moment = 1 : 2)

Solution:
Given: Let m1, m2 are the masses of truck and bike.
m1 = 4m2
Here kinetic energies of both truck and bike are same
=m2v22
Ratio of momenta: p1p2=m1v1m2v2=4m2m2⋅v12v1 = 2
P1 : P2 = 2 : 1.

Question 3.
“Wearing a helmet and fastening the seat belt is highly recommended for the safe journey”.Justify your answer using Newton’s laws of motion.

Answer:
During the motion of car and two wheelers, when the brakes are applied, the vehicles slow down but our body tends to continue in the same state of motion due to inertia. So this may cause injury to passengers. Hence they are advised to wear a helmet and seat belt.

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