10th Science Book Back Answer Unit 1 Laws of Motion
10th Standard Science Physics Laws of Motion Book Back Answers
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10th Standard Science Book Solution  Lesson 1 Laws of Motion
Important Formulae
Important Formulae  
Force : F  = m × a 
Linear Momentum : p  = m × v 
Torque : τ  = F × d 
Change in momentum : Δ𝑝  = 𝑃𝑓 − 𝑃𝑖 = 𝑚𝑣 − 𝑚𝑢 
Momentum of a couple : M  = F × S 
Impulse : J  = F × t = Δ𝑝 
Gravitational Force : F  = 𝐺𝑚^{1}𝑚^{2}/𝑟^{2} 
Acceleration due to gravity : g  = 𝐺𝑀/𝑅^{2} 
Weight : W  = m × g 
Kinetic Energy : 𝐸𝑘  = 1/2 𝑚𝑣^{2} = 𝑝^{2}/2𝑚 
Important Values to remember 
 Acceleration due to gravity on the surface of the Earth
on the surface of the Moon on the surface of the Moon Radius of Earth (R) Mass of Earth (M) Gravitational constant (G) 1N 1 kg f 1 g f 
= 9.8 m𝑠^{−2} = 1.625 ms^{2} = 6378km ≅ 6400km = 5.972 × 1024kg = 6.674×10−11 Nm2 kg2 = 1 kg m 𝑠−1 = 105 dyne = 9.8 N = 98 × 104 dyne = 9.8 × 10−3N= 980 dyne 
Important Principle
At equilibrium, the algebraic sum of the moments of all the individual forces about any point is equal to zero.
TEXTBOOK EVALUATION
I. Choose the correct answer
1) Inertia of a body depends on
 weight of the object
 acceleration due to gravity of the planet
 mass of the object
 Both a & b
Ans ; mass of the object
2) Impulse is equals to
 rate of change of momentum
 rate of force and time
 change of momentum
 rate of change of mass
Ans ; change of momentum
3) Newton’s III law is applicable
 for a body is at rest
 for a body in motion
 both a & b
 only for bodies with equal masses
Ans ; both a & b
4) Plotting a graph for momentum on the Yaxis and time on Xaxis. slope of momentumtime graph gives
 Impulsive force
 Acceleration
 Force
 Rate of force
Ans ; Force
5) In which of the following sport the turning of effect of force used
 swimming
 tennis
 cycling
 hockey
Ans ; cycling
6) The unit of ‘g’ is m s2. It can be also expressed as
 cms1
 Nkg1
 Nm2 kg1
 cm2 s2
Ans ; Nkg1
7) One kilogram force equals to
 8 dyne
 8 × 104 N
 98 × 104 dyne
 980 dyne
Ans ; 98 × 104 dyne
8) The mass of a body is measured on planet Earth as M kg. When it is taken to a planet of radius half that of the Earth then its value will be____kg
 4 M
 2M
 M/4
 M
Ans ; M/4
9) If the Earth shrinks to 50% of its real radius its mass remaining the same, the weight of a body on the Earth will
 decrease by 50%
 increase by 50%
 decrease by 25%
 increase by 300%
Ans ; decrease by 25%
10) To project the rockets which of the following principle(s) is /(are) required?
 Newton’s third law of motion
 Newton’s law of gravitation
 law of conservation of linear momentum
 both a and c
Ans ; both a and c
10th Science Book Back Answer Unit 1 Laws of Motion
II. Fill in the blanks
 To produce a displacement __________ is required. Ans ; Force
 Passengers lean forward when sudden brake is applied in a moving vehicle. This can be explained by __________ Ans ; Inertia of motion
 By convention, the clockwise moments are taken as __________ and the anticlockwise moments are taken as __________. Ans ; Negative, Positive
 __________ is used to change the speed of car. Ans ; Gears
 A man of mass 100 kg has a weight of __________ at the surface of the Earth. Ans ; 980 N
III. State whether the following statements are true or false. Correct the statement if it is false
 The linear momentum of a system of particles is always conserved. ( False )
 The linear momentum of a system of particles is always conserved only if no external force is applied.
 Apparent weight of a person is always equal to his actual weight. ( False )
 Apparent weight and actual weight is not equal during upward or downward motion.
 Weight of a body is greater at the equator and less at the polar region. ( False )
 Weight of a body is lesser at the equator and more at the polar region as g α 1/R2.
 Turning a nut with a spanner having a short handle is so easy than one with a long handle. ( False )
 Moment of force in longer handle is easy than one with a short handle.
 There is no gravity in the orbiting space station around the Earth. So the astronauts feel weightlesness. ( False )
 Apparent weight is zero. They are in the state of weightlessness.
10th Science Book Back Answer Unit 1 Laws of Motion
IV. Match the following
Column I Column II
 Newton’s I law – 1. propulsion of a rocket
 Newton’s II law – 2. Stable equilibrium of a body
 Newton’s III law – 3. Law of force
 Law of conservation of Linear momentum – 4. Flying nature of bird
Ans ; A – 2, B – 3, C – 4, D – 1
V. Assertion & Reasoning
 Mark the correct choice as
 If both the assertion and the reason are true and the reason is the correct explanation of assertion.
 If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
 Assertion is true, but the reason is false.
 Assertion is false, but the reason is true.
1. Assertion: The sum of the clockwise moments is equal to the sum of the anticlockwise momen.
Reason: The principle of conservation of momentum is valid if the external force on the system is zero.
Ans : (b) both the assertion and the reason are true, but the reason is not the correct explanation of the assertion
 Assertion: The value of ‘g’ decreases as height and depth increases from the surface of the Earth.
Reason: ‘g’ depends on the mass of the object and the Earth.
Ans : (c) Assertion is true, but the reason is false
VI. Answer briefly.
1. Define inertia. Give its classification.
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.
2. Classify the types of force based on their application.
The 2 types of forces are,
 Like parallel forces.
 Unlike parallel forces.
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.
Resultant Force = F2 – F1
= 15N – 5N
= 10N.
4. Differentiate mass and weight.
Mass
 Fundamental quantity
 It is the amount of matter contained in a body
 It’s unit is kilogram
 Remains the same
 It is measured using physical balance
Weight
 Derived quantity
 It is the gravitational pull acting on the body
 It is measured in newton
 Varies from place to place
 It is measured using a spring balance
5. Define moment of a couple.
The line of action of the two forces does not coincide. It does not produce any translatory motion since the resultant is zero. But a couple results in causes the rotation of the body. Rotating effect of a couple is known as moment of a couple.
Moment of couple = F × S
M = F × S. (S I Unit is Nm)
6. State the principle of moments.
When a number of like or unlike parallel forces act on a rigid body and the body is in equilibrium, then the algeberic sum of the moments in the clockwise direction is equal to the algebraic sum of the moments in the anticlockwise direction.
7. State Newton’s 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.
8. Why a spanner with a long handle is preferred to tighten screws in heavy vehicles?
The turning effect of a force is called moment of force.
Moment of Force = Force × Perpendicular distance = F × d
For the spanner with a long handle, ‘d’ is large. Therefore the moment of force is also large and hence it is easier to rotate the object (nut).
9. While catching a cricket ball the fielder lowers his hands backwards. Why?
In cricket, a fielder pulls back his hands while catching the ball. He experiences a smaller force for a longer interval of time to catch the ball, resulting in a lesser impulse on his hands.
10. How does an astronaut float in a space shuttle?
On the astronaut there is no external force on him due to planet or space ship. By the first law of motion the acceleration on him is zero. So he floats.
10th Science Book Back Answer Unit 1 Laws of Motion
VII. Solve the given problems
 Two bodies have a mass ratio of 3:4 The force applied on the bigger mass produces an acceleration of 12 ms2. What could be the acceleration of the other body, if the same force acts on it.
Let mass of the body A = 3 m
Mass of the body B = 4 m
Force applied = F
For body A
F = mass × acceleration
F = 3m × 12 ms–1 = 36 N.
For body B
F = mass × acceleration
Acceleration = Force / Mass
= 36N/4m = 9 ms2.
2. A ball of mass 1 kg moving with a speed of 10 ms1 rebounds after a perfect elastic collision with the floor. Calculate the change in linear momentum of the ball.
Mass of the ball  = 1 kg 
Initial speed  = 10 ms^{–1} 
Final speed  = –10ms^{–1} (rebounds) 
Change in momentum
ΔP ΔP
ΔP 
= mV – mu
= 1 × (–10) – 1 × 10 = –10 – 10 –20 kg m/s. 
3. A mechanic unscrew 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?
Equating the torque in both the cases.
F_{1} L_{1} 140N × 40cm L_{2} L_{2} 
= F_{2} L_{2}
= L_{2} × 40N = 140N 40 cm / 40N = 140 cm. = 1.4 m. 
4. The ratio of masses of two planets is 2:3 and the ratio of their radii is 4:7 Find the ratio of their accelerations due to gravity.
10th Science Book Back Answer Unit 1 Laws of Motion
VIII. Answer in detail
1. What are the types of inertia? Give an example for each type.
There are 3 types of Inertia. They are;
 Inertia at rest:
The resistance of a body to change its state of rest is called inertia of rest.
Example :
 When you vigorously shake the branches of a tree, some of the leaves and fruits are detached and they fall down (Inertia of rest).
 Inertia of motion:
The resistance of a body to change its state of motion is called inertia of motion.
Example :
 An athlete runs some distance before jumping because this will help him jump longer and higher.
 Inertia of direction :
The resistance of a body to change its direction of motion is called inertia of direction.
Example :
 When a bus turn towards right, the passangers are thrown towards left.
2. State Newton’s laws of motion?
Newton’s First Law :
This law states that 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.
Newton’s Second Law :
According to this 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.
Newton’s Third Law :
Newton’s third law states that for every action, there is an equal and opposite reaction. They always act on two different bodies.
3. Deduce the equation of a force using Newton’s second law of motion.
This law helps us to measure the amount of force. So it is also called as “law of 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 – m 
By Newton’s second law of motion, Force, F α rate of change of momentum.
F α change in momentum / time.
F α mv – mu / 4
F = K (m (vu)/t)
Here K is the proportionality constant. K=1 in all systems of units. Hence,
F since acceleration a Hence we have F Force 
= m (vu)/t
= change in velocity / time = vu/t = m × a = 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.
4. State and prove the law of conservation of linear momentum.
There is no change in the linear momentum of a system of bodies as long as no net external force acts on them.
Let us prove the law of conservation of linear momentum with the following illustration
Proof: Let two bodies A and B having masses m1 and m2 move with initial velocity u1 and u2 in a straight line. Let the velocity of the first body be higher than that of the second body. i.e., u1 > u2. During an interval of time ‘t’ second, they tend to have a colliusion. 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 
F_{A}  = F_{B} 
m_{1}(v_{1}–u_{1}) / t  = m_{2}(v_{2}–u_{2}) / t 
m_{1}v_{1 }+ m_{1}v_{1}  = m_{1}u_{1 }+ m_{1}u_{1} 
The above equation confirms in the absence of an external force, the algebraic sum of momentum after collision is numerically equal to the algebraic sum of the momentum before collision.
Hence the law of conservation of linear momentum is proved.
5. Describe rocket propulsion.
 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 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.
6. State the universal law of gravitation and derive its mathematical expression.
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 centers 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 α m^{1} × m^{2}
F α 1/r^{2}
On combining the above two expressions
F α G m_{1} m_{2}/r^{2}
F = G m1 m_{2}/r^{2}
Where G is the universal gravitational constant. Its value in SI unit is 6.674 × 10^{11} N m^{2} kg^{2}.
7. Give the applications of universal law gravitation.
 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.
 It helps in discovering new stars and planets.
 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.
 Helps to explain germination of roots is due to the property of geotropism which is the property of a root responding to the gravity.
 Helps to predict the path of the astronomical bodies.
IX. HOT Questions
 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: If 2 blocks are of mass 8 kg and 2 kg
Date  
m_{2} m_{2} m_{T}

= 8 kg
= 2 kg = Total mass = 8+2 = 10kg 
F 15 15 a 
= mT x a
= (8 + 2) x a = 10 x a = 15N/10kg = 3/2 ms1 
Force exerted by block 2 (2 kg)
So F Exerted Force 
= m x a
= 2 x 3/2 = 3N 
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 momenta = 1:2)
Solution : According to kinetic energy,
Date  
m_{1} m_{2} v_{1} v_{2} 
= Truck mass
= Bike mass = Truck Velocity = Bike Velocity 
Given K.E. are equal
m_{1} = 4m_{2}
ratio of momentum = ?
Since K.E are equal
1/2 m_{1}v_{1}^{2} m_{1/}m_{2} 4m_{1/}m_{2} v_{2}/ v_{1} Ratio of momentum Ratio of momentum 
= 1/2 m_{2}v_{2}^{2}
= v_{2}^{2 }/ v_{1}^{2} = v_{2}^{2 }/ v_{1}^{2} = 2 = m_{1}v_{1}/m_{2}v_{2} = 4m_{1}v_{1}/m_{2}v_{2} = 4 x 1/2 = 2/1 = 2:1 
3. “Wearing helmet and fastening the seat belt is highly recommended for safe journey” Justify your answer using Newton’s laws of motion.
Explanation :
 Wearing a helmet is strongly recommended for safe journy, because when a person fall from bike he exerts a force equal to product of mass of the persion and acceleration of the bike (Newton’s II law). According to Newton’s III law, inturn the ground offers and equal and opposite force on the persion, which will porduce large damage. In order to mimnse damages the persion must wear helmet.
 Fastening of seat belt will not allow a persion to meve from seat why the vechicle comes to rest suddengly by applying brake or by having some accidents.
 This is deu to inertia of motion. (Newton’s I law). When the speeding vechicle stops suddently the lower part in contact with the seat stops while the upperr patr of the body tends to maintain its uniform motion. Hence the persion will trun forward and obtain injuries. Inorder to avoid this, fastening of seat belt is important