Book Back Questions with Answers - Laws of Motion
I. Choose the correct answer
1) Inertia of a body depends on
- a) weight of the object
- b) acceleration due to gravity of the planet
- c) mass of the object
- d) Both a & b
2) Impulse is equals to
- a) rate of change of momentum
- b) rate of force and time
- c) change of momentum
- d) rate of change of mass
3) Newton’s III law is applicable
- a) for a body is at rest
- b) for a body in motion
- c) both a & b
- d) only for bodies with equal masses
4) Plotting a graph for momentum on the X-axis and time on Y-axis. slope of momentum-time graph gives
- a) Impulsive force
- b) Acceleration
- c) Force
- d) Rate of force
5) In which of the following sport the turning of effect of force used
- a) swimming
- b) tennis
- c) cycling
- d) hockey
6) The unit of ‘g’ is $m s^{-2}$. It can be also expressed as
- a) $cm s^{-1}$
- b) $N kg^{-1}$
- c) $N m^2 kg^{-1}$
- d) $cm^2 s^{-2}$
7) One kilogram force equals to
- a) 9.8 dyne
- b) $9.8 \times 10^4 N$
- c) $98 \times 10^4$ dyne
- d) 980 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
- a) 4 M
- b) 2M
- c) M/4
- d) M
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
- a) decrease by 50%
- b) increase by 50%
- c) decrease by 25%
- d) increase by 300%
10) To project the rockets which of the following principle(s) is/are required?
- a) Newton’s third law of motion
- b) Newton’s law of gravitation
- c) law of conservation of linear momentum
- d) both a and c
II. Fill in the blanks
- To produce a displacement force is required.
- Passengers lean forward when sudden brake is applied in a moving vehicle. This can be explained by inertia of motion.
- By convention, the clockwise moments are taken as negative and the anticlockwise moments are taken as positive.
- Gear is used to change the speed of car.
- A man of mass 100 kg has a weight of 980 N at the surface of the Earth.
III. State whether the following statements are true or false. Correct the statement if it is false:
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The linear momentum of a system of particles is always conserved.
False.
In the absence of external force, the linear momentum of a system of particle is always conserved. -
Apparent weight of a person is always equal to his actual weight.
False.
Both apparent weight and actual weight can be greater or lesser according to the movement of the person inside the lift. -
Weight of a body is greater at the equator and less at the polar region.
False.
Weight of the body is less at equator, more at polar region. -
Turning a nut with a spanner having a short handle is so easy than one with a long handle.
False.
Turning effect (i.e torque, $\tau$) depends on perpendicular distance of the line of action of the applied force $\tau = F \times d$. -
There is no gravity in the orbiting space station around the Earth. So the astronauts feel weightlessness.
False.
When space station and astronauts have equal acceleration, they are under free fall condition, so both astronaut and space station are in the state of weightlessness.
IV. Match the following
| Column I | Column II |
|---|---|
| a. Newton’s I law | Propulsion of a rocket |
| b. Newton’s II law | Stable equilibrium of a body |
| c. Newton’s III law | Law of force |
| d. Law of conservation of Linear momentum | Flying nature of bird |
Answer:
| Column I | Column II |
|---|---|
| a. Newton’s I law | Stable equilibrium of a body |
| b. Newton’s II law | Law of force |
| c. Newton’s III law | Flying nature of bird |
| d. Law of conservation of Linear momentum | Propulsion of a rocket |
V. Assertion & Reasoning
Mark the correct choice as:
- a) If both the assertion and the reason are true and the reason is the correct explanation of assertion.
- b) If both the assertion and the reason are true, but the reason is not the correct explanation of the assertion.
- c) Assertion is true, but the reason is false.
- d) Assertion is false, but the reason is true.
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Assertion: The sum of the clockwise moments is equal to the sum of the anticlockwise moments.
Reason: The principle of conservation of momentum is valid if the external force on the system is zero.
Answer: b) If 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.
Answer: c) Assertion is true, but the reason is false.
VI. Answer briefly
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1. Define inertia. Give its classification.
The inherent property of the objects 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".
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2. Classify the types of force based on their application.
There are two types of force:
(i) Contact force: A force which is applied by means of direct physical contact between two objects is known as "contact force". Examples: pushing or pulling an object, muscular force, frictional force, compressing a spring, kicking a football, etc.
(ii) Non-contact force: A force which is applied without any physical contact between the objects is known as "non-contact force". Examples: gravitational force, magnetic force, electromagnetic force, etc.
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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.
Let the forces be $F_1 = 5 \text{ N}$ and $F_2 = 15 \text{ N}$.
Since they act in opposite directions, we take one as negative.
Resultant Force, $R = F_1 + (- F_2$)
$R = 5 - 15 = -10 \text{ N}$
The resultant force is -10 N, and the negative sign indicates it acts in the direction of the larger force (15 N).
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4. Differentiate mass and weight.
Mass Weight 1. It is the quantity of matter contained in the body. 1. It is the gravitational force exerted on a body due to the earth's gravity alone. 2. Mass is a scalar quantity. 2. Weight is a vector quantity. 3. Its unit is kg (kilogram). 3. Its unit is N (newton). 4. Mass of a body remains the same at any point on the earth. 4. Weight of a body varies from one place to another place on the earth. 5. Mass can be measured using a physical balance. 5. Weight can be measured using a spring balance. -
5. Define moment of a couple.
Couple: Two equal and unlike parallel forces applied simultaneously at two distinct points constitute 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 causing the rotation of the body. Rotating effect of a couple is known as moment of a couple.
Examples: Turning a tap, winding or unwinding a screw, spinning of a top, etc.
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6. State the principle of moments.
Principle of moments states that 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 moments in clockwise direction is equals to the algebraic sum of moments in anticlockwise direction.
Moment in clockwise direction = Moment in anticlockwise direction
$$ F_1 \times d_1 = F_2 \times d_2 $$ -
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 force.
$$ F = m \times a $$Force = mass x acceleration
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8. Why a spanner with a long handle is preferred to tighten screws in heavy vehicles?
The turning effect of a body depends upon the distance of the line of action of the applied force from the axis of rotation. Larger the perpendicular distance, lesser is the force required to turn the body. So spanner with long handle is preferred.
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9. While catching a cricket ball the fielder lowers his hands backwards. Why?
When the fielder pulls back his hands he experiences a smaller force for a longer interval of time leading to less damage to his hands.
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10. How does an astronaut float in a space shuttle?
(i) Both the astronauts and the space craft experience the same gravitational force of the earth.
(ii) There is no weight of the astronauts exerted on the surface of the space craft.
(iii) In turn, there is no reaction force exerted by the surface of the space craft on the astronauts ($R = 0$). So they float.
VII. Solve the given problems
1. Two bodies have a mass ratio of 3:4. The force applied on 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.
Given:
Mass ratio is 3:4. Let's assume:
Mass of smaller body, $m_1 = 3 \text{ kg}$
Mass of bigger body, $m_2 = 4 \text{ kg}$
Acceleration of bigger body, $a_2 = 12 \text{ ms}^{-2}$
To Find: Acceleration of the smaller body, $a_1$.
Solution:
According to Newton's second law, $F = m \times a$.
Force on bigger body: $$F_2 = m_2 \times a_2 = 4 \times 12 = 48 \text{ N}$$
Since the same force acts on the smaller body, $F_1 = F_2 = 48 \text{ N}$.
$$F_1 = m_1 \times a_1$$
$$48 = 3 \times a_1$$
$$a_1 = \frac{48}{3} = 16 \text{ ms}^{-2}$$
The acceleration of the smaller body is $16 \text{ ms}^{-2}$.
2. A ball of mass 1 kg moving with a speed of 10 ms-1 rebounds after a perfect elastic collision with the floor. Calculate the change in linear momentum of the ball.
Given:
Mass, $m = 1 \text{ kg}$
Initial velocity, $u = 10 \text{ ms}^{-1}$
Final velocity, $v = -10 \text{ ms}^{-1}$ (rebounds in the opposite direction)
To Find: Change in linear momentum, $\Delta p$.
Solution:
Momentum before collision: $p_{\text{initial}} = mu = (1 \times 10) = 10 \text{ kg ms}^{-1}$
Momentum after collision: $p_{\text{final}} = mv = (1 \times -10) = -10 \text{ kg ms}^{-1}$
Change in momentum, $\Delta p = p_{\text{final}} - p_{\text{initial}}$
$$\Delta p = mv - mu = -10 - 10 = -20 \text{ kg ms}^{-1}$$
The change in linear momentum is $-20 \text{ kg ms}^{-1}$.
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?
Given:
Force $F_1 = 140 \text{ N}$ with length $L_1 = 40 \text{ cm} = 0.4 \text{ m}$
Force $F_2 = 40 \text{ N}$
To Find: Required length, $L_2$.
Solution:
By the principle of moments, the torque required is the same in both cases.
$$ \text{Torque}_1 = \text{Torque}_2 $$
$$ F_1 \times L_1 = F_2 \times L_2 $$
$$ 140 \times 0.4 = 40 \times L_2 $$
$$ 56 = 40 \times L_2 $$
$$ L_2 = \frac{56}{40} = 1.4 \text{ m} $$
The length of the spanner should be $1.4 \text{ m}$.
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.
Given:
Ratio of masses, $m_1:m_2 = 2:3$
Ratio of radii, $R_1:R_2 = 4:7$
To Find: Ratio of acceleration due to gravity, $g_1:g_2$.
Solution:
The formula for acceleration due to gravity is $g = \frac{GM}{R^2}$.
The ratio $\frac{g_1}{g_2}$ is:
$$ \frac{g_1}{g_2} = \frac{\frac{GM_1}{R_1^2}}{\frac{GM_2}{R_2^2}} = \left(\frac{M_1}{M_2}\right) \times \left(\frac{R_2^2}{R_1^2}\right) = \left(\frac{M_1}{M_2}\right) \times \left(\frac{R_2}{R_1}\right)^2 $$Substitute the given ratios:
$$ \frac{g_1}{g_2} = \left(\frac{2}{3}\right) \times \left(\frac{7}{4}\right)^2 $$ $$ \frac{g_1}{g_2} = \frac{2}{3} \times \frac{49}{16} = \frac{98}{48} = \frac{49}{24} $$The ratio of their accelerations due to gravity, $g_1:g_2$, is 49:24.
VIII. Answer in detail.
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1. What are the types of inertia? Give an example for each type.
(i) Inertia of rest: The resistance of a body to change its state of rest is called inertia of rest.
Ex: When you vigorously shake the branches of a tree some of the leaves and fruit 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.
Ex: 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.
Ex: When you make a sharp turn while driving a car, you tend to lean sideways.
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2. State Newton’s laws of motion?
(i) Newton's First law: Every object 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.
(ii) Newton's second law of motion: The force acting on an object is directly proportional to the rate of change of linear momentum of the object and the change in momentum takes place in the direction of force. This law helps us to measure the amount of force. So it is called as "law of force".
(iii) Newton's third law of motion: Newton's third law states that "for every action there is an equal and opposite reaction. They always act on two different bodies".
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3. Deduce the equation of a force using Newton’s second law of motion.
(i) Let '$m$' be the mass of a moving body, moving along a straight line with an initial speed '$u$'.
(ii) After a time interval of '$t$', the velocity of the body changes to '$v$' due to the impact of unbalanced external force F.
(iii) Initial momentum of the body, $P_i = mu$
(iv) Final momentum of the body, $P_f = mv$
(v) Change in momentum, $\Delta p = P_f - P_i = mv - mu$
By Newton's second law of motion, Force is proportional to the rate of change of momentum:
$$ F \propto \frac{\Delta p}{t} \implies F \propto \frac{mv - mu}{t} $$ $$ F = k \frac{m(v - u)}{t} $$Here, k is the proportionality constant, and its value is 1 in all systems of units. Hence,
$$ F = \frac{m(v - u)}{t} $$Since acceleration $a = \frac{v - u}{t}$, we substitute this into the equation:
$$ F = m \times a $$Force = mass × acceleration
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4. State and prove the law of conservation of linear momentum.
The principle of conservation of linear momentum states that "There is no change in the linear momentum of a system of bodies as long as no net external force acts on them".
Conservation of linear momentum Proof:
Let two bodies A and B having masses $m_1$ and $m_2$ move with initial velocity $u_1$ and $u_2$ in a straight line. Let $u_1 > u_2$. During a time interval $t$, they collide. After impact, they move with velocities $v_1$ and $v_2$ respectively.
Force on body B due to A: $$F_B = \frac{m_2(v_2 - u_2)}{t}$$
Force on body A due to B: $$F_A = \frac{m_1(v_1 - u_1)}{t}$$
By Newton’s III law, Action force = Reaction force:
$$ F_A = -F_B $$ $$ \frac{m_1(v_1 - u_1)}{t} = - \frac{m_2(v_2 - u_2)}{t} $$ $$ m_1v_1 - m_1u_1 = - (m_2v_2 - m_2u_2) $$ $$ m_1v_1 - m_1u_1 = -m_2v_2 + m_2u_2 $$ $$ m_1v_1 + m_2v_2 = m_1u_1 + m_2u_2 $$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 of linear momentum is proved.
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5. Describe rocket propulsion.
(i) Propulsion of rockets is based on law of conservation of linear momentum as well as Newton's III law of motion.
(ii) Rockets are filled with a fuel (either liquid or solid) in the propeller.
(iii) When the rocket is fired, this fuel is burnt and a hot gas is ejected with high speed from the back nozzle producing a huge momentum.
(iv) To balance this momentum, an equal and opposite reaction force is produced which makes the rocket project forward.
(v) While in motion, 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.
(vi) The mass of the rocket decreases with altitude that results gradual increase in velocity of the rocket.
(vii) 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.
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6. State the universal law of gravitation and derive its mathematical expression.
Newton’s Universal law of gravitation: 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 centers of these masses. The direction of the force acts along the line joining the masses.
Derivation:
Let $m_1$ and $m_2$ be the masses of two bodies A and B placed at a distance $r$ apart in space.
Gravitational force between two masses From the law, the force $F$ is:
$$ F \propto m_1 \times m_2 $$ $$ F \propto \frac{1}{r^2} $$Combining the above two expressions:
$$ F \propto \frac{m_1 m_2}{r^2} $$ $$ F = G \frac{m_1 m_2}{r^2} $$Where G is the universal gravitational constant. Its value in SI units is $6.674 \times 10^{-11} \text{ Nm}^2\text{kg}^{-2}$.
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7. Give the applications of universal law gravitation.
Application of Newton’s law of gravitation
(i) Dimensions of the heavenly objects can be measured using 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. Mass of the double stars can be calculated.
(iii) One of the irregularities in the motion of stars is called "Wobble" which leads to the disturbance in the motion of planet nearby. In this condition mass of the star can be calculated using law of gravitation.
(iv) Helps to explain germination of roots due to the property of geotropism, which is the property of root responding to the gravity.
(v) Helps to predict the path of the astronomical bodies.
IX. HOT Questions
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.
Given:
Mass of block 1, $m_1 = 8 \text{ kg}$
Mass of block 2, $m_2 = 2 \text{ kg}$
Total mass, $m = m_1 + m_2 = 8 + 2 = 10 \text{ kg}$
Applied Force, $F = 15 \text{ N}$
To Find: Force exerted on the 2 kg mass, $F_2$.
Solution:
First, find the acceleration of the combined system:
$$ F = m \times a \implies 15 = 10 \times a $$
$$ a = \frac{15}{10} = 1.5 \text{ ms}^{-2} $$
The force exerted on the 2 kg mass is the force required to accelerate it:
$$ F_2 = m_2 \times a = 2 \times 1.5 = 3 \text{ N} $$
The force exerted on the 2 kg mass is 3 N.
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)
Given:
Kinetic Energy of truck = Kinetic Energy of bike ($KE_t = KE_b$)
Mass of truck, $m_t = 4 \times$ Mass of bike, $m_b$
To Find: Ratio of their momenta, $p_t : p_b$.
Solution:
The relationship between Kinetic Energy (KE) and momentum (p) is $KE = \frac{p^2}{2m}$, which means $p = \sqrt{2m \cdot KE}$.
Ratio of momenta:
$$ \frac{p_t}{p_b} = \frac{\sqrt{2m_t \cdot KE_t}}{\sqrt{2m_b \cdot KE_b}} $$Since $KE_t = KE_b$, they cancel out:
$$ \frac{p_t}{p_b} = \sqrt{\frac{2m_t}{2m_b}} = \sqrt{\frac{m_t}{m_b}} $$Substitute $m_t = 4m_b$:
$$ \frac{p_t}{p_b} = \sqrt{\frac{4m_b}{m_b}} = \sqrt{4} = 2 $$So, $\frac{p_t}{p_b} = \frac{2}{1}$. This means the ratio of momenta (truck to bike) is 2:1. The question asks for the ratio of bike to truck, which would be 1:2.
3. “Wearing helmet and fastening the seat belt is highly recommended for safe journey” Justify your answer using Newton’s laws of motion.
(i) According to Newton’s second law, when you fall from a bike on the ground with a force equal to your mass and acceleration of the bike and according to Newton’s third law, an equal and opposite reacting force on the ground is exerted on your body or head. When you do not wear a helmet, this reacting force can cause fatal head injuries. So it is important to wear helmet for a safe journey.
(ii) Consider for instance of an unfortunate collision of a car with another car (or any obstacles in its path). Upon contact with an obstacle, an unbalanced force acts upon the car to abruptly decelerate it to rest. The passengers in the car would also be decelerated to rest if strapped to the seats by seat belts.
(iii) If not strapped to the seats, they no longer share the same state of rest as the car. According to Newton’s first law, the passenger in the car are more likely to maintain the same state of motion, which will result in banging the glass (wind shield) or being thrown forward by breaking the glass windshield. In case of the driver, he might bang himself on the steering wheel. So wearing seatbelts is highly recommended for a safe journey.
