[2 Mark Question Answer]

Question 12 Marks

What is the (a) force of gravity and (b) weight of a block of mass 10.5 kg ? Take $g =10 ms^{-2}$.

Answer

Mass, $m = 10.5 kg$
$G = 10 m/s^2$
(a) Force, $F = mg$
$F = (10.5) (10) = 105 N$
(b) Weight, $w = mg$
$w = (10.5) (10) = 105 N$

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Question 22 Marks

What is the importance of law of gravitation ?

Answer

The gravitational force of attraction is significant to explain the motion of heavenly bodies, e.g. motion of planets around the Sun, motion of the Moon around the Earth etc.

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Question 32 Marks

A man weighs $600 N$ on the Earth. What would be his approximate weight on the Moon? Give a reason for your answer?

Answer

Man's weight on Earth $= 600 N$
Man's weight on the Moon $= (1/6)$ man's weight on Earth;
Because the acceleration due to gravity on the Moon is $1/6^{th}$ that of Earth and w = mg.
Therefore, man's weight on Moon $= (600/6) = 100 N.$

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Question 42 Marks

The weight of a body on Earth is $98 N$ , where acceleration due to gravity is $9.8 m s ^{-2}$. What will be its $(a)$ mass and $(b)$ weight on the Moon, where acceleration due to gravity is $1.6 m s ^{-2} ?$

Answer

Weight of the body on Earth $=98 N$.
Acceleration due to gravity on Earth $=9.8 m / s ^2$.
Let ' $m$ ' be the mass of the body on Earth.
$m =W / g$
$m =(98 / 9.8)=10 kg$
Thus, the mass of the body is 10 kg , which always remains constant.
(a) Mass on moon = mass on Earth $=10 kg$
(b) Let weight on moon is $W ^{\prime}$.
$W ^{\prime}=$ mass $\times$ acceleration due to gravity on the Moon.
[Given, acceleration due to gravity on the Moon $=1.6 m / s ^2$ ]
$W' = 10 \times 1.6 =16 N.$

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Question 52 Marks

The weight of a body is 2.0 N . What is the mass of the body? $\left( g =10 m s ^{-2}\right)$

Answer

Weight, $W = 2.0 N$
$g = 9.8 m/s^2$
Let 'm' be the mass of the body.
$W = mg$
Or,$ m = W/g = (2/9.8) kg = 0.2 kg.$

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Question 62 Marks

State the magnitude and direction of the force of gravity acting on the body of mass $5 \ kg$. Take $g = 9.8 m s^{-2}.$

Answer

Mass $= 5 kg.$
$g = 9.8 m/s^2.$
Let $F$ be the force of gravity,
$F = mg.$
$F = (5) (9.8) = 49 N.$
Force of gravity always acts downwards.

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Question 72 Marks

Calculate the weight of a body of mass $10 kg$ in (a) kgf and (b) newton. Take $g = 9.8 m s^{-2}.$

Answer

Mass = 10 kg
(i) Weight (in kgf) = 10 x 1 kgf = 10 kgf
$[1 kgf = 9.8 N]$
(ii) Weight (in newton) $= 10 x 9.8 = 98 N.$

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Question 82 Marks

Write an expression for the gravitational force of attraction between two bodies of masses $m_1$ and $m_2$ separated by a distance r.

Answer

$F \propto \frac{ Gm _1 m _2}{ d ^2}$ Here G is a constant of proportionality called the universal gravitational constant.

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Question 92 Marks

Write the approximate weight of a body of mass $5 \ kg$. What assumption have you made?

Answer

Given: mass $= 5 kg$
Assumption made is value of acceleration due to gravity is $10 ms^{-2}$
Weight $= mg$
$W = 5 \times 10$
$W = 50 N$

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Question 102 Marks

The value of g at the centre of Earth is zero. What will be the weight of a body of mass m kg at the centre of the Earth?

Answer

W = mg
At the centre of Earth, g = 0.
Therefore, W = 0.

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Question 112 Marks

A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking $g =10$ $m s ^{-2}$ and speed of sound $=330 m s ^{-1}$. Find : The depth of water surface

Answer

Time $t =20 s$
$g = 10 m/s^2$
Let $'D$' be the depth of the well.
Using the second equation of motion,
$D = ut + (1/2) gt^2$
$D = 0 + (1/2)(10)(20)^2$
$D = 2000 m$

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Question 122 Marks

Distinguish between mass and weight.

Answer

Mass is a scalar quantity, but weight is a vector quantity. Mass is the measure of the quantity of matter contained in a body, but weight is the measure of force with which the Earth attracts the body. Mass of a body is always constant but weight varies from place to place.

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Question 132 Marks

Define the terms mass and weight.

Answer

Mass: The mass of a body is the quantity of matter it contains. Weight: The weight of a body is the force with which the Earth attracts it.

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Question 142 Marks

ball is thrown vertically upwards with an initial velocity of $49 m s ^{-1}$. Calculate: The time taken by it before it reaches the ground again. (Take $\left.g =9.8 m s ^{-2}\right)$.

Answer

Initial velocity, u = $49 m / s$
$
g =9.8 m / s ^2
$
Total time of flight is given by $t =2 u / g$
Or, $t =2(49) / 9.8$
Or, $t=10 s$

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Question 152 Marks

A body falls from the top of a building and reaches the ground $2.5 s$ later. How high is the building? (Take $g = 9.8 m s^{-2})$

Answer

Given time $t =2.5, g =9.8 m / s ^2$
Height, $H=(1 / 2) gt ^2$
Or, $H=(1 / 2)(9.8)(2.5)^2$
Or, $H =30.6 m$

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Question 162 Marks

A body is thrown vertically upwards with an initial velocity u. Write the expression for the maximum height attained by the body.

Answer

A body is thrown vertically upwards with an initial velocity $u$. Write the expression for the maximum height attained by the body.
$
v^2=u^2-2 g h
$
We - get,
$
h_{\max }=\frac{u^2}{2 g}
$

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Question 172 Marks

A body falls freely under gravity from rest and reaches the ground in time t. Write the expression for the height fallen by the body.

Answer

Initial velocity, $u =0$.
Time taken $= t$.
Acceleration due to gravity $= g$.
Let ' $h$ ' be the height fallen.
$
\begin{aligned}
& h = ut +\frac{1}{2} g t ^2 \\
& h =\frac{1}{2} g t ^2
\end{aligned}
$

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Question 182 Marks

What will be the velocity of the stone on reaching the ground? (Take $g=10 m s^{-2})$

Answer

Let ' $v$ ' be the velocity on reaching the ground.
Using the third equation of motion,
$
v^2-u^2=2 g H
$
or, $v^2-0=2(10)(80)$
or, $v^2=1600$
or, $v=40 m / s$

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Question 192 Marks

How long will a stone take to fall to the ground from the top of a building 80 m high

Answer

Height, $s=80 m$
$
g =10 m / s ^2
$
Using the second equation of motion,
$
S=u t+(1 / 2) g t^2
$
Or, $80=0+(1 / 2)(10)(t)^2$
Or, $( t )^2=16$
Or, $t=4 s$

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Question 202 Marks

A pebble is thrown vertically upwards with a speed of $20 m s ^{-1}$. How high will it be after 2 s ? (Take $\left.g =10 m s ^{-2}\right)$

Answer

Initial velocity, $u=20 m / s$
Time, $t =2 s$
$
g =10 m / s ^2
$
Maximum height reached in $2 s , H =(1 / 2) gt ^2$
Or, $H =(1 / 2)(10)(2)^2$
Or, $H =20 m$

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Question 212 Marks

An body is dropped from the top of a tower. It acquires a velocity $20 m s ^{-1}$ on reaching the ground. Calculate the height of the tower. (Take $g =10 m s ^{-2}$ )

Answer

Initial velocity $u =0$ Final velocity $=20 m / s g =10 m / s ^2$ Let ' $h$ ' be the height of the tower. Using the third equation of motion, $v ^2- u ^2=2 gs$ or, $(20)^2-0=2(10) h$ or, $h =20 m$

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Question 222 Marks

A ball is thrown vertically upwards. It goes to a height $20 m$ and then returns to the ground. Taking acceleration due to gravity g to be $10 ms^{-2}$, find:
the initial velocity of the ball

Answer

Given, maximum height reached, $s=20 m$ Acceleration due to gravity, $g=10 m / s ^2$ Let ' $u$ ' be the initial velocity.
At the highest point, velocity $=0$
Using the third equation of motion,
$v^2-u^2=2 g s$
or, $0-u^2=2(10)(20) m / s$
or, $u^2=-(400) m / s$ [Negative sign indicates that the motion is against gravity]
or, $u=20 m / s$

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Question 232 Marks

What force, in newton, your muscles need to apply to hold a mass of 5 kg in your hand? State the assumption.

Answer

Mass, $m=5 kg$
Force, $F = mg$
$
F =(5)(9.8)=49 N
$
Assumption: Value of acceleration due to gravity is $9.8 m / s ^2$

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Question 242 Marks

Write an expression for the force due to gravity on a body of mass m and explain the meaning of symbols used in it.

Answer

The force due to gravity on a body of mass m kept on the surface of Earth (mass=M and radius=R) is equal to the force of attraction between the Earth and that body. $F =\frac{ GMm }{ R ^2}$

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Question 252 Marks

To move a boat ahead in water, the boatman has to push the water backwards by his oar. Explain this statement.

Answer

To move a boat, the boatman pushes (action) the water backwards with his oar. In this response, the water exerts an equal and opposite force (reaction) in the forward direction on the boat due to which the boat moves ahead.

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Question 262 Marks

State Newton's third law of motion.

Answer

According to Newton's third law of motion, to every action there is always an equal and opposite reaction. The action and reaction act simultaneously on two different bodies.

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Question 272 Marks

Comment on the statement 'the sum of action and reaction on a body is zero'.

Answer

The given statement is wrong. Reason: According to Newton's third law of motion, the action and reaction act simultaneously on different bodies. Hence they do not cancel each other.

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Question 282 Marks

A light ball falling on ground, after striking the ground rises upwards. Explain the reason.

Answer

When a falling ball strikes the ground, it exerts a force on the ground. The ground exerts a force back at the ball in the opposite direction. This is the reason the ball rises upwards.

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Question 292 Marks

A force causes an acceleration of $10 m s ^{-2}$ in a body of mass $500 g$ . What acceleration will be caused by the same force in a body of mass $5 \ kg$ ?

Answer

Let the force be $F$.
Force $F$ causes an acceleration, $a =10 m / s ^2$ in a body of mass, $m =500 g$ or $0.5 kg$
Thus, $F = ma$
Or, $F=(0.5)(10)=5 N$
Let $a^{\prime}$ be the acceleration which force $F(=5 N )$ cause on a body of mass, $m ^{\prime}=5 kg$.
Then, $a ^{\prime}= F / m ^{\prime}$.
Or, $a ^{\prime \prime}=(5 / 5) ms ^{-2}$.
Or, $a ^{\prime}=1 ms ^{-2}$.

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Question 302 Marks

Write the mathematical form of Newton's second law of motion. State the conditions if any.

Answer

Mathematical expression of Newton's second law of motion is as shown below: Force = Mass × Acceleration Above relation holds for the following conditions: (i) When the velocity of the body is much smaller than the velocity of light. (ii) When the mass remains constant.

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Question 312 Marks

Figure shows the velocity-time graph of a particle of mass 100 g moving in a straight line. Calculate the force acting on the particle.
(Hint : Acceleration = Slope of the v-t graph)

Answer

Given mass, $m =100 g =0.1 kg$.
Slope of a velocity-time graph gives the value of acceleration.
$
\text { Acceleration }=\text { Slope }=\frac{20}{5}=4 m / s ^2 .
$
Force $=$ Mass $\times$ Acceleration.
Force $=0.1 \times 4=0.4 N$.

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Question 322 Marks

How does Newton's second law of motion differ from the first law of motion?

Answer

Newton's first law of motion gives the qualitative definition of force. It explains the force as the cause of acceleration only qualitatively but Newton's second law of motion gives the quantitative value of force. It states force as the product of mass and acceleration. Thus, it relates force to the measurable quantities such as acceleration and mass.

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Question 332 Marks

State the Newton's second law of motion. What information do you get from it?

Answer

According to Newton's second law of motion, the rate of change of momentum is directly proportional to the force applied on it and the change of momentum takes place in the direction in which the force is applied. It gives the quantitative value of force, i.e. it relates the force to the measurable quantities such as acceleration and mass.

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Question 342 Marks

A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate : The magnitude of the force

Answer

Mass, $m=100 kg$
Distance moved, $s =100 m$
Initial velocity, $u =0$
Force, $F = ma$
Or, $F =(100)$ (2) N.
Or, $F=200 N$.

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Question 352 Marks

A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate : The acceleration produced by the force

Answer

Mass, $m=100 kg$
Distance moved, $s =100 m$
Initial velocity, $u =0$
From Newton's third equation of motion,
$
v ^2- u ^2=2 as \text {. }
$
Or, $a=\left(v^2-u^2\right) / 2 s$.
Or, $a=\left[\left(20^2-0^2\right) / 2(100)\right] ms ^{-2}$.
Or, $a =2 ms ^{-2}$.

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Question 362 Marks

A force acts for $10 s$ on a stationary body of mass $100 kg$, after which the force ceases to act. The body moves through a distance of 100 m in the next $5 s$. Calculate : The velocity acquired by the body

Answer

Mass, $m = 100 kg$
Distance moved, $s = 100 m$
Initial velocity, $u = 0$
Because the body moves through a distance of $100 m$ in $5 s,$
Velocity of the body = Distance moved / time taken
Velocity $= (100/5) = 20 m/s^{-1}$

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Question 372 Marks

Calculate the magnitude of force which when applied on a body of mass $0.5 kg$ produces an acceleration of $5 m s^{-2}.$

Answer

Mass, $m = 0.5 kg.$
Acceleration, $a = 5 ms^{-2}$
Force, F = ma [ From Newton's second law]
Or, $F = (0.5) (5) N = 2.5 N.$

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Question 382 Marks

A force of 10 N acts on a body of mass 5 kg. Find the acceleration produced.

Answer

Force, $F=10 N$
Mass, $m=5 kg$
Acceleration, $a = F / m$ [ From Newton's second law]
Or, $a=(10 / 5) ms ^{-2}$
Or, $a =2 ms ^{-2}$

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Question 392 Marks

A force of 15 N acts on a body of mass 2 kg. Calculate the acceleration produced.

Answer

Force, $F=15 N$
Mass, $m =2 kg$
Acceleration, $a= F / m$ [ From Newton's second law]
Or, $a=(15 / 2) ms ^{-2}$
Or, $a =7.5 ms ^{-2}$

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Question 402 Marks

The linear momentum of a ball of mass $50 \ g$ is $0.5 kg m s ^{-1}$. Find its velocity.

Answer

Linear momentum $=0.5 kg m / s$
Mass, $m =50 g =0.05 kg$
Velocity = Linear momentum / mass
$=0.5 / 0.05 m / s$
$=10 m / s ^{-1}$

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Question 412 Marks

A body of mass $5 \ kg$ is moving with velocity $2 m s^{-1}.$ Calculate its linear momentum.

Answer

Mass of the body, $m=5 kg$
Velocity, $v =2 m / s$
Linear momentum $=m v=(5)(2) kg m / s$
$=10 kg m / s ^{-1}$

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Question 422 Marks

Use Newton's second law of motion to explain the following instance :
An athlete prefers to land on sand instead of hard floor while taking a high jump .

Answer

When an athlete lands from a height on a hard floor, his feet comes to rest instantaneously, so a very large force is exerted by the floor on his feet, but if he lands on sand, his feet push the sand for some distance; therefore, the time duration in which his feet comes to rest increases. As a result, the force exerted on his feet decreases and he is saved from getting hurt.

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Question 432 Marks

Use Newton's second law of motion to explain the following instance :

A cricketer pulls his hands back while catching a fast moving cricket ball .

Answer

We pull our hands back while catching a fast moving cricket ball, because by doing so, we increase the time of catch, i.e. increase the time to bring about a given change in momentum, and hence, the rate of change of momentum decreases. Thus, a small force is exerted on our hands by the ball.

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Question 442 Marks

Why does a glass vessel break when it falls on a hard floor, but it does not break when it falls on a carpet?

Answer

When a glass vessel falls from a height on a hard floor, it comes to rest almost instantaneously, i.e. in a very short time, so the floor exerts a large force on the vessel and it breaks. However, if it falls on a carpet, then the time duration, in which the vessel comes to rest, increases, so the carpet exerts less force on the vessel and it does not break.

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Question 452 Marks

Name the $S.I$. and $C.G.S$. units of force. How are they related?

Answer

The S.I. unit of force is newton and the C.G.S. unit of force is dyne. $1 N=10^5$ dyne.

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Question 462 Marks

What is the C.G.S. unit of force? How is it defined?

Answer

The C.G.S. unit of force is dyne.
One dyne is the force which acts on a body of mass 1 gramme and produces an acceleration of $1 cms ^{-2}$, i.e. 1 dyne $=1 g \times 1 cms ^{-2}$.

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Question 472 Marks

Name the S.I. unit of force and define it .

Answer

The S.I. unit of force is newton.
One newton is the force which acts on a body of mass $1 kg$ and produces an acceleration of $1 m / s ^2$, i.e. $1 N =1 kg \times 1 m / s ^2$.

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Question 482 Marks

Draw a graph to show the dependence of force on mass for a constant acceleration.

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Question 492 Marks

Draw a graph to show the dependence of acceleration on force for a constant mass.

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Question 502 Marks

How can Newton's first law of motion be obtained from the second law of motion?

Answer

From Newton's second law of motion, F = ma. If F = 0, then a = 0. This means that if no force is applied on the body, its acceleration will be zero. If the body is at rest, then it will remain in the state of rest and if it is moving, then it will remain moving in the same direction with the same speed. Thus, a body not acted upon by an external force, does not change its state of rest or motion. This is the statement of Newton's first law of motion.

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