##
*Gravitation*

*Gravitation*

Johannes Kepler - proposed laws of planetary motion |

**NCERT Q & A, Study Notes, Numerical Problems, Questions asked in CBSE Papers.****Q1: What was geocentric model? Who proposed this theory?**

Answer: Ptolemy a Greek mathematician and astronomer proposed geocentric model around 125 AD. According to this model, the earth is in the center and sun, other planets and stars move around it.

To explain retrograde motion of planets, he also proposed that those planet also move in small circle. The geocentric theory held valid till 1543 AD when Copernicus proposed heliocentric model.

**Q2: What was heliocentric theory? Who proposed this?**

Answer: Copernicus proposed heliocentric model around 1543 AD. he proposed that the sun is in the center and all planets move around it.

**Q3: What is retrograde motion of planets (say mars planet)?**

Answer: Retrograde motion is the apparent backward motion of planets (e.g. Mars) as seen from the earth against the background of stars.E.g. Mars planet when seen from the earth, appears to move in a loop. This had puzzled the astronomers for a long time till Copernicus proposed heliocentric model. Later Brahe and Kepler solved it by telling the world the planets move in elliptical path. (Click here to read more about this.)

**Q4:What are the Kepler Laws that govern the motion of the planets?**

Answer: Kepler three laws are:

- THE LAW OF ORBITS: All planets move in elliptical orbits. with the Sun at one focus.
- THE LAW OF AREAS: A line that connects a planet to the Sun sweeps out equal areas in the plane of the planet's orbit in equal time intervals. It implies that the planet will move most slowly when it is farthest from the Sun and most rapidly when it is nearest to the Sun.
- THE LAW OF PERIODS: The square of the orbital period (T) of any planet is proportional to the cube is of its orbit (r) i.e. T
^{2}∝ r^{3}

**Q5(NCERT/CBSE 2010): State the universal law of gravitation.**

Answer: Every object of the universe attracts other object. The force of attraction between the two objects is proportional to the product of their mass and inversely proportional to the square of the distance between them. The direction of this force is along the line joining the centres of the objects.

If m

_{1}and m

_{2}are the two objects separated by a distance

*d*, then force attraction between them is:

F = Gm

_{1}m

_{2}/d

^{2}

where G is the universal Gravitation constant = 6.67 × 10

^{-11}Nm

^{2}/kg

^{2}

**Q6**

**(NCERT)**

**: What is the importance of Universal Law of Gravitation?**

or

**Q**

**(CBSE 2010)**

**: State any one phenomena related to the Universal Law of Gravitation.**

Answer: The law is universal i.e. it is applicable to all bodies, whether the bodies are big or small, whether they are celestial or terrestrial. It successfully explains various phenomena:

- the force that binds us to the earth.
- the motion of the moon around the earth
- the motion of planets around the Sun
- the tides due to the moon and the Sun.
- it also helps us finding the masses of planets and stars.

**Q7(CBSE 2010): Why is G called the Universal Constant?**

Answer: At any place in the universe and at any time, the value of G is found to be constant for any two bodies. Thus G is called the universal constant of gravitation.

It is defined as force of attraction acting between two objects of mass 1 kg each placed at a distance of 1 m.

Mathematically, G = F.d

^{2}/m

_{1}m

_{2}

Thus the SI unit of the universal constant of gravitation is Nm

^{2}/kg

^{2}.

The value of G is 6.67 × 10

^{-11}Nm

^{2}/kg

^{2}

**Q8: Who measured the value of G first time, experimentally?**

Answer: Henry Cavendish

**Q9**

**(CBSE 2010/NCERT): What is the gravitational force between the Earth and a body called? In which direction does it occur?**

Answer: Weight of the body or gravity. It always acts towards the centre of the earth i.e. vertically downwards.

**Q10: An object moves in a circular motion due to centripetal force, acting towards the center. The moon also rotates around the earth? Does centripetal force acts on it? Why does the moon not fall on the earth?**

or

**Q(NCERT):If the moon attracts the earth, why does the earth not move towards the moon?**

Answer: Yes the centripetal force acts on the moon. This force is due to gravitational force between the earth and the moon and it acts on both the objects. Gravitational force is always attractive, still the moon does not fall on the earth, because the tangential speed (due to centripetal force) of the moon make it escape from the gravitational force and is enough to hold it in its orbital path.

**Q11: The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which**

the moon attracts the earth? Why?

the moon attracts the earth? Why?

Answer: The two objects attract each other with equal forces of attraction but in opposite directions.

**Q12: How gravitational force is different from or similar to magnetic force (between two magnets) or coulomb force (between two charged object)**

**?**

**Answer: Similarity: These three forces follow inverse square rule i.e. force between the two objects is inversely proportional to the square of distance between them.**

Difference: While gravitational force is always attractive, magnetic force and coulomb forces can be attractive or repulsive (e.g. North-North pole repel each other.). Among the three, gravitational force is the weakest.

**Q13**

**(NCERT): Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10**

^{24}kg and of the Sun = 2 × 10^{30}kg. The average distance between the two is 1.5 × 10^{11}m.Answer: Given, mass of the earth (m

_{E}) = 6 × 10

^{24}kg

mass of the sun (m

_{S}) = 2 × 10

^{30}kg

Distance between the bodies (d) = 1.5 × 10

^{11}m

Gravitational Constant (G) = 6.67 × 10

^{-11}Nm

^{2}/kg

^{2}

Applying formula of Newton's Law of Gravitation,

F = Gm

_{E}m

_{S}/d

^{2}= 6.67 × 10

^{-11}× 6 × 10

^{24}× 2 × 10

^{30}/ (1.5 × 10

^{11})

^{2}

F = 80.04 × 10

^{-11+24+30}/ 2.25 × 10

^{22}= 35.57 × 10

^{43-22}= 35.57 × 10

^{21}

F =

**3.56 × 10**

^{22}N**Q14(CBSE 2010): The Earth attracts an apple. Does the apple also attract the Earth? If it does, why does the Earth not move towards the apple?**

Answer: According to Newton's law of gravitation, both apple and the earth attract each other with equal force but in opposite directions. Also they they produce different accelerations depending on the mass of the object. It means apple also attract the Earth. Since the mass of the earth is extremely large as compared to mass of the apple, acceleration of the earth due to this force of attraction is very low (of order 10

^{-25}m/s

^{2}).

Mathematically, mass of an apple (m

_{apple}) = 150g = 0.15 kg

mass of the earth (m

_{E}) = 6 × 10

^{24}kg

acceleration acting on apple due to gravity (g) of earth = 9.8 m/s

^{2}

Weight of the apple = m

_{apple}

_{}× g = 0.15 × 9.8 = 1.47 N

⇒ Earth also experiences 1.47 N force.

i.e. m

_{E}× a = 1.47N

a = 1.47 / 6 × 10

^{24}= 2.45 × 10

^{-25}m/s

^{2}(extremely small value)

**Q15(CBSE 2010): How is gravitational force of attraction between two bodies affected if**

(i) Mass of both bodies is doubled.

(ii) Distance between them is halved.

(i) Mass of both bodies is doubled.

(ii) Distance between them is halved.

Answer: The gravitational force between two objects is directly proportional to the product of their mass and inversely proportional to the square of the distance between them.

(i) if mass of both bodies is doubled, the force of the attraction will increase by 4 time.

(ii) if distance between them is halved, the force of attraction will increase by 4 times.

In case above two cases occur in parallel, the force of attraction will increase by 16 times.

**Q16**

**(NCERT/CBSE 2011)**

**: What happens to the force between two objects, if**

(i) the mass of one object is doubled?

(i) the mass of one object is doubled?

**(ii) the distance between the objects is doubled and tripled?**

(iii) the masses of both objects are doubled?

(iii) the masses of both objects are doubled?

Answer: The gravitational force between two objects is directly proportional to the product of their mass and inversely proportional to the square of the distance between them.

F = Gm

_{1}m

_{2}/d

^{2}

(i) Doubled

If m

_{1}becomes 2m

_{1}

F

_{New}= G2m

_{1}m

_{2}/d

^{2}= 2(Gm

_{1}m

_{2}/d

^{2}) = 2F

(ii) One fourth and one-ninth

if distance between objects is doubled i.e. 2d, F

_{New}= Gm

_{1}m

_{2}/(2d)

^{2}= Gm

_{1}m

_{2}/4d

^{2}

F

_{New }= F/4

if distance is tripled i.e. 3d, F

_{New}= Gm

_{1}m

_{2}/(3d)

^{2}= Gm

_{1}m

_{2}/9d

^{2}

F

_{New }= F/9

(iii) four times

If both masses are doubled i,e, 2m

_{1 }and 2m

_{2}

F

_{New}= G2m

_{1}2m

_{2}/d

^{2}= 4(Gm

_{1}m

_{2}/d

^{2}) = 4F

**Q17: An object is placed at the surface of the earth and is at distance R from the center of the earth. The object weighs 180N. What will be its weight if it is at distance 3R from the center of the earth.**

Answer: 20N. [Hint: See logic in Q16-9(i)]

**Q18: Does the force of attraction between two objects depend on the properties of intervening medium?**

Answer: No.

**Q19: Does Law of gravitation obey Newton's third law of motion?**

Answer: Yes.

**Q20(NCERT): What is free fall?**

Answer: The falling of a body from height towards the earth under the influence of gravitation force of the earth alone only is called free fall.

**Q21: Define 'g' or acceleration due to gravity? Does it depend on the mass of the body (near the surface of the earth)experiencing 'g'?**

Answer: The constant acceleration experienced by a freely falling object towards the earth is called acceleration due to gravity (g). Its average value on the surface of the earth is 9.8 m/s

^{2}. It does not depend on the mass of the body experiencing 'g'.

i.e. g = Gm

_{E}/R

^{2}= 9.8 m/s

^{2}.

where m

_{E}is the mass of the earth, R is the radius of the earth and G is the universal constant.

**Q22: (a)(CBSE 2011) What is the relationship between g and G?**

**(b) Show mathematically whether g (acceleration due to gravity on the earth) depends on the mass of the object itself or not.**

**(c) Show that for a given location on the surface of the earth g is almost constant.**

Answer: Let M

_{E}is the mass of the earth,

m is the mass of the object on the surface of the earth.

R is the radius of the earth

g is the acceleration due to gravity and

G is the universal constant.

According to Newton's second law, Force on the body due to acceleration due to gravity is

F = mass(m) × acceleration due to gravity(g) = m×g ...(I)

Using Newton's law of gravitation, the force of attraction F = GmM

_{E}/R

^{2}... (II)

Since both these forces are equal, equating I and II,

mg = GmM

_{E}/R

^{2}

⇒ g = GM

_{E}/R

^{2}...(III)

The above equation (III) shows, acceleration due to gravity(g) is independent of its mass. It depends on the mass of the earth(or planet) and the distance between the two objects.

For a given location on the surface of the earth, M

_{E}and R are constant. ∴ for that location value of g is constant.

**Q23: How does value of 'g' vary on the earth? What are the factors responsible for this?**

Answer: The value of 'g' varies slightly from location to location. It is due to the following reasons:

- Earth is not sphere but ellipsoid. i.e. it is flattened at the poles. Equator radius is more than Polar radius. ∴ Value of g is maximum at the poles and minimum at equator.
- Earth's mass is not uniformly distributed. Since density of the earth crust varies region to region, g also varies.
- Earth is rotating on its axis.
- Altitude: As we move away from the earth, the distance increases, value of g also decreases.
- Depth: As we move towards, g increases to certain depth and after that it starts decreasing and is zero at the centre of the earth.

➽ The value of 'g' varies at different places, this property is used to detect mineral ores, mines and oil fields.

**Q24: What happens to 'g' experienced by a body when it moves up the earth's surface?**

Answer: When a body moves up the earth surface, the value of 'g' decreases because its distance from the centre of the earth increases.

Let h = height of the body from the earth's surface and R

_{E}is the radius of the earth. Acceleration at height 'h' will be:

g

_{h }= GM

_{E}/(R

_{E}+ h)

^{2}= GM

_{E}/R

_{E}

^{2}× R

_{E}

^{2}

_{}/(R

_{E}+ h)

_{}

^{2}

⇒ g

_{h }= gR

_{E}

^{2}/(R

_{E}+ h)

^{2}(where g = GM

_{E}/R

_{E}

^{2}= 9.81 m/s

^{2}.)

**Q25: At what height above the surface of the earth, the value of 'g' becomes 64% of its value at the surface of the earth. Take radius of the earth = 6400 km.**

Answer: Let g = acceleration due to gravity at the earth surface.

Given g

_{h }= acceleration due to gravity at height h = 0.64g

Because, g

_{h }= gR

_{E}

^{2}/(R

_{E}+ h)

^{2}

⇒ 0.64g = gR

_{E}

^{2}/(R

_{E}+ h)

^{2}

⇒ 0.64 (R

_{E}+ h)

^{2}= R

_{E}

^{2}

⇒ 0.8(R

_{E}+ h) = R

_{E}

⇒ 0.8h = (1 - 0.8)R

_{E}

⇒ h = 2 × 6400 / 8 =

**1600 km**

**Q26(NCERT): Why is the weight of an object on the moon 1/6th its weight on the earth?**

Answer:

Let 'm' be the mass of an object.

R

_{E }be radius of the earth = 6400km = 6.4 × 10

^{6}m

M

_{E}= mass of the earth = 5.98 × 10

^{24}kg

R

_{M}= radius of the moon. = 1.74 × 10

^{6}m

M

_{M}= mass of the moon = 7.36 × 10

^{22}kg

W

_{E}= weight of an object on the earth.

W

_{M}= weight of an object on the moon.

Using Newton's Law of gravitation,

W

_{E}= GM

_{E}m/R

_{E}

^{2}

W

_{M}= GM

_{M}m/R

_{M}

^{2}

⇒ W

_{M}/W

_{E}= M

_{M}/R

_{M}

^{2}× R

_{E}

^{2}/M

_{E}= M

_{M}R

_{E}

^{2}/M

_{E}R

_{M}

^{2}

⇒ W

_{M}/W

_{E}= (7.36 × 10

^{22})(6.4 × 10

^{6})

^{2}/(5.98 × 10

^{24})(1.74 × 10

^{6})

^{2}

⇒ W

_{M}/W

_{E}= 0.165 = 1/6

⇒ W

_{M}

_{}= W

_{E}/6

**Q27**

**(NCERT): Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?**

Answer: All objects fall under the influence of gravity and experience constant acceleration (g).

g = GM

_{E}/R

^{2}

Since g is constant for a location and independent of mass of the object, heavy objects do not fall faster than light objects.

**Q28**

**(NCERT): What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10**

^{24}kg and radius of the earth is 6.4 × 10^{6}m).Answer: Given,

R

_{E }radius of the earth = 6.4 × 10

^{6}m

M

_{E}= mass of the earth = 6 × 10

^{24}kg

G = Universal Gravitational Constant = 6.67 × 10

^{-11}Nm

^{2}/kg

^{2}

m = mass of object = 1kg

According to Newton's Law of gravitation, the gravitation force (F) is

F = GM

_{E}m/R

_{E}

^{2}

F = (6.67 × 10

^{-11})(6 × 10

^{24})(1)/(6.4 × 10

^{6})

_{}

^{2}

F =

**9.8N**

**Q29: Define mass.**

Answer: The mass of the body is the quantity of matter it contains. It is the measure of inertia of the

body. It is a scalar quantity. Its SI unit is kg. The mass of an object is constant and does not change from place to place.

**Q30(CBSE 2010): Define weight? How it is related to mass?**

Answer: The weight of the body is the force experienced by the body due to its gravity. It is a vector quantity and always directs towards the centre of the earth. The SI unit of weight is Newton (N). Its cgs unit is dyne.

1N = 10

^{5}dyne

Weight is also called force of gravity on the body.

Weight and mass are related as follows:

Weight = mass(m) × acceleration due to gravity(g) = mg

**Q31(CBSE 2010): Mass of an object is 10 kg. What is its weight on the Earth? (g = 9.8 m/s**

^{2}.)Answer: Weight = mass(m) × acceleration due to gravity(g) = mg

W = 10 × 9.8 =

**98N**

**Q32**

**(NCERT/CBSE 2011): What are the differences between the mass of an object and its weight?**

Answer:

Mass | Weight |
---|---|

1. It is the quantity of the matter contained in the body. | 1. It is the gravitational force of attraction on the body by the earth(planet). |

2. SI unit is kg | 2. SI unit is Newton(N) |

3. It is a scalar quantity i.e. has magnitude only. | 3. It is a vector quantity i.e. has both magnitude and direction. e.g. On the Earth, its direction is always towards the centre of the earth/planet. |

4. It is a measure of inertia of the body. | 4. It is the measure of the gravity. |

5. It is measured by physical or beam balance. | 5. It is measured by spring balance calibrated to read in Newton scale or Kg-Wt. |

6. It is constant for a body and does not change from place to place. | 6. It is not constant but varies as the value of gravity changes place to place. |

**Q33**

**: Why the value of g is greater at the poles than at the equator?**

Answer: The value of 'g' for an object on the surface of the earth can be expressed as:

g = GM

_{E}/R

^{2}

where R is radius of the earth, G is universal constant and M

_{E}is the mass of the earth.

Since G and M

_{E}are constant, g ∝ 1/R

^{2}

Because Equator radius (R

_{eq}) is more than Polar radius (R

_{P}). ∴ Value of g is greater at the poles than that of at equator.

Or The value of g increases as we move equator to poles.

**Q34**

**(NCERT)**

**: Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator].**

Answer: Let 'm' be the mass of the gold. Because weight = mass × acceleration = mg

∴ Weight of the gold at poles (W

_{P}) = mg

_{P}

and Weight of gold at equator (W

_{Eq}) = mg

_{Eq}

∵ g

_{P}> g

_{Eq}⇒ W

_{P}> W

_{Eq}

Or gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought.

**Q35**

**(NCERT)**

**: Gravitational force on the surface of the moon is only 1/6th as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth? What is the value of 'g' at the surface of the moon?**

Answer: weight = mass × acceleration = mg

Given mass of the object (m) = 10kg

Value of g on the earth = 9.8 m/s

^{2}.

Weight of the object on the earth (W

_{E}) = 10 × 9.8 =

**98N**.

Weight of the object on the moon (W

_{M}) = 1/6th of W

_{E}= 98/6 =

**16.34 N**

acceleration due to gravity at the moon's surface = W

_{M}/m = 16.34 / 10 =

**1.63 m/s**.

^{2}**Q36(CBSE 2010): Write the standard equations of motion when an object is thrown upwards with some initial velocity?**

Answer: When an object is thrown upwards with initial velocity (u), it reaches to certain height(h) and its velocity (v) becomes zero. Then it starts to fall. In this case, the acceleration due to gravity (g) is opposing the vertical upward motion and is taken as negative(-ve).

Equations become:

v = u + at ⇒ 0 = u - gt ⇒

**u = gt**

S = ut + ½at

^{2}⇒

**h = ut - ½gt**

^{2}v

^{2}- u

^{2}=2aS ⇒ 0 - u

^{2}=-2gh ⇒

**u**

^{2}= 2gh**Q37: Write the standard equations in the form when an object is dropped from a height?**

Answer: When an object is dropped from a height(h), its initial velocity(u = 0) is zero. It begins to free fall i.e. under the influence of gravity. It means the acceleration due to gravity(taken as positive), will accelerate the object and it reaches ground with some final velocity (v).

Equations of linear motion become:

v = u + at ⇒

**v = gt**

S = ut + ½at

^{2}⇒ h = 0 + ½gt

^{2}⇒

**h = ½gt**

^{2}v

^{2}- u

^{2}=2aS ⇒

**v**

^{2}= 2gh**Q38: Is acceleration due to gravity a vector or a scalar? Write its S.I. unit.**

Answer: It is a vector quantity and hence has both magnitude and direction. The SI unit is m/s

^{2}.

**Q39**

**(NCERT)**

**: A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate**

(i) the maximum height to which it rises.

(ii)the total time it takes to return to the surface of the earth.

(i) the maximum height to which it rises.

(ii)the total time it takes to return to the surface of the earth.

Answer: Given, initial velocity (u) = 49 m/s

final velocity the ball reaches max. height (v) = 0 m/s

acceleration due to gravity (g) = -9.8 m/s

^{2}.

g is -ve because it opposed the vertical motion.

height (h) ball will reach = ? m

Time of ascent i.e. time taken to reach max. height (t) = ? s

Using equation v

^{2}- u

^{2}=2aS,

⇒ 0

^{2}- 49

^{2}= -2(9.8)h

⇒ h = (49 × 49)/19.6 = 490/4 =

**122.5 m**

Using equation v = u + at i.e. 0 = 49 - 9.8t

⇒ t = 49/9.8 = 5s

Since Time of ascent = Time of descent.

Total time the ball will take to reach ground = 5 + 5 =

**10s**

**Q40**

**(NCERT): A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.**

Answer: Given initial velocity (u) = 0 m/s

Final velocity (v) = ?

height (distance) h = 19.6 m

acceleration due to gravity (g) = +9.8 m/s

^{2}.

g is taken as +ve because it will accelerate the vertical motion.

Using equation v

^{2}- u

^{2}=2aS, ⇒ v

^{2}- 0 = 2gh

⇒ v

^{2}= 2 × 9.8 × 19.6 = (19.6)

^{2}

⇒ v =

**19.6 m**/s

**Q41**

**(NCERT)**

**: A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s**

^{2}, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?Answer: Given, initial velocity (u) = 40 m/s

Final velocity when stone reaches at max. height (v) = 0 m/s

Acceleration due gravity in upward motion of stone g = -9.8 m/s

^{2}

g is taken -ve because it opposed the vertical motion.

max. height the stone will reach (h) = ?

Using equation v

^{2}- u

^{2}=2aS,

⇒ 0 - u

^{2}= -2gh

⇒ h = u

^{2}/(2g) = (40)

^{2}/(2 × 10) =

**80m**

It means the stone will reach upto 80m height and then it comes back to ground(original position).

Total Distance covered by stone = 2 × 80 =

**160m**

Net displacement = Upward Displacement + Downward Displacement = 80 + (-80) =

**0m**

**Q42**

**(NCERT)**

**: A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.**

Answer: Let us assume both stones meet at time 't'.

Height of the tower (h) = 100m

**Stone 1 dropped from the top of a tower:**

initial velocity of the stone (u) = 0 m/s

acceleration due to gravity (g) is +ve = +9.8 m/s

^{2}

At time t, it will cover distance (h

_{1}), then using equation S = ut + ½at

^{2}

⇒ h

_{1}= 0 + ½gt

^{2}= (½)(9.8)t

^{2}= 4.9t

^{2}

**Stone 2 thrown upwards:**

initial velocity (u) = 25 m/s

acceleration due to gravity (g) is negative = - 9.8 m/s

^{2}

At time t, 2nd stone reaches height (h

_{2}), then using equation S = ut + ½at

^{2}

⇒ h

_{2}= 25t - ½gt

^{2}= 25t - (½)(9.8)t

^{2}= 25t - 4.9t

^{2}

Because Total height h = 100 m = h

_{1}+ h

_{2}

⇒ 4.9t

^{2}+ 25t - 4.9t

^{2 }= 100

⇒ 25t - 100

⇒ t = 4s.

In 4 seconds, both stones will meet.

h

_{1}(height from top) = 4.9 × 16 = 78.4m or 100- 78.4 = 21.6m from ground.

Both stones will meet at

**21.6 m**from ground.

(

**Note**: If you take g = 10 m/s

^{2}, the height will come as 20m)

**Q43: A ball is thrown upwards from the ground of a tower with a speed of 20 m/s. There is a window in the tower at the height of 15m from ground. How many times and when will the ball pass the window? (g = 10**

**m/s**

^{2})

Answer: Given, initial velocity (u) = 20 m/s

Max height the ball will reach (h) = ?

Using equation u

^{2}= 2gh,

⇒ h = 20

^{2}/(2×10) = 20m

This means ball will reach the height of 20m and comes back. It will pass the window two times.

Let us calculate the time ball will take to reach 15 m height.

Using equation,

*h = ut - ½gt*

^{2}15 = 20t -

*½*

*×10t*

^{2}⇒ 5

*t*- 20t +15 = 0

^{2}⇒

*t*- 4t +3 = 0

^{2}⇒

*t*- t - 3t +3 = 0

^{2}⇒ t(t -1) -3(t -1) = 0

⇒ (t -1)(t - 3) = 0

⇒ t = 1s and 3s

Ball will pass the window at 1s and 3s respectively.

**Q44: A ball is thrown up and is caught by the player after 4 seconds.How high did it go and with what velocity was it thrown? How far below the ball will be from its highest point after 3 seconds from start?**

Answer: Since we know that, Time of Ascent = Time of Descent.

⇒ Time taken by ball to reach highest point = 4 /2 = 2s

Let u be the initial velocity the ball is thrown upwards.

At its highest point (h), final velocity (v) = 0 m/s

Using equation v = u + at

⇒ 0 = u - 2g

⇒ u = 2 × 9.8 =

**19.6 m/s**

The ball was thrown with initial velocity 19.6 m/s

Maximum height it will reach,

*h = ut - ½gt*

^{2}⇒ h = 19.6 × 2 -

*½*

*× 9.8 × 2*

*= 19.6 × 2 - 19.6 =*

^{2}**19.6 m**

After t = 3 seconds, the distance traveled will be

*h*= 19.6 × 3 -

_{3}= ut - ½gt^{2}*½*

*× 9.8 × 3*

*= 58.8 - 44.1 = 14.7 m wrt ground.*

^{2}Ball's position w.r.t. to its high point = 19.6 - 14.7 =

**4.9m**

**Q45: A ball is thrown with some velocity 'u' m/s. Show that under free fall, it will fall the ground with same velocity.**

Answer: When the ball is thrown upwards, it will reach certain height (h) and starts falling. At height 'h', the final velocity will be (v) = 0.

Max. height reached by the ball, (acceleration = -g)

Using equation v

^{2}- u

^{2}=2aS

⇒ 0

^{2}- u

^{2}= - 2gh

⇒ h = u

^{2}/2gh ... (i)

In second case when the ball starts to fall, the initial velocity = 0. It will accelerate due to gravity i.e. a = +g and reach ground with velocity (say

*v*)

_{2}Using equation v

^{2}- u

^{2}=2aS

⇒

*v*

_{2}^{2}- 0

^{2}=2gh

⇒

*v*

_{2}^{2}= 2g

*×*u

^{2}/2g (from eqn. i)

⇒

*v*

_{2}^{2}= u

^{2}

⇒

*v*= u

_{2}∴ the ball will reach ground with same velocity.

**Q46: Define thrust.**

Answer: Its a type of force which acts normal (perpendicular) to the surface.

**Q47: Two stones A and B are dropped from a multistoried building. A is dropped from 100 m and after some time B is dropped from 50 m height. Both of them reach the earth at the same time. Will they have equal velocity while reaching the ground ? Calculate and find out the answer (take g =10 m/s**

^{2}).Answer: For stone A, height (h) = 100m and a = g = 10m/s

^{2}(taken as +ve)

Initial velocity (u) = 0 m/s, Final velocity

*v*= ?

_{A}Using equation, v

^{2}- u

^{2}=2aS

⇒

*v*

_{A}^{2}- 0

^{2}=2gh = 2 × 10 × 100 = 2000

⇒

*v*

_{A}^{}=

*√(2000) =***44.72 m/s**

For stone B, initial velocity (u) = 0 m/s, height h = 50m,

Final velocity (

*v*) = ?

_{B}Using equation, v

^{2}- u

^{2}=2aS

⇒

*v*

_{B}^{2}- 0

^{2}= 2gh = 2 × 10 × 50 = 1000

⇒

*v*=

_{B}

*√(1000) =***31.6 m/s**

∴ Both stones will reach ground with different velocities.

**See Also:**

I have found a mistake - Look at answer of Q6. The last word of the fifth point is "stars" not "starts".

ReplyDeletefixed. thanks for pointing.

DeleteWelcome ! And great study material compilation. This has helped me a lot.

Deletethanks for the notes!!:)

ReplyDeleteVry good material

ReplyDeletethanku so much

ReplyDeletethanku so much

ReplyDeletethanks..........

ReplyDeletevery good and useful. thank u

ReplyDeletevery good and useful. thank u

ReplyDeletethanks a lot......

ReplyDeleteit is very useful .it help me a lot.....

This comment has been removed by the author.

ReplyDeleteThanks a lot for this because only because of these notes I have secured first in class in the test of this chapter .........whom so ever have posted this a very huge thanks to him or herðŸ˜˜ðŸ˜˜ðŸ˜˜ðŸ˜˜ðŸ˜˜ but this. Is. Only for himðŸ˜˜ðŸ˜˜ðŸ˜˜

ReplyDelete