Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 6:

What happens to the force between two objects, if

(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?

Answer:

According to the universal law of gravitation, the force of gravitation between two objects is given by:

(i) F is directly proportional to the masses of the objects. If the mass of one object is doubled, then the gravitational force will also get doubled.

(ii) F is inversely proportional to the square of the distances between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value.

Similarly, if the distance is tripled, then the gravitational force becomes one-ninth of its original value.

(iii) F is directly proportional to the product of masses of the objects. If the masses of both the objects are doubled, then the gravitational force becomes four times the original value.

Question 7:

What is the importance of universal law of gravitation?

Answer:

The universal law of gravitation proves that every object in the universe attracts every other object.

Question 8:

What is the acceleration of free fall?

Answer:

When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 m s⁻², which is constant for all objects (irrespective of their masses).

Question 9:

What do we call the gravitational force between the Earth and an object?

Answer:

Gravitational force between the earth and an object is known as the weight of the object.

Question 10:

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:

Weight of a body on the Earth is given by:

W = mg

Where,

m = Mass of the body

g = Acceleration due to gravity

The value of g is greater at poles than at the equator. Therefore, gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought.

Question 11:

Why will a sheet of paper fall slower than one that is crumpled into a ball?

Answer:

When a sheet of paper is crumbled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper.

Question 12:

Gravitational force on the surface of the moon is only 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?

Answer:

Weight of an object on the moon   Weight of an object on the Earth

Also,

Weight = Mass × Acceleration

Acceleration due to gravity, g = 9.8 m/s²

Therefore, weight of a 10 kg object on the Earth = 10 × 9.8 = 98 N

And, weight of the same object on the moon 

Question 13:

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.

Answer:

(i) 122.5 m (ii) 10 s

According to the equation of motion under gravity:

v² − u² = 2 gs

Where,

u = Initial velocity of the ball

v = Final velocity of the ball

s = Height achieved by the ball

g = Acceleration due to gravity

At maximum height, final velocity of the ball is zero, i.e., v = 0

u = 49 m/s

During upward motion, g = − 9.8 m s⁻²

Let h be the maximum height attained by the ball.

Hence,

Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:

v = u + gt

We get,

But,

Time of ascent = Time of descent

Therefore, total time taken by the ball to return = 5 + 5 = 10 s

Question 14:

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:

According to the equation of motion under gravity:

v² − u² = 2 gs

Where,

u = Initial velocity of the stone = 0

v = Final velocity of the stone

s = Height of the stone = 19.6 m

g = Acceleration due to gravity = 9.8 m s⁻²

∴ v² − 0² = 2 × 9.8 × 19.6

v² = 2 × 9.8 × 19.6 = (19.6)²

v = 19.6 m s^{− 1}

Hence, the velocity of the stone just before touching the ground is 19.6 m s⁻¹_.

Question 15:

A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s², find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Answer:

According to the equation of motion under gravity:

v² − u² = 2 gs

Where,

u = Initial velocity of the stone = 40 m/s

v = Final velocity of the stone = 0

s = Height of the stone

g = Acceleration due to gravity = −10 m s⁻²

Let h be the maximum height attained by the stone.

Therefore,

Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m

Net displacement of the stone during its upward and downward journey

= 80 + (−80) = 0

Question 16:

Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 10²⁴ kg and of the Sun = 2 × 10³⁰ kg. The average distance between the two is 1.5 × 10¹¹ m.

Answer:

According to the universal law of gravitation, the force of attraction between the Earth and the Sun is given by:

Where,

M_Sun = Mass of the Sun = 2 × 10³⁰ kg

M_Earth = Mass of the Earth = 6 × 10²⁴ kg

R = Average distance between the Earth and the Sun = 1.5 × 10¹¹ m

G = Universal gravitational constant = 6.7 × 10⁻¹¹ Nm² kg⁻²

Question 17:

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 the two stones meet after a time t.

(i) For the stone dropped from the tower:

Initial velocity, u = 0

Let the displacement of the stone in time t from the top of the tower be s.

Acceleration due to gravity, g = 9.8 m s⁻²

From the equation of motion,

(ii) For the stone thrown upwards:

Initial velocity, u = 25 m s⁻¹

Let the displacement of the stone from the ground in time t be s‘.

Acceleration due to gravity, g = −9.8 m s⁻²

Equation of motion,

The combined displacement of both the stones at the meeting point is equal to the height of the tower 100 m.

In 4 s, the falling stone has covered a distance given by equation (1) as

Therefore, the stones will meet after 4 s at a height (100 − 80) = 20 m from the ground

Question 18:

A ball thrown up vertically returns to the thrower after 6 s. Find

(a) the velocity with which it was thrown up,

(b) the maximum height it reaches, and

(c) its position after 4 s.

Answer:

Hence, it has taken 3 s to attain the maximum height.(a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey.

Final velocity of the ball at the maximum height, v = 0

Acceleration due to gravity, g = −9.8 m s⁻²

Equation of motion, v = u + gt will give,

0 = u + (−9.8 × 3)

u = 9.8 × 3 = 29.4 ms^{− 1}

Hence, the ball was thrown upwards with a velocity of 29.4 m s⁻¹.

(b) Let the maximum height attained by the ball be h.

Initial velocity during the upward journey, u = 29.4 m s⁻¹

Final velocity, v = 0

Acceleration due to gravity, g = −9.8 m s⁻²

From the equation of motion,

(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.

In this case,

Initial velocity, u = 0

Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s − 3 s = 1 s.

Equation of motion,   will give,

Total height = 44.1 m

This means that the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.