# Laws of Motion Exe-3E Numericals Answer Physics Class-9 ICSE Selina Publishers

Laws of Motion Exe-3E Gravitation Numericals Answer Type for Class-9 ICSE Concise Physics. There is the solutions of Numericals Answer type Questions of your latest textbook which is applicable in 2023-24 academic sessionVisit official Website CISCE for detail information about ICSE Board Class-9.

## Laws of Motion Exe-3E Gravitation Numericals Answer

(ICSE Class – 9 Physics Concise Selina Publishers)

 Board ICSE Class 9th Subject Physics Writer / Publication Concise selina Publishers Chapter-3 Laws of Motion Exe – 3E Gravitation Topics Solution of Exe-3(E) Numericals Answer Type Academic Session 2023-2024

### Exe-3E Gravitation Numericals Answer Type

Ch-3 Laws of Motion Physics Class-9 ICSE Concise

Page 89

#### Question 1. The force of attraction between two bodies at certain separation is 10 N. What will be the force of attraction between them if the separation between them is reduced to half?

Given the force of attraction between two bodies = 10 N

Now,  F=G Mm/R².

If the new distance R’= R/2, then let F’ be the force acting between the bodies. Then:

#### Question 2. Write the approximate weight of a body of mass 5 kg. What assumption have you made?

Weight = mg

W = (5) (9.8) = 50 N.

Assumption: Value of acceleration due to gravity = 9.8 m/s2.

#### Question 3. Calculate the weight of a body of mass 10 kg in (a) kgf and (b) newton. Take g = 9.8 m/ s

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.

#### Question 4. 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.

Mass = 5 kg.

g = 9.8 m/s2.

Let F be the force of gravity,

F = mg.

F = (5) (9.8) = 49 N.

Force of gravity always acts downwards.

#### Question 5. The weight of a body is 2.0 N. What is the mass of the body? (g = 10 m s-2)

Weight, W = 2.0 N

g = 9.8 m/s2

Let ‘m’ be the mass of the body.

W = mg

Or, m = W/g = (2/9.8) kg = 0.2 kg.

#### Question 6. 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?

Weight of the body on Earth = 98 N.

Acceleration due to gravity on Earth = 9.8 m/s2.

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’.

W’ = mass  acceleration due to gravity on the Moon.

[Given, acceleration due to gravity on the Moon = 1.6 m/s2]

W’ = 10  1.6 =16 N.

#### Question 7. A man weighs 600 N on the Earth. What would be his approximate weight on the Moon? Give a reason for your 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 1/6th that of Earth and w = mg.

Therefore, man’s weight on Moon = (600/6) = 100 N.

#### Question 8. What is the (a) force and (b) weight act on a mass of 10.5 kg being pulled by the Earth? Take g = 10 m s–

Mass, m = 10.5 kg

G = 10 m/s2

(a) Force, F = mg

F = (10.5) (10) = 105 N

(b) Weight, w = mg

w = (10.5) (10) = 105 N

(Laws of Motion Exe-3E Numericals Class-9 ICSE)

#### Question 9. A ball is released from a height and it reaches the ground in 3 s. If g= 9.8 m s-2, find the value for the following variables:

(a) The height from which the ball was released.

(b) The velocity with which the ball will strike the ground.

Let ‘S’ be the height.

Time taken, t = 3s; g = 9.8 m/s2

Initial velocity, u = 0 (because the body starts from rest)

##### (a) Using the second equation of motion,

S = ut + (1/2) gt2

We get,

S = 0 + (1/2) (9.8) (3) (3)

S = 44.1 m

##### (b) Let ‘v’ be the velocity with which the ball strikes the ground.

Using the third equation of motion,

v2  u2 = 2gs

or, v2 – 02 = 2(9.8) (44.1)

or, v2 = 864.36

or, v = 29.4 m/s

#### Question 10. What force, in newton, your muscles need to apply to hold a mass of 5 kg in your hand? State the assumption.

Mass, m = 5kg

Force, F = mg

F = (5) (9.8) = 49 N

Assumption: Value of acceleration due to gravity is 9.8 m/s2.

#### Question 11. 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 m s-2, find the following variables:

(a) The initial velocity of the ball

(b) The final velocity of ball on reaching the ground and

(c) The total time of journey of ball.

Given, maximum height reached, s = 20 m

Acceleration due to gravity, g = 10 m/s2

##### (i) Let ‘u’ be the initial velocity.

At the highest point, velocity = 0

Using the third equation of motion,

v  u= 2gs

or, 0  u= 2 (10) (20) m/s

or, u=  (400) m/s [Negative sign indicates that the motion is against gravity]

or, u = 20 m/s

##### (b) Let v’ be the final velocity of the ball on reaching the ground.

Considering the motion from the highest point to ground,

Velocity at highest point = 0 = Initial velocity for downward journey of the ball.

Distance travelled, s = 20m

Using the third equation of motion,

v u= 2gs

or, v 0 = 2 (10) (20) m/s

or, v= 400 m/s

or, v = 420 m/s

##### (c) Now total time for which the ball remains in air, t = 2u/g.

Or, t = 2 (20)/(10).

Or, t = 4s.

(Laws of Motion Exe-3E Numericals Class-9 ICSE)

#### Question 12. A 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)

Initial velocity u = 0

Final velocity = 20 m/s

g = 10 m/s2

Let ‘h’ be the height of the tower.

Using the third equation of motion,

v2  u2 = 2gs

or, (20)2  0 = 2 (10) h

or, h = 20 m

#### Question 13. A ball is thrown vertically upwards. It returns 6 s later. Calculate: (i) The greatest height reached by the ball and (ii) The initial velocity of the ball. (Take g = 10 m s-2)

Total time of journey = 6 s

g = 10 m/s2

#### (i) Let ‘H’ be the greatest height.

Time of ascent, t = 6/2 = 3 s,

For ascent, initial velocity, u = 0

Using the second equation of motion,

H = ut + (1/2) gt2

H = 0 + (1/2) (10) (3) 2

H = 45 m

#### (ii) Let u’ be the initial velocity.

Final velocity, v = 0

Using the third equation of motion,

v2  u2 = 2gH

or, v2  0 = 2(10) (45)

or, v2 = 900

or, v = 30 m/s

### Laws of Motion Exe-3E Numericals Class-9 ICSE

Ch-3 Gravitation Physics Class-9 ICSE Concise

Page 90

#### Question 14. A pebble is thrown vertically upwards with a speed of 20 m s-1. How high will it be after 2 s? (Take g = 10 m s-2)

Initial velocity, u = 20 m/s

Time, t = 2s

g = 10 m/s2

Maximum height reached in 2s, H = (1/2) gt2

Or, H = (1/2) (10) (2) 2

Or, H = 20 m

#### Question 15.

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

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

##### (a)

Height, s = 80m

g = 10 m/s2

Using the second equation of motion,

S = ut + (1/2) gt2

Or, 80 = 0+ (1/2) (10) (t) 2

Or, (t) 2 = 16

Or, t = 4s

##### (b)

Let ‘v’ be the velocity on reaching the ground.

Using the third equation of motion,

v2  u2 = 2gH

or, v2  0 = 2(10) (80)

or, v2 = 1600

or, v = 40 m/s

#### Question 16. 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)

Given time t = 2.5, g = 9.8 m/s2

Height, H = (1/2) gt2

Or, H = (1/2) (9.8) (2.5)2

Or, H = 30.6 m

#### Question 17. A ball is thrown vertically upwards with an initial velocity of 49 m s-1. Calculate: (i) The maximum height attained, (ii) The time taken by it before it reaches the ground again. (Take g = 9.8 m s-2).

Initial velocity, u = 49 m/s

g = 9.8 m/s2

##### (i) Let H be the maximum height attained.

At the highest point, velocity = 0.

Using the third equation of motion,

v2  u2 = 2gH

or,0  492 = 2(9.8) (H)

or, H = (492)/ 19.6

or, H = 122.5 m

##### (ii) Total time of flight is given by t = 2u/g

Or, t = 2(49)/ 9.8

Or, t = 10 s

#### Question 18. A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10m s-2, calculate the height of the tower.

Initial velocity u = 0

Time t = 4 s

g = 10 m/s2

Let ‘H’ be the height of the tower.

Using the second equation of motion,

H = ut + (1/2) gt2

Or, H = 0 + (1/2)(10)(4)2

Or, H = 80 m

#### Question 19. 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: (i) The depth of water surface and (ii) The time when echo is heard after the pebble is dropped.

(i) Time t =20 s

g = 10 m/s2

Let ‘D’ be the depth of the well.

Using the second equation of motion,

D = ut + (1/2) gt2

D = 0 + (1/2)(10)(20) 2

D = 2000 m

(ii) Speed of sound = 330 m/s

Depth of well = 2000 m

Time taken to hear the echo after the pebble reaches the water surface = Depth/speed

= (2000/330) s

= 6.1 s

Time taken for pebble to reach the water surface = 20 s.

Therefore, the total time taken to hear the echo after the pebble is dropped = 20 + 6.1 = 21.6 s.

(Laws of Motion Exe-3E Numericals Class-9 ICSE)

#### Question 20. A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6 m s-1. The ball reaches the ground after 5 s. Calculate: (i) The height of the tower, (ii) The velocity of the ball on reaching the ground. Take g= 9.8 ms-2.

(i) Let x be the height of the tower.

Let h be the distance from the top of the tower to the highest point as shown in the diagram.

Initial velocity u = 19.6 m/s

g = 9.8 m/s2

At the highest point, velocity = 0

Using the third equation of motion,

v2  u2 = 2gh

Or, (19.6) 2 = 2 (9.8) h

Or, h = 19.6 m

If the ball takes time t1 to go to the highest point from the top of building, then for the upward journey from the relation, v = u  gt,

0 = 19.6  (9.8) (t1)

Or, t1 = 2s

##### (ii) Let us consider the motion for the part (x+h)

Time taken to travel from highest point to the ground = (5  2) = 3s

Using the equation s = ut + (1/2) gt2

We get,

(x + h) = 0 + (1/2) (9.8) (3) 2

Or, (x + 19.6) = 44.1 m

Or, x = 44.1  19.6 = 24.5 m

Thus, height of the tower = 24.5 m

(iii) Let v be the velocity of the ball on reaching the ground.

Using the relation, v = u + gt

We get:

v = 0 + (9.8) (3)

Or, v = 29.4 m/s

—  : End of Laws of Motion Exe-3E Gravitation Numericals Answer Type  Solutions :–

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