Gravitation

Question (12) 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?

Answer

We know that acceleration due to gravity on earth = 9.8 m s^–2

As the gravitational force on the surface of moon is 1/6 of the gravitational force on earth.

Therefore, the acceleration due to gravity on moon

`=9.8xx1/6\ m//s^2 = 1.63\ m\ s^(-2)`

Given, the mass of object = 10kg

We know that, weight, W = m × g

Thus, weight of the given object on earth,

= 10 kg × 9.8 m s^–2

= 98 kg m s^–2

Thus, weight of object on earth = 98 N

Weight on moon = m x g

= 10 kg × 1.63 m s^–2

= 16.3 kg m s^–2

Thus, weight of object on moon = 16.3 N

Thus, weight of object on earth = 98 N and weight of the given object on moon = 16.3 N Answer

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

Given, initial velocity, u = 49 m/s

The final velocity, v = 0, when it achieves the maximum height

As the object is thrown vertically, hence, g = - 9.8 m/s²

(i) the maximum height to which it rises,

Let height = h

We know that,`v^2=u^2+2gh`

`=>0=(49\ m\ s^(-1))^2+2xx(-9.8m\ s^(-2))xxh`

`=>0=2401\ m^2\ s^(-2)-19.6\ m\ s^(-2)xx h`

`=>2401\ m^2\ s^(-2) = 19.6\ m\ s^(-2)\ h`

`=>h = (2401\ m^2\ s^(-2))/(19.6\ m\ s^(-2)`

`=>h = 122.5\ m`

The ball rises to a maximum height of 122.5 m Answer

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

Let, Time taken to reach the maximum height = t

Here, u = 49m/s

v =0

height, h = 122.5 m

Time, t =?

We know, that v = u + gt

⇒ 0 = 49m/s + (-9.8 m s^–2) × t

⇒ 49 m s^–1 = 9.8 m s^–2 × t

`=>t = (49m\ s^(-1))/(9.8\ m\ s^(-2))`

⇒ t = 5 s

We know that time taken to achieve the maximum height = time taken to reach at the ground.

Here, time taken to achieve the maximum height = 5 s

Therefore, total time taken to return on the ground = 5 s + 5 s = 10s

Total time taken to return at the surface of the earth = 10 s Answer

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

Given,

Height, h = 19.6 m

Initial velocity, u = 0

Final velocity, v =?

We know that,

We know that, `v^2=u^2+2gh`

⇒ v² = (0)² + 2 × 9.8 m s^–2 × 19.6 m

⇒ v²= 2 × 2 × 9.8 m s^–2 × 19.6 m

⇒ v² = 19.6 × 19.6 m² s^–2

`=>v = sqrt(19.6xx19.6\ m^2\ s^(-2))`

`=>v = 19.6\ m\ s^(-1)`

Hence, final velocity of stone will be 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

Given, Initial velocity, u = 40m/s

g = 10m/s²

Final velocity, v = 0

Height, h =?

Net displacement =?

Total distance covered =?

We know that, v² = u² + 2 g h

⇒ 0 = (40m/s)² + 2 × (–10m/s²) × h

⇒ 0 = 1600 m²/s² – 20 m/s² × h

⇒ 20 m/s² × h = 1600 m²/s²

`=>h = (1600\ m^2//s^2)/(20\ m//s^2)`

⇒ h = 80 m

As the stone returns on the ground, hence the total displacement = 0

Total distance covered by the stone,

= height × 2

= 80 m × 2 = 160 m

Hence,

Maximum height reached by the stone = 80 m

Displacement = 0

Total distance covered by the stone = 160 m

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

Given,

Mass of earth, M = 6 × 10²⁴ kg

Mass of sun, m = 2 × 10³⁰ kg

Distance between earth and sun, d = 1.5 × 10¹¹ m.

We know that the value of G = 6.67 × 10^–11 N m² kg²

We know that, `F=Gxx(Mxxm)/d^2`

`=6.67xx10^(-11)xx(6xx10^24xx2xx10^30)(1.5xx10^11)^2\ N`

`=(6.67xx6xx10^43xx2)/(1.5xx1.5xx10^11xx10^11)\ N`

`=(6.67xx6xx2xx10^21)/(1.5xx1.5)\ N`

`=3.57xx10^22\ N`

Hence, the force of gravitation between earth and sun = 3.57 × 10²² N Answer

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 a point, which is `p` meter above from the ground.

Given, height of tower = 100 m

Therefore,

Stone which is falling from the height will meet at `100-p` meter above the ground.

(i) For stone which is falling from the tower.

Initial velocity, u = 0

g = 9.8 m s^–2

Height, h = 100 – p

Let it reaches at the height of `100-p` meter in time `t` second.

We know that, `h=ut+1/2\ g\ t^2`

`=>100-p=0xxt+1/2xx9.8xxt^2`

`=>100-p=1/2x9.8xxt^2`

`=>100-p=4.9\ t^2` ------------- (i)

(ii) for stone which is thrown vertically

Height, = p

Initial velocity, u = 25m/s

g = –9.8 m/s²

Time = t

We know that, We know that, `h=ut+1/2\ g\ t^2`

`=>p = 25xxt+1/2xx(-9.8)xxt^2`

`=>p=25t-4.9 t^2` ---------- (ii)

By adding equation (i) and equation (ii) we get

`100-p+p=25t-4.9t^2+4.9t^2`

`=>100 = 25t`

`=>t = 100/25=4\ s`

By putting the value of t in equation (i), we get

`100-p=4.9xx(4)^2\ m`

`=>100-p=4.9xx16\ m`

`=>100-p=78.4\ m`

`=>p = 100-78.4\ m=21.6\ m`

Thus, both of the stone will meet at 21.6 m above the ground after 4 second. Answer

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

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

Given,

Final velocity, while reaching at the maximum height v =0

Time, t = 6/2 s = 3 s

g = – 9.8 m/s²

Initial velocity, u =?

We know that,

v = u + g t

⇒ 0 = u – 9.8 m/s² × 3s

⇒ u = 29. 4 m/s

Hence, the ball was thrown with a velocity of 29.4 m/s

(b) the maximum height it reaches, and

Initial velocity, u = 29.4m/s

Final velocity, v = 0

Time = 3 s

g = – 9.8 m/s²

h =?

We know that, `s=ut+1/2\ g\ t^2`

`=29.4\ m//s xx3s+1/2xx-9.8\ m//s^2xx(3s)^2`

`=29.4\ m//sxx3s-4.9\ m//s^2xx9s^2`

`=88.2\ m-44.1\ m`

`=>s=44.1\ m`

Hence, maximum height it reaches = 44.1 m

(c) Position after 4 s

It will reaches at the maximum height of 44.1 m in 3 s

Therefore, we have to calculate its position 1 s after start falling.

Hence,

Initial speed, u = 0

Time, t = 1 s

g = 9.8m/s²

h = ?

We know, `s=ut+1/2\ g\ t^2`

`=0xx1\ s+1/2xx9.8\ m//s^2xx(1x)^2`

`=4.9\ m//s^2xx1s^2`

`=>s=4.9\ m`

Position above the ground = total height ? height covered in 1 second

= 44.1 m– 4.9 m = 39.2 m

Hence, the position of ball after 4 second of throwing = 39.2 m above the ground.

Question (19) In what direction does the buoyant force on an object immersed in a liquid act?

Answer

When an object is immersed in a liquid the buoyant force acts in upward direction at the bottom of the object.

Question (20) Why does a block of plastic released under water come up to the surface of water?

Answer

Because the density of block of plastic is less than the water, so it comes up to the surface of water when released under water.

Alternate Answer

When a block of plastic is immersed in water, the weight of water displaced by it is more than the weight of block of plastic, hence it comes on the surface of water when released under water.

Question (21) The volume of 50 g of a substance is 20 cm³. If the density of water is 1 g cm^–3, will the substance float or sink?

Answer

Given, the volume of substance = 20cm³

Mass of substance = 50g

We know that, `text(Density) = {text(Mass)}/{text(Volume)}`

`=(50\ g)/(20\ cm^3)=2.5\ g\ cm^(-3)`

Thus, density of the substance = 2.5 g cm^–3

Since, the density of substance is more than the water, hence given substance will sink in the water.

Question (22) The volume of a 500 g sealed packet is 350 cm³. Will the packet float or sink in water if the density of water is 1 g cm^–3? What will be the mass of the water displaced by this packet?

Answer

Given, Mass of sealed packet = 500g

Volume of the sealed packet = 350 cm³

We know that, We know that, `text(Density) = {text(Mass)}/{text(Volume)}`

`=(500\ g)/(350\ cm^3)`

`=1.42\ g\ cm^(-3)`

Therefore, density of the substance = 1.42 g cm^–3

As the density of substance is more than the water, hence it will sink in the water.

The object which sinks in water displaces the volume of water equal to its volume. Hence, the packet will displace 350 cm³ of water.

We know that,

`text(Density of water)={text(Mass of water (M))}/{text(Volumr of water)}`

`=>1\ g\ cm^(-3)=M/(350\ cm^3)`

`=>M=1\ g\ cm^(-3)xx350\ cm^3`

`=>M=350\ g`

Hence, the sealed packet will displace 350g of water.