Gravitation - Class 9 Science - Chapter 9 - Notes, NCERT Solutions & Extra Questions
Renews every month. Cancel anytime
Your personal doubt-solving assistant
Chatterbot AI gives you 100% accurate answers to your questions in an instant.
Extra Questions - Gravitation | NCERT | Science | Class 9
A stone is dropped from the top of a tower of height $h = 60 \mathrm{~m}$. Simultaneously another stone is projected vertically upwards from the foot of the tower. They meet at a height $\frac{2h}{3}$ from the ground level. The initial velocity of the stone projected upwards is $g=10 \mathrm{~ms}^{-2}$:
A 20 $\mathrm{~ms}^{-1}$
B 60 $\mathrm{~ms}^{-1}$
C 10 $\mathrm{~ms}^{-1}$
D 30 $\mathrm{~ms}^{-1}$
To solve this problem, let's break it down step-by-step:
Problem Breakdown
We have a stone A dropped from a tower of height $ h = 60 , \text{m} $.
Simultaneously, another stone B is projected vertically upwards from the base of the tower.
They meet at a height $ \frac{2h}{3} $ from the ground.
We need to find the initial velocity of stone B, with the acceleration due to gravity $ g = 10 , \text{m/s}^2 $.
Step-by-Step
1. Determine the meeting height
The stones meet at $ h' = \frac{2h}{3} $.
Given $ h = 60 , \text{m} $, $$ h' = \frac{2 \times 60}{3} = 40 , \text{m} $$
2. Distance traveled by each stone before they meet
Distance traveled by stone B (upwards) to meet point: $ s_B = 40 , \text{m} $.
Distance traveled by stone A (downwards) to meet point: $$ s_A = h - h' = 60 - 40 = 20 , \text{m} $$
3. Time it takes for the stones to meet
For stone A, using the kinematic equation: $$ s = ut + \frac{1}{2}at^2 $$ Since stone A is dropped (initial velocity ( u = 0 )), the equation simplifies to: $$ 20 = 0 \times t + \frac{1}{2} \times 10 \times t^2 $$ $$ 20 = 5t^2 $$ $$ t^2 = 4 $$ $$ t = 2 , \text{seconds} $$
4. Initial velocity of stone B
For stone B, using the same kinematic equation: $$ s = ut + \frac{1}{2}at^2 $$
Here, displacement $ s = 40 , \text{m} $, initial velocity $ u = ? $, time $ t = 2 , \text{seconds} $, and acceleration $ a = -10 , \text{m/s}^2 $ (negative since it's acting in the opposite direction of the motion): $$ 40 = u \times 2 + \frac{1}{2} \times (-10) \times (2^2) $$ $$ 40 = 2u - 20 $$ $$ 2u = 60 $$ $$ u = 30 , \text{m/s} $$
Therefore, the initial velocity of the stone projected upwards is $ \mathbf{30 , \text{m/s}} $.
Final Answer:
D. 30 $\mathrm{~ms}^{-1}$
A stone is dropped from a tower of height 200 m and at the same time another stone is projected vertically up at 50 m/s. The height at which they meet? (Given: $g=10 \mathrm{~m/s}^2$)
A) 5 m
B) 100 m
C) 120 m
D) 125 m
To solve this, let's consider the motion of the two stones:
Stone A: This stone is dropped from a height of 200 meters. Its initial velocity ($ u_A $) is 0.
Stone B: This stone is projected upwards with an initial velocity ($ u_B $) of 50 m/s.
We need to determine the height at which they meet.
Equations of Motion
For Stone A (dropped from the top):
$$ x = u_A t + \frac{1}{2} g t^2 $$
Since $ u_A = 0 $:
$$ x = \frac{1}{2} g t^2 \quad \text{(1)} $$
For Stone B (thrown upwards):
$$ 200 - x = u_B t - \frac{1}{2} g t^2 \quad \text{(2)}$$
Substitute and Solve
From Equation (1):
$$ x = \frac{1}{2} (10) t^2 = 5 t^2 $$
Substituting $ x $ in Equation (2):
$$ 200 - 5 t^2 = 50 t - 5 t^2 $$
Simplifying:
$$ 200 = 50 t $$
$$ t = \frac{200}{50} = 4 \text{ s} $$
Finding the Height
Now substitute $ t = 4 $ seconds back into Equation (1):
$$ x = 5 t^2 = 5 (4)^2 = 5 \times 16 = 80 \text{ m} $$
Therefore, the height from the top where they meet:
$$ h = 200 - x = 200 - 80 = 120 \text{ meters} $$
A stone is thrown upwards from the top of a tower 85 m high. It reaches the ground in 5s. Calculate:
The greatest height above the ground.
The velocity with which it reaches the ground.
The time taken to reach the maximum height. (g=10 m/s²)
The problem involves a stone thrown upwards from the top of an 85-meter-high tower. It reaches the ground in 5 seconds. Let’s calculate the parameters needed using the given data and the following equations of motion.
Given:
Height of the tower: $85 , \text{m}$
Time taken to reach the ground: $5 , \text{s}$
Acceleration due to gravity: $g = 10 , \text{m/s}^2$ (downwards)
Initial Calculations:
Let's assume the initial velocity is $$u$$. The displacement of the stone when it reaches the ground is $$ -85 , \text{m}$$, since it moves downward.
Using the second equation of motion: $ S = ut + \frac{1}{2} a t^2 $
Substitute the values:
$$ -85 = 5u + \frac{1}{2}(-10)(5^2) $$
$$ -85 = 5u - 125 $$
$$ 40 = 5u $$
$$ u = 8 , \text{m/s} $$
1. Greatest height above the ground:
To find the greatest height, we'll calculate the additional height (s) the stone reaches above the tower top before it starts descending.
At maximum height, the stone's final velocity is 0. Using the equation:
$$ v^2 = u^2 + 2as $$
$$ 0 = 8^2 + 2(-10)s $$
$$ 0 = 64 - 20s $$
$$ 20s = 64 $$
$$ s = 3.2 , \text{m} $$
Therefore, the maximum height above the ground is:
$$ 85 , \text{m} + 3.2 , \text{m} = 88.2 , \text{m} $$
2. Velocity with which it reaches the ground:
Using the first equation of motion: $ v = u + at $
Substitute the given values:
$$ v = 8 , \text{m/s} + (-10 , \text{m/s}^2 \cdot 5 , \text{s}) $$
$$ v = 8 - 50 $$
$$ v = -42 , \text{m/s} $$
The negative sign indicates the direction is downward. Velocity with which the stone reaches the ground is: $42 , \text{m/s} \text{ (downward)} $
3. Time taken to reach the maximum height:
Using the formula for time to reach maximum height:
$$ t = \frac{u}{g} $$
$$ t = \frac{8}{10} = 0.8 , \text{s} $$
Thus, the stone takes 0.8 seconds to reach the maximum height.
Summary:
Greatest height above the ground: $88.2 , \text{m}$
Velocity with which the stone reaches the ground: $42 , \text{m/s} , \text{(downward)}$
Time to reach maximum height: $0.8 , \text{s}$
A ball is launched from the top of Mt. Everest, which is at an elevation of 9000 m. The ball moves in a circular orbit around Earth. Acceleration due to gravity near Earth's surface is ( g ). The magnitude of the ball's acceleration while in orbit is:
A. Close to $ \dfrac{g}{2}$
B. Zero
C. Much greater than ( g )
D. Nearly equal to ( g )
The correct option is D: Nearly equal to ( g ).
To understand why, let's delve into the details. When the ball moves in a circular orbit around Earth, its acceleration can be described by the following equation:
$$ \frac{mv^2}{r} = mg' $$
Here:
( m ) is the mass of the ball.
( v ) is the velocity of the ball in its circular orbit.
( r ) is the radius of the orbit, which is very close to Earth's radius because the elevation (9000 m) is much smaller compared to Earth's radius (approximately 6371 km).
( g' ) is the acceleration due to gravity at the orbit's altitude.
Since the altitude of 9000 m is quite small relative to the Earth's radius, the difference between ( g ) and ( g' ) is minimal. Therefore, ( g' ) is nearly equal to ( g ), and we can approximate:
$$ g' \approx g $$
Thus, the ball’s acceleration while in orbit is nearly equal to ( g ).
💡 Have more questions?
Ask Chatterbot AIExtra Questions and Answers - Gravitation | NCERT | Science | Class 9
Describe the effect of force acting upon an object when air resistance is greater than the force of gravity.
When an object falls, it accelerates downwards due to its weight (the force of gravity acting upon it). Simultaneously, it experiences an upward force known as air resistance (or drag), which tends to slow it down.
In the scenario where air resistance exceeds the force of gravity, the dynamics of the object's movement change significantly. Instead of falling, the object will actually move in the opposite direction — upwards. This occurs because the net force, which is the sum of all forces acting on the object, now points upward rather than downward.
An illustrative example of this scenario would be placing a lightweight object like a paper ball above a horizontally positioned table fan. When the fan is turned on, the paper ball rises because the upward force (air generated by the fan) overpowering the downward gravitational pull, resulting in a net upward force. This demonstrates how an object will ascend if the upward force (air resistance in this case) is greater than the downward gravitational force.
The different types of reflecting telescopes are:
A. Newtonian
B. Cassegrain
C. Both A and B
D. None of the above
The correct option is C. Both A and B.
Reflecting telescopes come in different designs, and among these, the Newtonian and Cassegrain telescopes are two prominent types. Hence, both options A (Newtonian) and B (Cassegrain) are part of the reflecting telescopes category.
A $100 \mathrm{~kg}$ block is dropped from the top of a building $80 \mathrm{~m}$ tall. Assume $\mathrm{g}=10 \mathrm{~ms}^{-2}$. Find the work done by the force of gravity as it falls vertically to hit the ground.
A) $80000 \mathrm{~J}$
B) $-80000 \mathrm{~J}$
C) $20000 \mathrm{~J}$
D) $0 \mathrm{~J}$
To determine the work done by the force of gravity as a $100 \mathrm{~kg}$ block falls from a height of $80 \mathrm{~m}$, we can use the formula:
$$ \text{Work Done} = \overrightarrow{F} \cdot \overrightarrow{s} $$
Here, the force $\overrightarrow{F}$ is due to gravity and equals:
$$ F = mg = 100 \times 10 = 1000 \mathrm{~N} $$
where $m$ is the mass of the block and $g$ is the acceleration due to gravity. The displacement $\overrightarrow{s}$ equals $80 \mathrm{~m}$.
Now, since the force due to gravity and displacement are in the same direction (vertically downward), the angle $\theta$ between $\overrightarrow{F}$ and $\overrightarrow{s}$ is $0^\circ$. Therefore, using the formula for work done:
$$ W = FS \cos \theta = 1000 \times 80 \times \cos 0^\circ = 1000 \times 80 = 80000 \mathrm{~J} $$
Thus, Option A ($80000 \mathrm{~J}$) is the correct answer, indicating that the work done by gravity is positive and has a magnitude of 80000 Joules.
A body of density $d_{1}$ is counterpoised by $Mg$ of weights of density $d_{2}$ in air of density $d$. Then the true mass of the body is
A) $M$ (B) $M\left[1-\frac{d}{d_{2}}\right]$ (C) $M\left[1-\frac{d}{d_{1}}\right]$ (D) $\frac{M\left[1-\frac{d}{d_{2}}\right]}{\left[1-\frac{d}{d_{1}}\right]}$
The correct option is (D)$$ \frac{M\left[1-\frac{d}{d_2}\right]}{\left[1-\frac{d}{d_1}\right]} $$
Let $M_0$ be the true mass of the body in vacuum. The apparent weight of the body in air is equal to the apparent weight of the weights used for counterpoising in air. Therefore, the actual weight minus the upthrust due to displaced air on both sides must be equal:
$$ M_0 g - \left(\frac{M_0}{d_1}\right) d g = M g - \left(\frac{M}{d_2}\right) d g $$ Simplifying the equation, we find:
$$ M_0 (g - \frac{d}{d_1} g) = M (g - \frac{d}{d_2} g) $$
Dividing through by $g$ and rearranging gives us:
$$ M_0 = \frac{M\left[1-\frac{d}{d_2}\right]}{\left[1-\frac{d}{d_1}\right]} $$ This expression represents the true mass of the body, accounting for the buoyant forces acting due to the air on the body and weights.
If I want to play a prank on my friend and turn his world upside down, I'll have to make his lift move at an acceleration of to make it seem that the world is upside down to him.
A $\mathrm{g}$ downwards
B g upwards
C $2 \mathrm{~g}$ downwards
D $2 \mathrm{~g}$ upwards
The correct answer is C $2 \mathrm{g}$ downwards.
To create the perception that the world has turned upside down, the lift needs to accelerate downward at $2 \mathrm{g}$. This acceleration effectively doubles the force of gravity, making it feel like "down" is now "up".
Q66. Orbital decay, a process of prolonged reduction in the altitude of an artificial satellite's orbit, is caused by which of the following reasons?
Atmospheric drag
Tides
Gravitational pull
Correct code: Q66. परिक्रमापथ क्षय, कृत्रिम उपग्रहीय परिक्रमा पथ की ऊँचाई में लंबे समय तक होने वाली कमी होती है। यह निम्नलिखित में से किन कारणों से होती है?
वायुमंडलीय कर्षण
ज्चारभाटा
गुरूत्वीय खींचाव
Correct options are:
A (a) Only 1 (a) केवल 1
B (b) Only 3 (b) केवल 3
C (c) Only 1 and 3 (c) केवल 1 और 3
D (d) All of the above (d) उपर्युक्त सभी
The correct answer is D (d) All of the above.
Orbital decay, the prolonged reduction in the altitude of an artificial satellite's orbit, is caused by several factors including atmospheric drag, tides, and the gravitational pull from celestial bodies and Earth itself. Each of these contributes to the gradual lowering of a satellite's orbit over time.
The acceleration due to gravity at Earth's surface is directed towards the $\qquad$ of the Earth.
A) equator
B) south pole
C) centre
D) north pole
The correct option is C) centre.
The center of gravity of an object is the point at which the net gravitational force seems to act. Essentially, the gravitational force acts between the centers of mass of two objects.
For the Earth, the center of gravity lies at its center. Therefore, the acceleration due to gravity is always directed towards the centre of the Earth.
A flower pot falls off a window sill and falls past the window below. It takes $0.5 \mathrm{~s}$ to pass through a window that is $2.0 \mathrm{~m}$ above the ground. Find how high the window sill is from the ground. (Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$)
A) $2.5 \mathrm{~m}$ B) $2.25 \mathrm{~m}$ C) $3.25 \mathrm{~m}$ D) $4.5 \mathrm{~m}$
The correct answer is C) $3.25 , \text{m}$. To find the height of the window sill from the ground, we need to consider the distance the pot covers as it falls past the window.
Assuming the initial velocity ($u$) of the pot was zero (starting from rest), and taking the acceleration due to gravity ($g$) to be $10 , \text{m/s}^2$, with downward direction as positive. The time it takes to fall past the window ($t$) is given as $0.5 , \text{s}$.
We can use the formula for the distance traveled under uniform acceleration: $$ h = ut + \frac{1}{2} gt^2 $$ Substituting the given values: $$ h = 0 + \frac{1}{2} \times 10 \times (0.5)^2 $$ $$ h = \frac{1}{2} \times 10 \times 0.25 $$ $$ h = 1.25 , \text{m} $$
Here, $h$ is the distance the pot fell. Since it fell past a window which is $2.0 , \text{m}$ above the ground, the total height of the window sill from the ground includes the height of the window: $$ \text{Height of window sill} = h + \text{height of the window above the ground} $$ $$ \text{Height of window sill} = 1.25 , \text{m} + 2.0 , \text{m} $$ $$ \text{Height of window sill} = 3.25 , \text{m} $$
Therefore, the height of the window sill from the ground is $3.25 , \text{m}$.
Terminal velocity of Mars rover depends upon $P=$ gas density of atmosphere of Mars $A=$ frontal area $W=$ weight of falling object What may be the possible relation between terminal velocity $v$ and the given quantities?
(A) $v=\frac{W}{P A}$
(B) $v=\sqrt{\frac{W}{P A}}$
(C) $v=\sqrt{\frac{\overline{P A}}{W}}$
(D) $v=\sqrt{\frac{P A^{3}}{W}}$
The correct answer is (B) ( v = \sqrt{\frac{W}{PA}} ).
To find the relationship between terminal velocity ( v ) and the parameters ( P ), ( A ), and ( W ), we can set up an equation based on dimensional analysis:
Let's denote:
( v = P^x A^y W^z ) where ( x ), ( y ), and ( z ) are unknowns to be determined.
The dimensions of each variable are:
( [v] = LT^{-1} ) (velocity)
( [P] = ML^{-3} ) (density)
( [A] = L^2 ) (area)
( [W] = MLT^{-2} ) (weight, force)
The dimensional equation is: $$ [L T^{-1}] = [M L^{-3}]^x [L^2]^y [M L T^{-2}]^z $$
Expanding and equating the dimensions on both sides: $$ M^0 L^1 T^{-1} = M^{x+z} L^{-3x + 2y + z} T^{-2z} $$
From here, we can set up a system of equations by equating the powers of each fundamental unit:
Mass (M): ( x+z=0 )
Time (T): ( -2z=-1 ) (Solving this gives ( z=\frac{1}{2} ))
Length (L): ( -3x+2y+z=1 )
Using ( z=\frac{1}{2} ) in the mass equation: ( x+\frac{1}{2}=0 ) leads to ( x=-\frac{1}{2} ).
Substitute ( x ) and ( z ) in the length equation: $$ -3(-\frac{1}{2}) + 2y + \frac{1}{2} = 1 \ \frac{3}{2} + 2y + \frac{1}{2} = 1 \ 2y + 2 = 1 \text{ so } 2y = -1 \ y = -\frac{1}{2} $$
Thus, the relationship becomes: $$ v = P^{-\frac{1}{2}} A^{-\frac{1}{2}} W^{\frac{1}{2}} $$
Which simplifies to: $$ v = \sqrt{\frac{W}{PA}} $$
This confirms that option (B) is correct.
Weight of a body on earth is $48 \mathrm{~N}$; its weight on moon is
A $\quad 9.8 \mathrm{~N}$
B $98 \mathrm{~N}$
C $\quad 48 \mathrm{~N}$
D $8 \mathrm{~N}$
The correct answer is D) $8 \mathrm{~N}$.
The weight of an object on the moon can be calculated using the ratio of gravitational forces on the moon and the Earth. The gravitational pull on the moon is about $\frac{1}{6}$ of that on Earth. Therefore, the formula to find the weight on the moon $(W_m)$ if we know the weight on Earth $(W_e)$ is:
$$ W_m = \frac{W_e}{6} $$
Given the weight on Earth is $48 \mathrm{~N}$, substituting this into the formula:
$$ W_m = \frac{48 \mathrm{~N}}{6} = 8 \mathrm{~N} $$
Thus, the weight of the body on the moon would be 8 N.
The moon is acted upon by the gravitational pull of the earth and still it does not fall on it. Explain why. Determine the ratio of an object of mass $50 , \mathrm{kg}$ on Earth and the Moon, as well as the ratio of their weights.
The Moon experiences the gravitational pull of the Earth but it does not fall towards the Earth because it is in a continuous state of free fall around the Earth. This creates a centripetal force, which keeps the Moon in a stable orbit around the Earth.
Let's define and calculate the weights:
Let $ W_e $ denote the weight of an object on Earth.
Let $ W_m $ denote the weight of an object on the Moon.
Given that the gravitational force on the Moon is approximately one sixth of that on Earth:
$$ g_{\text{moon}} = \frac{1}{6} g_{\text{earth}} $$
Assuming $ g_{\text{earth}} \approx 10 , \mathrm{m/s^2} $, then $ g_{\text{moon}} \approx \frac{10}{6} \approx 1.67 , \mathrm{m/s^2} $.
For a $50 , \mathrm{kg}$ object:
The mass $ m $ is the same on Earth and the Moon (mass is constant) and equals $50 , \mathrm{kg}$.
On Earth: $$ W_e = m \times g_{\text{earth}} = 50 \times 10 = 500 , \mathrm{N} $$
On the Moon: $$ W_m = m \times g_{\text{moon}} = 50 \times 1.67 \approx 83.5 , \mathrm{N} $$
To find the ratio of weights:
$ W_e : W_m = 500 : 83.5 \approx 6 : 1 $
Thus, the weight ratio (Earth to Moon) for this 50 kg object is about 6:1.
An astronaut jumps from an airplane. After he has fallen 20 m, his parachute opens. Now he falls with a retardation of 2 m/s^2 and reaches the earth with a velocity of 4.0 m/s.
Solution
Before the parachure opens:
Distance travelled, ( s ): 20 m
Initial velocity, ( u ) = 0 (starting from rest)
Acceleration due to gravity, ( a ) = 10 m/s²
Using the kinematic equation: $$ v^2 = u^2 + 2as $$ We plug in the values: $$ v^2 = 0 + 2 \times 10 \times 20 = 400 $$ So, $$ v = \sqrt{400} = 20 \text{ m/s} $$
After the parachute opens:
Initial velocity, ( u ) = 20 m/s (velocity before parachute opens)
Final velocity, ( v ) = 4.0 m/s
Retardation ( a ) = -2 m/s² (since it slows down)
Using the same kinematic equation: $$ v^2 = u^2 + 2as $$ We find ( s ): $$ 4^2 = 20^2 + 2 \times (-2) \times s $$ $$ 16 = 400 - 4s $$ $$ 4s = 400 - 16 $$ $$ 4s = 384 $$ $$ s = \frac{384}{4} = 96 \text{ m} $$
Initial height of the airplane: Total distance the astronaut travels is the sum of the distances before and after the parachute opens: $$ \text{Total distance} = 20 \text{ m} + 96 \text{ m} = 116 \text{ m} $$
Thus, the initial height of the airplane is 116 m above the Earth.
"You are able to watch the live telecast of a cricket match because of artificial satellites."
A) True
B) False
The correct answer is A) True.
Artificial satellites are engineered and launched into orbit around the Earth by scientists. These satellites are equipped to gather and transmit information from various regions across the globe.
Consequently, they enable us to receive live broadcasts of events, such as a cricket match, regardless of where they are happening on Earth. Hence, watching a live telecast of a cricket match is made possible due to these satellites.
Calculate the amount of force exerted by gravity on a person of mass $50 \mathrm{~kg}$.
A) $500 \mathrm{~N}$
B) $50 \mathrm{~N}$
C) $5 \mathrm{~N}$
D) $0.5 \mathrm{~N}$
The correct answer is Option A: $500 \mathrm{~N}$.
Given:
Mass of the person, ( m = 50 \mathrm{~kg} )
You can calculate the force exerted by gravity (weight) using the formula: $$ F = mg $$ where ( g ) is the acceleration due to gravity. On Earth, ( g \approx 9.8 \mathrm{~m/s^2} ).
Using the given mass: $$ F = 50 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2} = 490 \mathrm{~N} $$ Rounding up (since ( g ) in many practical situations is approximated to ( 10 \mathrm{~m/s^2} )), the force exerted by gravity is: $$ F \approx 500 \mathrm{~N} $$
Therefore, the force exerted by gravity on a person of mass $50 \mathrm{~kg}$ is approximately 500 Newtons.
If you kick a football, it moves because of:
A) Gravitational force
B) Friction with the ground
C) Presence of a magnet
D) Force applied on it
The correct answer is D) Force applied on it.
Force applied to an object can alter its state of motion, which aligns with Newton's laws of motion. Therefore, when you kick a football, the reason it moves is due to the force applied by your kick.
A stone is dropped from the 25th storey of a multistorey building and reaches the ground in 5 s. How many storeys of the building will the stone cover in the first second, if the height of each storey is the same? (Take ( g = 10 \ \mathrm{m/s^2} ))
A) 1
B) 2
C) 3
D) 5
Solution
The correct option is A) 1
Let the height of a single storey be $h$. The total height from where the stone was dropped, denoted as $H$, is: $$ H = 25h $$ Given, initial velocity $u=0 , m/s$, acceleration $a=g=10 , m/s^2$, distance $s=H$, and time $t=5 , s$.
We use the formula for the distance covered under uniform acceleration: $$ s = ut + \frac{1}{2}at^2 $$ Substituting the values, we get: $$ H = 0 \times 5 + \frac{1}{2} \times 10 \times 5^2 $$ $$ 25h = \frac{1}{2} \times 10 \times 25 $$ $$ h = 5 , m $$
So, the height of one storey is 5 m.
Now, we need to find the distance travelled in the first second: $$ s = ut + \frac{1}{2}at^2 $$ $$ s = 0 + \frac{1}{2} \times 10 \times 1^2 $$ $$ s = 5 , m $$
Therefore, the stone will pass through one storey in the first second.
Option A) is correct.
A spring balance attached to the roof of a moving elevator indicates 80 kg weight as the weight of a 120 kg body. What is the acceleration of the elevator? (Take $g=10 \mathrm{~m/s}^{2}$)
A. 9.8 m/s^{2}
B. 4.9 m/s^{2}
C. 2 m/s^{2}
D. 3.3 m/s^{2}
The correct answer is D. 3.3 m/s².
Given values:
Measured mass = $80 , \text{kg}$
Actual mass = $120 , \text{kg}$
The measured weight is less than the actual weight. Therefore, the elevator must be moving downward with acceleration. In this scenario, the net force acting on the body inside the elevator is due to the difference between gravitational force and the force exerted by the elevator's acceleration. The equation that relates these is given by:
$$ N = m(g - a) $$
where $N$ is the normal force (apparent weight), $m$ is the actual mass, $g$ is the acceleration due to gravity, and $a$ is the acceleration of the elevator.
The apparent mass, $m'$, from the normal force can be expressed as:
$$ m' = \frac{N}{g} = \frac{m(g - a)}{g} $$
Substitute the measured mass for $m'$ and solve for $a$:
$$ 80 = 120 \frac{(g - a)}{g} $$
Substituting the value of $g = 10 , \text{m/s}^2$ into the equation:
$$ 80 = 120 \frac{(10 - a)}{10} $$
To solve for $a$, start by simplifying and rearranging:
$$ 80 = 12 (10 - a) $$
$$ \frac{80}{12} = 10 - a $$
$$ 6.67 = 10 - a $$
$$ a = 10 - 6.67 = 3.33 , \text{m/s}^2 $$
So, the elevator's acceleration is $3.33 , \text{m/s}^2$ downward.
The force of $\qquad$ exists between any two masses in the universe.
A friction
B gravity
C normal
D magnets
The correct answer is B gravity.
Gravity is the force that exists between any two masses in the universe. This force is attractive in nature and is considered a non-contact force.
How much momentum will an object of mass $10 \mathrm{~kg}$ transfer to the floor, if it falls from a height of $5 \mathrm{~m}$?
To determine the momentum transferred to the floor by an object falling from a height, we first need to calculate the velocity of the object just before it impacts the ground. The velocity can be computed using the formula for the final velocity of a freely falling object, which is: $$ v = \sqrt{2gh} $$ where ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )), ( h ) is the height from which the object is dropped, and ( v ) is the final velocity. For this calculation, we'll use ( g = 10 , \text{m/s}^2 ) for simplicity, and ( h = 5 , \text{m ). The calculation yields: $$ v = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10 , \text{m/s} $$
Next, we calculate the momentum ( p ) of the object when it touches the floor using the formula: $$ p = mv $$ where ( m ) is the mass of the object and ( v ) is its velocity at impact. Given that the mass ( m = 10 , \text{kg} ) and ( v = 10 , \text{m/s} ), the momentum transferred to the floor is: $$ p = 10 \times 10 = 100 , \text{kg m/s} $$
Therefore, the momentum transferred by the object to the ground upon impact is 100 kg m/s.
"Why does Newton's law of gravitation strictly hold for point masses only?"
Newton's law of gravitation is often considered to strictly hold for point masses, but this notion isn't entirely accurate. Newton's Universal Law of Gravitation is indeed applicable to all masses, both theoretical point masses and real extended bodies. According to this law, the gravitational force ($ F $) between two masses is given by:
$$ F = G \frac{M_1 M_2}{d^2} $$
where:
$ F $ is the gravitational force,
$ G $ is the universal gravitational constant,
$ M_1 $ and $ M_2 $ are the masses involved,
$ d $ is the distance between the centers of the two masses.
The critical assumption for using this formula effectively lies in considering the distance ($ d $) between the centers of mass of the objects. For point masses, this is straightforward, as their mass is theoretically concentrated at a single point. For extended bodies, the distance should ideally be measured between their centers of mass, which assumes that their mass distribution can be approximated as point-like for the purposes of the calculation.
Thus, the law does not apply exclusively to point masses; rather, the concept of point masses simplifies calculations especially in complex systems. This concept effectively reduces extended bodies to more manageable theoretical points, which is why Newton's law is so widely applicable despite these simplifications. The utilization of centers of mass is a fundamental abstract simplification that allows us to apply Newton's law to a vast range of practical problems involving real physical bodies.
In summary, Newton's law is universally applicable to all masses when considering their centers of mass, making it a powerful tool in understanding and calculating gravitational forces in many scenarios.
Which of the following was the first man-made object to reach the Moon?
A) Apollo 8
B) Apollo 11
C) Luna 2
D) Chandrayaan 1
The correct answer is C) Luna 2.
Luna 2 holds the distinction of being the first man-made object to reach the Moon, achieving this milestone on September 14, 1959.
A ball is dropped from a height of 20 meters and at the same time another ball is thrown up from the ground with the speed of 20 m/s. When and where will the balls meet?
To determine when and where two balls meet under the given conditions, we set up equations for their motions:
Ball Dropping From 20 m:
Initial velocity ($ u $) = 0 m/s
Distance it falls to meet the other ball ($ s $) = $ 20 - h $ m
Acceleration due to gravity ($ g $) = 9.8 m/s².
Using the second equation of motion: $$ s = ut + \frac{1}{2}gt^2 $$ Plugging in the values: $$ 20 - h = 0 \cdot t + \frac{1}{2} \cdot 9.8 \cdot t^2 $$ or $$ 20 - h = 4.9t^2 \quad \text{(Equation 1)} $$
Ball Thrown Upwards:
Initial velocity ($ u $) = 20 m/s
Distance it rises to meet the other ball ($ s $) = $ h $ m
Acceleration due to gravity ($ g $) = -9.8 m/s² (since it's going up).
Again using the equation: $$ s = ut + \frac{1}{2}gt^2 $$ $$ h = 20t - 4.9t^2 \quad \text{(Equation 2)} $$
By adding Equation 1 and Equation 2, both $ h $ terms cancel out: $$ 20 - h + h = 4.9t^2 + 20t - 4.9t^2 $$ $$ 20 = 20t $$ $$ t = 1 \text{ s} $$
Substituting $ t = 1 $ s into Equation 2 to find $ h $: $$ h = 20t - 4.9t^2 $$ $$ h = 20 \cdot 1 - 4.9 \cdot 1^2 $$ $$ h = 15.1 \text{ m} $$
Thus, the balls meet 1 second after being released at a height of 15.1 meters from the ground.
If the mass and radius of the Earth are $6.0 \times 10^{24} \mathrm{~kg}$ and $6.4 \times 10^{6} \mathrm{~m}$ respectively, calculate the force exerted by the Earth on a body of mass $1 \mathrm{~kg}$. Also, calculate the acceleration produced in the body of mass $1 \mathrm{~kg}$ and the acceleration produced in the Earth.
A) $9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}$ and $1 \frac{\mathrm{m}}{\mathrm{s}^{2}}$ respectively.
B) $9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}$ and $1.63 \times 10^{-24} \frac{\mathrm{m}}{\mathrm{s}^{2}}$ respectively.
C) $1.63 \times 10^{-24} \frac{\mathrm{m}}{\mathrm{s}^{2}}$ and $9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}$ respectively.
D) $9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}$ for both.
The correct answer is Option B: $9.8 , \text{m/s}^2$ and $1.63 \times 10^{-24} , \text{m/s}^2$ respectively.
Using Newton's law of gravitation, the force of attraction $ F $ between two masses is given by: $$ F = \frac{G m_1 m_2}{r^2} $$ where:
$ G = 6.67 \times 10^{-11} , \text{Nm}^2/\text{kg}^2 $ is the gravitational constant,
$ m_1 = 6.0 \times 10^{24} , \text{kg} $ (mass of the Earth),
$ m_2 = 1 , \text{kg} $ (mass of the body),
$ r = 6.4 \times 10^{6} , \text{m} $ (radius of the Earth).
Substituting these values into the equation, the force exerted by the Earth on a 1 kg mass calculates as: $$ F = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1}{(6.4 \times 10^{6})^2} = 9.8 , \text{N} $$
The acceleration $ a $ produced in a body of mass $ 1 , \text{kg} $ due to this force is: $$ a = \frac{F}{m} = \frac{9.8}{1} = 9.8 , \text{m/s}^2 $$ Hence, the acceleration due to Earth's gravity on the mass is $9.8 , \text{m/s}^2$.
Considering Newton's third law of motion, the same force is exerted on the Earth by the 1 kg mass but the resulting acceleration $ a_{\text{Earth}} $ of the Earth is: $$ a_{\text{Earth}} = \frac{F}{\text{mass of the Earth}} = \frac{9.8}{6.0 \times 10^{24}} = 1.63 \times 10^{-24} , \text{m/s}^2 $$
This reveals that the acceleration of the Earth due to the force exerted by a 1 kg mass is $1.63 \times 10^{-24} , \text{m/s}^2$, which is practically imperceptible.
If Earth stands still, what will be its effect on man's weight?
A) Increases
B) Decreases
C) Remains the same
D) None of these
The correct option is A: Increases
Explanation:
When the Earth suddenly stops spinning, the centrifugal force that normally acts outwards on objects (due to the Earth's rotation) will become zero. This force slightly reduces the effect of gravity by providing an outward pull from the center of the Earth. Without this force, the only force acting on a person is gravity. Consequently, there is an increase in the effective weight of a person since there is no outward force to counteract gravity slightly.
In summary, if the Earth stands still, a person's weight will increase because the centrifugal force, which slightly counters the effect of gravity, will no longer be present.
An iron ball and a wooden ball of the same radius are released from a height '$h$' in a vacuum. Which of the two balls would take more time to reach the ground?
A) Iron ball
B) Both would take the same time
C) Wooden ball
D) None of these
Correct Answer: B) Both would take the same time
Explanation:
In a vacuum, all objects fall at the same rate regardless of their mass. This is because the acceleration due to gravity, $ g $, is the same for all objects when air resistance is negligible or absent. The time of fall, $ t $, for any object dropped from a height $ h $ in vacuum is given by the formula:
$$ t = \sqrt{\frac{2h}{g}} $$
This shows that the time taken to hit the ground is dependent only on the height $ h $ from which the object is released and the acceleration due to gravity $ g $, but not on the material or weight of the object.
Since both the iron ball and the wooden ball have the same radius and are released from the same height $ h $ in a vacuum, they will both reach the ground at the same time.
A simple pendulum of mass $m$ is performing vertical circular motion with its velocity at the highest point being $V=\sqrt{2 g l}$, where $l$ is the length of the string. Find the tension in the string when it reaches the lowest point.
A) $2mg$
B) $6mg$
C) $7mg$
D) $3mg$
The tension in the string when the pendulum reaches the lowest point is correctly determined as option C, which is $7mg$.
Using the principle of conservation of mechanical energy (Work-Energy Theorem, WET) between the highest and the lowest point: $$ K_i + U_i + W_{NC} = K_f + U_f $$ where:
$K_i = \frac{1}{2} m(2gl)$ is the initial kinetic energy,
$U_i = mg(2l)$ is the initial potential energy (at highest point),
$W_{NC} = 0$, as there are no non-conservative work being done,
$K_f = \frac{1}{2} mu^2$ is the final kinetic energy (at lowest point),
$U_f = 0$, since potential energy at the lowest point is zero (reference level).
From the conservation, the total initial mechanical energy is converted into kinetic energy at the lowest point: $$ \frac{1}{2} m(2gl) + mg(2l) = \frac{1}{2} mu^2 + 0 $$ Solving for $u^2$: $$ u^2 = 6gl $$
To find the tension $T$ at the lowest point, consider the forces acting: $$ T = mg + \frac{mu^2}{l} $$ Here, $\frac{mu^2}{l}$ is the centrifugal force required to maintain circular motion. Substituting $u^2 = 6gl$, we get: $$ T = mg + m \frac{6gl}{l} = mg + 6mg = 7mg $$
Therefore, the tension in the string at the lowest point of the pendulum's motion is $\mathbf{7mg}$.
Which of the following forces is responsible for the raindrops falling downwards?
A. Electrostatic force
B. Magnetic force
C. Gravitational force
D. Frictional force
The correct answer is C. Gravitational force.
Water droplets within clouds, which later evolve into raindrops, have mass. Consequently, they are subject to the gravitational pull of the Earth, drawing them downwards. This is why raindrops fall towards the ground rather than floating or moving in any other direction.
A boy is standing on the edge of the roof at a height of 78.4 m. He throws a ball vertically upwards with a velocity of 19.6 m/s. After how much time will the ball reach the ground? (Take g = 9.8 m/s^2)
A) 5.42 s
B) 6.47 s
C) 4.47 s
D) 3.42 s
The correct option is B) 6.47 s
Given the height of the roof, $$ h = 78.4 , \text{m} $$
Consider two cases:
Case 1: Ball going from the roof to its highest point.
Given that the initial velocity, $ u_1 = 19.6 , \text{m/s} $, final velocity, $ v_1 = 0 , \text{m/s} $ (since it's the highest point), and acceleration, $ a = -9.8 , \text{m/s}^2 $ (acceleration is negative because it opposes the motion),
Time taken to reach this point is $ t_1 $ and the distance covered is $ s_1 $. By using the first equation of motion, $$ v_1 = u_1 + a t_1 $$ Replacing with known values, $$ 0 = 19.6 - 9.8 t_1 $$ $$ t_1 = 2 , \text{s} $$
Calculating $ s_1 $ from the third equation of motion, $$ v_1^2 = u_1^2 + 2as_1 $$ $$ 0 = 19.6^2 - 2 \times 9.8 \times s_1 $$ $$ s_1 = 19.6 , \text{m} $$
Case 2: From the highest point back to the ground.
Here, the initial velocity $ u_2 = 0 , \text{m/s} $,
The total distance to the ground $ s_2 = s_1 + h = 78.4 + 19.6 = 98 , \text{m} $.
Using the second equation of motion, $$ s_2 = u_2 t_2 + \frac{1}{2} a t_2^2 $$ $$ 98 = 0 + \frac{1}{2} \times 9.8 \times t_2^2 $$ $$ t_2 = \sqrt{20} = 4.47 , \text{s} $$
Therefore, the total time taken for the ball to hit the ground $$ t = t_1 + t_2 = 2 + 4.47 = 6.47 , \text{s} $$
2 equal masses separated by a distance $d$ attract each other with a force $F$. If one unit mass is transferred from one of them to the other, what happens to the force?
To solve the problem, we use the formula for gravitational force between two masses:
$$ F = \frac{G \cdot m_1 \cdot m_2}{d^2} $$
Here, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two bodies, and $d$ is the distance between them. Initially, since the masses are equal, let's denote each mass as $m$. Therefore, the initial force of attraction is calculated as:
$$ F = \frac{G \cdot m \cdot m}{d^2} = \frac{G \cdot m^2}{d^2} $$
Now, when one unit of mass is transferred from one mass to the other, the masses change to $m-1$ and $m+1$. The new force of attraction, $F'$, becomes:
$$ F' = \frac{G \cdot (m-1) \cdot (m+1)}{d^2} $$
Using the difference of squares, it simplifies to:
$$ F' = \frac{G \cdot (m^2 - 1)}{d^2} $$
This new force $F'$ can be rewritten relative to the original force $F$:
$$ F' = \frac{G \cdot m^2}{d^2} - \frac{G}{d^2} = F - \frac{G}{d^2} $$
Based on pure calculation from moving forward from the solution, it is determined that $F'$ was erroneously calculated initially, and instead should rely on more basic algebraic evaluations:
The correct reduction in force is:
$$ F' = F - \frac{G}{d^2} $$
Therefore, the resultant force decreases by $\frac{G}{d^2}$, assuming $G$ and $d^2$ remain constants. The force calculation earlier using $F' = F - \frac{4G}{d^2}$ was incorrect as my initial calculations simplified wrongly. We simply have to remember that transferring mass in such a way decreases the product $m_1 \cdot m_2$ slightly due to the unit difference squared being subtracted, hence decreasing the force by $\frac{G}{d^2}$.
"If we release all the vacuum from the breaker and put an iron nail and feather in it, which of these will fall faster and why?"
Solution:
In a vacuum, both the iron nail and the feather will fall at the same rate. This occurs because, without air, the only significant force acting on the objects is gravity. Normally, air resistance affects the rate at which objects with different masses and surface areas fall. However, removing the air eliminates this resistance.
Here's how the situation breaks down mathematically:
The net force acting on an object in a vacuum is simply due to gravity: $$ F = ma $$ Where:
$ m $ is the mass of the object,
$ a $ is the acceleration.
However, the force due to gravity can also be expressed as: $$ F = mg $$ Where:
$ g $ is the acceleration due to gravity.
Setting the forces equal, we get: $$ ma = mg \ a = g $$
This shows that the acceleration ($ a $) of both the iron nail and the feather is equal to the acceleration due to gravity ($ g $), independent of their mass. Thus, they will fall at the same rate when placed in a vacuum.
Identify free fall situations from the following:
A. A man landing with a parachute.
B. An apple falling on the Moon.
C. A rocket propelling into space.
D. A block falling from a cliff.
The situations constituting free fall from the provided scenarios are:
B. An apple falling on the Moon.
This represents a true free fall scenario as the apple is primarily influenced by the Moon's gravitational pull and there isn't any considerable atmospheric drag due to the extremely thin atmosphere on the Moon.
D. A block falling from a cliff.
Similarly, a block falling from a cliff on Earth is primarily influenced by Earth's gravity. The effect of air resistance is often considered small enough to be negligible in comparison to the gravitational force, thus classifying this as a free fall.
Analysis of Other Options:
A. A man landing with a parachute.
This is not a free fall because the parachute creates significant air resistance or drag, which counteracts the gravitational pull, slowing down the descent.
C. A rocket propelling into space.
Rockets are propelled by the force generated through the expulsion of exhaust gases, and they must work against Earth’s gravity and atmospheric drag. This shows multiple forces are acting simultaneously, distancing this scenario from a free fall.
In summary, scenarios B and D depict free fall conditions, where the motion is predominantly governed by gravitational forces, minimizing the impacts of other forces.
What is the difference between mass and weight?
Mass and weight are two fundamental but different physical concepts often used interchangeably in everyday language. Here are the distinct differences between them:
Mass is the amount of matter contained within a body, depicting how much stuff is there. Weight, on the other hand, is the force exerted on an object due to the gravitational attraction of another body like Earth, the sun, etc.
Mass is an intrinsic property of a body, meaning it does not change regardless of location or environment. Weight is an extrinsic property, which changes depending on the gravitational field the object is in.
The mass of a body remains consistent throughout the universe; however, the weight of a body varies depending on the local acceleration due to gravity.
The SI unit for mass is the kilogram (kg), while since weight is a force, its SI unit is the Newton (N).
Mass can be measured using instruments such as beam balance or pan balance. Weight is measured using instruments that rely on spring mechanisms, such as spring balance or weighing machines.
An object's mass can never be zero, while its weight can be zero if it is in a location with no gravitational pull (e.g., far space).
Mass is a measure of an object's inertia, and weight is a measure of gravitational force acting on the object.
Understanding these differences is crucial in fields such as physics and engineering, where precise measurements and distinct distinctions matter.
A body of mass $500 \mathrm{~g}$ is accelerated at $2 \mathrm{~m} / \mathrm{sec}^2$. Find magnitude of accelerating force.
To solve for the magnitude of the accelerating force acting on a body, you can use Newton's second law of motion, which states that the force exerted on an object is equal to the mass of the object multiplied by its acceleration. This can be expressed mathematically as: $$ F = ma $$ where:
$F$ is the force,
$m$ is the mass of the object,
$a$ is the acceleration.
Given in the problem:
Mass, $m = 500 , \text{g}$ (which needs to be converted to kilograms to use SI units, $1 \text{kg} = 1000 \text{g}$),
Acceleration, $a = 2 , \text{m/s}^2$.
First, convert the mass from grams to kilograms: $$ m = 500 , \text{g} = 0.5 , \text{kg} $$
Now, plug the values into Newton's second law formula: $$ F = 0.5 , \text{kg} \times 2 , \text{m/s}^2 = 1 , \text{kg m/s}^2 $$ Also known as $1 , \text{Newton}$, since $1 , \text{N} = 1 , \text{kg m/s}^2$.
Thus, the magnitude of the force accelerating the body is $1 , \text{Newton}$.
4
An astronaut standing on the surface of the moon throws a ball upwards. The ball would (a) directly fall down from the point it is released. (b) hang in space. (c) go up and then come back to the surface of the moon. (d) keep going up never to come back.
The correct behavior of the ball when thrown upwards by an astronaut on the moon is that it will ascend and then descend back to the surface. This is because, although the gravitational pull on the moon is weaker than on Earth, it still exerts a force that causes the ball to eventually return after reaching its peak height.
Hence, the correct answer is:
(c) go up and then come back to the surface of the moon.
A rocket has a mass of $1000 \mathrm{~kg}$ with $50,000 \mathrm{~kg}$ of fuel being stored on the rocket. The nozzle of the rocket lets out the exhaust gases at a speed of $500 \mathrm{~m} / \mathrm{s}$. If the rocket consumes fuel at the rate of $100 \mathrm{~kg} / \mathrm{s}$, what is the acceleration of the rocket after 5 minutes in a gravity-free space?
(A) $10 \mathrm{~m} / \mathrm{s}^{2}$
B) $3.95 \mathrm{~m} / \mathrm{s}^{2}$
C) $2.38 \mathrm{~m} / \mathrm{s}^{2}$
(D) $1.56 \mathrm{~m} / \mathrm{s}^{2}$
Solution The correct answer is (C) $2.38 , \text{m/s}^2$.
Given:
Mass of the rocket, $m_r = 1000 , \text{kg}$
Mass of fuel, $m_f = 50000 , \text{kg}$
Total initial mass of the rocket and fuel, $m = m_r + m_f = 51000 , \text{kg}$
Rate of fuel consumption, $\frac{dm}{dt} = 100 , \text{kg/s}$
Exhaust velocity, $u = 500 , \text{m/s}$
Time elapsed, $T = 5 , \text{minutes} = 300 , \text{seconds}$
From Newton's third law and considering the momentum conservation principle, the thrust ($F$) provided by the rocket engines due to the expelled gases can be represented by: $$ F = u \frac{dm}{dt} $$
Substituting the given values into the thrust formula gives: $$ F = 500 \times 100 = 50000 , \text{N} $$
Since the mass of the fuel decreases as it's consumed, the overall mass of the rocket at any time $t$ is: $$ m_t = m - \frac{dm}{dt} \times t $$
Plug in the time elapsed $T = 300 , \text{s}$: $$ m_{300} = 51000 - 100 \times 300 = 21000 , \text{kg} $$
The acceleration $a$ of the rocket can be found by dividing the thrust by the changing mass: $$ a = \frac{F}{m_{300}} = \frac{50000}{21000} \approx 2.38 , \text{m/s}^2 $$
The acceleration of the rocket after 5 minutes in gravity-free space is $2.38 , \text{m/s}^2$. Thus, the correct option is (C).
An aeroplane is rising vertically with acceleration $f$. Two stones are dropped from it at an interval of time $t$. The distance between them at time $t'$ after the second stone is dropped will be:
A) $\frac{1}{2}(g+f) t t'$
B) $\frac{1}{2}(g+f)\left(t+2 t'\right) t$
C) $\frac{1}{2}(g+f)\left(t-t'\right)^{2}$
D) $\frac{1}{2}(g+f)\left(t+t'\right)^{2}$
The correct answer is Option B: $$ \frac{1}{2}(g+f)\left(t+2 t^{\prime}\right) t $$
Explanation:
When the first stone is released, it initially shares the velocity of the aeroplane and then starts accelerating due to gravity, but with an additional acceleration equal to the aeroplane's upward acceleration, $f$. The displacement of the first stone relative to the aeroplane after $t$ seconds can be calculated using:
$$ h_1 = \frac{1}{2} (g + f) t^2 $$
Here, $\mathbf{g}$ denotes the acceleration due to gravity. At this time, the velocity of the aeroplane is $u + ft$ and the velocity of the first stone is $u - gt$, where $u$ is the initial velocity of the aeroplane when the first stone was dropped.
The second stone is dropped $t$ seconds later, thus it starts with a velocity equal to the aeroplane's at that time, i.e., $u + ft$. The relative velocity between the first and second stone immediately upon the second stone's release is:
$$ (u + ft) - (u - gt) = (g + f) t $$
This relative speed is maintained as both stones continue to accelerate downwards due to gravity, compounded by the aeroplane's acceleration. The relative displacement between the first and second stone after an additional duration $t'$ can be computed using:
$$ h_2 = (g + f) t t' $$
Combining the displacements from $h_1$ and $h_2$ to find the total displacement between the two stones at $t + t'$ seconds since the first stone was dropped gives us:
$$ h_1 + h_2 = \frac{1}{2} (g + f) t^2 + (g + f) t t' = \frac{1}{2} (g + f) \left(t + 2 t'\right) t $$
This confirms that Option B is indeed the correct choice.
The gravitational unit of force is
A $\mathrm{Ns}$
B $\mathrm{kg}$ wt
C $\mathrm{ms}^{-2}$
D (N) newton
The correct answer is B $\mathbf{kg , wt}$.
Kilogram-weight ($\mathbf{kg , wt}$) refers to the gravitational force exerted by a mass of one kilogram under Earth's gravity. This unit specifically measures force in terms of its gravitational component, thus is the gravitational unit of force.
Consider the statements A and B and choose the appropriate option from those given below:
A: The position of the center of mass of a body is independent of the acceleration due to gravity. B: The center of mass of a body always lies within the boundaries.
A) A true, B false
B) A false, B true
C) A true, B true
D) A false, B false
The correct answer is C) A true, B true.
Statement A posits that the position of the center of mass of a body is independent of the acceleration due to gravity. This is indeed true because the center of mass is determined by the distribution of mass within an object, not by external forces such as gravity.
Statement B claims that the center of mass of a body always lies within the boundaries of that body. This statement is generally true for most practical purposes and regular objects, although there are cases (like certain shaped objects or systems of particles) where the center of mass can lie outside the physical boundaries.
Hence, both statements are true, making option C the correct choice.
Identify the free fall situations from the following:
A. A man landing through a parachute
B. An apple falling on the moon
C. A rocket propelling into space
D. A heavy stone falling from a cliff
The correct options are:
B. An apple falling on the moon
D. A heavy stone falling from a cliff
The concept of free fall refers to situations where gravitation is the only force acting on an object. Here's why options B and D qualify:
In option D, a heavy stone falling from a cliff is under the influence of Earth's gravity. Assuming negligible air resistance due to the stone’s density, this represents a free fall scenario.
Option B involves an apple falling on the Moon, which is also a free fall. The Moon's gravitational pull is the sole force acting on the apple, and importantly, the Moon lacks an atmosphere, which means no air resistance.
For the other options:
Option A. A man landing with a parachute is not in free fall. The parachute creates a significant drag force (air resistance), which counteracts gravity.
Option C. In the case of a rocket propelling into space, it is driven by the upward thrust from the gases it expels, which is a key force besides gravity during its ascent. Thus, it does not fall under the category of free fall.
A boy, while playing under a mango tree, saw a mango falling on his head. This is due to the force of:
A) gravity
B) friction
C) magnetic force
D) can't be said
The correct answer is A) gravity.
Gravity is the force responsible for pulling objects towards the center of the Earth. In the scenario described, when the mango detaches from the tree, it is pulled downwards by the force of gravity. This natural phenomenon occurs because the Earth exerts a gravitational pull on all objects, causing them to fall towards its center once they are no longer supported. Hence, the mango falls on the boy’s head due to gravity, a fundamental force acting on all matter on Earth.
A small steel ball of radius $r$ is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity $\eta$. After some time the velocity of the ball attains a constant value known as terminal velocity $v_{T}$. The terminal velocity depends on (i) the mass of the ball $m$, (ii) $\eta$, (iii) $r$, and (iv) acceleration due to gravity $g$. Which of the following relations is dimensionally correct?
$v_{T} \propto \frac{m g}{\eta r}$
$v_{T} \propto \frac{\eta r}{mg}$
$v_{T} \propto \eta rrmg$
$v_{T} \propto \frac{mgr}{n}$
The correct option is (A): $$ v_{T} \propto \frac{mg}{\eta r} $$
To determine this, we'll perform a dimensional analysis.
Dimensional Analysis
The dimension of terminal velocity $v_{T} $ is: $$ [v_{T}] = [L T^{-1}] $$
From the formula of drag force $ F_D $: $$ F_D = 6 \pi \eta r v_{T} $$
This can be rewritten using dimensional terms: $$ [\eta] = \frac{[F_D]}{[6 \pi r v_{T}]} $$
Given: $$ [F_D] = [M L T^{-2}] $$
Therefore: $$ [\eta] = \frac{[M L T^{-2}]}{[L] [L T^{-1}]} = [M L^{-1} T^{-1}] $$
Verifying Option (a)
For Option (a): $$ v_{T} \propto \frac{mg}{\eta r} $$
Let's break this down dimensionally: $$ \frac{[mg]}{[\eta r]} = \frac{[M L T^{-2}]}{[M L^{-1} T^{-1} \times L]} = [L T^{-1}] $$
From our analysis, this matches the dimensions of $ v_{T} $: $$ [L T^{-1}] $$
Hence, option (a) is dimensionally correct.
Derivation for acceleration due to gravity on its effects of height and depth. Explain.
Acceleration Due to Gravity
1) On the Earth’s Surface
Definition: The acceleration due to gravity of an object is the rate at which its velocity changes due to the Earth's gravitational pull, directed towards the center of the Earth.
The gravitational force acting on a body of mass $m$ on the Earth's surface is given by: $$ F = \frac{GMm}{R^2} \tag{1} $$ where:
$R$ is the Earth's radius,
$M$ is the Earth's mass,
$G$ is the gravitational constant.
From Newton’s Second Law of Motion: $$ F = mg \tag{2} $$ Here, $g$ is the acceleration due to gravity.
Equating (1) and (2), we get: $$ mg = \frac{GMm}{R^2} \implies g = \frac{GM}{R^2} \tag{3} $$
Another expression for $g$ is derived given the mean density of Earth ($\rho$): $$ M = \frac{4}{3} \pi R^3 \rho $$ Thus, $$ g = G \left( \frac{4}{3} \pi R^3 \rho \right) / R^2 \implies g = \frac{4}{3} \pi R \rho G \tag{4} $$
2) Variation of Acceleration Due to Gravity with Height
At a height $h$ from the Earth's surface, the gravitational force on an object of mass $m$ is: $$ F = \frac{GMm}{(R+h)^2} $$ Here, $(R + h)$ is the distance between the object and the Earth's center.
Let $g_1$ be the gravitational acceleration at height $h$. We have: $$ mg_1 = \frac{GMm}{(R+h)^2} \implies g_1 = \frac{GM}{(R+h)^2} \tag{5} $$
On the Earth's surface: $$ g = \frac{GM}{R^2} $$
Taking the ratio: $$ \frac{g_1}{g} = \frac{R^2}{(R+h)^2} \implies g_1 = g \left( \frac{R}{R+h} \right)^2 \implies g_1 = g \left( 1 - \frac{h}{R} \right)^2 \approx g \left( 1 - \frac{2h}{R} \right) \tag{6} $$
Thus, as altitude $h$ increases, the value of $g_1$ decreases.
3) Variation of Acceleration Due to Gravity with Depth
Let a body of mass $m$ be at a point $A$ at a depth $h$ beneath the Earth’s surface. The distance from the center of the Earth is $(R - h)$.
Mass of the inner sphere at depth $h$: $$ M' = \frac{4}{3} \pi (R-h)^3 \rho $$
At point $A$, the gravitational force is: $$ F = \frac{G M' m}{(R-h)^2} = G \left[ \frac{4}{3} \pi (R-h)^3 \rho \right] / (R-h)^2 \implies F = \frac{4}{3} \pi (R-h) \rho m G $$
Let $g_2$ be the acceleration due to gravity at this depth: $$ g_2 = \frac{F}{m} = \frac{4}{3} \pi (R-h) \rho G \tag{7} $$
On the Earth's surface: $$ g = \frac{4}{3} \pi R \rho G $$
Taking the ratio: $$ \frac{g_2}{g} = \frac{(R-h)}{R} \implies g_2 = g \left( 1 - \frac{h}{R} \right) \tag{8} $$
Thus, as depth $h$ increases, the value of $g_2$ decreases.
Comparison of Variation Equations
At height $h$ from the Earth's surface: $$ g_1 = g \left(1 \ - \ \frac{2h}{R}\right) \tag{6} $$
At depth h from the Earth's surface: $$ g_2 = g \left(1 \ - \ \frac{h}{R} \right) \tag{8} $$
Key observations:
Both $g_1$ and $g_2$ are less than $g$ on the Earth's surface.
$g_1 < g_2$, indicating the acceleration due to gravity decreases more significantly with an increase in height compared to depth.
💡 Have more questions?
Ask Chatterbot AINCERT Solutions - Gravitation | NCERT | Science | Class 9
How does the force of gravitation between two objects change when the distance between them is reduced to half?
The force of gravitation between two objects is given by Newton's law of universal gravitation, which states that the gravitational force ($F$) between two masses ($m_1$ and $m_2$) is directly proportional to the product of their masses and inversely proportional to the square of the distance ($r$) between their centers:
$$F = G \cdot \frac{m_1 \cdot m_2}{r^2}$$
where $G$ is the gravitational constant.
If the distance between two objects is reduced to half, the new distance between them is $\frac{r}{2}$. Plugging this into the equation, we get:
$$ F' = G \cdot \frac{m_1 \cdot m_2}{\left(\frac{r}{2}\right)^2} = G \cdot \frac{m_1 \cdot m_2}{\frac{r^2}{4}} = 4 \cdot G \cdot \frac{m_1 \cdot m_2}{r^2}$$
Comparing the new force ($F'$) to the original force ($F$):
$$ F' = 4 \cdot F $$
This means that the gravitational force becomes four times stronger when the distance between two objects is reduced to half.
Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
The gravitational force acting on an object is indeed proportional to its mass, as given by Newton's law of universal gravitation:
$$ F = G \frac{m_1 m_2}{r^2}$$
where $F$ is the force between the masses, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the objects, and $r$ is the distance between the centers of the two masses.
When an object falls towards the Earth, the Earth's mass ($m_1$) and the object's mass ($m_2$) determine the gravitational force. However, according to Newton's second law of motion:
$$F = ma$$
where $F$ is the force applied to the object, $m$ is the mass of the object, and $a$ is the acceleration of the object.
When we set the gravitational force equal to the mass times acceleration ($F = ma$), we have:
$$G \frac{m_1 m_2}{r^2} = m_2 a$$
If we solve for acceleration ($a$), we see that the mass of the falling object ($m_2$) cancels out:
$$a = G \frac{m_1}{r^2}$$
This shows that the acceleration due to gravity is independent of the mass of the falling object and is only dependent on the mass of the Earth and the distance to the center of the Earth. Therefore, in the absence of air resistance, all objects fall at the same rate of acceleration, which is approximately $9.81 \, \text{m/s}^2$ on the surface of the Earth. That is why a heavy object does not fall faster than a light object.
What is the magnitude of the gravitational force between the earth and a $1 \mathrm{~kg}$ object on its surface? (Mass of the earth is $6 \times 10^{24} \mathrm{~kg}$ and radius of the earth is $6.4 \times 10^{6} \mathrm{~m}$.
The gravitational force between two objects, according to Newton's law of universal gravitation, is given by the formula:
$$F = \frac{G m_1 m_2}{r^2}$$
where $F$ is the gravitational force, $G$ is the gravitational constant ($6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2$), $m_1$ is the mass of the first object, $m_2$ is the mass of the second object, and $r$ is the distance between the centers of the two objects.
Given:
Mass of the Earth ($m_1$): $6 \times 10^{24} \, \text{kg}$
Mass of the object ($m_2$): $1 \, \text{kg}$
Radius of the Earth ($r$): $6.4 \times 10^{6} \, \text{m}$
Plugging in these values:
$$F = \frac{(6.674 \times 10^{-11} \, \text{m}^3/\text{kg} \cdot \text{s}^2)(6 \times 10^{24} \, \text{kg})(1 \, \text{kg})}{(6.4 \times 10^{6} \, \text{m})^2}$$
Solving this, we can find the magnitude of the gravitational force between the Earth and a $1 \, \text{kg}$ object on its surface. The magnitude of the gravitational force between the Earth and a $1 \, \text{kg}$ object on its surface is approximately $9.776 \, \text{N}$ (newtons).
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 Earth and the Moon attract each other with forces that are equal in magnitude but opposite in direction. This is a consequence of Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. In the context of gravity, this means that if the Earth exerts a gravitational force on the Moon, then the Moon also exerts an equal gravitational force on the Earth.
The law of universal gravitation, formulated by Sir Isaac Newton, further clarifies this interaction with the equation:
$$F = G \frac{m_1 m_2}{r^2}$$
where:
- $F$ is the magnitude of the gravitational force between the two masses,
- $G$ is the gravitational constant,
- $m_1$ and $m_2$ are the masses of the two objects,
- $r$ is the distance between the centers of the two masses.
According to this equation, the force depends only on the masses and the distance between them, not on the individual properties of either object. That is why the force with which the Earth attracts the Moon is the same as the force with which the Moon attracts the Earth. However, the effects of these forces are different due to the different masses of the Earth and Moon: Earth has a much larger mass than the Moon, so the acceleration the Moon experiences due to Earth's gravitational pull is much larger than the acceleration the Earth experiences due to the Moon's pull.
If the moon attracts the earth, why does the earth not move towards the moon?
The Earth and the Moon do exert gravitational forces on each other, and both bodies actually move in response to this force. However, the movement is not as simple as the Earth moving towards the Moon because both the Earth and the Moon are also in orbit around their common center of mass.
Here's why the Earth does not simply move towards the Moon:
Mutual Gravitation: Both the Earth and the Moon exert a gravitational pull on each other. This force is governed by Newton's law of universal gravitation, which states that the force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Center of Mass: The Earth and the Moon orbit around their common center of mass, also known as the barycenter. Even though Earth is much more massive than the Moon, the location of the barycenter is still within Earth's volume but not at its center. The Earth does move in a small circle around this point as the Moon orbits Earth.
Earth's Inertia: Earth has a much larger mass compared to the Moon, and due to its inertia, the amount of movement produced by the gravitational pull of the Moon is much less than that of the Moon itself.
Orbital Motion: Both the Earth and the Moon are in constant motion due to their orbital momentum. The Moon's orbital motion around Earth is the result of its velocity in space being balanced by the gravitational attraction between Earth and the Moon. If the Earth didn't move, the Moon would spiral in and collide with Earth, but because of the Earth's movement, the Moon instead maintains its orbit.
Tidal Forces: The gravitational interaction causes tides in Earth's oceans. Interestingly, it's this interaction that is slowly changing both the Earth's rotation and the Moon's orbit. Over very long periods of time, the Moon is gradually moving away from Earth due to the transfer of Earth's rotational energy to the Moon's orbital energy via tidal forces.
In summary, while the Moon’s gravity does affect Earth, both bodies are in motion around a common center of mass, and the Earth's large mass means it only moves slightly in response to the Moon's pull. This movement is part of a delicate balance that results in the stable, nearly elliptical orbits we observe.
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?
The force between two objects is determined by Newton's law of universal gravitation, which states that the force ($F$) between two objects with masses $m_1$ and $m_2$ is directly proportional to the product of their masses and inversely proportional to the square of the distance ($r$) between them. The gravitational force can be calculated using the following formula:
$$F = G \frac{m_1 m_2}{r^2}$$
where $G$ is the gravitational constant.
Now let's analyze what happens in each of the given scenarios:
(i) If the mass of one object is doubled:
If the mass of one of the objects (say $m_1$) is doubled, the new force ($F'$) becomes:
$$F' = G \frac{2m_1 m_2}{r^2}$$
This means that the new force is double the original force ($F' = 2F$).
(ii) If the distance between the objects is doubled and tripled:
If the distance is doubled ($r$ becomes $2r$), the new force ($F''$) would be:
$$F'' = G \frac{m_1 m_2}{(2r)^2} = G \frac{m_1 m_2}{4r^2}$$
This means the force is reduced to a quarter ($\frac{1}{4}$) of the original force.
If the distance is tripled ($r$ becomes $3r$), the new force ($F'''$) would be:
$$F''' = G \frac{m_1 m_2}{(3r)^2} = G \frac{m_1 m_2}{9r^2}$$
This means the force is reduced to a ninth ($\frac{1}{9}$) of the original force.
(iii) If the masses of both objects are doubled:
If the masses of both objects are doubled, the new force ($F''''$) will be:
$$F'''' = G \frac{2m_1 \times 2m_2}{r^2} = 4G \frac{m_1 m_2}{r^2}$$
This means the new force is four times ($4F$) the original force.
In summary, doubling one mass doubles the force, doubling the distance reduces the force to one-fourth, tripling the distance reduces the force to one-ninth, and doubling both masses quadruples the force.
What is the importance of universal law of gravitation?
The universal law of gravitation is highly important for various reasons:
Explains Planetary Motion: It explains the motion of planets around the sun as observed by Kepler and others. The law provides the mathematical basis for gravity, which explains the elliptical orbits of planets.
Describes Tidal Effects: The gravitational pull of the moon and the sun on the Earth's oceans results in tides. Understanding this force helps us predict and understand tidal phenomena.
Predicts Celestial Events: By using the law, scientists can predict celestial events like eclipses, the return of comets, and the positions of stars and planets in the sky.
Aids in Space Missions: The law is fundamental in calculating trajectories, ensuring satellites orbit correctly, and in the planning and execution of space missions.
Part of the Fundamental Forces: Gravitation is one of the four fundamental forces of nature. It acts on matter at all scales, from subatomic particles to galaxies, and thus plays a role in the large-scale structure and evolution of the cosmos.
Contributes to Astrophysics and Cosmology: The law assists in understanding the lifecycle of stars, galaxy formation, the behaviour of black holes, and other phenomena.
Supports General Relativity: Newton's law was superseded by Einstein's theory of General Relativity, yet the law is still used for practical calculations on scales where relativistic effects are negligible.
Enables Technological Applications: Knowledge of gravitational effects is essential for technologies such as GPS and other systems that require precise measurements of the Earth's gravitational field.
Overall, the universal law of gravitation is a cornerstone in our understanding of the physical universe, providing insights into both celestial mechanics and terrestrial phenomena.
What is the acceleration of free fall?
The acceleration of free fall, also known as standard gravity, is approximately \(9.81 , \text{m/s}^2\) on the surface of the Earth. This value can slightly vary depending on altitude and location due to the Earth's rotation and the distribution of its mass.
What do we call the gravitational force between the earth and an object?
The gravitational force between the Earth and an object is commonly referred to as the object's weight. Weight is a measure of the force of gravity acting on an object's mass. The weight of an object can be calculated by using the formula:
$$F = m \times g$$
where $F$ is the force of gravity (or the weight), $m$ is the mass of the object, and $g$ is the acceleration due to gravity near the surface of the Earth, which is approximately $9.81$ m/s$^2$ (meters per second squared). It's important to note that while mass is an intrinsic property of an object and does not change, the weight can vary depending on the gravitational field where the object is located.
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.]
The weight of an object is the force with which it is pulled towards the center of the Earth due to gravity, and it is calculated using the formula:
$$ \text{weight} = m \times g$$
where
$m$ is the mass of the object and
$g$ is the acceleration due to gravity.
The acceleration due to gravity ($g$) is not the same everywhere on Earth; it varies slightly depending on latitude and elevation. At the poles, Earth's radius is smaller compared to the equator due to its slightly flattened shape (oblate spheroid). Since gravitational force is inversely proportional to the square of the distance from the center of the Earth, the gravitational pull is stronger at the poles where you are closer to Earth's center.
So, at the poles, where the value of $g$ (gravitational acceleration) is greater, the weight of the gold will be more than its weight at the equator, where $g$ is less. The mass ($m$) of the gold remains constant, as it is a measure of the amount of matter, which does not change unless gold is removed or added.
If Amit measures the gold in grams at the poles (which is a measure of mass), this value will not change when he moves to the equator. However, if he measures the weight in units of force (such as newtons), he will find that the weight of the gold (the force) is less at the equator than at the poles due to the difference in $g$.
Therefore, if his friend is expecting the same weight (force) rather than the same mass (grams), he might be surprised to find that the weight is slightly less at the equator due to the lower value of $g$.
Why will a sheet of paper fall slower than one that is crumpled into a ball?
A sheet of paper falls slower than one that is crumpled into a ball due to air resistance and surface area effects.
When a sheet of paper is spread out, it has a large surface area in comparison to its mass. As it falls through the air, this large surface area interacts with a lot of air molecules. This interaction with the air creates a form of friction known as air resistance or drag. The large surface area of the flat sheet presents more resistance to the air through which it is falling, causing it to fall more slowly.
On the other hand, when the paper is crumpled into a ball, the surface area is greatly reduced. This smaller surface area means that there is less air resistance acting against the motion of the falling ball of paper. With less air resistance, the crumpled ball of paper falls faster than the uncrumpled sheet.
Additionally, the flat sheet of paper tends to flutter and may even catch air currents that keep it aloft longer, while the ball shape is more aerodynamic and will generally not experience such effects.
In the absence of air (in a vacuum), both the flat sheet and the crumpled ball of paper would fall at the same rate due to gravity. However, in Earth's atmosphere, air resistance affects the rate at which objects with different shapes and surface areas fall.
Gravitational force on the surface of the moon is only $\frac{1}{6}$ as strong as gravitational force on the earth. What is the weight in newtons of a $10 \mathrm{~kg}$ object on the moon and on the earth?
The weight of a $10\,\text{kg}$ object on the Earth is calculated using the formula:
$$ \text{Weight} = \text{mass} \times \text{acceleration due to gravity}$$
For Earth, the acceleration due to gravity $g$ is approximately $9.8\,\text{m/s}^2$. Thus, the weight $W$ on Earth is:
$$W = 10\,\text{kg} \times 9.8\,\text{m/s}^2 = 98\,\text{N}$$
Since the gravitational force on the surface of the moon is $\frac{1}{6}$ of that on Earth, the weight of the same object on the moon is:
$$\frac{98\,\text{N}}{6} \approx 16.33\,\text{N}$$
Therefore, a $10\,\text{kg}$ object weighs approximately $98\,\text{N}$ on Earth and $16.33\,\text{N}$ on the moon.
A ball is thrown vertically upwards with a velocity of $49 \mathrm{~m} / \mathrm{s}$. Calculate
(i) the maximum height to which it rises,
(ii) the total time it takes to return to the surface of the earth.
To calculate the maximum height $(h)$ and the total time $(t)$ it takes for the ball to return to the surface of Earth after being thrown vertically upwards with a velocity of $49 \, \text{m/s}$, we can use kinematic equations for uniformly accelerated motion.
Let's start with (i) the maximum height to which it rises. We'll use the following equation which relates the initial velocity $(v_0)$, the final velocity $(v)$, acceleration $(a)$, and distance $(h)$:
$$v^2 = v_0^2 + 2ah$$
When the ball reaches its maximum height, its velocity $(v)$ will be $0 \, \text{m/s}$, since it momentarily stops before falling down again. The acceleration $(a)$ is due to gravity, which we'll take as $-9.81 \, \text{m/s}^2$, the negative sign indicating that it's directed downwards, opposite to the initial velocity of the ball.
Plugging in the numbers we get:
$$(0 \, \text{m/s})^2 = (49 \, \text{m/s})^2 + 2(-9.81 \, \text{m/s}^2)h$$
Solving for $(h)$, we find:
$$h = \frac{(49 \, \text{m/s})^2}{2 \cdot 9.81 \, \text{m/s}^2}$$
For (ii) the total time it takes to return to the surface, we'll use the equation that relates the initial velocity $(v_0)$, acceleration $(a)$, and time $(t)$:
$$v = v_0 + at$$
Again, final velocity $(v)$ at maximum height is $0 \, \text{m/s}$, and we can solve for the time $(t)$ it takes to reach that height:
$$0 \, \text{m/s} = 49 \, \text{m/s} + (-9.81 \, \text{m/s}^2)t$$
Solving for $(t)$ we get:
$$t = \frac{-49 \, \text{m/s}}{-9.81 \, \text{m/s}^2}$$
This gives us the time to reach the maximum height. The total time to return
(ii) The total time it takes for the ball to return to the surface of Earth is approximately $9.99$ seconds.
A stone is released from the top of a tower of height $19.6 \mathrm{~m}$. Calculate its final velocity just before touching the ground.
To calculate the final velocity of a stone released from the top of a tower, we can use the equation for the final velocity of an object under constant acceleration due to gravity. The formula is:
$$v^2 = u^2 + 2as$$
where:
$v$ is the final velocity,
$u$ is the initial velocity (which is $0$ since the stone is released),
$a$ is the acceleration (which is $g = 9.8 \, \mathrm{m/s^2}$, the acceleration due to gravity on the surface of the Earth),
$s$ is the distance the object has fallen.
Since the stone is released from rest, the initial velocity $u$ is $0$. We can thus simplify the equation as follows:
$$v^2 = 0 + 2as = 2 \cdot g \cdot s$$
Plugging in the values for $g$ and $s$:
$$v^2 = 2 \cdot 9.8 \, \mathrm{m/s^2} \cdot 19.6 \, \mathrm{m}$$
$$v^2 = 2 \cdot 9.8 \, \mathrm{m/s^2} \cdot 19.6 \, \mathrm{m} = 19.6 \cdot 19.6 \, \mathrm{m^2/s^2}$$
$$v^2 = 384.16 \, \mathrm{m^2/s^2}$$
To find $v$, we take the square root:
$$v = \sqrt{384.16 \, \mathrm{m^2/s^2}} = 19.6 \, \mathrm{m/s}$$
So, the final velocity of the stone just before touching the ground is $19.6 \, \mathrm{m/s}$.
A stone is thrown vertically upward with an initial velocity of $40 \mathrm{~m} / \mathrm{s}$. Taking $g=10 \mathrm{~m} / \mathrm{s}^{2}$, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
To find the maximum height reached by the stone, we'll use the kinematic equation for motion under constant acceleration (gravity, in this case):
$$ v^2 = u^2 + 2 g h$$
where:
- $v$ is the final velocity ($0 \ m/s$ at the maximum height),
- $u$ is the initial velocity ($40 \ m/s$),
- $g$ is the acceleration due to gravity ($-10 \ m/s^2$, negative because it is directed downward),
- $h$ is the maximum height reached.
Solving for $h$ will give us the maximum height. Let's rearrange the equation and insert the values:
$$0 = u^2 - 2gh$$
Now, solving for the height $h$:
$$h = \frac{u^2}{2g}$$
Inserting the given values:
$$h = \frac{40^2}{2\times (-10)} = \frac{1600}{-20} = -80 \ m$$
The calculation mistake made here should be corrected. Since we take gravity as negative, we should use its absolute value in the formula for finding height, as it represents a magnitude in this context. So correctly:
$$h = \frac{40^2}{2\times 10} = \frac{1600}{20} = 80 \ m$$
Therefore, the maximum height reached by the stone is 80 meters.
Next, for the net displacement: the stone returns to the starting point, so the net displacement is 0.
For the total distance covered, we have to consider the stone going up and down. The distance it travels upwards is the maximum height of 80 meters. Since it falls back to the starting point, it covers the same distance downward; thus, the total distance is:
$$80 \ m \ (\text{up}) + 80 \ m \ (\text{down}) = 160 \ m$$
The total distance covered by the stone is 160 meters.
Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth $=6 \times 10^{24} \mathrm{~kg}$ and of the Sun $=2 \times 10^{30} \mathrm{~kg}$. The average distance between the two is $1.5 \times 10^{11} \mathrm{~m}$.
The gravitational force between the Earth and the Sun, given the masses and the distance you've provided, is approximately $3.56 \times 10^{22}$ newtons (N). This calculation is based on Newton's law of universal gravitation, which is expressed by the formula:
$$F = G \frac{m_1 m_2}{r^2}$$
where $F$ is the gravitational force, $G$ is the gravitational constant ($6.674 \times 10^{-11}$ m\(^3\)kg\(^{-1}\)s\(^{-2}\)), $m_1$ and $m_2$ are the masses of the two bodies, and $r$ is the distance between their centers of mass.
A stone is allowed to fall from the top of a tower $100 \mathrm{~m}$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25 \mathrm{~m} / \mathrm{s}$. Calculate when and where the two stones will meet.
Let's solve the system of equations representing the positions of the stones as functions of time. Setting the acceleration due to gravity ($g$) to approximately $9.8 \, \mathrm{m/s}^2$, we have two equations:
For the first stone falling from the tower:
$$ h_1 = 100 \, \mathrm{m} - \frac{1}{2} g t^2 $$
For the second stone projected upwards:
$$ h_2 = 25 t - \frac{1}{2} g t^2 $$
To find when and where they will meet, we need to set $h_1 = h_2$. This gives us the equation:
$$ 100 \, \mathrm{m} - \frac{1}{2} g t^2 = 25 t - \frac{1}{2} g t^2 $$
Simplifying the equation by canceling out the second term on both sides and solving for $t$ gives:
$$ 25 t = 100 $$
$$ t = 4 \, \mathrm{s} $$
Now, plugging $t$ back into one of the original equations (for example, $h_2$) will give us the height at which they meet:
$$ h_2 = 25 \times 4 - \frac{1}{2} \times 9.8 \times 4^2 $$
$$ h_2 = 100 - \frac{1}{2} \times 9.8 \times 16 $$
$$ h_2 = 100 - 78.4 $$
$$ h_2 = 21.6 \, \mathrm{m} $$
So, the stones will meet after 4 seconds at a height of 21.6 meters above the ground.
A ball thrown up vertically returns to the thrower after $6 \mathrm{~s}$. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after $4 \mathrm{~s}$.
To solve this problem, we can use the equations of uniformly accelerated motion. Since the ball is thrown vertically and returns to the thrower, we will assume that the only acceleration acting on it is due to gravity ($g$), which we will take as $-9.8 \, \text{m/s}^2$ (negative because it's directed downwards). Moreover, we assume no air resistance.
(a) Initial Velocity ($u$)
The ball takes a total of 6 seconds for the entire trip, which means it reaches its maximum height in half the time, i.e., 3 seconds.
We use the equation of motion which relates initial velocity ($u$), acceleration ($a$), and time ($t$):
$$ v = u + at $$
When the ball reaches the maximum height, its final velocity ($v$) is 0 m/s. Solving for initial velocity ($u$) gives us:
$$ 0 = u + (-9.8 \, \text{m/s}^2)(3 \, \text{s}) $$
$$ u = 9.8 \, \text{m/s}^2 \times 3 \, \text{s} $$
$$ u = 29.4 \, \text{m/s} $$
So the initial velocity with which the ball was thrown up is $29.4 \, \text{m/s}$.
(b) Maximum Height ($h_{\text{max}}$)
Using the second equation of motion which relates initial velocity, final velocity, acceleration, and distance ($s$) traveled:
$$ v^2 = u^2 + 2as $$
With the final velocity at the maximum height being 0 m/s, we have:
$$ 0 = (29.4 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2)s $$
$$ s = \frac{(29.4 \, \text{m/s})^2}{2 \times 9.8 \, \text{m/s}^2} $$
$$ s = \frac{864.36 \, \text{m}^2/\text{s}^2}{19.6 \, \text{m/s}^2} $$
$$ s = 44.1 \, \text{m} $$
Thus, the maximum height reached by the ball is $44.1 \, \text{m}$.
(c) Position after 4 seconds ($s_4$)
We will utilize the first equation of motion to find the position after 4 seconds:
$$ s = ut + \frac{1}{2}at^2 $$
Using $u = 29.4 \, \text{m/s}$, $a = -9.8 \, \text{m/s}^2$, and $t = 4 \, \text{s}$, we find:
$$ s = (29.4 \, \text{m/s})(4 \, \text{s}) + \frac{1}{2}(-9.8 \, \text{m/s}^2)(4 \, \text{s})^2 $$
$$ s = 117.6 \, \text{m} + \frac{1}{2}(-39.2 \, \text{m/s}^2)(16 \, \text{s}^2) $$
$$ s = 117.6 \, \text{m} - 313.6 \, \text{m} $$
$$ s = -196 \, \text{m} $$
Given that the displacement is negative, this means that after 4 seconds, the ball is $196 \, \text{m}$ away from the maximum height point in the downward direction.
Keep in mind that our starting point is the thrower's hand. We have calculated the upward and downward motion separately, so to find the position relative to the thrower after 4 seconds, we need to look at the ball's return journey. Since the total time to the top is 3 seconds, after 4 seconds, the ball has been falling for 1 second. We can recalculate the displacement for this 1-second fall, which will be its position above the thrower's hand:
$$ s_{\text{fall}} = ut + \frac{1}{2}at^2 $$
$$ s_{\text{fall}} = 0 \, \text{m/s} \times 1 \, \text{s} + \frac{1}{2}(9.8 \, \text{m/s}^2)(1 \, \text{s})^2 $$
$$ s_{\text{fall}} = 0 \, \text{m} + 4.9 \, \text{m} $$
$$ s_{\text{fall}} = 4.9 \, \text{m} $$
So, after 4 seconds, the ball is $4.9 \, \text{m}$ above the thrower's hand on its way down.
In what direction does the buoyant force on an object immersed in a liquid act?
The buoyant force on an object immersed in a liquid acts in the upward direction. This force is also known as buoyancy and it is caused by the pressure differences in the fluid due to the presence of the object. The fluid exerts a greater pressure on the bottom surface of the object than on the top surface, resulting in a net upward force that opposes the weight of the object. This is described by Archimedes' principle, which states that the buoyant force on an object in a fluid is equal to the weight of the fluid that the object displaces.
Why does a block of plastic released under water come up to the surface of water?
A block of plastic will rise to the surface when released underwater because of the principle of buoyancy, which is described by Archimedes' principle. This principle states that any object submerged in a fluid experiences an upward force equal to the weight of the fluid that is displaced by the object. The behavior of an object in a fluid, whether it sinks or floats, depends on the density of the object relative to the density of the fluid.
Here's a step-by-step explanation:
Density Comparison: Density is the mass per unit volume of a substance. If the density of the plastic block is less than the density of water, the block will float.
Buoyant Force: When the block is submerged, it displaces a volume of water. The weight of this displaced water creates an upward force known as the buoyant force.
Weight of the Block: At the same time, the block itself has weight due to the gravitational force acting on its mass.
Net Force: If the buoyant force is greater than the weight of the plastic block, there will be a net upward force causing the block to rise to the surface.
Equilibrium: Once the plastic block reaches the surface, it will displace its own weight in water, and it will float partially submerged until the upward buoyant force balances out the downward gravitational force.
In summary, a block of plastic rises to the surface of the water because its density is less than that of water, and therefore, the water's buoyant force on the block is greater than the gravitational force pulling it down.
The volume of $50 \mathrm{~g}$ of a substance is $20 \mathrm{~cm}^{3}$. If the density of water is $1 \mathrm{~g} \mathrm{~cm}^{-3}$, will the substance float or sink?
To determine if the substance will float or sink in water, we need to compare its density with that of water.
The density ($\rho$) of a substance can be calculated using the formula:
$$ \rho = \frac{\text{mass}}{\text{volume}}$$
Given the mass and volume of the substance, we can calculate its density:
$$\rho_{\text{substance}} = \frac{50 \text{ g}}{20 \text{ cm}^3} = 2.5 \text{ g/cm}^3$$
Since the density of water is $1 \text{ g/cm}^3$, and the density of the substance is $2.5 \text{ g/cm}^3$, the substance is denser than water.
Therefore, the substance will sink in water.
The volume of a $500 \mathrm{~g}$ sealed packet is $350 \mathrm{~cm}^{3}$. Will the packet float or sink in water if the density of water is $1 \mathrm{~g}$ $\mathrm{cm}^{-3}$ ? What will be the mass of the water displaced by this packet?
To determine whether the packet will float or sink, we need to compare the density of the packet to the density of water.
The density of an object is given by the formula:
$$\text{Density} = \frac{\text{Mass}}{\text{Volume}}$$
Given the mass ($m = 500\ g$) and the volume ($V = 350\ cm^3$), the density ($\rho$) of the packet can be calculated as:
$$\rho = \frac{m}{V} = \frac{500\ g}{350\ cm^3} \approx 1.43\ g/cm^3$$
Now, since the density of water is $1\ g/cm^3$, we compare the two:
- If the density of the packet is greater than the density of water, it will sink.
- If the density of the packet is less than the density of water, it will float.
Here, $\rho_{\text{packet}} = 1.43\ g/cm^3$ is greater than $\rho_{\text{water}} = 1\ g/cm^3$, so the packet will sink.
When the packet is completely submerged in water, the mass of the displaced water will be equal to the volume of the packet since the density of water is $1\ g/cm^3$. The volume of the packet is $350\ cm^3$, so the mass of the displaced water $m_{\text{displaced}}$ will also be $350\ g$.
💡 Have more questions?
Ask Chatterbot AINotes - Gravitation | Class 9 NCERT | Science
Concepts, Laws, and Important Examples
Introduction to Gravitation
Gravitation is a natural phenomenon by which all things with mass are brought towards one another. It is a crucial topic in the Class 9 science curriculum, providing foundational knowledge for understanding celestial mechanics and various physical phenomena on Earth.
Newton's Universal Law of Gravitation
Isaac Newton formulated the law of universal gravitation, which states that every object in the universe attracts every other object with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematical Formulation:[ F = G \frac{{M \cdot m}}{{d^2}} ]
where:
( F ) is the gravitational force between two objects
( G ) is the universal gravitational constant (6.673 × 10^-11 N m²/kg²)
( M ) and ( m ) are the masses of the objects
( d ) is the distance between the centers of the two objects
Calculating Gravitational Force
To calculate the gravitational force between two objects:
Identify the masses of both objects (( M ) and ( m )).
Measure the distance (( d )) between their centers.
Plug these values into the formula.
Example Problem: Calculate the gravitational force between Earth (mass ( 6 \times 10^{24} ) kg) and the Moon (mass ( 7.4 \times 10^{22} ) kg), separated by a distance of ( 3.84 \times 10^8 ) meters.
Solution:[ F = 6.673 \times 10^{-11} \frac{(6 \times 10^{24}) \cdot (7.4 \times 10^{22})}{(3.84 \times 10^8)^2} ]
[ F \approx 2.02 \times 10^{20} \text{ N} ]
Applications of the Universal Law of Gravitation
Binding force of the Earth: Keeps us and everything else anchored to the surface.
Motion of Planets: Governs the orbits of planets around the Sun.
Tides: Caused by the gravitational pull of the Moon and Sun on Earth’s water.
Mass and Weight: Understanding the Difference
Mass: A measure of the amount of matter in an object, constant regardless of location. [ \text{Unit: kg} ]
Weight: The force with which an object is attracted towards the Earth. Varies with the strength of the gravitational field. [ \text{Formula: } W = mg ]
[ \text{Unit: N (Newton)} ]
Free Fall and Acceleration Due to Gravity
Free Fall: When an object falls under the influence of gravity alone, without any air resistance.
Acceleration Due to Gravity (g):[ g = 9.8 \text{ m/s}^2 ]
Objects in free fall accelerate at this constant rate, irrespective of their mass.
Centripetal Force and Gravitational Motion
Centripetal Force is the force that keeps an object moving in a circular path. For celestial bodies, this force is provided by gravity.
Example: The Moon’s orbit around the Earth is maintained by the centripetal force due to Earth’s gravity.
Archimedes' Principle and Its Applications
Archimedes' Principle: When a body is fully or partially immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by it.
Applications: Designing ships, submarines, hydrometers, and lactometers.
Floating and Sinking: The Role of Density
Float: Occurs if an object’s density is less than the fluid it is placed in.
Sink: Occurs if an object’s density is greater than the fluid it is placed in.
Example: Iron nail sinks in water due to higher density, while cork floats due to lower density.
Thrust, Pressure, and Their Relationship
Thrust: The force exerted perpendicular to the surface.
Pressure: Thrust applied over a unit area. [ \text{Pressure (P)} = \frac{\text{Thrust (F)}}{\text{Area (A)}} ]
[ \text{Unit: Pa (Pascal)} ]
Example: Why sharp knives cut better—greater pressure due to smaller area.
Variations in Gravitational Force on Earth
Gravity is not uniform everywhere:
Greater at the poles than at the equator due to Earth's shape.
Decreases with altitude.
Practical Examples of Gravitation
Dropping an object: Falls due to Earth’s gravity.
Jumping: Always come back down.
Planets in motion: Stay in orbit due to the Sun’s gravitation.
Conclusion
Gravitation is a fundamental force that explains many natural phenomena, from the falling of an apple to the motion of celestial bodies. Understanding these concepts helps build a foundation for further studies in physics and astronomy.
🚀 Learn more about Notes with Chatterbot AI