Motion under Gravity

11. In a vacuum, a five-rupee coin, a feather of a sparrow bird and a mang

In a vacuum, a five-rupee coin, a feather of a sparrow bird and a mango are dropped simultaneously from the same height. The time taken by them to reach the bottom is t1, t2 and t3 respectively. In this situation, we will observe that

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”t<sub>1</sub>
”t<sub>1</sub>
”t<sub>3</sub>
”t<sub>1</sub>
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UPSC NDA-2 – 2017
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In a vacuum, there is no air resistance. According to the principles of physics, in the absence of air resistance, all objects dropped from the same height will fall with the same acceleration due to gravity, regardless of their mass, size, or shape. This means they will take the same amount of time to reach the ground. Therefore, the five-rupee coin, the feather, and the mango will all reach the bottom simultaneously, meaning $t_1 = t_2 = t_3$.
– In a vacuum, acceleration due to gravity is constant for all objects ($g$).
– Absence of air resistance.
– Time taken to fall from the same height is independent of mass.
This principle was first hypothesized by Galileo Galilei and later demonstrated convincingly in experiments, including the famous experiment conducted by astronaut David Scott on the Moon during the Apollo 15 mission, where a hammer and a feather were dropped simultaneously and hit the lunar surface at the same time. On Earth, air resistance significantly affects the fall time of objects, especially those with large surface area relative to mass like a feather.

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12. Two balls, A and B, are thrown simultaneously. A vertically upward wit

Two balls, A and B, are thrown simultaneously. A vertically upward with a speed of 20 m/s from the ground and B vertically downward from a height of 40 m with the same speed and along the same line of motion. At what points do the two balls collide by taking acceleration due to gravity as 9.8 m/s²?

The balls will collide after 3s at a height of 30·2 m from the ground
The balls will collide after 2s at a height of 20·1 m from the ground
The balls will collide after 1s at a height of 15·1 m from the ground
The balls will collide after 5s at a height of 20 m from the ground
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UPSC NDA-2 – 2016
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The correct answer is C) The balls will collide after 1s at a height of 15·1 m from the ground.
Let the origin be the ground level, with the upward direction as positive.
For ball A (thrown upward from ground):
Initial position, y₀_A = 0
Initial velocity, u_A = +20 m/s
Equation of motion: y_A(t) = y₀_A + u_A*t – (1/2)gt² = 0 + 20t – (1/2)(9.8)t² = 20t – 4.9t²

For ball B (thrown downward from 40 m):
Initial position, y₀_B = 40 m
Initial velocity, u_B = -20 m/s
Equation of motion: y_B(t) = y₀_B + u_B*t – (1/2)gt² = 40 – 20t – (1/2)(9.8)t² = 40 – 20t – 4.9t²

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Collision occurs when y_A(t) = y_B(t):
20t – 4.9t² = 40 – 20t – 4.9t²
Add 20t and 4.9t² to both sides:
40t = 40
t = 1 second

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Substitute t = 1s into either equation to find the height:
y_A(1) = 20(1) – 4.9(1)² = 20 – 4.9 = 15.1 m
y_B(1) = 40 – 20(1) – 4.9(1)² = 40 – 20 – 4.9 = 15.1 m

The balls collide after 1 second at a height of 15.1 m from the ground.

– Use equations of motion under constant acceleration (gravity).
– Define a consistent coordinate system (origin and positive direction).
– Set the positions of the two objects equal to find the time of collision.
– Use the time of collision to find the position (height) of collision.
– The acceleration due to gravity (g) is taken as 9.8 m/s² downwards.
– The relative velocity approach could also be used for the time of collision: v_rel = v_A – v_B. Initial relative velocity = 20 – (-20) = 40 m/s. The relative acceleration is g – g = 0. The initial separation is 40m. Time to collide = separation / relative velocity = 40m / 40m/s = 1s. This confirms the time calculation.

13. A ball is thrown vertically upward from the ground with a speed of 25-

A ball is thrown vertically upward from the ground with a speed of 25-2 m/s. The ball will reach the highest point of its journey in

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5·14 s
3·57 s
2·57 s
1·29 s
This question was previously asked in
UPSC NDA-2 – 2016
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Using the kinematic equation $v = u + at$, where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration, and $t$ is time.
Given initial velocity $u = 25.2 \text{ m/s}$ upwards, and acceleration due to gravity $a = -9.8 \text{ m/s}^2$ (assuming upward is positive). At the highest point, the velocity $v = 0 \text{ m/s}$.
Substituting the values into the equation: $0 = 25.2 + (-9.8)t$.
Solving for $t$: $9.8t = 25.2$, so $t = 25.2 / 9.8 = 252 / 98 = 126 / 49 = 18 / 7 \approx 2.57$ seconds.

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14. A stone is thrown horizontally from the top of a 20 m high building wi

A stone is thrown horizontally from the top of a 20 m high building with a speed of 12 m/s. It hits the ground at a distance R from the building. Taking g = 10 m/s² and neglecting air resistance will give:

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R=12m
R=18m
R=24m
R=30m
This question was previously asked in
UPSC NDA-1 – 2023
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The correct option is C.
For projectile motion, the horizontal and vertical components of motion are independent (neglecting air resistance). The time it takes for the stone to hit the ground is determined by the vertical distance and the acceleration due to gravity. The horizontal distance (range) is then calculated by multiplying the horizontal velocity by the time of flight.
The vertical motion is governed by $y = v_{y0}t + \frac{1}{2}gt^2$. Since the stone is thrown horizontally, $v_{y0} = 0$. The height is $y = 20$ m and $g = 10$ m/s². So, $20 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \implies 20 = 5t^2 \implies t^2 = 4 \implies t = 2$ s (time of flight). The horizontal motion is uniform with velocity $v_x = 12$ m/s. The range $R$ is given by $R = v_x \cdot t = 12 \text{ m/s} \cdot 2 \text{ s} = 24$ m.

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15. A ball is thrown vertically upward with a speed of 40 m/s. The time ta

A ball is thrown vertically upward with a speed of 40 m/s. The time taken by the ball to reach the maximum height would be approximately

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2 s
3 s
4 s
5 s
This question was previously asked in
UPSC NDA-1 – 2022
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The time taken by the ball to reach the maximum height would be approximately 4 s.
When an object is thrown vertically upward, its velocity decreases due to the acceleration due to gravity ($g$) acting downwards. At the maximum height, the velocity of the object momentarily becomes zero. We can use kinematic equations to find the time taken.
Given initial velocity $u = 40$ m/s. At maximum height, final velocity $v = 0$ m/s. The acceleration due to gravity is $a = -g$. Taking $g \approx 10$ m/s² (a common approximation for simplicity), $a = -10$ m/s². Using the kinematic equation $v = u + at$: $0 = 40 + (-10)t$. Solving for $t$: $10t = 40$, so $t = \frac{40}{10} = 4$ s. Using $g \approx 9.8$ m/s² would give a time slightly less than 4s, but 4s is the closest approximation among the options.

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16. A man weighing 70 kg is coming down in a lift. If the cable of the lif

A man weighing 70 kg is coming down in a lift. If the cable of the lift breaks suddenly, the weight of the man would become

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70 kg
35 kg
140 kg
Zero
This question was previously asked in
UPSC NDA-1 – 2016
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The weight of the man would become zero.
Weight is the force of gravity acting on an object’s mass (W = mg). Apparent weight is the normal force exerted by the supporting surface on the object. When the lift cable breaks, the lift and everything inside it are in free fall, accelerating downwards at the acceleration due to gravity (g). In a state of free fall, there is no supporting force counteracting gravity. Therefore, the apparent weight of the man (the force exerted by the lift floor on him) becomes zero. This is the sensation of weightlessness. His actual weight (mass * g) remains the same, but his apparent weight is zero.
If the lift were accelerating upwards, the apparent weight would be greater than his actual weight. If the lift were accelerating downwards at a rate less than g, his apparent weight would be less than his actual weight but greater than zero. If the lift were stationary or moving at constant velocity, his apparent weight would equal his actual weight.

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17. A body has a free fall from a height of 20 m. After falling through a

A body has a free fall from a height of 20 m. After falling through a distance of 5 m, the body would

lose one-fourth of its total energy
lose one-fourth of its potential energy
gain one-fourth of its potential energy
gain three-fourth of its total energy
This question was previously asked in
UPSC NDA-1 – 2016
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Let the total height be H = 20 m. The initial height is H = 20 m. After falling through a distance of 5 m, the body is at a height of h = 20 m – 5 m = 15 m from the ground.
We use the principle of conservation of mechanical energy, assuming no air resistance. Total Mechanical Energy (TE) = Potential Energy (PE) + Kinetic Energy (KE).
Initial state (at 20 m height):
Initial PE = mgH = mg(20).
Initial KE = 0 (free fall starts from rest).
Initial TE = mg(20) + 0 = 20mg.

After falling 5 m (at 15 m height):
Potential Energy at 15m = PE’ = mgh = mg(15).
The loss in potential energy is Initial PE – PE’ = 20mg – 15mg = 5mg.
The fraction of the *initial* potential energy lost is (Loss in PE) / (Initial PE) = (5mg) / (20mg) = 1/4.

By conservation of energy, the energy lost from potential energy is gained as kinetic energy.
KE gained = Loss in PE = 5mg.
Kinetic Energy at 15m = KE’ = 5mg.
Total Energy at 15m = PE’ + KE’ = 15mg + 5mg = 20mg. The total energy remains constant.

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Let’s evaluate the options:
A) lose one-fourth of its total energy: Incorrect, total energy is conserved.
B) lose one-fourth of its potential energy: The initial potential energy was 20mg. The loss is 5mg, which is indeed one-fourth (1/4) of the initial potential energy. Correct.
C) gain one-fourth of its potential energy: Incorrect, potential energy decreases as the body falls.
D) gain three-fourth of its total energy: Incorrect, total energy is conserved.

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Note: If the question meant “lose one-fourth of its *remaining* potential energy”, that would be different, but the phrasing “lose one-fourth of its total potential energy” usually refers to the initial maximum potential energy.

In free fall (assuming no air resistance), mechanical energy is conserved. As potential energy decreases, kinetic energy increases by an equal amount. Potential energy is proportional to height.
The loss of potential energy is equal to the gain in kinetic energy. After falling 5m from 20m, the body is at 15m. The initial PE was proportional to 20m, the PE at 15m is proportional to 15m. The drop in PE is proportional to 5m. 5/20 = 1/4, meaning 1/4 of the *initial* PE is lost, and 1/4 of the initial PE is converted into KE.

18. Statement-I: A body weighs less on a hill top than on earth’s surface

Statement-I: A body weighs less on a hill top than on earth’s surface even though its mass remains unchanged.
Statement-II: The acceleration due to gravity of the earth decreases with height.

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Both the statements are individually true and Statement II is the correct explanation of Statement I
Both the statements are individually true but Statement II is not the correct explanation of Statement I
Statement I is true but Statement II is false
Statement I is false but Statement II is true
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UPSC NDA-1 – 2015
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The correct option is A. Both statements are individually true, and Statement II is the correct explanation of Statement I.
Statement I is true because weight (W) is defined as mass (m) multiplied by the acceleration due to gravity (g), W = mg. A body’s mass remains constant regardless of location.
Statement II is true because the acceleration due to gravity (g) decreases with increasing height above the Earth’s surface. The gravitational force, and hence ‘g’, depends on the distance from the center of the Earth. At a higher altitude like a hill top, this distance is greater than on the surface at sea level. The relationship is approximately g’ = g(R/(R+h))², where R is Earth’s radius and h is height. As h increases, g’ decreases.
Since weight is directly proportional to ‘g’ (with constant mass), a decrease in ‘g’ at a hill top compared to the Earth’s surface directly causes the body to weigh less. Therefore, Statement II provides the correct reason for Statement I.

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19. Which one of the following statements is not correct for a freely fall

Which one of the following statements is not correct for a freely falling object?

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It accelerates.
Its momentum keeps on changing.
Its motion is affected only by the gravity.
Its motion is affected both by the gravity as well as by the air resistance.
This question was previously asked in
UPSC Geoscientist – 2022
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In physics, a “freely falling object” is ideally defined as an object that is accelerating due to gravity only, with no other forces acting on it (specifically, negligible air resistance).
– A) It accelerates: Correct. It accelerates downwards due to gravity (acceleration due to gravity, g).
– B) Its momentum keeps on changing: Correct. As it accelerates, its velocity changes, and momentum (mass x velocity) therefore changes.
– C) Its motion is affected only by the gravity: Correct. This is the definition of free fall in an ideal scenario (e.g., vacuum).
– D) Its motion is affected both by the gravity as well as by the air resistance: This statement is not correct for a *freely falling object* as defined in physics. Free fall explicitly excludes significant air resistance. While real objects falling in an atmosphere experience both, the term “freely falling object” usually implies the ideal case where air resistance is negligible.
– Free fall is motion under the sole influence of gravity.
– Air resistance is typically ignored in the definition of free fall.
– A freely falling object accelerates and its momentum changes.
In practical scenarios, objects falling through air do experience air resistance, which opposes the motion and increases with speed. Eventually, a terminal velocity is reached when air resistance equals gravity. However, the term “freely falling” in the context of fundamental physics principles often refers to the ideal case without air resistance.

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20. A stone of mass 1 kg and initially at rest is dropped from a tower of

A stone of mass 1 kg and initially at rest is dropped from a tower of height 40 m. When it reaches a height of 10 m from the ground level, what will be the values of its potential energy (PE) and kinetic energy (KE)? (Acceleration due to gravity is 10 m/s²)

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PE = 300 J, KE = 100 J
PE = 200 J, KE = 200 J
PE = 100 J, KE = 300 J
PE = 100 J, KE = 200 J
This question was previously asked in
UPSC Geoscientist – 2021
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The total mechanical energy (Potential Energy + Kinetic Energy) of the stone is conserved, assuming no air resistance.
Initial state (at height 40 m, at rest):
Initial PE = mgh₁ = 1 kg * 10 m/s² * 40 m = 400 J
Initial KE = ½ mv₁² = ½ * 1 kg * (0 m/s)² = 0 J
Total Energy = Initial PE + Initial KE = 400 J + 0 J = 400 J.
Final state (at height 10 m):
Final PE = mgh₂ = 1 kg * 10 m/s² * 10 m = 100 J.
Since Total Energy is conserved:
Total Energy = Final PE + Final KE
400 J = 100 J + Final KE
Final KE = 400 J – 100 J = 300 J.
So, at a height of 10 m, PE = 100 J and KE = 300 J.
In the absence of non-conservative forces (like air resistance), the total mechanical energy of a system remains constant. Energy is transformed between potential and kinetic forms.
The velocity of the stone at 10 m height can also be calculated from KE = ½ mv², giving 300 J = ½ * 1 kg * v², so v² = 600 m²/s², and v = √600 ≈ 24.5 m/s. Alternatively, one could use kinematic equations (v² = u² + 2as) to find the velocity and then calculate KE. The energy conservation method is often simpler for problems involving height and speed changes.

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