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Question

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 t 1, t 2and t 3respectively. In this situation, we will observe that

The correct answer is

t 1= t 2=t 3

Understanding Free Fall in a Vacuum

The question asks about the time taken for different objects – a five-rupee coin, a feather, and a mango – to reach the bottom when dropped simultaneously from the same height in a vacuum. The times are denoted as $t_1$, $t_2$, and $t_3$ respectively.

Let's first understand what happens when objects fall in a vacuum. A vacuum is a space completely devoid of matter, including air. In the absence of air, there is no air resistance or drag force acting on the falling objects. The only significant force acting on the objects is gravity.

Gravity and Acceleration in Vacuum

Gravity is the force of attraction between objects with mass. On Earth, gravity pulls objects towards the center of the Earth. The acceleration caused by gravity is denoted by 'g'. Near the Earth's surface, the value of 'g' is approximately $9.8 \, m/s^2$. This acceleration is nearly constant for all objects, regardless of their mass or shape.

When objects fall in a vacuum, because there is no air resistance to oppose their motion, the only force is gravity. According to Newton's second law of motion, force equals mass times acceleration ($F=ma$). In this case, the force is gravity, which can be expressed as $F_g = mg$, where 'm' is the mass of the object. So, $mg = ma$. This means the acceleration 'a' of the object is equal to 'g'.

Thus, all objects in a vacuum accelerate downwards at the same rate, 'g', irrespective of their mass, size, or shape.

Calculating Time of Fall in Vacuum

For objects dropped from rest (initial velocity $u=0$) from a certain height 'h' under constant acceleration 'g', we can use the kinematic equation:

\[s = ut + \frac{1}{2}at^2\]

Here, 's' is the distance covered (which is the height 'h'), 'u' is the initial velocity ($0$), 'a' is the acceleration (which is 'g' in vacuum), and 't' is the time taken. Substituting these values, we get:

\[h = (0)t + \frac{1}{2}gt^2\] \[h = \frac{1}{2}gt^2\]

We can rearrange this equation to find the time taken 't':

\[t^2 = \frac{2h}{g}\] \[t = \sqrt{\frac{2h}{g}}\]

From this equation, we can see that the time taken to fall from a height 'h' in a vacuum depends only on the height 'h' and the acceleration due to gravity 'g'. It does not depend on the mass, density, or shape of the object.

Applying to the Given Scenario: Coin, Feather, Mango

In this problem, the five-rupee coin, the feather of a sparrow bird, and the mango are dropped simultaneously from the same height in a vacuum. Let the height be 'h'.

  • Time taken by the five-rupee coin ($t_1$) depends on 'h' and 'g'.
  • Time taken by the feather ($t_2$) depends on 'h' and 'g'.
  • Time taken by the mango ($t_3$) depends on 'h' and 'g'.

Since they are all dropped from the same height 'h' and are falling under the same acceleration 'g' (in a vacuum), the time taken for each object to reach the bottom will be the same.

Therefore, $t_1 = t_2 = t_3$.

Comparison: Vacuum vs. Air

It is important to note the difference between falling in a vacuum and falling in air. In air, there is air resistance, which is a force that opposes the motion of the object through the air. Air resistance depends on factors such as the object's speed, shape, size, and the density of the air. Objects with larger surface areas (like a feather) experience more significant air resistance than compact objects (like a coin or mango) at the same speed. This is why, in air, a coin or mango falls faster than a feather.

However, in a vacuum, air resistance is absent. The motion is purely governed by gravity, which affects all objects equally in terms of acceleration.

Condition Forces Acting Effect on Different Objects Time to Fall from Same Height
In Air Gravity, Air Resistance Air resistance varies with object properties (shape, size, speed). Lighter/less dense objects or those with large surface area experience relatively larger air resistance, slowing them down more. Different times (heavier/denser objects with less relative air resistance fall faster).
In Vacuum Gravity Only Gravity causes the same acceleration 'g' for all objects, regardless of their properties. Same time for all objects.

Conclusion

Because the five-rupee coin, the feather, and the mango are dropped simultaneously from the same height in a vacuum, where there is no air resistance, they all experience the same acceleration due to gravity. Consequently, they will take the same amount of time to reach the bottom.

The correct observation is $t_1 = t_2 = t_3$.

Revision Table: Free Fall Concepts

Concept Description Key Factor in Vacuum
Free Fall Motion under the influence of gravity only. Pure free fall occurs in vacuum (no air resistance).
Acceleration due to Gravity (g) Rate at which gravity accelerates objects. Approximately $9.8 \, m/s^2$ near Earth's surface. Constant for all objects regardless of mass or shape.
Air Resistance Force opposing motion through air. Depends on speed, shape, size. Absent in a vacuum.
Time of Fall (from height h, in vacuum) Time taken to reach the ground. Given by $t = \sqrt{\frac{2h}{g}}$, independent of object's mass/shape.

Additional Information: Historical Context

The idea that objects of different masses fall at the same rate in the absence of air resistance was famously demonstrated by Galileo Galilei. While the story of him dropping objects from the Leaning Tower of Pisa might be apocryphal, his experiments and reasoning led to the understanding that all objects accelerate equally under gravity if air resistance is negligible or absent. This principle was later formalized by Newton's laws of motion and universal gravitation.

Modern demonstrations, like dropping a feather and a bowling ball in a large vacuum chamber, clearly show them hitting the bottom at the exact same moment, confirming the principle of free fall in a vacuum.

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Important Questions from Gravitation

  1. Which one of the following statement is true for the relation, \(F= \frac{{G{m_1}{m_2}}}{{{r^2}}}\) ?

    (All symbols have their usual meanings)
  2. The free-fall acceleration g increases as one proceeds, at sea level, from the equator toward either pole. The reason is

  3. A planet has a mass M 1and radius R 1. The value of acceleration due to gravity on its surface is g 1. There is another planet 2, whose mass and radius both are two times that of the first planet. Which one of the following is the acceleration due to gravity on the surface of planet 2?

  4. Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M each are placed R/2 distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be

  5. Suppose the force of gravitation between two bodies of equal masses is F. If each mass is doubled keeping the distance of separation between them unchanged, the force would become

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