/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 A glass marble dropped from a ce... [FREE SOLUTION] | 91Ó°ÊÓ

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A glass marble dropped from a certain height above the horizontal surface reaches the surface in time \(t\) and then continues to bounce up and down. The time in which the marble finally comes to rest is? (Note: The coefficient of restitution is \(e\) ) (A) \(e^{n} t\) (B) \(e^{2} t\) (C) \(t\left[\frac{1+e}{1-e}\right]\) (D) \(t\left[\frac{1-e}{1+e}\right]\)

Short Answer

Expert verified
(C) \(t\left[\frac{1+e}{1-e}\right]\)

Step by step solution

01

Calculate the time taken for the 1st bounce

Since the marble reaches the horizontal surface in time \(t\), we have the time taken for the first half (descent) of the bounce as \(t\).
02

Calculate the time taken for the 2nd and subsequent bounces

For the 2nd and subsequent bounces, the time taken for each half (ascent and descent) will be multiplied by the coefficient of restitution \(e\). So for the second bounce, the time taken for the ascent (and similarly, for the descent) will be \(et\). And for the third bounce, it will be \(e^2 t\), and so on.
03

Find the total time taken for all the bounces

To find the total time until the marble comes to rest, we need to sum the time taken for all the bounces (each ascent and descent). This gives us a geometric series with the first term as \(t\) and the common ratio as \(e\): \( T_{total} = t + 2(et + e^2t + e^3t + e^4t + \cdots) \)
04

Use the geometric series sum formula

Since the given series is a geometric series, we can use the formula for the sum of an infinite geometric series: \( S = \frac{a_1}{1 - r} \) Where \(S\) is the sum, \(a_1\) is the first term, and \(r\) is the common ratio. In our case, \(a_1 = et\) and \(r = e\). Plugging these into the formula, we get: \( S = \frac{et}{1 - e} \) Now, substituting the value of \(S\) in the equation for \(T_{total}\), we get: \( T_{total} = t + 2\left(\frac{et}{1 - e}\right) \)
05

Simplify and find the answer

Now, let's simplify the expression for the total time taken: \( T_{total} = t\left(1 + \frac{2e}{1 - e}\right) = t\left[\frac{1 - e + 2e}{1 - e}\right] = t\left[\frac{1 + e}{1 - e}\right] \) Looking at the given options, the correct answer is: (C) \(t\left[\frac{1+e}{1-e}\right]\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Restitution
The coefficient of restitution (denoted by the letter e) is a measure of how much energy is conserved in a collision or bounce. This concept is crucial in understanding the dynamics of collisions for objects in motion. In simple terms, it's a way to quantify how bouncy an object is when it hits another surface.

When an object like a glass marble collides with a surface, some of its kinetic energy is converted into other forms of energy, like heat or sound, and some is conserved. The coefficient of restitution tells us the ratio of the speed post-collision to the speed pre-collision along the line of impact. It has a value that ranges from 0 (perfectly inelastic collision, where the object does not bounce at all) to 1 (perfectly elastic collision, where no kinetic energy is lost in the bounce).

Mathematically, if we consider a collision where an object strikes a surface with a speed v and bounces off with a speed ev, then e is the coefficient of restitution. This figure is applied repeatedly to problems involving bounces, especially in the context of competitive exams like the JEE MAIN.
Geometric Series in Physics
A geometric series in physics often models scenarios where there's exponential growth or decay, and this is particularly pertinent to motion with repeated events such as bouncing or oscillations. In the context of our exercise, the times taken for the successive bounces of a marble form a geometric series, where each term after the first is found by multiplying the previous term by a common ratio.

The sum of an infinite geometric series S is given by the formula S = a_1 / (1 - r), where a_1 is the first term and r is the common ratio. This formula is pivotal when we want to calculate the total time taken for an object to come to rest after it has been dropped, as long as the bounces continue infinitely and the ratio between successive times is constant. In competitive exams, knowing how to apply this formula quickly and accurately is often the key to solving complex physics problems efficiently.
Mechanics for Competitive Exams
Mechanics is a fundamental section in the syllabus of physics for competitive exams such as JEE MAIN. It encompasses various concepts, including motion, forces, energy, and momentum, which are all foundational to the study of physics.

Students preparing for such exams need to be adept at solving problems that involve these principles. One of the hallmarks of competitive exams is the application of these concepts to multi-step problems, which can include a combination of idealized systems like perfectly elastic collisions or perfect simple harmonic motion, and real-world scenarios like the bouncing marble problem.

The ability to understand and apply the coefficient of restitution in collision problems, geometric series in repeated motion scenarios, and principles of mechanics like conservation laws are all crucial. Good problem-solving skills in mechanics involve not just memorizing formulas, but also understanding the underlying physical principles which allow for the application of these formulas in various contexts. Therefore, an emphasis on conceptual clarity, alongside practice with varied and complex problem sets, is essential for students aiming to excel in competitive exams.

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Most popular questions from this chapter

A particle of mass \(m\) describes a circle of radius \(r\). The centripetal acceleration of the particle is \(4 / r^{2}\). The momentum of the particle is (A) \(\frac{4 m}{r}\) (B) \(\frac{2 m}{r}\) (C) \(\frac{4 m}{\sqrt{r}}\) (D) \(v\)

A ball of mass \(m\) approaches a wall of mass \(M(>>m)\) with speed \(4 \mathrm{~m} / \mathrm{s}\) along the normal to the wall. The speed of wall is \(1 \mathrm{~m} / \mathrm{s}\) towards the ball. The speed of the ball after an elastic collision with the wall is (A) \(5 \mathrm{~m} / \mathrm{s}\) away from the wall. (B) \(9 \mathrm{~m} / \mathrm{s}\) away from the wall. (C) \(3 \mathrm{~m} /\) s away from the wall. (D) \(6 \mathrm{~m} / \mathrm{s}\) away from the wall.

A bomb of mass \(3 m \mathrm{~kg}\) explodes into two pieces of mass \(m \mathrm{~kg}\) and \(2 \mathrm{~m} \mathrm{~kg}\). If the velocity of \(m \mathrm{~kg}\) mass is \(16 \mathrm{~m} / \mathrm{s}\), the total energy released in the explosion is (A) \(192 \mathrm{~mJ}\) (B) \(96 \mathrm{~m} \mathrm{~J}\) (C) \(384 \mathrm{~m} \mathrm{~J}\) (D) \(768 \mathrm{~m} \mathrm{~J}\)

A bomb of mass \(12 \mathrm{~kg}\) is dropped by a fighter plane moving horizontally with a speed of \(100 \mathrm{~ms}^{-1}\) from a height of \(1 \mathrm{~km}\) from the ground. The bomb exploded after 10 s into two pieces of masses in the ratio \(1: 5\). If the small part started moving horizontally with a speed of \(600 \mathrm{~ms}^{-1}\) the speed of bigger part will be \(\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)\) (A) \(100 \mathrm{~ms}^{-1}\) (B) \(10 \sqrt{65} \mathrm{~ms}^{-1}\) (C) \(120 \mathrm{~ms}^{-1}\) (D) \(100 \sqrt{2} \mathrm{~ms}^{-1}\)

A ball of mass \(1 \mathrm{~kg}\) strikes a wedge of mass \(4 \mathrm{~kg}\) horizontally with a velocity of \(10 \mathrm{~m} / \mathrm{s}\). Just after collision velocity of wedge becomes \(4 \mathrm{~m} / \mathrm{s}\). Friction is absent everywhere and collision is elastic. Then (A) the speed of ball after collision is \(6 \mathrm{~m} / \mathrm{s}\). (B) the speed of ball after collision is \(8 \mathrm{~m} / \mathrm{s}\). (C) the speed of ball after collision is \(4 \mathrm{~m} / \mathrm{s}\). (D) the speed of ball after collision is \(10 \mathrm{~m} / \mathrm{s}\).

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