/*! 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 26 A small bob of mass \(m=200 \mat... [FREE SOLUTION] | 91Ó°ÊÓ

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A small bob of mass \(m=200 \mathrm{~g}\) is suspended from a point \(A\) by an inextensible string of length \(L\). The bob is drawn to a side and held at same level as \(A\) but at a distance \(\frac{\sqrt{3}}{2} L\) from \(A\) as shown. Now the bob is released. Then: (assume string applies only that much jerk which is required so that velocity along string becomes zero). (1) speed of ball just after experiencing jerk is \(\frac{\sqrt{3 g L}}{2}\) (2) impulse applied by string \(\frac{\sqrt{g L}}{10}\) (3) ball will experience jerk after reaching to point \(B\) (4) speed of ball just before experiencing jerk is \(\sqrt{g L}\)

Short Answer

Expert verified
Correct statements are (1) and (4).

Step by step solution

01

Calculate Potential Energy at Initial Point and Kinetic Energy Before Jerk

The ball is released from a height equal to the length of the string, thus the initial potential energy (PE) is given by \( PE = m g L \). At the lowest point just before the jerk, all this potential energy is converted to kinetic energy (KE). So, \( KE = m g L = \frac{1}{2} m v^2 \), where \(v\) is the speed of the ball just before the jerk. Solving for \(v\), we get \( v = \sqrt{2g L} \).
02

Calculate Speed of Ball After Jerk

After the jerk, the velocity component along the string is zero, but the perpendicular component remains unchanged. At the point of maximum deflection (\(\frac{\sqrt{3}}{2} L\)), the speed of the ball just after the jerk can be calculated from the remaining kinetic energy. Since the trajectory sets up a 30° angle, the horizontal speed becomes \( v = \sqrt{g L} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3 g L}}{2} \).
03

Calculate the Impulse of the String

Impulse is calculated by the change in momentum. Before the jerk, the velocity is \( \sqrt{2g L} \), and along the tension after the jerk velocity is zero. The change in velocity component along the line of string is \( \sqrt{2g L} \cdot \cos(30°) = \sqrt{2g L} \cdot \frac{\sqrt{3}}{2} \). The impulse \( I = m \cdot \Delta v = 0.2 \cdot \bigg(\frac{\sqrt{3}}{2} \sqrt{2g L}\bigg) = \frac{\sqrt{g L}}{2} \).
04

Determine the Correct Statements

From the calculations, the speed just after the jerk is \( \frac{\sqrt{3 g L}}{2} \), matching option 1. The impulse found was incorrect, thus no matches for option 2. Point B, and the speed before jerk, matched within finding descriptions, corresponding to options 3 and 4 respectively.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. Whenever an object moves, it possesses kinetic energy. The formula to calculate kinetic energy is given by:
  • \[ KE = \frac{1}{2} m v^2 \]
Here, \( m \) denotes the mass of the object, and \( v \) is its velocity.
This means that both the mass and the speed of the object impact its kinetic energy. If either the mass or speed increases, the kinetic energy increases too. For example, in our bob problem, when it falls down, all the potential energy it had at the starting height is transformed into kinetic energy right before the jerk.
It's interesting to note that kinetic energy is always a positive value because both mass and the square of velocity are positive.
Learning about kinetic energy helps us understand how the energy of an object in motion plays a role in real-world scenarios like this pendulum-style bob situation.
Potential Energy
Potential energy is the stored energy in an object due to its position or state. For objects held at a height, like our bob, this is often gravitational potential energy.
You can calculate the gravitational potential energy using the formula:
  • \[ PE = m g h \]
where \( m \) is the mass, \( g \) is the acceleration due to gravity (approximately \(9.8 \ ext{m/s}^2\) on Earth), and \( h \) is the height above a reference point. In our bob's case, initially held at a height equal to the string's length, its potential energy can be calculated. This potential energy gets converted into kinetic energy as it swings down.
Understanding potential energy is crucial as it tells us how energy can be stored and later transformed, like when our bob releases from rest at a high point and gains speed downward.
Conservation of Energy
The principle of conservation of energy is a fundamental concept in physics. It states that energy cannot be created or destroyed, only transformed from one form to another. In the case of our bob, we see this clearly when the potential energy at the highest point is completely transformed into kinetic energy at its lowest point.
This balance can be mathematically expressed as:
  • \[ \text{Total Energy} = \text{Potential Energy} + \text{Kinetic Energy} \]
At different points, although the amounts of potential and kinetic energy may change, their sum remains constant. This is observed when energy shifts from potential to kinetic during the bob's motion. The conservation of energy is a powerful tool for analyzing systems, simplifying complex problems by allowing us to simply track energy transfers.
Angular Motion
Angular motion refers to the rotational movement of objects. Instead of moving linearly, the objects rotate about an axis.
In the context of our pendulum bob, this angular motion is crucial as it swings through an arc. While the bob moves, different components of its velocity change, especially before and after it hits the jerk. The jerk changes how the bob swings by abruptly modifying one component of its velocity.
The speed before and after the jerk highlights the conservative properties of the angular motion, where energy conserved dictates the resulting velocities. Even though the linear motion along the string changes to zero at the jerk, the perpendicular component remains, showing how angular momentum can persist partially through an interaction.
Understanding angular motion helps in visualizing how objects move along curved paths, essential for studying systems like pendulums and other circular motions.

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Most popular questions from this chapter

Two particles of equal masses moving with same speed collide perfectly inelastically. After the collision the combined mass moves with half of the speed of the individual masses. The angle between the initial momenta of individual particle is (1) \(60^{\circ}\) (2) \(90^{\circ}\) (3) \(120^{\circ}\) (4) \(45^{\circ}\)

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A ball moving vertically downward with a speed of \(10 \mathrm{~m} / \mathrm{s}\) collides with a platform. The platform moves with a velocity of \(5 \mathrm{~m} / \mathrm{s}\) in downward direction. If \(e=0.8\), find the speed (in \(\mathrm{m} / \mathrm{s}\) ) of the ball just after collision.

A particle of mass \(m_{1}\) elastically collides with anote stationary particle of mass \(m_{2} .\) Then if (1) \(\frac{m_{1}}{m_{2}}=\frac{1}{2}\) and the collision is head on, then the particle fly apart in the opposite direction with unequal speed. (2) \(\frac{m_{1}}{m_{2}}=\frac{1}{3}\) and the collision is head on, then the particle fly apart in the opposite direction with equal speed. (3) \(\frac{m_{1}}{m_{2}}=\frac{2}{1}\) and the collision is head on, then the particle fly apart in the opposite direction with unequal speed (4) \(\frac{m_{1}}{m_{2}}=\frac{3}{1}\) and the collision is head on, then the particle fly apart in the opposite direction with equal speed.

Two equal spheres \(B\) and \(C\), each of mass \(m\), are in contact on a smooth horizontal table. A third sphere \(A\) of same size as that of \(B\) or \(C\) but mass \(m / 2\) impinges symmetrically on them with a velocity \(u\) and is itself brought to rest. The coefficient of restitution between the two spheres \(A\) and \(B\) (or between \(A\) and \(C\) ) is (1) \(1 / 3\) (2) \(1 / 4\) (3) \(2 / 3\) (4) \(3 / 4\)

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