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Which of the following statements is correct? (1) Kinetic energy of a system can be changed without changing its momentum. (2) Kinetic energy of a system cannot be changed without changing its momentum. (3) Momentum of a system cannot be changed without changing its kinetic energy. (4) A system cannot have energy without having momentum.

Short Answer

Expert verified
(1) is correct: Kinetic energy can change without changing momentum.

Step by step solution

01

Understand Kinetic Energy and Momentum

Kinetic energy (\( KE \) is given by \( KE = \frac{1}{2}mv^2 \), whereas momentum (\( p \) is given by \( p = mv \). Both terms depend on velocity (\( v \)) and mass (\( m \)) of an object. While both depend on mass and velocity, they are not the same thing.
02

Analyzing the Possibility of Changing Kinetic Energy Without Changing Momentum

To change kinetic energy without changing momentum, consider applying an external force that alters velocity's magnitude, not direction. This is possible because kinetic energy is a square of velocity whereas momentum is linear. Thus, changing velocity (not direction) can change kinetic energy and keep momentum constant.
03

Checking Whether Momentum Without Changing Kinetic Energy is Possible

Altering just the direction of velocity (vector change) can keep kinetic energy constant while changing momentum since momentum has direction unlike kinetic energy which is a scalar. Thus, changing direction without changing speed alters momentum with constant kinetic energy.
04

Investigating Energy Without Momentum

A system can have internal energy (e.g., potential energy) without momentum. For instance, a book on a table has gravitational potential energy but no momentum. Thus, energy does not mandate momentum.
05

Identify the Correct Statement

Statement (1) is correct because kinetic energy can be changed without changing momentum, by changing the magnitude of velocity while keeping its direction.

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Key Concepts

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

Physics JEE ADVANCED
Physics JEE Advanced explores deep concepts in physics. These exercises prepare students for competitive exams and help them understand various physical phenomena. One key area in these exercises is understanding kinetic energy and momentum, critical elements in mechanics.

To grasp these concepts, students learn that kinetic energy (\( KE = \frac{1}{2}mv^2 \)) is a scalar quantity dependent on mass and the square of velocity. Momentum (\( p = mv \)), however, is a vector quantity, dependent on mass and velocity direction.
  • Scalar quantities have magnitude but no direction, like kinetic energy.
  • Vector quantities have both magnitude and direction, such as momentum.
This distinction helps students differentiate how energy and momentum individually contribute to system dynamics. Students aiming for JEE Advanced need to master these subtleties to tackle physics problems effectively.
Conservation of Momentum
The principle of conservation of momentum is fundamental in mechanics. It states that for a closed system, the total momentum remains constant, provided no external forces act on it. This principle helps explain various phenomena, from simple collisions to complex systems.

In practice, this means that if two objects collide, their combined momentum before impact will equal their combined momentum after impact.
  • Internal forces between objects do not affect the total momentum.
  • Only external forces can change the total system momentum.
Therefore, in events like collisions, you can predict the outcome by applying this conservation law. Understanding this principle is crucial for solving problems in mechanics and is heavily tested in examinations like JEE Advanced.
Mechanics of Systems
Mechanics of systems delves into the behavior of multiple interacting objects. This branch of physics combines principles like forces, energy, and momentum to understand how systems operate.

When analyzing a system, it’s important to consider:
  • The interaction between objects, like gravitational or frictional forces.
  • The transfer of energy between parts of the system.
  • How internal and external forces impact system behavior.
Unlike single-object mechanics, systems can have kinetic energy without momentum due to internal energy types such as potential energy. For instance, an unbalanced vibrant system can store energy due to height (potential) yet have no momentum if stationary.
This conceptual understanding is critical in fields such as engineering and physics, offering insights into everything from star systems dynamics to simple machine operations.

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Most popular questions from this chapter

The potential energy of a particle is determined by the expression \(U=\alpha\left(x^{2}+y^{2}\right)\), where \(\alpha\) is a positive constant. The particle begins to move from a point with coordinates \((3,3)\), only under the action of potential field force. Then its kinetic energy \(T\) at the instant when the particle is at a point with the coordinates \((1,1)\) is (1) \(8 \alpha\) (2) \(24 \alpha\) (3) \(16 \alpha\) (4) Zero

A man pushes a block of \(30 \mathrm{~kg}\) along a level floor at a constant speed with a force directed at \(45^{\circ}\) below the horizontal. If the coefficient of friction is \(0.20\), then match the following. $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Work done by all forces exerted by the surface on the block in \(20 \mathrm{~m}\)} & \text { a. } \text{Zero} \\ \hline \text { ii. } \text{Work done by the force of gravity} & \text { b. } \text{\(-1500 \mathrm{~J}\)} \\ \hline \text { iii. } \text{Work done by the man on the block in pushing it through \(10 \mathrm{~m}\)} & \text { c. } \text{\(750 \mathrm{~J}\)}\\\ \hline \text { iv. } \text{Net force on the block} & \text { d. } \text{\(30 \mathrm{~J}\)} \\ \hline \end{array} $$

A block of mass \(2 \mathrm{~kg}\) is hanging over a smooth and light pulley through a light string. The other end of the string is pulled by a constant force \(F=40\) \(\mathrm{N}\). The kinetic energy of the particle increase \(40 \mathrm{~J}\) in a given interval of time. Then: \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (1) tension in the string is \(40 \mathrm{~N}\) (2) displacement of the block in the given interval of time is \(2 \mathrm{~m}\) (3) work done by gravity is \(-20 \mathrm{~J}\) (4) work done by tension is \(80 \mathrm{~J}\)

In which of the following cases can the work done increase the potential energy? (1) Both conservative and non-conservative forces (2) Conservative force only (3) Non-conservative force only (4) Neither conservative nor non-conservative forces.

A block of mass \(m\) lies on a wedge of mass \(M\). The wedge in tum lies on a smooth horizontal surface. Friction is absent everywhere. The wedge-block system is released from rest. All situations given in Column I are to be estimated in duration the block undergoes a vertical displacement \(h\) starting from rest. Match the statements in Column I with the results in Column II. ( \(g\) is acceleration due to gravity.) $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Work done by normal reaction acting on the block is} & \text { a. } \text{positive} \\ \hline \text { ii. } \text{Work done by normal reaction (exerted by block) acting on the wedge is} & \text { b. } \text{negative} \\ \hline \text { iii. } \text{The sum of work done by normal reaction on the block and work done by normal on wedge} & \text { c. } \text{Zero}\\\ \hline \text { iv. } \text{Net work done by all forces on the block is} & \text { d. } \text{less than \(m g h\) in magnitude} \\ \hline \end{array} $$

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