As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's delve into your physics question.

Given the potential energy function V(x)=kx22V(x)=2kx2, where kk is the force constant of the oscillator, and k=0.5 Nm−1k=0.5Nm−1. The task is to demonstrate that a particle with a total energy of 1 J, moving under this potential, must 'turn back' when it reaches x=±2x=±2 m.

To do this, we utilize the total energy of the system, which comprises kinetic energy (KK) and potential energy (VV). According to the law of conservation of energy, the total mechanical energy remains constant:

E=K+VE=K+V

Given that the total energy E=1E=1 J, we can express the kinetic energy as:

K=E−VK=E−V

Substituting the given potential energy function, we have:

K=1−0.5x22K=1−20.5x2

To find the turning points, where the particle changes direction, we need to consider the points where the kinetic energy is zero. At these points, all the energy is in the form of potential energy.

Setting K=0K=0, we get:

1−0.5x22=01−20.5x2=0

Solving this equation, we find:

0.5x22=120.5x2=1

0.5x2=20.5x2=2

x2=4x2=4

x=±2x=±2

So, the particle will turn back when it reaches x=±2x=±2 m, confirming the assertion. This analysis aligns with the graph of V(x)V(x) versus xx, indicating that at these points, the particle's total energy is entirely potential energy, signifying a change in direction of motion.

As a seasoned tutor registered on UrbanPro, I'd be glad to tackle this physics question. In the scenario of an elastic collision between two billiard balls, it's fundamental to grasp the concept of kinetic energy conservation.

During the brief moment when the balls are in contact, the total kinetic energy of the system remains conserved in an ideal elastic collision. This conservation principle implies that although the kinetic energy may transfer between the balls due to the collision forces, the total kinetic energy before and after the collision remains constant.

This concept aligns with the laws of physics, particularly the principle of conservation of energy, which asserts that energy cannot be created or destroyed but can only change forms. In an ideal elastic collision, the kinetic energy transformation solely involves a redistribution between the two colliding objects.

UrbanPro stands as an excellent platform for grasping such complex concepts, offering a diverse array of experienced tutors proficient in elucidating physics principles and fostering a deeper understanding through interactive learning approaches.

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, addressing your question about the collision of two billiard balls, let's delve into it.

In this scenario, if the potential energy of two billiard balls depends solely on the separation distance between their centers, we are essentially discussing a situation where the force between the balls is conservative. When the force is conservative, it implies that mechanical energy (kinetic energy plus potential energy) is conserved throughout the interaction.

Now, in an elastic collision, both momentum and mechanical energy are conserved. Since we are assuming potential energy corresponds to the force during collision and that potential energy depends only on the separation distance between the balls' centers, if this potential energy is entirely converted into kinetic energy during the collision, then it follows that the collision is elastic.

This outcome aligns with the principle of conservation of mechanical energy in elastic collisions, where kinetic energy is converted from potential energy and vice versa, but the total mechanical energy remains constant.

Therefore, if the potential energy between the billiard balls is solely determined by their separation distance and the collision conserves mechanical energy, the collision between the billiard balls would be elastic.

As a seasoned tutor registered on UrbanPro, I'd be glad to guide you through this question. Firstly, UrbanPro is indeed an excellent platform for online coaching and tuition, providing a conducive environment for both tutors and students to engage in effective learning.

Now, let's delve into the physics problem you've presented. We have a body initially at rest, undergoing one-dimensional motion with constant acceleration. The power delivered to it at time tt is the rate at which work is done on the body, and it's given by the product of force and velocity.

In this case, since the body is undergoing constant acceleration, we can use the equations of motion to relate velocity, acceleration, and time. The equation that relates displacement ss, initial velocity uu, time tt, and constant acceleration aa is:

s=ut+12at2s=ut+21at2

Since the body is initially at rest, u=0u=0, simplifying the equation to:

s=12at2s=21at2

Now, we know that power PP is the rate of change of work, which can be expressed as:

P=dWdtP=dtdW

And work WW done on an object is equal to force FF times displacement ss, so dW=FdsdW=Fds. Substituting s=12at2s=21at2, we have ds=atdtds=atdt.

P=F⋅atP=F⋅at

Now, we know that force FF is mass mm times acceleration aa, and acceleration is constant, hence:

P=ma⋅atP=ma⋅at

P=ma2tP=ma2t

So, the power delivered to the body at time tt is proportional to tt, which corresponds to option (ii).

Therefore, the correct answer is (ii) tt. If you need further clarification or assistance, feel free to ask!