Figure 8-73 a shows a molecule consisting of two atoms of...

Figure $8-73 a$ shows a molecule consisting of two atoms of masses $m$ and $M$ (with $m \ll M)$ and separation $r$. Figure $8-73 b$ shows the potential energy $U(r)$ of the molecule as a function of $r$. Describe the motion of the atoms (a) if the total mechanical energy $E$ of the two-atom system is greater than zero (as is $\left.E_{1}\right)$, and (b) if $E$ is less than zero $\left(\right.$ as is $\left.E_{2}\right) .$ For $E_{1}=1 \times 10^{-19} \mathrm{~J}$ and $r=0.3 \mathrm{~nm}$, find $(\mathrm{c})$ the potential energy of the system, (d) the total kinetic energy of the atoms, and (e) the force (magnitude and direction) acting on each atom. For what values of $r$ is the force (f) repulsive, $(g)$ attractive, and (h) zero?

Figure $8-73 a$ shows a molecule consisting of two atoms of masses $m$ and $M$ (with $m \ll M)$ and separation $r$. Figure $8-73 b$ shows the potential energy $U(r)$ of the molecule as a function of $r$. Describe the motion of the atoms (a) if the total mechanical energy $E$ of the two-atom system is greater than zero (as is $\left.E_{1}\right)$, and (b) if $E$ is less than zero $\left(\right.$ as is $\left.E_{2}\right) .$ For $E_{1}=1 \times 10^{-19} \mathrm{~J}$
and $r=0.3 \mathrm{~nm}$, find $(\mathrm{c})$ the potential energy of the system, (d) the total kinetic energy of the atoms, and (e) the force (magnitude and direction) acting on each atom. For what values of $r$ is the force (f) repulsive, $(g)$ attractive, and (h) zero?

Key Concepts

Mechanical Energy in Molecular Systems

This concept refers to the total energy of a system, which is the sum of kinetic and potential energies. In molecular systems, a positive total mechanical energy indicates that the atoms or molecules are not bound together and can move apart, resulting in unbound or scattering states. Conversely, a negative total mechanical energy signifies that the system is in a bound state, where the atoms remain confined due to the attractive forces summarized in the potential energy function.

Interatomic Potential Energy

Interatomic potential energy describes the energy stored in the system as a function of the distance between atoms. It typically features a minimum at a characteristic separation, indicating the equilibrium bond length where the attractive and repulsive interactions balance. The shape of this potential energy curve determines how the force between atoms varies with distance, which is essential in predicting molecular behavior and stability.

Kinetic Energy Distribution in Two-Body Systems

In a system consisting of two interacting particles, the total kinetic energy arises from the motion of both atoms. The distribution of kinetic energy depends on the masses of the atoms and their velocities. When analyzing such systems, it is often useful to consider concepts like the center-of-mass frame or reduced mass to simplify the understanding of how energy is partitioned and conserved during motion.

Force as Derivative of Potential Energy

The force acting on an atom in a molecular system is related to the spatial derivative of the potential energy with respect to the interatomic distance. Mathematically, this is expressed as the negative gradient of the potential energy. This relationship implies that by knowing the potential energy function, one can determine both the magnitude and the direction of the force acting on each atom, providing insights into the dynamic behavior of the system.

Attractive and Repulsive Interaction Regions

The regions of attractive and repulsive interactions are determined by the sign of the force, which in turn is derived from the gradient of the potential energy curve. When the force is directed to decrease the separation between atoms, it is considered attractive; conversely, if the force increases the separation, it is repulsive. Additionally, the point where the force is zero corresponds to the equilibrium separation where the net force on each atom vanishes, indicating a stable configuration.

Transcript

00:01 In this question we have given a pulley whose radius is given to us that is 10 semi and moment of inertia is given to us and it is told that the planes are frictionless and we have to calculate the acceleration of block this that is 4 kg block.

00:22 Now i assume that the mass of this block is m1 so m1 is given to us that is 2 kg and m2 is 4 kg we have given the moment of inertia of this pulley that is i is equal to 0 .5 kg meter square and radius of the pulley is 0 .1 meter so i'm going to show the forces acting on the system one is the component of this weight in this direction that is m1g sine theta and there also the component of weight in this direction is m2g sine theta which causes the acceleration of the block in this direction and tension in this string is let that is t1 and tension in this string is t2 now due to the motion of this block there is a rotation in this pulley that the angular acceleration is given by alpha and due to and now the force acting in this block is m2a and on this block this is m1a which causes this block to move in upward direction and this causes to move block in this direction now i'm going to solve this problem so from the fbd we can write that from free body diagram we can say that t1 minus m1g sine theta that is net force is equal to m1a this is our first this is the for block m1 similarly for pulley i can write t2 minus t1 is equal to i into a by r square this is our equation second and for block m2 i can write that m2g sine theta minus t 2 is equal to m2a this is our equation third now adding equation 1 and third so add equation 1 and 3rd we get m2g sine theta plus this is minus t 2 plus t 1 is a minus m 1g sine theta this equals to m1 plus m2 into a now simplify this equation so this is m2 minus m1 g sine theta is equal to sorry this is plus minus i upon r square into a is equals to m1 plus m2 into from here we get the value of acceleration a that is m2 minus m1 g sine theta upon this is m1 plus m2 plus i upon r square so this is the generalized equation now put the values in this equation we have all the known information so m2 is 4 kg m1 is 2 kg into this is 10 sine angle is 45 upon 4 plus sorry this is 2 plus 4 plus i is given to that is 0 .5 upon this is 0 .1 whole square after solving this we get acceleration that is 0 .248 or i can write it 0 .25 meter per second square...

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