Two disks are initially at rest, each of mass M, connected by a string between their centers, as shown in Figure 9.55. The disks slide on low-friction ice as the center of the string is pulled by a string with a constant force F through a distance d. The disks collide and stick together, having moved a distance b horizontally.

(a) What is the final speed of the stuck-together disks? (b) When the disks collide and stick together, their temperature rises Calculate the increase in internal energy of the disks assuming that the process is so fast that there is insufficient time for there to be much transfer of energy to the ice due to a temperature difference. (Also ignore the small amount of energy radiated away as sound produced in the collisions between the disks)

The given data can be listed below,

• The mass of the disk is$M$ .
• The string is pulled by a string with a constant force is $F$.
• The string is pulled by string at constant force at a distance is $d$.
• The disk collide and stick together move a distance is$b$ .

The term "translate" in physics and mathematics refers to moving from one point to another. Translational motion is the most common type of motion we've studied at so far.

The total work done on disk is given as,

$w=Fd$

The translational kinetic energy of the disk is given as,

$K{E}_{trans}=Fb$

When the disk is collide and sticks together then the mass of the disk is together is 2M.

2M

So, the translational kinetic energy is given as,

$K{E}_{trans}=\frac{1}{2}{M}_t{v}^2$

Here${}^{M_t}$ is the mass of the disk together; whose value is 2M.

Now equating translational kinetic energy, which is given as:

$$\begin{gathered} \frac{1}{2}\left(2M\right)v^2=Fb \\ {v}^2=\frac{Fb}{M} \\ v=\sqrt{\frac{Fb}{M}} \end{gathered}$$

Thus, the expression for the speed of the disk is stuck together is $\sqrt{\frac{Fb}{M}}$.

Using conservation of energy, the equation is given as,

$∆E_{sys}=W+Q$

Here${}^{∆E_{sys}}$is the energy of system, Q is the heat energy due to temperature difference.

Now,

$$\begin{gathered} K{E}_{trans}+∆E_{int}=W \\ ∆E_{int}=W-K{E}_{trans} \end{gathered}$$

Substitutefor Fd for W, and Fb for $K{E}_{trans}$in the above equation.

$$\begin{gathered} ∆E_{int}=Fd-Fb \\ =F\left(d-b\right) \end{gathered}$$

Thus, the expression for the increase in internal energy of the disk is${}^{F\left(d-b\right)}$.