Question

4.there are two weights of the same mass of 2.0 kg attached to a string looped over a braked pulley. the left weight is lying on the floor, the right one is suspended 1.0 m above the floor. there is another weight of mass 0.5 kg attached to the string 20 cm above the right weight. the string except the part around the pulley is stretched vertically. we eventually released the pulley. 1. what is the speed at which the lower right weight hits the floor? 2. what is the height that the left weight rises to after the lower right weight falls inelastically to the floor? 3. what is the maximum height that the lower right weight rises back to? neglect the mass of both the string and the pulley, deformation of the string by tension, and friction.

Explanation:

Step1: Determine the total mass of the system

The total mass of the system $M_{total}=2M + m$, where $M = 2.0\ kg$ and $m=0.5\ kg$. So $M_{total}=2\times2 + 0.5=4.5\ kg$.

Step2: Apply conservation of mechanical - energy for the first part (finding the speed when the lower right - weight hits the floor)

The initial potential energy of the system is due to the lower right - weight of mass $M$ at height $h_1 = 1.0\ m$ and the upper right - weight of mass $m$ at height $h_1 + d$ (where $d = 0.2\ m$). The initial kinetic energy is $K_i = 0$. The final kinetic energy is $K_f=\frac{1}{2}M_{total}v^2$ and the final potential energy is $U_f$ (which we can set to 0 at the floor level for the lower right - weight).
The initial potential energy $U_i = Mgh_1+mg(h_1 + d)$.
By conservation of mechanical energy $U_i=K_f$, so $Mgh_1+mg(h_1 + d)=\frac{1}{2}(2M + m)v^2$.
Substitute $M = 2\ kg$, $m = 0.5\ kg$, $h_1 = 1\ m$, $d = 0.2\ m$, and $g = 9.8\ m/s^2$:
\[

$$\begin{align*} 2\times9.8\times1+0.5\times9.8\times(1 + 0.2)&=\frac{1}{2}(2\times2 + 0.5)v^2\\ 19.6+0.5\times9.8\times1.2&=\frac{1}{2}\times4.5v^2\\ 19.6 + 5.88&=2.25v^2\\ 25.48&=2.25v^2\\ v&=\sqrt{\frac{25.48}{2.25}}\approx3.37\ m/s \end{align*}$$

\]

Step3: Analyze the situation after the lower right - weight hits the floor inelastically

When the lower right - weight hits the floor inelastically, the mass of the moving system becomes $m + M$ (where $M = 2\ kg$ and $m = 0.5\ kg$). The initial kinetic energy of the remaining moving system (the upper right - weight and the left - weight) is $K_{i2}=\frac{1}{2}(M + m)v^2$ (with the $v$ found above), and this kinetic energy will be converted into potential energy $U_{f2}=(M + m)gh_2$ as the left - weight rises.
\[

$$\begin{align*} \frac{1}{2}(M + m)v^2&=(M + m)gh_2\\ h_2&=\frac{v^2}{2g} \end{align*}$$

\]
Substitute $v\approx3.37\ m/s$ and $g = 9.8\ m/s^2$: $h_2=\frac{3.37^2}{2\times9.8}\approx0.58\ m$.

Step4: Analyze the situation for the maximum height the lower right - weight rises back

After the left - weight reaches its maximum height and starts to fall, the system (left - weight and upper right - weight) will have kinetic energy again. When the string becomes taut again, the lower right - weight will start to move up.
We can use conservation of mechanical energy again. The initial kinetic energy of the system (left - weight and upper right - weight) when the string becomes taut is converted into potential energy of the whole system.
The maximum height the lower right - weight rises back is the same as the height it fell from initially, which is $h_1 = 1.0\ m$ (because of conservation of energy in the absence of non - conservative forces).

Answer:

1. The speed at which the lower right weight hits the floor is approximately $3.37\ m/s$.
2. The height that the left weight rises to after the lower right weight falls inelastically to the floor is approximately $0.58\ m$.
3. The maximum height that the lower right weight rises back to is $1.0\ m$.