Question

Two objects are connected by a light string passing over a light, frictionless pulley as shown in Figure P8.7. The object of mass ${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{}\mathbf{kg}$is released from rest at a height $\mathit{h}\mathbf{=}\mathbf{4}\mathbf{}\mathbf{m}$above the table. Using the isolated system model, (a) determine the speed of the object of mass ${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{3}\mathbf{}\mathbf{kg}$ just as the 5 kg object hits the table and (b) find the maximum height above the table to which the 3 kg object rises.

Short answer

(a) The speed of the object of mass $m_2=3\mathrm{kg}$ just as the 5 kg object hits the table is 4.43 m/s.

(b) The maximum height above the table to which the 3 kg object rises is 5 m.

Step-by-step solution

Given data:

Mass of the first block,$m_1=5\mathrm{kg}$

Mass of the second block,$m_2=3\mathrm{kg}$

Initial height of the first block, h = 4 m

Potential and kinetic energy and equation of motion:

The potential energy of a mass m at a height h from the ground is,

P = mgh .....(I)

Here, g is the acceleration due to the gravity of value$\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$

The kinetic energy of a mass m moving with a velocity v is

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$ .......(II)

From the second equation of motion, the maximum distance traveled upwards by an object with an initial velocity v is

$\mathbf{S}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{2}\mathbf{g}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(\mathrm{III}\right)$

(a) Determining the speeds of the objects:

Initially the two objects are at rest. So, their initial kinetic energies are zero.

The 3 kg object is initially on the table which can be considered the zero state. So, its initial potential energy is zero.

Thus, the initial energy of the system is,

$E=m_{1}gh$

The final potential energy of the 5 kg object is zero.

Let the final velocities of the objects when the 5 kg object hits the ground be v.

From the conservation of energy and equations (I) and (II) you get,

$$\begin{gathered} m_{1}gh=m_{2}gh+\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2 \\ {v}^2=\frac{2\left(m_1-m_2\right)gh}{\left(m_1+m_2\right)} \\ v=\sqrt{\frac{2\left(m_1-m_2\right)gh}{\left(m_1+m_2\right)}} \end{gathered}$$

Substitute the values to get

$$\begin{gathered} v=\sqrt{\frac{2\left(5\mathrm{kg}-3\mathrm{kg}\right)\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 4\mathrm{m}}{\left(5\mathrm{kg}+3\mathrm{kg}\right)}} \\ =\sqrt{19.6{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =4.43\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the required velocity is 4.43 m/s.

(b) Determining the maximum height reached by the 3 kg  object:

From equation (III), the maximum height reached by the 3 kg object after the 5 kg object hits the ground is

$$\begin{gathered} S=\frac{{\left(4.43\mathrm{m}/\mathrm{s}\right)}^2}{2\times 9.8\mathrm{m}/{\mathrm{s}}^2} \\ =1\mathrm{m} \end{gathered}$$

Thus, the maximum height reached above the table is

$$\begin{gathered} h\text{'}=h+S \\ =4\mathrm{m}+1\mathrm{m} \\ =5\mathrm{m} \end{gathered}$$

Hence, the maximum height reached is 5 m.