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 is released from rest at h height above the table. Using the isolated system model,(a) determine the speed of m₂ just as m₁ hits the table and (b) find the maximum height above the table to which m₂ rises.

Short answer

(a) The speed of just as object hits the table is $\sqrt{\frac{2\left(m_1-m_2\right)gh}{\left(m_1+m_2\right)}}$

(b) The maximum height above the table to which rises is $\frac{2m_1}{\left(m_1+m_2\right)}h$

Step-by-step solution

Given data:

Mass of the first block is m₁

Mass of the second block is m₂

Initial height of the first block is h

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.........................................................................1

Here, g is the acceleration due to the gravity of value 9.8m/s²

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

$K=\frac{1}{2}m{v}^2$

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

$S=\frac{v^2}{2g}$

(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=m₁gh

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 conservation of energy and equations (I) and (II)

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

Thus, the required velocity is $\sqrt{\frac{2\left(m_1-m_2\right)gh}{\left(m_1+m_2\right)}}$

(b) Determining the maximum height reached by m2 :

From equation (III), the maximum distance traveled upwards by m₂ object after m₁ hits the ground is

$$\begin{gathered} S=\frac{v^2}{2g} \\ =\frac{\frac{2\left(m_1-m_2\right)gh}{\left(m_1+m_2\right)}}{2g} \\ =\frac{\left(m_1-m_2\right)}{\left(m_1+m_2\right)}h \end{gathered}$$

Thus, the maximum height reached above the table is

$$\begin{gathered} h\text{'}=h+S \\ =h+\frac{\left(m_1-m_2\right)}{\left(m_1+m_2\right)}h \\ =\frac{\left(m_1+m_2+m_1-m_2\right)}{\left(m_1+m_2\right)}h \\ =\frac{2m_1}{\left(m_1+m_2\right)}h \end{gathered}$$

Hence, the maximum height reached is$\frac{2m_1}{\left(m_1+m_2\right)}h$