NLM O28

1. Analysis of the System at Terminal Velocity

Let the terminal velocity be $v_t$. Since the balls have equal volumes, they displace the same amount of air. Therefore, the buoyant force $B$ acting on each ball is identical. Similarly, since the drag force depends on the velocity and the size/shape (which are identical), the viscous drag force $f_v$ acting on each ball is the same.

Considering the system of two balls together as a single unit, the internal tension $T$ cancels out. At terminal velocity, the net acceleration is zero. The equilibrium equation for the whole system is:

$$ \text{Total Upward Forces} = \text{Total Downward Forces} $$ $$ 2f_v + 2B = (m_1 + m_2)g $$

From this, we can solve for the combined resistive force per ball ($F_{resistive} = f_v + B$):

$$ f_v + B = \frac{(m_1 + m_2)g}{2} $$

2. Analysis of the Lighter Ball ($m_1$)

Now, let’s isolate the lighter ball $m_1$ (which is being pulled down by the heavier mass $m_2$). The forces acting on $m_1$ are:

• Gravitational force ($m_1g$) acting downwards.
• Tension ($T$) acting downwards.
• Resistive forces ($f_v + B$) acting upwards.

Since the acceleration is zero, we can write the force balance equation for $m_1$:

$$ f_v + B = m_1g + T $$

3. Calculating Tension

Substitute the value of $(f_v + B)$ derived in Step 1 into the equation from Step 2:

$$ \frac{(m_1 + m_2)g}{2} = m_1g + T $$

Rearranging for Tension $T$:

$$ T = \frac{(m_1 + m_2)g}{2} – m_1g $$ $$ T = \frac{m_1g + m_2g – 2m_1g}{2} $$ $$ T = \frac{m_2g – m_1g}{2} $$