Part (a)

To solve for the maximum height $d$ that the second ball reaches, we need to consider the principle of conservation of momentum and energy.

1. Momentum Conservation:
   The collision between the two balls is assumed to be perfectly inelastic, meaning that after the collision, the first ball falls straight down, and the second ball moves horizontally. By the conservation of linear momentum:

   $m_1 V_1 = m_2 V_2$

   Solving for $V_2$, the velocity of the second ball immediately after the collision:

   $V_2 = \frac{m_1 V_1}{m_2}$

2. Energy Conservation:
   After the collision, the second ball will convert its kinetic energy into potential energy as it swings upward. The kinetic energy after the collision is:

   $\frac{1}{2} m_2 V_2^2$

   At the highest point of the swing, all of this kinetic energy is converted into gravitational potential energy:

   $m_2 g d$

   Equating the kinetic energy and potential energy:

   $\frac{1}{2} m_2 V_2^2 = m_2 g d$

   Canceling out $m_2$ from both sides:

   $\frac{1}{2} V_2^2 = g d$

   Substituting $V_2 = \frac{m_1 V_1}{m_2}$ into the equation:

   $\frac{1}{2} \left( \frac{m_1 V_1}{m_2} \right)^2 = g d$

   Solving for $d$:

   $d = \frac{ \left( m_1 V_1 \right)^2 }{2 m_2^2 g}$

Thus, the expression for the closest distance from the ceiling that the second ball will reach is:

$d = \frac{ \left( m_1 V_1 \right)^2 }{2 m_2^2 g}$

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Part (b)

Now, let's find the angle that the string makes with the vertical at the highest point of the swing.

1. Height $d$:
   From part (a), the vertical distance $d$ the ball rises is:

   $d = \frac{ \left( m_1 V_1 \right)^2 }{2 m_2^2 g}$

   Using the given values:
   $m_1 = 0.27 \, \text{kg}$,
   $m_2 = 0.93 \, \text{kg}$,
   $V_1 = 4.5 \, \text{m/s}$,
   $g = 9.8 \, \text{m/s}^2$,
   we can calculate $d$:

   $d = \frac{(0.27 \times 4.5)^2}{2 \times (0.93)^2 \times 9.8}$

   Let's compute that.

$d = \frac{(1.215)^2}{2 \times 0.8649 \times 9.8} = \frac{1.476225}{16.92402} \approx 0.0872 \, \text{m}$

2. Angle $\theta$:
   The maximum height reached corresponds to a vertical distance of $d$. The string's length $L = 1.45 \, \text{m}$ forms a right triangle with $L$ as the hypotenuse and $L - d$ as the adjacent side to the angle $\theta$.

   Using trigonometry:

   $\cos \theta = \frac{L - d}{L}$

   Substituting the values of $L = 1.45 \, \text{m}$ and $d = 0.0872 \, \text{m}$:

   $\cos \theta = \frac{1.45 - 0.0872}{1.45} = \frac{1.3628}{1.45} \approx 0.939$

   Taking the inverse cosine:

   $\theta = \cos^{-1}(0.939) \approx 20.7^\circ$

Thus, the angle $\theta$ is approximately $20.7^\circ$.

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Would you like further clarification on any of these steps?

Related Questions:

1. How does the length of the string $L$ affect the maximum angle $\theta$?
2. What assumptions are made about the collision between the two balls?
3. How would the result change if the collision was perfectly elastic?
4. What is the impact of air resistance on the final height $d$?
5. How would changing the mass of the second ball $m_2$ affect the outcome?

Tip:

When solving momentum and energy problems, always identify the types of collisions and energy transformations involved.