**Problem 6:**  A baseball of mass m1m1 is thrown at another ball hanging from the ceiling by a length of string LL. The second ball of mass m2m2 is initially at rest while the baseball has an initial horizontal velocity of V1V1. After the collision the first baseball falls straight down (no horizontal velocity).


**Part (a)** Write an expression for the magnitude of the closest distance from the ceiling the second ball will reach dd.
**Expression**   :
**dd** = __________________________________________

Select from the variables below to write your expression. Note that all variables may not be required.
**g**, **L**, **m1**, **m2**, **V1**

**Part (b)** Let m1=0.27 kgm1=0.27 kg, L=1.45 mL=1.45 m, m2=0.93 kgm2=0.93 kg, and V1=4.5 m/sV1=4.5 m/s. What is the angle that the string makes with the vertical at the highest point of travel in degrees?
**Numeric**   : A numeric value is expected and not an expression.
**θθ** = __________________________________________∘∘


### 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:

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

---

### 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 \( 20.7^\circ \).

---

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.

A baseball of mass m1 is thrown at another ball hanging from the ceiling by a length of string L. The second ball of mass m2 is initially at rest while the baseball has an initial horizontal velocity of V1. After the collision, the first baseball falls straight down (no horizontal velocity). 
Part (a) Write an expression for the magnitude of the closest distance from the ceiling the second ball will reach d.
Part (b) Let m1=0.27 kg, L=1.45 m, m2=0.93 kg, and V1=4.5 m/s. What is the angle that the string makes with the vertical at the highest point of travel in degrees?

Explore the solution to a physics problem involving momentum and energy conservation where a baseball is thrown at a hanging ball. Learn how to calculate the closest distance the second ball reaches to the ceiling and the angle made with the vertical at its highest point. Suitable for Grades 11-12.

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