A 15.0 kg package is moving at a speed of 10.0 m/s vertically upward along a y axis when it explodes into three fragments: a 2.00 kg fragment is shot upward with an initial speed of 20.0 m/s and a 3.00 kg fragment is shot in the positive direction of a horizontal x axis with an initial speed of 5.00 m/s. Find (a) the speed of the third fragment right after the explosion and (b) the total kinetic energy provided by the explosion.

Answer

To solve the problem, we will break it down into two parts: (a) finding the speed of the third fragment after the explosion and (b) calculating the total kinetic energy provided by the explosion.

### Part (a): Finding the Speed of the Third Fragment

1. **Initial Momentum Calculation**:
- The initial momentum of the system (the package) before the explosion can be calculated using the formula:
\[ p_{\text{initial}} = m_{\text{initial}} \cdot v_{\text{initial}} = 15 \, \text{kg} \cdot 10 \, \text{m/s} = 150 \, \text{kg m/s} \, \hat{j} \]
- This momentum is directed vertically upward along the y-axis.

2. **Final Momentum Calculation**:
- After the explosion, the package breaks into three fragments:
- Fragment 1: \( m_1 = 2 \, \text{kg} \), \( v_1 = 20 \, \text{m/s} \) (upward)
- Fragment 2: \( m_2 = 3 \, \text{kg} \), \( v_2 = 5 \, \text{m/s} \) (horizontal, x-direction)
- Fragment 3: \( m_3 = 10 \, \text{kg} \) (unknown speed \( v_3 \))

3. **Conservation of Momentum**:
- According to the law of conservation of momentum, the total momentum before the explosion equals the total momentum after the explosion:
\[ p_{\text{initial}} = p_{\text{final}} \]
- The final momentum can be expressed as:
\[ p_{\text{final}} = m_1 v_1 \hat{j} + m_2 v_2 \hat{i} + m_3 v_3 \]
- Substituting the values:
\[ 150 \hat{j} = (2 \cdot 20) \hat{j} + (3 \cdot 5) \hat{i} + (10 \cdot v_3) \]
- This simplifies to:
\[ 150 \hat{j} = 40 \hat{j} + 15 \hat{i} + 10 v_3 \]

4. **Separating Components**:
- For the y-component:
\[ 150 = 40 + 10 v_{3y} \implies 10 v_{3y} = 110 \implies v_{3y} = 11 \, \text{m/s} \]
- For the x-component:
\[ 0 = 15 + 10 v_{3x} \implies 10 v_{3x} = -15 \implies v_{3x} = -1.5 \, \text{m/s} \]

5. **Magnitude of the Velocity of the Third Fragment**:
- The speed of the third fragment \( v_3 \) can be calculated using the Pythagorean theorem:
\[ v_3 = \sqrt{v_{3x}^2 + v_{3y}^2} = \sqrt{(-1.5)^2 + (11)^2} = \sqrt{2.25 + 121} = \sqrt{123.25} \approx 11.1 \, \text{m/s} \]

### Part (b): Total Kinetic Energy Provided by the Explosion

1. **Initial Kinetic Energy**:
- The initial kinetic energy \( KE_{\text{initial}} \) of the package is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m_{\text{initial}} v_{\text{initial}}^2 = \frac{1}{2} \cdot 15 \cdot (10)^2 = \frac{1}{2} \cdot 15 \cdot 100 = 750 \, \text{J} \]

2. **Final Kinetic Energy**:
- The final kinetic energy \( KE_{\text{final}} \) of the fragments is:
\[ KE_{\text{final}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + \frac{1}{2} m_3 v_3^2 \]
- Substituting the values:
\[ KE_{\text{final}} = \frac{1}{2} \cdot 2 \cdot (20)^2 + \frac{1}{2} \cdot 3 \cdot (5)^2 + \frac{1}{2} \cdot 10 \cdot (11.1)^2 \]
- Calculating each term:
- For fragment 1: \( \frac{1}{2} \cdot 2 \cdot 400 = 400 \, \text{J} \)
- For fragment 2: \( \frac{1}{2} \cdot 3 \cdot 25 = 37.5 \, \text{J} \)
- For fragment 3: \( \frac{1}{2} \cdot 10 \cdot 123.21 \approx 616.05 \, \text{J} \)
- Thus:
\[ KE_{\text{final}} = 400 + 37.5 + 616.05 \approx 1053.55 \, \text{J} \]

3. **Total Kinetic Energy Provided by the Explosion**:
- The change in kinetic energy \( \Delta KE \) is:
\[ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 1053.55 - 750 = 303.55 \, \text{J} \]

### Final Answers:
- (a) The speed of the third fragment is approximately **11.1 m/s**.
- (b) The total kinetic energy provided by the explosion is approximately **303.55 J**.