9. A 1050 kg car has a velocity of $$ 2,65 \mathrm{~m} \cdot \mathrm{~s}^{-1} $$ north. The car hits the rear of a stationary truck, and the bumpers lock together. The velocity of the car-truck system immediately after the collision is $$ 0,78 \mathrm{~m} \cdot \mathrm{~s}^{-1} $$ north. Calculate the mass of the truck? 10. A $$ 0,25 \mathrm{~kg} $$ volleyball is thrown horizontally at $$ 2 \mathrm{~m} \cdot \mathrm{~s}^{-1} $$ west. It strikes a $$ 0,62 \mathrm{~kg} $$ stationary basketball. The volleyball rebounds east at $$ 0,79 \mathrm{~m} . \mathrm{s}^{-1} $$. Calculate the velocity of the basketball immediately after the collision?

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UpStudy AI · Accepted Answer

Let's solve both problems step-by-step using the principles of **conservation of momentum**.

---

### **Problem 9**

**Given:**
- Mass of the car, $$ m_{\text{car}} = 1050 \, \text{kg} $$
- Initial velocity of the car, $$ v_{\text{car}} = 2.65 \, \text{m/s north} $$
- Initial velocity of the truck, $$ v_{\text{truck initial}} = 0 \, \text{m/s} $$ (stationary)
- Final velocity of the combined car-truck system, $$ v_{\text{final}} = 0.78 \, \text{m/s north} $$
- Mass of the truck, $$ m_{\text{truck}} = ? $$

**Approach:**
Apply the **conservation of momentum**:
$$
\text{Total initial momentum} = \text{Total final momentum}
$$
$$
m_{\text{car}} \cdot v_{\text{car}} + m_{\text{truck}} \cdot v_{\text{truck initial}} = (m_{\text{car}} + m_{\text{truck}}) \cdot v_{\text{final}}
$$
Since the truck is initially stationary:
$$
m_{\text{car}} \cdot v_{\text{car}} = (m_{\text{car}} + m_{\text{truck}}) \cdot v_{\text{final}}
$$
Solving for $$ m_{\text{truck}} $$:
$$
m_{\text{truck}} = \frac{m_{\text{car}} \cdot v_{\text{car}}}{v_{\text{final}}} - m_{\text{car}}
$$
$$
m_{\text{truck}} = 1050 \cdot \left( \frac{2.65}{0.78} - 1 \right)
$$
$$
m_{\text{truck}} = 1050 \cdot (3.403 - 1) = 1050 \cdot 2.403 = 2523.15 \, \text{kg}
$$

**Answer:**
The mass of the truck is approximately **2523  kg**.

---

### **Problem 10**

**Given:**
- Mass of the volleyball, $$ m_{\text{volleyball}} = 0.25 \, \text{kg} $$
- Initial velocity of the volleyball, $$ v_{\text{vol initial}} = 2 \, \text{m/s west} $$
- Final velocity of the volleyball, $$ v_{\text{vol final}} = 0.79 \, \text{m/s east} $$
- Mass of the basketball, $$ m_{\text{basketball}} = 0.62 \, \text{kg} $$
- Initial velocity of the basketball, $$ v_{\text{basket initial}} = 0 \, \text{m/s} $$
- Final velocity of the basketball, $$ v_{\text{basket final}} = ? $$

**Approach:**
Choose **east** as the positive direction and **west** as negative.

Apply the **conservation of momentum**:
$$
m_{\text{volleyball}} \cdot v_{\text{vol initial}} + m_{\text{basketball}} \cdot v_{\text{basket initial}} = m_{\text{volleyball}} \cdot v_{\text{vol final}} + m_{\text{basketball}} \cdot v_{\text{basket final}}
$$
Substitute the known values:
$$
0.25 \cdot (-2) + 0.62 \cdot 0 = 0.25 \cdot 0.79 + 0.62 \cdot v_{\text{basket final}}
$$
$$
-0.5 = 0.1975 + 0.62 \cdot v_{\text{basket final}}
$$
Solve for $$ v_{\text{basket final}} $$:
$$
-0.5 - 0.1975 = 0.62 \cdot v_{\text{basket final}}
$$
$$
-0.6975 = 0.62 \cdot v_{\text{basket final}}
$$
$$
v_{\text{basket final}} = \frac{-0.6975}{0.62} \approx -1.124 \, \text{m/s}
$$
The negative sign indicates the direction is **west**.

**Answer:**
Immediately after the collision, the basketball moves west at approximately **1.12 m/s**.
**Problem 9:** The mass of the truck is approximately **2523 kg**. **Problem 10:** Immediately after the collision, the basketball moves west at approximately **1.12 m/s**.

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