Check point 8 In Figure 1.48 , an 8001 g car travels at $$ 20 \mathrm{~ms} s^{\prime} $$ east. It colldes with a 3000 kg truck travelling at $$ 12 \mathrm{~ms} s^{-1} $$ wot. The collision lasts $$ 0.5 s $$. After the collision, the truck moves at $$ 4 \mathrm{~ms} s^{1} $$ west

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To analyze the collision between the car and the truck, we can use the principles of conservation of momentum. Here are the known conditions extracted from the problem:

1. Mass of the car, $$ m_c = 8001 \, \text{g} = 8.001 \, \text{kg} $$
2. Velocity of the car before the collision, $$ v_{c_i} = 20 \, \text{m/s} $$ (east)
3. Mass of the truck, $$ m_t = 3000 \, \text{kg} $$
4. Velocity of the truck before the collision, $$ v_{t_i} = -12 \, \text{m/s} $$ (west, negative direction)
5. Duration of the collision, $$ t = 0.5 \, \text{s} $$
6. Velocity of the truck after the collision, $$ v_{t_f} = -4 \, \text{m/s} $$ (west, negative direction)

### Step 1: Calculate the initial momentum of the system

The total initial momentum $$ p_i $$ of the system (car + truck) can be calculated using the formula:

$$
p_i = m_c \cdot v_{c_i} + m_t \cdot v_{t_i}
$$

### Step 2: Calculate the final momentum of the system

The total final momentum $$ p_f $$ of the system after the collision can be calculated as:

$$
p_f = m_c \cdot v_{c_f} + m_t \cdot v_{t_f}
$$

Where $$ v_{c_f} $$ is the final velocity of the car after the collision, which we need to find.

### Step 3: Apply the conservation of momentum

According to the conservation of momentum:

$$
p_i = p_f
$$

### Step 4: Solve for the final velocity of the car

We can rearrange the equation to solve for $$ v_{c_f} $$:

$$
m_c \cdot v_{c_i} + m_t \cdot v_{t_i} = m_c \cdot v_{c_f} + m_t \cdot v_{t_f}
$$

Now, let's calculate the initial momentum and set up the equation to find $$ v_{c_f} $$.

Let's perform the calculations.
Calculate the value by following steps:
- step0: Calculate:
  $$8001\times 20+3000\left(-12\right)$$
- step1: Multiply the numbers:
  $$160020+3000\left(-12\right)$$
- step2: Multiply the numbers:
  $$160020-36000$$
- step3: Subtract the numbers:
  $$124020$$
Solve the equation $$ 8001 * 20 + 3000 * (-12) = 8001 * v_c_f + 3000 * (-4) $$.
Solve the equation by following steps:
- step0: Solve for $$v_{c}$$:
  $$8001\times 20+3000\left(-12\right)=8001v_{c}+3000\left(-4\right)$$
- step1: Simplify:
  $$124020=8001v_{c}-12000$$
- step2: Swap the sides:
  $$8001v_{c}-12000=124020$$
- step3: Move the constant to the right side:
  $$8001v_{c}=124020+12000$$
- step4: Add the numbers:
  $$8001v_{c}=136020$$
- step5: Divide both sides:
  $$\frac{8001v_{c}}{8001}=\frac{136020}{8001}$$
- step6: Divide the numbers:
  $$v_{c}=\frac{45340}{2667}$$
Calculate or simplify the expression $$ 45340/2667 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{45340}{2667}$$
### Step 1: Calculate the initial momentum

The initial momentum $$ p_i $$ of the system is calculated as follows:

$$
p_i = 8001 \cdot 20 + 3000 \cdot (-12) = 124020 \, \text{kg m/s}
$$

### Step 2: Set up the equation for final momentum

Using the conservation of momentum, we set up the equation:

$$
124020 = 8001 \cdot v_{c_f} + 3000 \cdot (-4)
$$

### Step 3: Solve for the final velocity of the car

Rearranging the equation gives:

$$
8001 \cdot v_{c_f} = 124020 + 12000
$$

$$
8001 \cdot v_{c_f} = 136020
$$

Now, solving for $$ v_{c_f} $$:

$$
v_{c_f} = \frac{136020}{8001}
$$

Calculating this gives:

$$
v_{c_f} \approx 17.0 \, \text{m/s}
$$

### Conclusion

The final velocity of the car after the collision is approximately $$ 17.0 \, \text{m/s} $$ east.
After the collision, the car moves at approximately 17.0 m/s east.

Check point 8 In Figure 1.48 , an 8001 g car travels at \( 20 \mathrm{~ms} s^{\prime} \) east. It colldes with a 3000 kg truck travelling at \( 12 \mathrm{~ms} s^{-1} \) wot. The collision lasts \( 0.5 s \). After the collision, the truck moves at \( 4 \mathrm{~ms} s^{1} \) west

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