Question

Two students in bumper cars want to prove conservation of momentum. They collide their bumper cars. The first bumper car has a mass of 120 kg
and was moving with a velocity of 4.0 m/s
before the collision and with a velocity of −2.0 m/s
after the collision. The second bumper car has a mass of 90 kg
and was moving at a velocity of −5.0 m/s
before the collision. To prove conservation of momentum, what must the velocity of the second bumper car be after the collision?(1 point)
Responses

1.0 m/s
1 point 0 meters per seconds

2.3 m/s
2 point 3 meters per seconds

3.0 m/s
3 point 0 meters per seconds

2.7 m/s

Answer 1

To prove the conservation of momentum, we can use the principle that the total momentum before the collision must equal the total momentum after the collision.

The formula for momentum (p) is:

\[ p = m \cdot v \]

Where:
- \( m \) is the mass
- \( v \) is the velocity

Let's denote the variables:
- Mass of the first bumper car, \( m_1 = 120 \, \text{kg} \)
- Initial velocity of the first bumper car, \( v_{1i} = 4.0 \, \text{m/s} \)
- Final velocity of the first bumper car, \( v_{1f} = -2.0 \, \text{m/s} \)

- Mass of the second bumper car, \( m_2 = 90 \, \text{kg} \)
- Initial velocity of the second bumper car, \( v_{2i} = -5.0 \, \text{m/s} \)
- Final velocity of the second bumper car, \( v_{2f} \) (what we want to find)

### Step 1: Calculate the total momentum before the collision
\[
\text{Total momentum before} = p_{1i} + p_{2i}
\]

\[
p_{1i} = m_1 \cdot v_{1i} = 120 \, \text{kg} \cdot 4.0 \, \text{m/s} = 480 \, \text{kg m/s}
\]
\[
p_{2i} = m_2 \cdot v_{2i} = 90 \, \text{kg} \cdot (-5.0 \, \text{m/s}) = -450 \, \text{kg m/s}
\]

Total momentum before:
\[
\text{Total momentum before} = 480 \, \text{kg m/s} - 450 \, \text{kg m/s} = 30 \, \text{kg m/s}
\]

### Step 2: Calculate the total momentum after the collision
\[
\text{Total momentum after} = p_{1f} + p_{2f}
\]
\[
p_{1f} = m_1 \cdot v_{1f} = 120 \, \text{kg} \cdot (-2.0 \, \text{m/s}) = -240 \, \text{kg m/s}
\]
\[
p_{2f} = m_2 \cdot v_{2f} = 90 \, \text{kg} \cdot v_{2f}
\]

### Step 3: Set total momentum before equal to total momentum after
\[
30 \, \text{kg m/s} = -240 \, \text{kg m/s} + 90 \, \text{kg} \cdot v_{2f}
\]

### Step 4: Solve for \( v_{2f} \)
Rearranging gives:
\[
90 \, \text{kg} \cdot v_{2f} = 30 \, \text{kg m/s} + 240 \, \text{kg m/s}
\]
\[
90 \, \text{kg} \cdot v_{2f} = 270 \, \text{kg m/s}
\]
\[
v_{2f} = \frac{270 \, \text{kg m/s}}{90 \, \text{kg}} = 3.0 \, \text{m/s}
\]

Thus, the velocity of the second bumper car after the collision must be **3.0 m/s**.

Final answer: **3.0 m/s** (response 3).