Question

Cart 1 is 4.0 kg and traveling 6.0 m/s to the right, colliding with Cart, which has a mass of 2.5 kg and is traveling 3.0 m/s to the left. What is the total momentum of the system? (Do not include units in the answer.)
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7. In the problem above, if Cart 1 continues traveling to the right at 4.5 m/s, what is the velocity of cart 2 after the elastic collision? (Do not include units in the answer.)

Answer 1

To find the total momentum of the system, we can use the formula for momentum: \( p = m \cdot v \).

For Cart 1:
- Mass \( m_1 = 4.0 \, \text{kg} \)
- Velocity \( v_1 = 6.0 \, \text{m/s} \) (to the right, we consider it positive)

For Cart 2:
- Mass \( m_2 = 2.5 \, \text{kg} \)
- Velocity \( v_2 = -3.0 \, \text{m/s} \) (to the left, we consider it negative)

Now we calculate the total momentum:

\[
\text{Total Momentum} = m_1 \cdot v_1 + m_2 \cdot v_2
\]

Substituting in the values:

\[
\text{Total Momentum} = (4.0 \, \text{kg} \cdot 6.0 \, \text{m/s}) + (2.5 \, \text{kg} \cdot -3.0 \, \text{m/s})
\]
\[
= 24.0 \, \text{kg m/s} - 7.5 \, \text{kg m/s}
\]
\[
= 16.5 \, \text{kg m/s}
\]

The total momentum of the system is **16.5**.

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Now, to find the velocity of Cart 2 after the elastic collision, we will use the principles of conservation of momentum and conservation of kinetic energy.

Let \( v_{2f} \) be the final velocity of Cart 2 after the collision.

Using the conservation of momentum:
\[
m_1 v_1 + m_2 v_2 = m_1 v_{1f} + m_2 v_{2f}
\]
Where:
- \( v_{1f} = 4.5 \, \text{m/s} \) (final velocity of Cart 1)

Plugging in the initial values into the momentum equation:

\[
(4.0)(6.0) + (2.5)(-3.0) = (4.0)(4.5) + (2.5)v_{2f}
\]
\[
24.0 - 7.5 = 18.0 + 2.5v_{2f}
\]
\[
16.5 = 18.0 + 2.5v_{2f}
\]
Subtracting 18.0 from both sides:
\[
16.5 - 18.0 = 2.5v_{2f}
\]
\[
-1.5 = 2.5v_{2f}
\]
Dividing by 2.5:
\[
v_{2f} = \frac{-1.5}{2.5} = -0.6 \, \text{m/s}
\]

The final velocity of Cart 2 after the elastic collision is **-0.6**.