A 2 kg cart moving at 4 m/s collides and sticks to a 3 kg cart at rest. What is the final velocity?

• Total initial momentum: \( 2 \cdot 4 = 8 \, \text{kg m/s} \)
• Total mass: \( 2 + 3 = 5 \, \text{kg} \)
• Final velocity: \( \frac{8}{5} = 1.6 \, \text{m/s} \)

Two carts are on a frictionless track. Cart A has a mass of 2 kg and is moving to the right at 4 m/s. Cart B has a mass of 1 kg and is initially at rest. The two carts collide elastically.

Question: What are the final velocities of both carts after the collision?

Assume the velocity of cart A to be \(v_A\) and the velocity of cart B to be \(v_B\).

Momentum Conservation:

Initial momentum = \(2 \times 4 + 0 = 8 \, \text{kg} \cdot \text{m/s}\)

Final momentum = \(2v_A + v_B = 8\)

⇒ \(v_B = 8 - 2v_A\)

Kinetic Energy Conservation:

Initial energy = \(\frac{1}{2} \times 2 \times 4^2 = 16 \, \text{J}\)

Final energy = \(\frac{1}{2} \times 2 \times v_A^2 + \frac{1}{2} \times 1 \times v_B^2 = v_A^2 + 0.5 v_B^2\)

Substitute \(v_B = 8 - 2v_A\) into the energy equation:

\[ v_A^2 + 0.5(8 - 2v_A)^2 = 16 \]

Expand the square:

\[ v_A^2 + 0.5(64 - 32v_A + 4v_A^2) = 16 \]

Now simplify:

\[ v_A^2 + 32 - 16v_A + 2v_A^2 = 16 \]

\[ 3v_A^2 - 16v_A + 32 = 16 \]

\[ 3v_A^2 - 16v_A + 16 = 0 \]

Solving the Quadratic:

Use the quadratic formula:

\[ v_A = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 3 \cdot 16}}{2 \cdot 3} = \frac{16 \pm \sqrt{256 - 192}}{6} = \frac{16 \pm \sqrt{64}}{6} = \frac{16 \pm 8}{6} \]

So, \(v_A = \frac{24}{6} = 4\) or \(v_A = \frac{8}{6} = \frac{4}{3}\)

If \(v_A = 4\), then \(v_B = 8 - 2 \cdot 4 = 0\) — but this is the initial state, so it’s not the result after collision.

So the physical solution is:

• \(v_A = \frac{4}{3} \, \text{m/s}\)
• \(v_B = 8 - 2 \cdot \frac{4}{3} = \frac{16}{3} \, \text{m/s}\)

Answer: After the elastic collision:

• Cart A moves at \( \frac{4}{3} \, \text{m/s} \)
• Cart B moves at \( \frac{16}{3} \, \text{m/s} \)