Question

two carts collide inelastically and stick together. cart 1 has a mass of 6 kg and an initial velocity of 5 m/s. cart 2 has a mass of 4 kg and is initially at rest. what is their final velocity after collision? a 2.0 m/s b 1.5 m/s c 2.4 m/s d 3.0 m/s e 1.67 m/s

Explanation:

Step1: Recall Conservation of Momentum

For inelastic collision, momentum is conserved: \( m_1v_1 + m_2v_2=(m_1 + m_2)v_f \)
Here, \( m_1 = 6\,\text{kg} \), \( v_1 = 5\,\text{m/s} \), \( m_2 = 4\,\text{kg} \), \( v_2 = 0\,\text{m/s} \) (at rest).

Step2: Substitute Values into Formula

Substitute into \( m_1v_1 + m_2v_2=(m_1 + m_2)v_f \):
\( 6\times5 + 4\times0=(6 + 4)v_f \)
\( 30 + 0 = 10v_f \)

Step3: Solve for \( v_f \)

Divide both sides by 10: \( v_f=\frac{30}{10}=3.0\,\text{m/s} \)? Wait, no—wait, recalculate: Wait, \( 6\times5 = 30 \), \( m_1 + m_2 = 10 \), so \( v_f=\frac{30}{10}=3 \)? But wait, no, wait the options: Wait, no, wait I made a mistake? Wait, no, cart 1 is 6kg, 5m/s; cart 2 is 4kg, 0. So total momentum initial is \( 6*5 + 4*0 = 30 \). Total mass is \( 6 + 4 = 10 \). So final velocity \( v_f = 30 / 10 = 3.0 \, \text{m/s} \). So option D.

Wait, but let me check again. Yes, conservation of momentum: initial momentum = final momentum. Inelastic collision, so they stick together, so final mass is sum of masses. So \( m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \). Plugging in: \( 65 + 40 = (6 + 4)v_f \) → \( 30 = 10v_f \) → \( v_f = 3.0 \, \text{m/s} \). So the correct answer is D.

Answer:

D. 3.0 m/s