(1) pLAY O sTop 15 A 9300 kg railroad car traveling at a velocity of \( 5 \mathrm{~m} / \mathrm{s} \) strikes a second boxcar that weighs 5000 kg at rest. At what speed will the second boxcar move? A \( 18.6 \mathrm{~m} / \mathrm{s} \) B \( 1.86 \mathrm{~m} / \mathrm{s} \) C \( 7.5 \mathrm{~m} / \mathrm{s} \) D \( 9.3 \mathrm{~m} / \mathrm{s} \) III >

To solve this problem, we can use the principle of conservation of momentum. The total momentum of the system before the collision must equal the total momentum after the collision. Before the collision, the momentum of the first boxcar (A) is: \( p_A = m_A \cdot v_A = 9300 \, \text{kg} \cdot 5 \, \text{m/s} = 46500 \, \text{kg m/s} \) The second boxcar (B) is at rest, so its momentum is: \( p_B = 5000 \, \text{kg} \cdot 0 \, \text{m/s} = 0 \, \text{kg m/s} \) Total initial momentum: \( p_{initial} = p_A + p_B = 46500 + 0 = 46500 \, \text{kg m/s} \) Let \( v_B \) be the velocity of boxcar B after the collision. Total momentum after the collision is: \( p_{final} = m_A \cdot v_A + m_B \cdot v_B \) Setting the initial momentum equal to the final momentum: \( 46500 = 9300 \cdot 0 + 5000 \cdot v_B \) Solving for \( v_B \): \( 46500 = 5000 \cdot v_B \) \( v_B = \frac{46500}{5000} = 9.3 \, \text{m/s} \) So the answer is: **D \( 9.3 \mathrm{~m/s} \)**. Railroad cars, which provide an example here, have quite the history! They revolutionized transportation in the 19th century, reducing travel times and boosting economies. This made it easier for goods and people to move, fueling industrial growth and changing the way cities developed. If you’re ever in a physics class, be wary of two common pitfalls: forgetting to account for the direction of velocities and mistakenly using kinetic energy instead of momentum for inelastic collisions! Always remember that momentum is conserved, even if kinetic energy isn’t.