Question

1. a 1000 kg railroad car, a, traveling at a speed of 24.0 m/s strikes an identical car, b, at rest. if the cars lock together as a result of the collision, their common speed after collision is (a) 12.5 m/s (b) 15 m/s (c) 6 m/s (d) 13 m/s (e) none of the above

Explanation:

Step1: Apply conservation of momentum

The formula for conservation of momentum is $m_1v_1 + m_2v_2=(m_1 + m_2)v_{final}$. Here, $m_1 = 1000$ kg, $v_1=24.0$ m/s, $m_2 = 1000$ kg and $v_2 = 0$ m/s.

Step2: Substitute values into formula

Substituting the values we get $(1000\times24.0)+(1000\times0)=(1000 + 1000)v_{final}$. Which simplifies to $24000=2000v_{final}$.

Step3: Solve for $v_{final}$

Dividing both sides by 2000, we have $v_{final}=\frac{24000}{2000}=12$ m/s. Since 12 m/s is not in the given options.

Answer:

(e) none of the above