Question

two identical train cars are on the same track. the first train car is moving with a speed of 10 m/s toward the stationary second train car. if they get hooked together after contact, how fast will the newly connected trains be moving? view available hint(s) the connected trains will be moving 5 m/s. the speed of the connected trains will stay constant. the connected trains will be moving faster. the connected trains will be moving slower.

Explanation:

To solve this, we use the principle of conservation of momentum. Let the mass of each train car be \( m \). The initial momentum of the system is the momentum of the first train (since the second is stationary). Initial momentum \( p_i = m \times 10 + m \times 0 = 10m \). After collision, the mass is \( m + m = 2m \) and let the final velocity be \( v \). So final momentum \( p_f = 2m \times v \). By conservation of momentum, \( p_i = p_f \), so \( 10m = 2m \times v \). Solving for \( v \), we divide both sides by \( 2m \) ( \( m
eq 0 \) ), giving \( v = \frac{10m}{2m} = 5 \, \text{m/s} \). Also, since mass increases after collision (two cars instead of one) while momentum is conserved, the velocity should decrease (move slower than 10 m/s, here 5 m/s which is slower than 10 m/s).

Step1: Define variables and initial momentum

Let mass of each train be \( m \). Initial momentum: \( p_i = m \times 10 + m \times 0 = 10m \)

Step2: Define final momentum

Final mass: \( 2m \), final velocity \( v \). So \( p_f = 2m \times v \)

Step3: Apply conservation of momentum

\( 10m = 2m \times v \)

Step4: Solve for \( v \)

Divide both sides by \( 2m \): \( v = \frac{10m}{2m} = 5 \, \text{m/s} \). Also, since mass increased, velocity is less than initial (10 m/s), so moving slower.

Answer:

The correct options are:
A. The connected trains will be moving 5 m/s.
D. The connected trains will be moving slower.

(Note: Assuming the options are labeled as A, B, C, D in order. If the last option is D, then A and D are correct. The first option gives the correct speed, and the last option correctly states they move slower than the initial 10 m/s.)