Question

Just solve A and B please!

Velocity addition A train with proper length L' travels with velocity v_t. It takes the conductor in the train the time t_c^' to move through the train from front to back (and the same time to move back again). Things that are in the way will be kicked out of the train at different angles that_t^', but all at the same speed S_o^1 (in the trains frame). What is the speed of the conductor with respect to an observer 0 at rest outside the train when the conductor is moving from front to back (s_b) or from back to front (s_f) in the train? What are the magnitude of the velocity w and the angles theta_i objects flying out of the train (in 0's frame and with respect to the train's direction).

Answer 1

a) First, we have to find the conductor speed wheen he is moving trough the train, in the trains frame, which mean that we are going to find that speed beeing inside the train with him. This is:

$V_{c}^{'} = \frac{L^{'}}{t_{c}^{'}}$

Now, lets consider that we're off the train. We see the train traveling at a speed equals Vt, and we want to determine the speed of the conductor when he is walking from front to back and from back to front. If in an initial moment when we see the train drive away from us, we see the conductor walking from:

• From back to front: then his speed respect us (observer O) is equals: $V_{cO} = \frac{L^{'}}{t_{c}^{'}} + Vt$

• From front to back: then his speed respect us (observer O) is equals: $V_{cO} = Vt - \frac{L^{'}}{t_{c}^{'}}$

b) Now imagine that the conductor kicked all that things out of the train. This makes objects describe a parabolic path relative to the floor of the train at different angles as we know. If we see as the observer O (out of the train) all those objects flying, we will see than they described another parabolic path because of the train lineal speed (Vt). We can say then, that we have an horizontal speed component and a vertical speed component, which is used to find "w" (speed relative O's frame) with the Pythagoras theorem:

$W = \sqrt{Vt^{2} + S_{o}^{'}^{2}}$

Finally, we can also say that, respect an observer O, the horizontal speed component is Vt (adyacent side) and the vertical speed component is So (opossite side). by using trigonometry:

$\tan \theta _{i} = \frac{S_{o}^{'}}{Vt} \rightarrow \arctan \frac{S_{o}^{'}}{Vt} = \theta _{i}$