5.1 Exercises, examples & solutions

First we make a new sketch, now showing the two observers $S$ and $S'$ and their axis. We have made the velocity of the object red, the color of $S$. And we have given the coordinates of the front and back of the train in grey as these are specified according to $S'$. We do this, as it is is crucial to realize that we have ‘mixed’ information.

The velocity of the object is $u$ according to $S$. The observer in the train, $S'$, sees a different velocity.
The observer in the train will denote the position of the front of the train by $x_f' = L$ and of the back $x_b' = -L$. Both are, according to $S'$, fixed values. But $S$ will see that differently.

According to $S'$, the object moves with velocity $u' = u-V$. Note that this is a negative value, otherwise the object will not hit the back of the train.

$S'$ will describe the trajectory of the object by: $x'(t) = x'_0 +u' t$ with $x'_0=0$. Thus, the object will hit the back of the train at:

What does $S$ observe? It will write for the trajectory of the object $x_o(t) = u t$ (where we used that the object was released in the middle of the train at $t=0$ and both observers chose that as their origin).
According to $S$ also the back of the train is moving. It follows a trajectory $x_b = -L + Vt$, since at $t=0$ the back of the train was at position $x = -L$ according to $S$. The two will collide when

Hence we have $T$ and $T'$ as times of collision. But we already found $u' = u-V$. If we substitute this in $T'$ we get

Thus, indeed both observers see the collision at the same moment.

Sneak Preview: much to our surprise, when we enter the world of Special Relativity, this will no longer be the case!

The particle will ‘collide’ with the bump. This might cause the particle to start moving to the left. How to analyse this situation?

Perhaps it is easier when we view this from the point of view of an observer moving with the table.

Now we have a situation of a particle moving over a friction less table with velocity $v$. If we use conservation of energy, we can write down:

where we have taken $h$ as the height above the table and denote the velocity of $m$ at some point by $u$. The initial height is zero and the initial velocity is $v$.

So, if the initial velocity is such that $\frac{1}{2}mv^2 > mgH$, the particle will go over the bump and come back to height $h=0$. It will thus pass the bump and then continue moving with velocity $v$. For the original observer this means: the bump will pass the particle and after passing the particle is again laying still (but not at the same position!).

If, on the other hand $v$ is such that $\frac{1}{2}mv^2 < mgH$, the particle will not reach the top of the bump: it has insufficient kinetic energy. In stead it will stop at some height $h^* = \frac{v^2}{2g}$ and then fall of the bump again. It will continue with velocity $-v$ at the flat part of the table. To the original observer this means that $m$ first climbs the bump and returns to get a velocity $-2v$ on the flat part of the table.

The final possibility is $\frac{1}{2}mv^2 = mgH$. In that case the particle will exactly reach the top of the bump and stop there.

N.B. We have assumed that the bump is not too steep, because in such a case the particle will have a real collision with the bump. Think, for instance, of the bump as a sudden step. Then no matter how fast the particle is moving, it will not end up on the step, but bounce back.

First, a sketch:

It is a 1-dimensional problem, so an $x$-axis will do. We denote the velocity of your opponent (as seen by you) by $v_o$ and of the snowball $v_s$. The inertial system of you is $S$ and you are sitting in the origin $\mathcal{O}$. Similarly, you opponents inertial system is $S'$ with origin $\mathcal{O'}$ and finally the snowball has inertial system $S"$ and the snowball sits in the origin $\mathcal{O"}$.

1. Your perspective
   

   $$x_s(t) = v_s t$$

   (9)
   $$x_o(t) = L + v_o t$$

   (10)

   require: $x_s(t^*) = x_o(t^*)$

   $$\rightarrow t^* = \frac{L}{v_s - v_o} = 0.4 \; \mathrm{s} \rightarrow x^* = v_s t^* = 4 \; \mathrm{m}$$

   (11)

2. Your opponent’s perspective
   

   $$v'_s = v_s - v_0 =5 \; \mathrm{m/s}$$

   (12)

   require: $x'_s(t'^*) = 0$ since $S'$ is in $x'=0$. Thus

Same time of course. Position: your opponent concludes she is not moving and this she is hit at $x'=0$.

3. The snowballs perspective.
   According to the snowball $v"_o = v_o - v_s = -5\; \mathrm{m/s}$. Thus,

   $$x"_o = L + v"_o t$$

   (14)

   require: $x"_o(t"^*) = 0$

And, again the snowball will conclude that it all happened in its origin.

Galilean Transformation
We now have three different time/place coordinates for the event ‘snowball hits opponent’.

We could have found this directly from a GT.
a. from $S$ to $S'$: we need to take into account that at $t=0$ the origins do not coincide. Instead $\mathcal{O'}$ is shifted over a distance L w.r.t. $\mathcal{O}$

Thus: $x'_h = x_h - L - v_o t_h = 0$ and we get indeed $(x'_h,t'_h) = (0\; \mathrm{m}, 0.4\; \mathrm{s})$

b. We do a similar exercise for $S$ to $S"$:

Thus: $x"_h = x_h - v_s t_h = 0$ and we get $(x"_h,t"_h) = (0\; \mathrm{m}, 0.4\; \mathrm{s})$