Since the collision is perfectly elastic, the ball's velocity goes from -80
km/h from the train's reference point (negative being towards the train) to
+80 km/h from the train's reference point, a speed increase of 160 km/h.
For a stationery observer, therefore, the velocity goes from -30 km/h
(towards the train) to 130 km/h, an increase of 160 km/h.
The situation you describe only applies if the train isn't moving.
I made an error in units below, assuming the train is moving at 50 km per second (not hour) and that the ball is moving at 30 km per second (not hour), am I'm too lazy to correct it. The general principle, however, still applies.
The confusion may occur because we're ignoring the train's loss of momentum,
which means the train is going slower after the collision, and moving
backwards in its own frame of reference. A slightly more detailed
calculation:
Suppose the the train has mass
M
kg and the ball has massm
kg.From the "fixed" observer's point of view, the initial momentum (in
Newtons) is:
$rho =50 M-30 m$
and the initial kinetic energy (in Joules) is:
$e=450 m+1250 M$
- Let $v_t$ and $v_b$ be the train's and ball's velocities in
meters/second after the collision. Since perfectly elastic
collisions preserve both momentum and kinetic energy, we have:
$m v_b+M v_t=50 M-30 m$
$frac{m v_b^2}{2}+frac{M v_t^2}{2}=450 m+1250 M$
There are only two solutions to the equation above, one of which is
the initial conditions. The other is:
$
left{{v_b}to -frac{10 (3 m-13 M)}{m+M},{v_t}to -frac{10 (11 m-5
M)}{m+M}right}
$
Plugging in 0.0585 kg for the mass of a tennis ball and 640000 kg for
the train, this becomes:
${{v_b}to 129.9999854,{v_t}to 49.99998538}$
effectively confirming the calculation.
I'm not convinced this is a good analogy, however. Gravitational boost
occurs when a planet's gravity almost captures a spacecraft, thus
nearly making it a satellite, and giving it the same revolution
velocity around the Sun as the planet itself. The wikipedia analogy
bears only a passing resemblance to this.
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