The equipment in our laboratory consists of an infinitely long inclined plane at 45 degrees to the vertical and a ball, which bounces with no loss in energy.
The ball is dropped from some height above the plane, bounces on its way down the plane (Boing! Boing!).
The laboratory investigator records:
T(N) = the time between the Nth and (N+1)th bounce, and
H(N) = the maximum distance between the ball and the plane between the Nth and (N+1)th bounce.
Write an expression for
H(N+1)T(N+1)/( H(N)T(N) ) as a function of N.
The ratio is always unity, regardless of N. Further, H and T will each be constant on each bounce as well. (This is also regardless of the angle of the incline). The easy way to see this, is that upon the first bounce, the ball, in the frame of the plane, in one orthogonal direction is bouncing in the "normal" direction up and down in a reduced gravitational acceleration field: a_N=g sin (theta). In a frictionless world it keeps bouncing forever with the same period and reaching the same (normal) height. At the same time, in the other orthogonal direction, the ball experiences a constant transverse acceleration w.r.t. the plane of a_T = g cos(theta), moving faster and landing further, and soon going relativistic. It's like a bouncing ball experiencing a constant sideways acceleration.
Back in the Newtonian world, the dropped ball hits with vertical velocity v_0 and bounces with components v_N = v_0 sin(theta) and v_T = v_0 cos(theta).
Recall that the displacement of a ball thrown straight upward is
H = vt - 1/2 gt^2 Setting the normal displacement to 0,
0=v_N T - 1/2 a_N T^2
gives the period:
T=2 v_0/g (independent of theta, the inclination angle)
The maximum distance from the plane occurs at half that time, and putting that into the displacement formula gives
H=[v_0^2/(2g)]sin(theta)
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(BTW, If your lab can afford an infinite plane, then, does it get infinite funding? And, going downward, where does that plane lead?)
Edited on May 18, 2019, 12:05 am
attempt at better edition of soln text
