JC kinematics

Bigcable22 SG

The value of G can be measured very accurately by timing the vertical motion of ball as it rises past a pair of light gates separated by a known vertical distance-h.

If a ball spent a total time T1 above the lower light gate and a total time T2 above the upper light gate,show that g=8h / [ (T1)^2 - (T2)^2 ].

Audioboxing SG

Because of the symmetry of motion in the problem, the time taken to go from 1 to 2 is,

T1/2 - T2/2

hence, where u is the velocity at 1

h = u(T1/2 - T2/2) - 1/2 g (T1/2 - T2/2)^2

at the maximum height v=0 => u=g(T1/2)

Thus you will get,

g = 8h / [ (T1)^2 - (T2)^2 ]

ALTERNATIVELY,

you can use the work-energy principle, and since the mass is constant the equation is reduced to

1/2 [ g (T2/2) ]^2 - 1/2 [ g (T1/2) ]^2 = -gh

then solve for g in terms of h and T1 and T2 you get,

g = 8h / [ (T1)^2 - (T2)^2 ]

which is exactly the same thing!

DIAGRAM

------- max height , v=0 => u=g(T1/2)

------- sensor 2 , v= g (T2/2)

-------- sensor 1 , v=u

-------- origin / "ground" v=unknown, but no need to care

eagle SG

well done audioboxing! Only thing is the way of presentation needs to be a bit clearer... I got confused a little initially by your solution until I went to do it myself...

The confusing line is "at the maximum height, u=g(T1/2)"

u is a constant in its usage above, so there's no need to state "at max height". Instead, we should phrase it like

v = u - g(T1/2)

At max height, v = 0
Therefore, u=g(T1/2)

I think it would be clearer this way, to the readers, to markers, and most importantly to the students seeking an understandable solution.

Thanks! :D

Audioboxing SG

Ah...yes I made a mistake in the phrasing over there...initially wanted to type at max height v is 0 hence u= ...(mind move faster than fingers on the keyboard, lol)

Sorry if I confused anyone! :D