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Let the retarding force be F.
Suppose A having mass ‘m’ was initially moving with velocity ‘u’. It comes to rest in time ‘t’ after traveling a distance ‘x’.
Retardation of this mass is, a = F/m
And B having mass ‘M’ was initially moving with velocity ‘U’. It comes to rest in time ‘T’ after traveling a distance ‘X’.
Retardation of this mass is, A = F/M
A/Q,
t = 2T
x = X/3
Also,
F = ma = MA
For A,
0 = u – at
=> a = u/t
Similarly,
A = U/T
So,
a/A = uT/(Ut)
=> a/A = uT/(U × 2T)
=> a/A = u/(2U) ……………….(1)
Again for A,
02 = u2 – 2ax
=> a = u2/(2x)
A = U2/(2X)
a/A = [u2/(2x)]/[U2/(2X)]
=> a/A = (u2/U2)(X/x)
=> a/A = (u2/U2)[X/(X/3)]
=> a/A = 3(u2/U2) ………………(2)
(1) and (2) => u/(2U) = 3(u2/U2)
=> ½ = 3 u/U
=> u/U = 1/6
Thus, ratio of initial velocity of A to initial velocity of B is = 1 : 6
Regards
Arun (askIITians forum expert)
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