Two unequal masses Pand Q moving along a straight lines are brought to rest by applying equal retarding forces. if P moves twice the time q but goes only {1/3}rd of the distance covered by q before coming to rest , the ratio of their velocity is-

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)

Similarly,

A = U2/(2X)

So,

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)