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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-
Dear Ayush 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/mAnd 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/MA/Q,t = 2Tx = X/3Also,F = ma = MAFor A,0 = u – at=> a = u/tSimilarly,A = U/TSo,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 = 1/6Thus, ratio of initial velocity of A to initial velocity of B is = 1 : 6 RegardsArun (askIITians forum expert)
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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