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Dear ashok,
according to your que. the speed imparted to the car will be greater in the second case
first case,
intially whole system is at rest so total momentum is 0. when both person jumps off the car first person get a momentum of mu towards left and second person gets a momentum of mu towards right so by conservation of momentum
0=mu+MV-mu
V=0
second case,
when left person jumps off the car then
0=-mu+(M+m)V
V=mu/(M+m)
when right person jumps off the car then
0=mu-MV/
V/=mu/M
so net velocity= V-V/
=m2u/M(M+m)
case I => when both jumps simultaneously ...
let car is at origin & man jumps along +ve x axis ...
after jump let velocity 0f car is V then final momantam of system will be
P = momentam of man both men wrt ground after jump + momentam of car wrt ground after jump
momentam of car after jump is MV
momantam of man after jump is m(u-v) relative to ground
P = m(u-V) + m(u-V)i - M(V) ...........1
initially momantam was 0 so final should be 0 coz no external force acts on the system along horizontal direction..
2m(u-V) - MV = 0
V = 2mu/M+2m .................2
now if m<<<<<M then m/M can be neglected so eq 2 can be written as
V = 2m/M(1+2m/M) = 2(m/M)u ................3
case II ->
after 1 person jumps , mometam of system is given by
P = m(u-V) - (M+m)V = 0 ( 1 person remains in car)
V = mu/2m+M ...........1
now we our system consists of 1 man & plank moving with velocity V ...
let this man jumps with u & after jump velocity of plank is V1 then
final momantam = m(u-V1) -MV1 ....................2
initial momentam = (m+M)V = (m+M)mu/(2m+M) .............3
equating 2 & 3 we get
V1 = m2u/(m+M)(m+2M) .............4
now if m <<<M then m/M can be neglected eq 4 can be writtent by
V1 = m2u/M2 (m/M +1)(m/M + 2) = (m/M)2u/2 .................5
now comparing both results we can say that in first case speed of car will be more ....
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