vikas askiitian expert
Last Activity: 14 Years ago
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 ....