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two men each of mass m, stand on the edge of a stationary boggy of mass M. Assuming friction to be negligible, when both men jump off simultaneously with the same horizontal velocity u relative to the boggy then after their jump the velocity of the boggy will be

two men each of mass m, stand on the edge of a stationary boggy of mass M. Assuming friction to be negligible, when both men jump off simultaneously with the same horizontal velocity u relative to the boggy then after their jump the velocity of the boggy will be

Grade:12th pass

1 Answers

Arun
25750 Points
3 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
 
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
 
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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