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2 men, each with mass m, stand on a the edge of a stationary car of mass and jump off with velocity "u" relative to the car,first simultaneously and then one after another.If friction is negligible,in which case is greater speed imparted to the car.

2 men, each with mass m, stand on a the edge of a stationary car of mass and jump off with velocity "u" relative to the car,first simultaneously and then one after another.If friction is negligible,in which case is greater speed imparted to the car.

Grade:11

3 Answers

Vinay Goyal IITB
37 Points
10 years ago

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)

vikas askiitian expert
509 Points
10 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) -MV            ....................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 ....

Ankita
19 Points
3 years ago
Hmm...nice.......
The spelling of momentum is wrong everytime but still the answer seems to be impressively correct........

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