N women, each of mass m, stand on a railway flatcar of mass M. They jump off one end of the flatcar with velocity u relative to the car. The car rolls in the opposite direction without friction. (a) What is the final velocity of the flatcar if all the women jump at the same time? (b) What is the final velocity of the flatcar if the women jump off one at a time? (c) Which answer, either (a) or (b), is greater? Try to make physical and/or intuitive sense of this answer and comment on your thoughts.

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nuuk nuuk · Answer 1 · 2019-09-16T07:51:40+02:00

Answer:

Explanation:

Given

Velocity of women relative to the car is u

Let v be the velocity of car after women jump off

Therefore women velocity relative to the ground is v-u

Conserving momentum

Mv+Nm(v-u)=0

Mv+Nmv=Nmu

[tex]v=\frac{Nm}{M+Nm}[/tex]

(b)Let [tex]v_r[/tex] be the velocity of car after r women has jumped and [tex]v_N[/tex] be the final velocity of the car after N women have jumped.

After r women jumped and N-r on cart momentum is given by

[tex]P_r=\left [ M+\left ( N-r\right )m\right ]v_r[/tex]

After next woman jumps car has velocity of [tex]v_{r+1}[/tex] while the woman has velocity is [tex]v_{r+1}-u[/tex](relative to the ground)

Total momentum

[tex]P_{r+1}=\left [ M+\left ( N-r-1\right )m\right ]v_{r+1}+m\left ( v_{r+1}-u\right )[/tex]

Since total momentum in horizontal direction is conserved then

[tex]P_{r+1}=P_r[/tex]

[tex]v_{r+1}-v_r=\frac{mu}{\left ( M+\left ( N-r\right )m\right )}[/tex]

Summing the above expression from r=0 to r=N-1 we get

[tex]\sum_{j=1}^{j=N}\frac{mu}{\left ( M+jm\right )}[/tex]

(c)Answer in part b is greater because j\leq N[/tex]

Thus Velocity in part b is greater.

Intuitively if you jump after an another person you will impart extra momentum to the cart compared to when all persons jumped off simultaneously.