IIT JEE , AIEEE Forum - Physics , Mechanics - Centre of mass: Verma

anandghegde

verma pg 161 problem 28

2 persons each of mass m are standing at the 2 extremes of a railroad car of mass M. The person left jumps to the left with a speed u w.r.t the state of the car before the jump. Thereafter, the other person jumps to the right with a speed u w.r.t the state of the car before his jump. Find the final velocity of the car.

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Rates for everyone who replies.....

anandghegde

ramyani

whicz 1 is correct ?

anandghegde

I dont know

anandghegde

Want help from u ppl

ramyani

I checked. my ans was m² u / M (m + M )

shubham_sachdeva

mu = (m + M)v

v = mu / m + M towards right.......

so v with respect to ground after second jump should be = [mu /m + M ]+ u

= u { [m / m + M] + 1}

conserving momentum again we get

mu {[m/m+M ] + 1} + Mv' = mu

so we get

v' = - m^2 . u/m+M

- sign indicates left side.....

hope u got it!!!

cheers!!!

P.S. :- don't rate me!!!!!!

ramyani

2 persons each of mass m are standing at the 2 extremes of a railroad car of mass M resting on a smooth track.

First, the person on left jumps to left with vel u. Then

-- mu + ( m + M ) V = 0

or, V = mu / (M +m )     is the vel of train plus man towards right.

Now, when the man on the right jumps, the vel of it wrt car is u.

0 = mu - MV¹

or, V¹ = mu / M    towards left.

V is the change of vel of the platform. When the platform itself is taken as ref frame, assuming car to be at rest the above is derived.

Net vel towards left ( i,e, vel of car wrt earth )
                 = mu / M -- mu / (M +m ) = m² u / M (m + M )

whats wrong ?

edited.

shubham_sachdeva

after first man jumps the vel. of cart + 1 man = mu/m+M okk???? now the second man jumps in the same direction as that cart+man was moving so velocities will get added...so, mu/m+M + u....i didn't understand ur concept..???wats the prob.?

shubham_sachdeva

arrey dekho...while conserving momentum second time we will have to concentrate on both the three masses....this qn. basically tests the change of frame....while conserving momentum second time first term on lhs is for the second man , second term is for the cart and the term on rhs is for the first man.......hope its clear now!!!

pujasinghtwinkle

ans given in verma is wrong
it's

m ² u
-----------------------
(M+2m)(M+m)

anandghegde

anybody please explain which answer is right........

anandghegde

karthik or waterdemon
please solve and tell me which answer is right

karthik2007

Should be m²u/M(M+m).

Here is my solution anyway:

When the first man jumps out :

Initial momentum of the system is zero, so conserving momentum, we get:

mu = (M+m)v

which gives v = mu/(M+m)
The trolley moves with this speed to the right. (Whatever they are standing on X__X )

Now, at this stage, the momentum of the system is mu.

The man jumps with a velocity u with respect to the state of the car. So we have to find the absolute velocity of the man, which is u - (mu/M+m)

which gives v_absolute as Mu/(m+M).

Now, conserving momentum, we have (after the second man jumps out)

initial momentum of the system = mu.

Final momentum =
mMu(m+M) + Mv''.

So, mu = mMu/(m+M) + Mv''

which gives v'' = m²u/M(M+m)

Note : The direction will be towards the right. I have not shown the sign convention here. If you take the right handside to be the direction corresponding to the positive velocity vector, and vice versa for the left handside direction, you will get the final velocity to be negative, which means that the trolley moves in the left direction finally.

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