IIT JEE , AIEEE Forum - Physics , Mechanics

amitp91

A man of mass m stands on a long flat car of mass M, moving with velocity V . IF he now begins to run with a velocity u with respect to the car , in the same direction as V , the velocity of the car will be
a)V - mu/M
b)V - mu/(m+M)
c)V + mu/(m+M)
d)none

karthik2007

is it d)?

anchitsaini

amitp91

i dont know the answer but it is not d

anchitsaini

initial momentum = (m+M)V

final momentum = m(u+v1) + Mv1 where v1 is the new velocity of M
hence
(m+M)V=mu + mv1 + Mv1=mu+(m+M)v1
thus
v1=V-mu/(m+M)

karthik2007

hmm... thats what i did but i am getting d)!!!

edit : made error in simplification... ans is b) only. Anchit is right

konichiwa2x

There is no external force acting on the system hence the net linear momentum will be conserved.

----(1)

Let

be the velocity of the car.
The velocity of the man with respect to the car

then velocity of man with respect to ground

.

----(2)

equate (1) and (2) to get,

Option (b) is correct.

anchitsaini

see ,
let new velocity of car be v1
hence
relative velocity = u = v2-v1 (as its in same direction)
where v2 is velocity of man
hence v2=u+v1

now we can conserve momentum

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