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Q. A railroad flat car of mass M can roll without friction along a straight horizontal track. Initially a man of mass m is standing on the car which is moving to the right with speed v0 . What is da change in velocity of the car if da man runs to da left so that his speed relative to rhe car is v just before he jumps offf at the left end ?
Comments (8)
@ Ramyani
Firstly am amazed tht U didn't get this one......its more or less a sitter (basic problem).....personal advice.....take it cool!!Don't stress out over difficult ones, or U begin to mess up sitters like this one......No ofeence intended
As For The Solution:
Let M=mass of car
m=mass of man
Vi=Inital velocity of system
Vf=Final velocity of car
V=Velocity of man
Simple conservation of momentum
(M+m)Vi=MVf +m(Vf-V)
So (M+m)Vi +mV=(M+m)Vf
So ((M+m)Vi +mV)/M+m =Vf
Now we want the change in velocity which is simply Vf-Vi =((M+m)Vi +mV)/M+m -Vi
=mV/m+M
hope this helps
cheers!!
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Answer:
No external force (friction) acting , hence :
Momentum conserved :
Initial momentum = final momentum
(M+m)v0 = Mv1 - m(v1+v)
v1 = [Mv0+m(v0+v)]/(M-m)
If change is required , find (v0-v1) [Magnitude]
Hope its helpful.........