Five hoboes, each with mass m, stand on a railway flatcar of mass M. They jump o
ID: 1963861 • Letter: F
Question
Five hoboes, each with 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 men jump at the same time?
Data: m = 67 kg; M = 250 kg; u = 3.0 m/s.
(b) What is the final velocity of the flatcar if they jump off one at a time? Does case (a) or case (b) yield the largest final velocity of the flat car? Can you give a simple physical explanation for your answer?
Explanation / Answer
a. Pinit= Pfinal
0 = MV-5m(v) or,5mv=MV
67(5)(3) = 250V
V=4.02m/s
b. After the 1st man jumps off,
mu=(4m+M)v1
v1=m/(4m+M)*u=0.39m/s
after the 2nd man jumps off,
(4m+M)v1=-m(u-v1)+(3m+M)v2
v2=0.84m/s
after the 3rd man jumps off,
(3m+M)v2=-m(u-v2)+(2m+M)v3
v3=1.36m/s
after the 4th man jumps off,
(2m+M)v3=-m(u-v3)+(m+M)v4
v4=1.99m/s
after the 5th man jumps off,
(m+M)v4=-m(u-v4)+(M)v5
v5=2.79m/s
Explnation:
case (a) yeilds the largest final velocity. Physically it means that as u is the relative speed of the body wrt to the car so as people jump off,in case (b), their absolute velocities decrease(for example for 1st hobo the absolute speed is 3m/s,for 2nd one it is (3-v1),for 3rd it is(3-v2) and so on...)so the final combined momentum of the hobos will be less in part(b)as in part(a) where all have velocities 3m/s as a result of which the momentum imparted to the empty flatcar will be less which gives less velocity for the second part.
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