Question #146014

Ted and his ice-boat (combined mass = 240 kg) rest on the frictionless surface of a frozen lake. A heavy rope (mass of 80 kg and length of 100 m) is laid out in a line along the top of the lake. Initially, Ted and the rope are at rest. At time t=0, Ted turns on a wench which winds 0.5 m of rope onto the boat every second.

a) What is Ted’s velocity just after the wench turns on?

b) What is the velocity of the rope at the same time?

c) What is the Ted’s speed just as the rope finishes?

d) How far did the center-of-mass of Ted+boat+rope move

e) How far did Ted move?

f) How far did the center-of-mass of the rope move?


1
Expert's answer
2020-11-24T05:59:29-0500

Given,

mass of the ted and frozen (m)=240kg(m)=240kg

Mass of the rope(M)=80kg(M)=80kg

Length of the rope (l)=100m(l)=100m

velocity v=0.5m/sv=0.5m/s

Applying conservation of momentum


a)(M+m)V=mv(M+m)V=mv


V=mM+m×0.5=8080+240×0.5\Rightarrow V=\frac{m}{M+m}\times 0.5=\frac{80}{80+240}\times 0.5


V=0.54=0.125m/s\Rightarrow V = \frac{0.5}{4}=0.125m/s


b) For the velocity of rope

MV+mv1=0MV+mv_1=0

v1=Mm×0.125v_1=\frac{-M}{m}\times 0.125

v1=24080×0.125\Rightarrow v_1=\frac{-240}{80}\times 0.125

v1=0.375m/s\Rightarrow v_1=-0.375 m/s

c) Ted's speed just after the rope finish, will be zero because net momentum always be conserve so at the ends of the rope the velocity of ted becomes zero.


d) There is no external force working on system (rope+boat+ted), hence the center of mass remains unchanged. Hence net displacement will be 0.

e) Distance traveled by the ted =0.125×100=12.5m=0.125\times 100 = 12.5 m

f) Distance traveled by the center of mass of the rope =0.375×100=37.5m=0.375\times 100 =37.5 m


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Comments

Fidelity Banda
14.11.21, 16:06

i love this

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