SOLUTION: A tractor of mass 5*10^3kg is used to Low a car of mass 2.5*10^3kg. The tractor moved with a speed of 3.0m/sec just before the towing rope becomes taut. Calculate (1) speed of th

Algebra ->  Formulas -> SOLUTION: A tractor of mass 5*10^3kg is used to Low a car of mass 2.5*10^3kg. The tractor moved with a speed of 3.0m/sec just before the towing rope becomes taut. Calculate (1) speed of th      Log On


   

Question 1012422: A tractor of mass 5*10^3kg is used to Low a car of mass 2.5*10^3kg. The tractor moved with a speed of 3.0m/sec just before the towing rope becomes taut. Calculate
(1) speed of the tractor immediately the rope become taut.
(2) loss in K. E of the system just after the car has started moving.
(3) impulse in the rope when it jerks the car into motion.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Conservation of momentum.
m%5B1%5Dv%5B1%5D=%28m%5B1%5D%2Bm%5B2%5D%29v%5B2%5D
5000%283%29=%285000%2B2500%29v%5B2%5D
15000=7500v%5B2%5D
v%5B2%5D=2m%2Fs
.
.
Initially,
E%5B1%5D=%281%2F2%29%285000%29%283%29%5E2
E%5B1%5D=22500J
Then,
E%5B2%5D=%281%2F2%29%287500%29%282%29%5E2
E%5B2%5D=15000J
So,
DELTA%2AE=E%5B2%5D-E%5B1%5D
DELTA%2AE=15000-22500
DELTA%2AE=-7500J
.
.
.
Impulse equals the momentum change of the car.
I=m%28v%5B2%5D-0%29
I=2500%282%29
I=5000%28kg%2Am%29%2Fs