- #1
Ariano AnnaG
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- Homework Statement
- HALLIDAY CHAPTER 9, momentum and impulse
Statement of the problem:
Two vessels are proceeding in the same direction through a river. Their velocities are respectively
10 km/h and 20 km/h. While the second boat is getting over the first one, a men is throwing 1000kg of charcoal per minute from the slower the faster.
Which variation of the force has to be applied to the motor of the slower boat and which to the motor of the faster, in order to keep their velocities constant? This, considering that the transfer of the charcoal doesn't interfere with the motion of the vessels and that the friction force doesn’t depend on the charge of the boats.
Solution given by the book: (a) 46N (b) none
DATAS:
V1i= 10km/h = 10/3,6 m/s velocity of the first boat
V2i= 20km/h =20/3,6 m/s velocity of the second boat
k= 1000kg/min = 1000kg/60s quantity of charcoal moved by the men in a minute
- Relevant Equations
- impulse-momentum theorem general equation: FΔt=Δq
1)In the first case the mass changes, in this manner:
m1= m1i - k Δt
impulse-momentum theorem:
F1Δt=Δq
FΔt=m1V1-m1iV1i
I now can replace m1 with its definition and obtain:
F1Δt= (m1i - k Δt)V1 -m1iV1i = m1iV1i - kΔtV1- m1iV1i = - kΔtV1
It is possible to remove Δt as it appears in both members of the equation:
a) F1= -kV1= -1000kg/60s x 10/3,6 m/s= -46 N2)In the second case the mass changes, in this manner:
m2= m2i +k Δt
impulse-momentum theorem:
F2Δt=Δq
F2Δt=m2V2 -m2iV2i
I can now replace m1 with its definition and obtain:
F2Δt= (m2i + k Δt) V2 - m2iV2i = m2iV2i + kΔtV2- m2iV2i = +kΔtV2
It is possible to remove Δt as it appears in both members of the equation:
b) F2= kV2= +1000kg/60s x 20/3,6 m/s= +92,6 N ≅ 93 N
I thought I had to use the impulse-momentum theorem in the second case, like I did before to solve the first request of the problem, but my result (93 N) has nothing to do with the solution (b) of the book, (0 N, there isn’t a force).
What am I not considering or forgetting? Thank you in advance for your disposability and patience
m1= m1i - k Δt
impulse-momentum theorem:
F1Δt=Δq
FΔt=m1V1-m1iV1i
I now can replace m1 with its definition and obtain:
F1Δt= (m1i - k Δt)V1 -m1iV1i = m1iV1i - kΔtV1- m1iV1i = - kΔtV1
It is possible to remove Δt as it appears in both members of the equation:
a) F1= -kV1= -1000kg/60s x 10/3,6 m/s= -46 N2)In the second case the mass changes, in this manner:
m2= m2i +k Δt
impulse-momentum theorem:
F2Δt=Δq
F2Δt=m2V2 -m2iV2i
I can now replace m1 with its definition and obtain:
F2Δt= (m2i + k Δt) V2 - m2iV2i = m2iV2i + kΔtV2- m2iV2i = +kΔtV2
It is possible to remove Δt as it appears in both members of the equation:
b) F2= kV2= +1000kg/60s x 20/3,6 m/s= +92,6 N ≅ 93 N
I thought I had to use the impulse-momentum theorem in the second case, like I did before to solve the first request of the problem, but my result (93 N) has nothing to do with the solution (b) of the book, (0 N, there isn’t a force).
What am I not considering or forgetting? Thank you in advance for your disposability and patience