Conservation of momentum of cannon balls

[overside]

EDIT: MASS OF CANOE IS ACTUALLY 50G I typed it wrong the 1st time :/

Homework Statement

50kg girl sits in a 50kg canoe at rest on the water. She holds two 10 kg cannon balls. she throws them over the stern of her canoe one at a time, each ball leaving her hands at a velocity of 5.0m/s relative to the canoe. what is the final velocity of the canoe, neglecting friction between boat&water?

Homework Equations

ΣP = ΣP'
P = MV

The Attempt at a Solution

I don't really understand, probably misunderstood key concepts but I tried anyway.

1) finding momentum of 1st ball...
P = 5.0m/s * 10kg
=50 N*s

2) speed of boat after that 1st ball...
V = P/m
=50 N*S / (50kg + 50kg + 10kg)
=0.45m/s

3) here it gets tricky i think, and where I am lost. Because the speed is relative to canoe the speed of the 2nd ball should be

V= 5.0 m/s + 0.45 m/s = 5.45m/s ??

therefore

momentum of 2nd ball = 5.45m/s * 10kg = 54.5 N.s

and for the canoe's final velocity..

V = P/m
= 54.5 N.s / (50kg + 50kg)
= 0.54 m/smy answer books gives me 0.87 m/s and I am lost, so help pleeease ;D

Homework Statement

Homework Equations

The Attempt at a Solution

 

[xts]

My Lord!
Once again a school exercise forcing pupils to assume that dynamics of canoe floating on the water is the same as canoe on ice (frictionless move - frozen lake).

The real physical answer is: 'it is too dificult to model the flow of water around the canoe, as it moves. to predict it theoretically'.

Friction has really no significance. But in order to move the canoe, you must also move some water (displaced by the canoe). And it is extremely difficult to solve theoretically physical problem, depending on canoe shape, water depth, etc. In practice - even shipyar engineers do not solve it theoretically - they check it in practice, on ship models dragged in a pool.

 

[overside]

My Lord!
Once again a school exercise forcing pupils to assume that dynamics of canoe floating on the water is the same as canoe on ice (frictionless move - frozen lake).

The real physical answer is: 'it is too dificult to model the flow of water around the canoe, as it moves. to predict it theoretically'.

Friction has really no significance. But in order to move the canoe, you must also move some water (displaced by the canoe). And it is extremely difficult to solve theoretically physical problem, depending on canoe shape, water depth, etc. In practice - even shipyar engineers do not solve it theoretically - they check it in practice, on ship models dragged in a pool.

Well, I highly doubt my teacher will let me walk away if I told him what you said even if its true ...

you know how to solve this problem the way it is intented though? for the sake of me passing my next test, pleeease tell me how anyway D:

 

[ehild]

1) finding momentum of 1st ball...
P = 5.0m/s * 10kg
=50 N*s

2) speed of boat after that 1st ball...
V = P/m
=50 N*S / (50kg + 50kg + 10kg)
=0.45m/s

You need the momentum of the ball with respect to the ground, but you calculated it with respect to the boat.

Think: The ball has to accelerate to its final speed, 5 m/s with respect to the boat, before it is thrown. As the girl starts to move the ball the boat accelerates backwards. When the ball leaves the girl's hand, the boat has a forward velocity v, and the ball moves in the opposite direction, with velocity u=v-5. The total momentum is zero. Find the velocity of the boat after the first ball has been thrown.
Apply the same method for the second ball, with the new momentum of the boat.

(Taking Xts' comment into account, pretend that the boat is really a sledge on ice )

ehild

 

[xts]

ehild already helped you to pass the test ;)

But... as you finally pass it - ask your teacher to explain how real boat behaves in such situation. Press him for details about water flow around the canoe. And what happens when the canoe suddenly gets lighter by 10kg - as the girl throws out the cannonball. That is really something terribly difficult to describe theoretically, but such effects are overwhelming to "sledge dynamics" of two-body momentum conservation, you were asked to solve.

Maybe, next time, he'll give to you (and other schoolchildren) more realistic examples...

 

[overside]

You need the momentum of the ball with respect to the ground, but you calculated it with respect to the boat.

Think: The ball has to accelerate to its final speed, 5 m/s with respect to the boat, before it is thrown. As the girl starts to move the ball the boat accelerates backwards. When the ball leaves the girl's hand, the boat has a forward velocity v, and the ball moves in the opposite direction, with velocity u=v-5. The total momentum is zero. Find the velocity of the boat after the first ball has been thrown.
Apply the same method for the second ball, with the new momentum of the boat.

(Taking Xts' comment into account, pretend that the boat is really a sledge on ice )

ehild

arghh this is just making no sense to me, in fact I think i may be able to understand if you could just explain 1 point to me:

this is my teacher's answers key: http://fizics.com/physics30/answers/pdf/Lesson02.pdf
(Scroll down to #12, Girl part)

My teacher labled the V1 as 5.0m/s + V2(speed of boat) which i really don't understand why, if you can enlighten me on this part, then I will probably figure the problem out :D

 

[overside]

NVM BROS i figured it out, so simple in fact. Empty stomach does things to you !

 

[ehild]

I can not see your link.

The relative velocity of the ball is 5 m/s, then V(ball) = V(boat)+5 is correct, but in this case you get negative value for the velocity of the boat.

If you have a body moving with velocity v and an other one moving with V, then the relative velocity of the first body with respect to the second one is defined as v(rel)=v-V. If the relative velocity is given, v=V+v(rel)

I chose the relative velocity negative as the ball moves in opposite direction as the boat. My method gives a positive number for the velocity of the boat.

ehild