Homework: conservation of momentum

AI Thread Summary
The discussion focuses on calculating the initial speeds of two boats exchanging a 50kg parcel while conserving momentum. The user initially set up a momentum equation but struggled with the kinetic energy aspect, mistakenly assuming it was conserved. Participants clarified that this is not an elastic collision, meaning kinetic energy is not conserved, but momentum is. They provided guidance on formulating the correct equations to express the relationship between the two speeds, ultimately leading to the correct values of v1 and v2. The conversation emphasizes the importance of understanding momentum conservation in inelastic collisions.
Michael Kantor
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Hi,
could you please help me with following:
----------
two boats traveling in opposite direction with speed v1 and v2

when they are passing by they interchange a parcel of same mass M=50kg

as a result second boat stops and first boat continue traveling in same direction with speed u1=8.5m/s

calculate v1 and v2 when you know masses (including M) m1=1000kg and m2=500kg

------------

I wrote
m1*v1+m2*v2=m1*u1

but I am not able to find out second equation. I tried kinetic energy but
I think it is not conserved.

Thank you
Michael
 
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Welcome to PF,
What do you mean when you say;

Michael Kantor said:
when they are passing by they interchange a parcel of same mass M=50kg

~H
 
Sorry my English is very poor.

one boat is going East and the other West, when they meet
an object of mass 50kg is handed over from boat1 to boat2 and at the same time other object of same mass is handed from boat2 to boat1Thank you
Michael
 
No problem. I would say for this question you have to assume that kinetic energy is conserved. You correctly set up you momentum equation, if you then setup a kinetic energy equation, you have two equations (one for kinetic energy and one for momentum) and two unknowns (v1 and v2), you can then solve both equations simultaneously to obtain v1 and v2.

~H
 
thank you, but ...

Hootenanny said:
No problem. I would say for this question you have to assume that kinetic energy is conserved. You correctly set up you momentum equation, if you then setup a kinetic energy equation, you have two equations (one for kinetic energy and one for momentum) and two unknowns (v1 and v2), you can then solve both equations simultaneously to obtain v1 and v2.

~H

I wrote two equations

v1*m1+v2*m2 = u1*m1

and

(m1*v1^2)/2 + (m2*v2^2)/2 = (m1*u1^2)/2

the result is not M dependent ?
when I solve the equations I get wrong answers.
(correct is v1=9m/s and v2=-1m/s)

Or when and why should M in kinetic energy equation be ?

Thank you very much

Michael
 
Michael Kantor said:
I wrote two equations

v1*m1+v2*m2 = u1*m1

and

(m1*v1^2)/2 + (m2*v2^2)/2 = (m1*u1^2)/2

the result is not M dependent ?
when I solve the equations I get wrong answers.
(correct is v1=9m/s and v2=-1m/s)

Or when and why should M in kinetic energy equation be ?
This is not an elastic collision. Energy is not conserved. Only momentum is conserved. You know that the 50 kg mass traveling at v1 when caught by the second boat causes the second boat to lose all its momentum. Work out the formula for that. That should give you the ratio of the two speeds and will enable you to express v1 in terms of v2.

Then work out the equation for conservaton of momentum for the other boat and substitute your value for v1 in terms of v2. That will give you the answer.

AM
 
I've just worked through the calculations now (I've only just got a spare minute), and I made an incorrect assumtion that the collision was elastic (hence kinetic energy being conserved). If you consider the momentum of the package traveling at v1, you should be able to obtain a ratio of the two velcoities (one will be negative).

~H

Edit: AM got there before me.
 
Last edited:
Andrew Mason said:
This is not an elastic collision. Energy is not conserved.

Just a small niggling point (I'm sure you know, it's just that others might get the wrong idea) - energy (the sum total of all forms) is *always* conserved. Kinetic energy is not conserved (in this case). :smile:
 
Andrew Mason said:
This is not an elastic collision. Energy is not conserved. Only momentum is conserved. You know that the 50 kg mass traveling at v1 when caught by the second boat causes the second boat to lose all its momentum. Work out the formula for that. That should give you the ratio of the two speeds and will enable you to express v1 in terms of v2.

Then work out the equation for conservaton of momentum for the other boat and substitute your value for v1 in terms of v2. That will give you the answer.

AM

The two equations are

M*v1 + (m2-M)*v2 = 0

and

(m1-M)*v1 + M*v2 = m1*u1 ?

I hope so. At least they give correct answer, I use them.:smile:

Thank you
Michael
 
  • #10
Yup, they look correct to me.

~H
 
  • #11
Michael Kantor said:
The two equations are

M*v1 + (m2-M)*v2 = 0

and

(m1-M)*v1 + M*v2 = m1*u1 ?
You could also use:

m_1v_1 + m_2v_2 = m_1u_1

which expresses the conservation of momentum of the whole system.

AM
 
Last edited:

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