Homework: conservation of momentum

In summary: 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.
  • #1
Michael Kantor
4
0
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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  • #2
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
 
  • #3
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
 
  • #4
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
 
  • #5
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
 
  • #6
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
 
  • #7
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:
  • #8
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:
 
  • #9
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:

[tex]m_1v_1 + m_2v_2 = m_1u_1[/tex]

which expresses the conservation of momentum of the whole system.

AM
 
Last edited:

1. What is the definition of conservation of momentum?

Conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant over time. This means that the total momentum before and after a collision or interaction between objects will be the same.

2. How is conservation of momentum related to Newton's Third Law?

Newton's Third Law states that for every action, there is an equal and opposite reaction. In terms of momentum, this means that the total momentum of a system is conserved because the momentum lost by one object in a collision is equal to the momentum gained by the other object.

3. Can conservation of momentum be violated?

No, conservation of momentum is a fundamental law of physics and cannot be violated. However, in certain situations, it may appear that momentum is not conserved due to external factors such as friction or air resistance.

4. How is conservation of momentum applied in real-world situations?

Conservation of momentum is applied in a variety of real-world situations, such as collisions between objects, rocket propulsion, and sports. It is also used in designing safety features in vehicles and understanding the motion of objects in space.

5. How does the conservation of momentum principle relate to the conservation of energy?

The conservation of momentum principle is closely related to the conservation of energy. In fact, momentum is a component of kinetic energy, which is a form of energy that is also conserved in a closed system. This means that if the momentum is conserved, the energy will also be conserved.

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