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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: n1z on July 30, 2012, 09:18:22 pm

Title: Help! Momentum
Post by: n1z on July 30, 2012, 09:18:22 pm: 1.Does anyone know the formula for conservation of momentum.
2. Can anyone explain the principles of conservation of momentum.

Thanks
Title: Re: Help! Momentum
Post by: DisaFear on July 30, 2012, 09:32:00 pm: What is momentum?

As a formula, p=mv

p=momentum (kg m/s)
m=mass (kg)
v=velocity (m/s)

I don't know if this is correct, but I like to think of momentum as 'how hard it is to stop a moving object?'

Conservation of momentum?

Let's say we have a system, that is isolated, there are no outside forces messing up our system. There's no friction, or no silly child pushing or poking at it, as silly children do.

That means, whatever momentum we calculate for this system, it will be CONSTANT! If somehow the mass increases, then surely, to keep the momentum CONSTANT, the velocity must decrease!

Initial momentum = final momentum

Hope I haven't derped anywhere, reply if you need more help, I can provide some examples, esp with two objects heading towards each other, and stuff. These conservation principles can take a while to get used to ^-^
Title: Re: Help! Momentum
Post by: n1z on July 30, 2012, 09:58:19 pm: I'm still a little bit confused conservation wise.
Thanks for the formula
Title: Re: Help! Momentum
Post by: Lasercookie on July 30, 2012, 10:05:54 pm: Can you elaborate more on what's confusing you? Feel free to ramble.
Title: Re: Help! Momentum
Post by: n1z on July 30, 2012, 10:14:59 pm: Please tell me if I'm right. Lets say if a car collides with another car the momentum of car 2 is the same momentum as it was with car 1.
Title: Re: Help! Momentum
Post by: Bhootnike on July 30, 2012, 10:53:50 pm: Quote from: n1z on July 30, 2012, 09:58:19 pm
I'm still a little bit confused conservation wise.
Thanks for the formula

In a closed/isolated system - i.e. a situation where NO external forces act, the momentum before a collision = momentum after a collision.
Title: Re: Help! Momentum
Post by: Lasercookie on July 30, 2012, 11:10:58 pm: Quote from: n1z on July 30, 2012, 10:14:59 pm
Please tell me if I'm right. Lets say if a car collides with another car the momentum of car 2 is the same momentum as it was with car 1.
Not necessarily.

So car 1's momentum will be given by $p = mv$ , let's set m_1 to be it's mass, and u to be it's initial velocity. $p_1_{initial} = m_1u_1$

Car 2's momentum will be given by $p_2_{initial} = m_2u_2$

The principle of conservation of momentum is that the total momentum of the system will remain constant, that is:
$\sum p_{initial} = \sum p_{final}$

$p_1 + p_2 = p_1 + p_2$

Momentum isn't going to escape from our system.

So the total initial momentum in this system is given by
$\sum p_{initial} = p_1 + p_2 = m_1u_1 + m_2u_2$

So then the collision happens, and the velocity of the vehicles change etc. The new velocities are denoted 'v'.

$\sum p_{final} = p_1 + p_2 = m_1v_1 + m_2v_2$

So we can go back to the principle of conservation of momentum, and conclude that:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

If you add up p1 and p2 at any time, the total value should remain the same.

If you think in terms of Newton's Third Law, that is if you exert a force on an object, it will exert a force with equal magnitude in the opposite direction. We have our cars moving with a whatever driving force and then they collide.

$F = \frac{dp}{dt}$ (force is the rate of change of momentum). If force is increased, momentum is increased etc.

Car 1 exerts a force upon Car 2. Let's say this causes Car 2 to be pushed forward, i.e. it has speed up a bit. So the momentum of Car 2 increases by some amount. Note that the momentum $p_2_{final}$ does not become the value of $p_1_{initial}$ , since $p \not= F$ , force is equal to the rate of change of momentum.

$p_1_{initial} = m_1u_1$

$p_2_{final} = p_1_{initial} = m_1u_1$

Newton's Third Law. During the reaction, Car 2 exerts force, equal in magnitude but in an opposing direction, upon Car 1. The momentum of Car 1 decreases by the amount that momentum of car 2 increased. That is, Car 1 slows down a bit after the reaction. Note that this doesn't imply that the objects will have equal accelerations, the changes in velocity may very well be unequal.

If you have more than 2 objects in your system, you have to adjust slightly for that. If you're not yet convinced of Newton's laws, you might want to spend some time thinking about them. Also note that what'll happen (increases, decreases) will depend on the situation. Try playing around with assigning values.

Let's say take what you said and try and apply it:
Quote from: n1z on July 30, 2012, 10:14:59 pm
Please tell me if I'm right. Lets say if a car collides with another car the momentum of car 2 is the same momentum as it was with car 1.
What you've said is that $p_1_{initial} = p_2_{final}$

So due to conservation of momentum, we know that these two statements will be true:
$p_{1initial} + p_{2initial} = p_{1final} + p_{2final}$
$p_{1initial} + p_{2initial} - p_{1final} - p_{2final} = 0$ or more concisely stated as $\Delta p_{total} = 0$
This is also only in an isolated system, where there are no other forces to mess with things.

So what you've said is $p_{1initial} + p_{2initial} = p_{1final} + p_{1initial}$

$p_{1initial} + p_{2initial} - p_{1final} - p_{1initial} = 0$

$p_{2initial} = p_{1final}$

We can see that this value won't always be equal to zero, which means that conservation of momentum is violated. This will be when $p_{2initial}$ or $p_{1final}$ has a value, which it will if car 2 is moving at the start or if car 1 is still moving by the end (no violation if car 2 is initially stationary, or if car 1 is finally stationary).