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Consider two particles moving in the same direction on the same line, $A$ and $B$, with mass $m_A$ and $m_B$, respectively. They also have velocies $u_A$ and $u_B$. They collide. After the collision A and B have velocities $v_A$ and $v_B$.

I want to prove the equation for the Principle of Conservation of Momentum:

$m_Au_A+m_Bu_B=m_Av_A+m_Bv_B$.

So I begin:

Force is the rate of change of momentum, so

$F=\dfrac{\Delta mv}{\Delta t}$, so

$Ft=\Delta mv$.

Since the impulse on a particle is its change in momentum,

$I=\Delta mv=\Delta p$.

So taking partice B:

$I_B=\Delta p_B=m_Bv_B-m_Bu_B$.

And now A:

$I_A=\Delta p_A=m_Av_A-m_Au_A$.

I feel as if I'm on the right track to proving the equation, but how can I go from here?

I apologise if this seems like a homework question and doesn't belong here.

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  • $\begingroup$ You mean $u_A$, I think. $\endgroup$ Jul 3 '13 at 17:54
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The key component you're missing is Newton's Third Law. The impulse on a particle is $I=\Delta p=F\Delta t$. The interaction time is the same but the forces are equal in magnitude and opposite in direction for the two particles: $$F_A=-F_B\text{ so }I_A=-I_B.$$ With that all you need is some algebra.

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Suppose the collision between particles A and B lasts some time $\tau$, and during the collision the force between the particles is some complicated function of time, $F(t)$.

Consider particle A. The total impulse on it is the force times time, or more generally the integral of the force over the collision time:

$$ I_a = \int_0^\tau F_a(t)dt $$

And the change of momentum of particle A is just the impulse, $I_a$. Exactly the same applies to particle B, so:

$$ I_b = \int_0^\tau F_b(t)dt $$

But we know that action and reaction are equal and opposite, so $F_a = -F_b$, and therefore $I_a = -I_b$.

The total change of momentum is $I_a + I_b$, and since we've just shown that $I_a = -I_b$ the change is $-I_b + I_b = 0$. So the total change of momentum is zero i.e. momentum is conserved.

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