# Trigonometry in Newtonian Dynamics HARD

• August 26th 2009, 05:38 AM
TweedyTL
Trigonometry in Newtonian Dynamics HARD
A particle of mass $m\sb{1}$ is travelling along the x-axis with velocity $(v\sb{1},0,0)$. It collides elastically with a stationary particle of mass $m\sb{2}$. After the collision, the particle of mass $m\sb{1}$ is travelling with speed $v\sb{1}'$ at angle $a$ to the x-axis, and the particle of mass $m\sb{2}$ is travelling with speed $v\sb{2}'$ at angle $b$ to the x-axis.

The conservation of momentum laws for this collision are:
(i) $m\sb{1}v\sb{1}'sin(a)+m\sb{2}v\sb{2}'sin(b)=0$
(ii) $m\sb{1}v\sb{1}'cos(a)+m\sb{2}v\sb{2}'cos(b)=m\sb{1 }v\sb{1}$

The conservation of energy equation is:
(iii) $m\sb{1}(v\sb{1}')\sp{2}+m\sb{2}(v\sb{2}')\sp{2}=m\ sb{1}(v\sb{1})\sp{2}$

I have to show that these 3 equations imply:
$sin\sp{2}(a+b)=sin\sp{2}(b)+ksin\sp{2}(a)$, where $k=\frac{m\sb{1}}{m\sb{2}}$
• August 26th 2009, 11:24 AM
Opalg
Quote:

Originally Posted by TweedyTL
A particle of mass $m\sb{1}$ is travelling along the x-axis with velocity $(v\sb{1},0,0)$. It collides elastically with a stationary particle of mass $m\sb{2}$. After the collision, the particle of mass $m\sb{1}$ is travelling with speed $v\sb{1}'$ at angle $a$ to the x-axis, and the particle of mass $m\sb{2}$ is travelling with speed $v\sb{2}'$ at angle $b$ to the x-axis.

The conservation of momentum laws for this collision are:
(i) $m\sb{1}v\sb{1}'\sin(a)+m\sb{2}v\sb{2}'\sin(b)=0$
(ii) $m\sb{1}v\sb{1}'\cos(a)+m\sb{2}v\sb{2}'\cos(b)={\co lor{red}m_1}v\sb{1}$

The conservation of energy equation is:
(iii) $m\sb{1}(v\sb{1}')\sp{2}+m\sb{2}(v\sb{2}')\sp{2}=m\ sb{1}(v\sb{1})\sp{2}$

I have to show that these 3 equations imply:
$\sin\sp{2}(a+b)=\sin\sp{2}(b)+k\sin\sp{2}(a)$, where $k=\frac{m\sb{1}}{m\sb{2}}$

First step is to solve equations (i) and (ii) for $v_1'$ and $v_2'$. Multiply (i) by $\cos a$, multiply (ii) by $\sin a$, and subtract. That gives $m_2v_2'\sin(a-b)=m_1v_1\sin a$ (using the trig formula $\sin a\cos b-\cos a\sin b = \sin(a-b)$). Therefore $v_2' = \frac{kv_1\sin a}{\sin(a-b)}$. A similar procedure (multiplying (i) by $\cos b$ and (ii) by $\sin b$) gives $v_1' = -\frac{v_1\sin b}{\sin(a-b)}$.

Now square both of those expressions, and substitute those formulas for $v_1'^{\,2}$ and $v_2'^{\,2}$ into (iii). You'll find that after a bit of simplification and cancellation, it reduces to $\sin^2b+k\sin^2a = \sin^2(a-b)$. Note: $\sin^2(a-b)$, not $\sin^2(a+b)$ as (wrongly) stated in the question. Note also the missing $m_1$ in equation (ii).