Physics problem, stuck in algebra.

Question

I end up with the equations;

$$u=u_1' \cos(a)+u_2' \cos(b)$$ $$u_1' \sin(a)=u_2' \sin(b)$$ $$u^2=u_1'^2+u_2'^2$$

I have to show that $$a+b=\frac{\pi}{2}$$

$x'$ isn't the derivative of $x$, it's a convenient way not to mix variables in physics.

I have no approach. Tried substituting the trigonometrics as $x,y$ etc but it became more confusing and I couldn't do any trig identities easily.

Answer 1

Disclaimer

I did some changes in variable names, for brevity. Namely $u_1' = x, u_2' = y$

Solution

$$ u = x \cos a + y \cos b \\ x \sin a = y \sin b \\ u^2 = x^2 + y^2 $$ Square first equation and use second equation $$ u^2 = x^2(1 - \sin^2 a) + y^2 (1 - \sin^2 b) + 2xy \cos a \cos b = x^2 + y^2 - 2x^2 \sin^2 a + 2xy \cos a \cos b $$ Now use third equation, so $$ x \sin^2 a = y \cos a \cos b \implies y \sin a \sin b = y \cos a \cos b \implies \tan a = \cot b \ = \tan (\pi/2 - b) $$ Within the range of periodicity it means that $$ a + b = \frac \pi 2 $$

Answer 2

Hint: square the first identity, susbtitute in the third identity, and remember

$$ \cos^2 x + \sin^2 x = 1 $$

and

$$\cos(a+b) = \cos a \cos b - \sin a \sin b$$

Answer 3

Let $x=u_1' , y=u_2'$.

$$u=x \cos a+y \cos b \tag{1}$$ $$0=x \sin a-y \sin b \tag{2}$$ $$u^2=x^2+y^2 \tag{3}$$

$(1)^2+(2)^2 \implies $

$$u^2=x^2+y^2+2xy\cos (a+b) \tag{4}$$

From (4) and (3) we have:

$$ xy\cos (a+b)=0 \tag{5}$$

If $x\not =0$ and $y\not =0$, then $a+b=\pm k \pi+\pi/2$

Answer 4

Here is a vector-based approach. Your first two equations are saying that the (velocity) vectors $\mathbf{u}',\mathbf{u}_1',\mathbf{u}_2'$ satisfy $\mathbf{u}'=\mathbf{u}_1'+\mathbf{u}_2'$ (eqs. 1 and 2 express vector components with $\mathbf{u}'$ along the $x$-axis.) Then $$u'^2=|\mathbf{u}'|^2=|\mathbf{u}_1'+\mathbf{u}_2'|^2=u_1'^2+2\mathbf{u}_1'\cdot \mathbf{u}_2'+u_2^2=u_1'^2+2u_1 u_2 \cos\theta+u_2'^2$$

where $\theta$ is the angle between the two scattered particles. But your third equation will then only be satisfied if $\cos\theta=0$ which in this context only works if $\theta=\pi/2$---and what is $\theta$ in terms of $a$ & $b$?