X and Y collide elastically, find angle of Y

Question

A particle X moves with constant velocity collides elastically with an indentical particle Y which is stationary. After collision, X moves at an angle of 30° to its intitial direction of motion. What angle does Y make with the initial direction of X

The answer is 60°.

I used the principle of conservation of momentum and conservation of momentum, but failed to get the value. 

Answer 1

Let the initial speed be $u$. Let the final speeds be $v_1$ and $v_2$, with angle $\theta_1$ and $\theta_2$, respectively.

By conservation of energy, we have:

$$\frac{1}{2}mu^2=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2$$

$$\implies v_1^2+v_2^2=u^2$$

By conservation of momentum, we have:

$$\left\{\begin{array}{ccc}mu=mv_1\cos\theta_1+mv_2\cos\theta_2 &\implies& v_1\cos\theta_1+v_2\cos\theta_2=u \\ mv_1\sin\theta_1=mv_2\sin\theta_2 &\implies& v_1\sin\theta_1=v_2\sin\theta_2\end{array}\right.$$

$$u^2=\left(v_1\cos\theta_1+v_2\cos\theta_2\right)^2=v_1^2+v_2^2$$

$$v_1^2+v_2^2=v_1^2\cos^2\theta_1+v_2^2\cos^2\theta_2+2v_1v_2\cos\theta_1\cos\theta_2$$

$$v_1^2\sin^2\theta_1+v_2^2\sin^2\theta_2=2v_1v_2\cos\theta_1\cos\theta_2$$

$$2v_2^2\sin^2\theta_2=\frac{2v_2^2\sin\theta_2\cos\theta_1\cos\theta_2}{\sin\theta_1}$$

$$\sin\theta_1\sin\theta_2=\cos\theta_1\cos\theta_2$$ $$\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2=0$$ $$\cos\left(\theta_1+\theta_2\right)=0$$ $$\theta_1+\theta_2=90^\circ$$

Hence if given that $\theta_1=30^\circ$, then $$\theta_2=60^\circ.$$

Answer 2

It can be shown easily by geometry but I don't know how to draw figures here. So let me use vector algebra.

So let the initial velocity be $u\hat{x}$. Then the center of mass (CM) has velocity $\frac{u}{2}\hat{x}$. In the CM frame, we have two identical particles approaching each other with speed $u/2$.

The total momentum in the CM frame is, of course. zero. Therefore the final momentum should also be zero. Hence in the CM frame, one will see the particles moving away from each other with the same speed $v$ in opposite directions, making an angle $\theta$ with the $x$-axis.

By conservation of KE, $v=u/2$. And hence the final velocities are $$\frac{u}{2}\cos\theta\hat{i}+\frac{u}{2}\sin\theta\hat{j}$$ and $$\frac{u}{2}\cos(\theta+\pi)\hat{i}+\frac{u}{2}\sin(\theta+\pi)\hat{j}=-\frac{u}{2}\cos\theta\hat{i}-\frac{u}{2}\sin\theta\hat{j}$$.

Go back to the lab frame, we have velocities $$\frac{u}{2}(1+\cos\theta)\hat{i}+\frac{u}{2}\sin\theta\hat{j}$$ and $$\frac{u}{2}(1-\cos\theta)\hat{i}-\frac{u}{2}\sin\theta\hat{j}$$

Their dot product is hence $$\frac{u^2}{4}(1-\cos^2\theta-\sin^2\theta)=0$$

So if one particle makes $30^\circ$ with the original direction, the other angle will be $60^\circ$.