Determining final speed of a billiard ball after elastic collision

Question

A billiard ball traveling at 3.00 m/s collides perfectly elastically with an identical billiard ball initially at rest on the level table. The initially moving billiard ball deflects 30.0° from its original direction. What is the speed of the initially stationary billiard ball after the collision?

I set up two relationships involving momentum as follows

Let $m$ be the mass of both identical billiard balls, $v_1$ be speed of the initially moving billiard ball after the collision, $v_2$ be speed of the initially stationary billiard ball after the collision, and $\theta$ be the angle between the velocity vector of the second billiard ball after the collision and the direction of the initial momentum vector

Based on conservation of momentum:

$$1) \quad(3m/s)m=mv_1\cos(30^\circ)+mv_2\cos(\theta)$$ $$(3m/s)=v_1\cos(30^\circ)+v_2\cos(\theta)$$ $$2) \quad v_1\sin(30^\circ)=v_2\sin(\theta)$$

However this is as far as I got and I don't know how to solve for $v_2$. Am I missing something?

Btw I know that the answer is 1.5m/s.

Answer 1

If the question is about an elastic collision then no energy is lost in the collision, kinetic energy in = kinetic energy out. This should give you another equation to solve your problem - at the moment you have 3 unknowns and 2 equations - conserve energy and you will have 3 equations and should be able to solve.

The three unknowns are $v_1$, $v_2$ and $\theta$, of course, and the two equations you have are linear momentum conservation in the 2 dimensions of the 2D plane the collision occurs in - add conservation of energy and you should be able to solve the problem.

Answer 2

Momentum and energy must be conserved.

$v_1 \cos 30^\circ + v_2 \cos \theta = 3\\ v_1 \sin 30^\circ + v_2 \sin \theta = 0\\ v_1^2 + v_2^2 = 9$

The first two equations describe the momentum of the system, the last describes the energy.

$\theta = \arcsin \frac {v_1}{2v_2}\\ v_1 \frac {\sqrt 3}{2} + v_2 \sqrt {1-\frac {v_1^2}{4v_2^2}} = 3\\ v_1 \sqrt 3+ \sqrt {4v_2^2-v_1^2} = 6\\ \sqrt {36 - 5v_1^2} = 6 - v_1\sqrt 3\\ 36 - 5v_1^2 = 36 - v_112\sqrt 3 + 3v_1^2\\ 12\sqrt 3 v_1 = 8v_1^2\\ v_1 = 0 \text { or } v_1 = \frac {3\sqrt 3}{2}$

If $v_1 = 0$ then it isn't rolling away at a $30^\circ$ deflection.

$v_2 = \sqrt {9 - \left(\frac{3\sqrt 3}{2}\right)^2} = 1.50$