Answer to Question #118111 in Classical Mechanics for Lizwi

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?
Hint: What does "perfectly elastically" imply for the kinetic energy before and after the collision?
1 : 2.00 m/s
2 : 0.866 m/s
3 : 1.50 m/s
4 : 2.59 m/s
5 : 0.750 m/s

Answer 1

The correct answer is option 3.

Let $m_1,,m_2,u,v_1,v_2$ be the mass of first ball,initial velocity of the first ball,final velocity of first ball ans final velocity of second ball respectively.

Now, from the above figure ,apply the momentum conservation along horizontal direction and vertical direction we get,

$$m_1u=m_1v_1\cos(\theta_1)+m_2v_2\cos(\theta_2) \hspace{1cm}(1)\\ 0=m_1v_1\sin(\theta_1)-m_2v_2\sin(\theta_2)\hspace{1cm}(2)$$

Now, we are interested to find $v_2$ ,so using $m_1=m_2$ , and solving equation $(1)\&(2)$ , we get

$$v_2=\frac{u}{\cos(\theta_2)+\sin(\theta_2)\cot(\theta_1)}$$

By, plugin all the values we get,

$$v_2=\frac{3}{\frac{1}{2}+\frac{\sqrt{3}}{2}\sqrt{3}}\implies v_2=1.5ms^{-1}$$