Question

A pool ball moving 1.33 m/s strikes an identical ball at rest. Afterward, the first ball moves 0.750 m/s at a 33.30 angle. What is the velocity of the other ball?

Answer 1

Explanation:

We need to apply the conservation law of linear momentum to two dimensions:

Let $p_{1}$ = momentum of the 1st ball

$p_{2}$ = momentum of the 2nd ball

In the x-axis, the conservation law can be written as

$(p_{1} \cos \theta_{1})_{i} + (p_{2} \cos \theta_{2})_{i} = (p_{1} \cos \theta_{1})_{f} + (p_{2} \cos \theta_{2})_{f}$

or

$(m_{1}v_{1})_{i}= (m_{1}v_{1}\cos \theta_{1})_{f} + (m_{2}v_{2}\cos \theta_{2})_{f}$

Since we are dealing with identical balls, all the m terms cancel out so we are left with

$(v_{1})_{i} = (v_{1})_{f}\cos \theta_{1} + (v_{2})_{f}\cos \theta_{2}$

Putting in the numbers, we get

$1.33 = (0.750) \cos(33.30) + (v_{2})_{f} \cos \theta_{2}$

$= > (v_{2})_{f} \cos \theta_{2} = 0.703$

In the y-axis, there is no initial y-component of the momentum before the collision so we can write

$0 = (v_{1}\sin \theta_{1})_{f} + (v_{2}\sin \theta_{2})_{f}$

or

$= > (v_{2})_{f} \sin \theta_{2} = (0.750) \sin(33.30) = 0.412$

Taking the ratio of the sine equation to the cosine equation, we get

$\frac{ \sin \theta _{2}}{ \cos \theta_{2} } = \tan \theta_{2} = \frac{0.412}{0.703} = 0.586$

or

$\theta_{2} = { \tan}^{ - 1} (0.586) = 30.4$

Solving now for $(v_{2})_{f}$,

$(v_{2})_{f} = \frac{0.412}{ \sin(30.4) } = 0.815 \: \frac{m}{s}$