# How do you solve this physics problem ?

## there are two A and B balls on a pool table at rest, one next to the other. After the impulse, the A-ball moves with an acceleration of 12 m/s^2 and the B ball with an acceleration of 24 m/s^2. If the angle formed between the two balls is 60°, what will be the distance, in cm, between the two balls after one second, considering that neither of them has fallen into the hole? THANK YOU

Jan 25, 2018

First we find the distance covered by each ball, then the distance between them. They are $2080$ $c m$ apart after 1 second.

#### Explanation:

For both balls, the initial velocity $u = 0$ $m {s}^{-} 1$, so:

$s = u t + \frac{1}{2} a {t}^{2}$

reduces to:

$s = \frac{1}{2} a {t}^{2}$

For Ball A:

$s = \frac{1}{2} \cdot 12 \cdot {1}^{2} = 6$ $m$

For Ball B:

$s = \frac{1}{2} \cdot 24 \cdot {1}^{2} = 12$ $m$

We can't simply subtract one from the other, though, because they are at a ${60}^{o}$ angle to each other. We have a triangle with sides $6$ and $12$ $m$ and a ${60}^{o}$ angle between them. We can use the cosine law to find the third side of the triangle:

${a}^{2} = {b}^{2} + {c}^{2} - 2 b c \cos A$

$= {12}^{2} + {24}^{2} - 2 \cdot 12 \cdot 24 \cdot \cos {60}^{o}$

$= 144 + 576 - 576 \cdot 0.5 = 432$

Take the square root of both sides:

$a = 20.8$ $m$

We were asked for the answer in $c m$, so it is $2080$ $c m$.

(side note: this is clearly not a standard size pool or billiard table!)