# 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

#### Answer:

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!)